Wikipedia:Reference desk/Mathematics

Wikipedia:Reference desk/headercfg

October 4

NOT A QUESTION

Earlier this week I asked questions on a homework assignment and one user deleted the questions and told me to do my own homework. I became mad 'cussed out' the user. I want to aplogize to anyone I may offended, especially David Eppstein, whom I have sent a message on thier talk page. If anone was wodering, I ended up getting an 88 on the test. If soem Wikipedian believes this needs to be deleted because it's not what the desk is for, then do it, I jsut didn't see a more practical way to apologize. schyler 00:35, 4 October 2007 (UTC)

It's okay, we forgive you. I think this is a good case of WP:IAR to let you post this here.--Mostargue 00:53, 4 October 2007 (UTC)

Thanks for getting back to us - most don't bother. And, 88 ain't bad! - hydnjo talk 02:12, 4 October 2007 (UTC)

No problem schyler, although it's definitely good to avoid insults and name-calling, you have every right to object if you feel someone was being less-than-helpful with their answer. Sometimes people worry more about "enforcement" than about helping people figure out how to properly use Wikipedia and get familiar with all the "rules". This is not to "blame" anyone, just to say we can all do our part to help avoid misunderstandings.

In the future, if you want to make a post like this, (a general-purpose comment instead of a question), feel free to post it at Wikipedia talk:Reference desk. Just like with Wikipedia articles, the reference desk also has a talk page. (See also, Wikipedia:Talk page guidelines and Wikipedia:No personal attacks for more information, both of those have "quick summaries" so you don't have to read through the whole thing if you're busy).

Thanks again for your original question and for your apology. Regards. dr.ef.tymac 06:30, 4 October 2007 (UTC)

The original outburst appeared here, directed at David, so apologies both here and personally to David are fine.

It is not just "good" to avoid profanity and attacks, it is mandatory. Bans can be imposed if civility is flagrantly ignored.

That said, I understand that tests can be stressful. Shoe on the other foot, David is a university professor with years of experience dealing with students, which can also try the patience. And surely he has other stresses in his life as well.

In the end, your willingness to acknowledge that you behaved badly, and to apologize to all concerned, is a testament to your character. It suggests humility and consideration and courage. Thank you; and congratulations on your test showing. --KSmrq^T 09:14, 4 October 2007 (UTC)

Indeterminate form

I know that in early calculus, 0/0, (undefined)/(undefined), and (undefined)-(undefined) are indeterminate forms. Are there any other indeterminate forms that are suitable for newbies (to calculus)? --Kushal^{Click me!} write to me 03:43, 4 October 2007 (UTC)

0^0, 0^∞, 1^∞.--Mostargue 05:20, 4 October 2007 (UTC)

And $0\cdot \infty$, and $\infty -\infty$. – b_jonas 08:43, 4 October 2007 (UTC)

Also ∞/∞ and ∞/0. And (−∞)^L, in which L can be basically any limit, is in general not well defined. Actually, if you want to be exhaustive, −∞ and +∞ should be treated separately.  --Lambiam 12:27, 4 October 2007 (UTC)

Is 0^∞ really indeterminate ? Surely

$\lim _{x\rightarrow 0}(\lim _{n\rightarrow \infty }(x^{n}))=0$

$\lim _{n\rightarrow \infty }(\lim _{x\rightarrow 0}(x^{n}))=0$

so 0^∞ = 0 whichever way you look at it (and so it is different from 0⁰ and 1^∞, which certainly are indeterminate). Gandalf61 12:41, 4 October 2007 (UTC)

Even more general, if $x_{n},y_{n}$ are sequences of positive numbers such that $x_{n}\to 0$ and $y_{n}\to \infty$, then $x_{n}\leq 1$ and $y_{n}\geq 1$ for almost all n, so $0\leq x_{n}^{y_{n}}\leq x_{n}\to 0,$ so all relevant limits are 0, not just the iterated limits considered by Gandalf61. Kusma (talk) 13:42, 4 October 2007 (UTC)

Although there is a slight problem: lim_x→∞−exp(−x) = 0, lim_x→∞exp(x) = ∞, yet I'd say that lim_x→∞(−exp(−x))^exp(x) is undefined rather than 0.  --Lambiam 17:43, 4 October 2007 (UTC)

I went for "positive numbers" to avoid this problem of $(-0)^{\infty }$. The problem is not in the limit, but in defining arbitrary powers of negative numbers, though -- any useful definition should have $|x^{y}|=|x|^{y}$, and then the argument from above carries through to show that the limit is still zero. Kusma (talk) 11:19, 5 October 2007 (UTC)

Win some, lose some.--Mostargue 13:50, 4 October 2007 (UTC)

Interesting, (although I do not understand all of it.) Thanks guys! (this thread is still open I guess) --Kushal^{Click me!} write to me 17:29, 4 October 2007 (UTC)

ah, I see what my problem was. I meant $\infty ^{0}$.--Mostargue 11:07, 5 October 2007 (UTC)

Isn't that just 1 ? StuRat 15:46, 5 October 2007 (UTC)

Also, for indeterminate forms, couldn't we just look at the value of a discontinuous function at the discontinuity, such as tan(90°) (that is $+\infty$ and/or $-\infty$) ? StuRat 15:54, 5 October 2007 (UTC)

I am off a tangent with this question, but is $\infty ^{0}$ = 1? --Kushal^{Click me!} write to me 20:09, 5 October 2007 (UTC)

StuRat [edit - and Kushal]: let x_n be 2ⁿ, and y_n be 1/n. Then lim_n→∞x_n^y_n = 2. You can easily modify this to get any kind of limit you want. [moved down to answer Kushal as well. Tesseran 00:03, 6 October 2007 (UTC)

partial derivative

Is ðy/ðx=1/(ðx/ðy). —Preceding unsigned comment added by 218.248.2.51 (talk) 05:52, 4 October 2007 (UTC)

Generally, no. See Partial derivative.

• ${\frac {\partial y}{\partial x}}={\frac {\partial }{\partial x}}y$
• ${\frac {\partial x}{\partial y}}={\frac {\partial }{\partial y}}x$

Basically, it's because ${\frac {\partial }{\partial x}}$ and ${\frac {\partial }{\partial y}}$ are operators. --Mostargue 06:02, 4 October 2007 (UTC)

Also see Differential operator.--Mostargue 06:06, 4 October 2007 (UTC)

Rather than "no", I'd say that both expressions are not meaningful at the same time. If the expression ∂y/∂x is meaningful, then y is an expression depending on several variables, including x. In that case x is not an expression depending on several variables, including y, so then the expression ∂x/∂y is not meaningful. (If "several" can include the case of just one dependent variable, so that you can write just dy/dx, and the relationship is such that x can be considered as functionally depending on y equally well, then indeed dy/dx = 1/(dx/dy) with some caveats such as that thou shalt not divide by zero.)  --Lambiam 06:35, 4 October 2007 (UTC)

Lamb-"Both Guns Blazing"-Iam. More questions I answer, the more of my own questions are answered.--Mostargue 06:49, 4 October 2007 (UTC)

outcircles?

I've just looked up incircle and was surprised to see the definition given for excircles. What is the circle called that goes through the three corners of a triangle? Do we have an article on it? -- SGBailey 06:24, 4 October 2007 (UTC)

Circumscribed circle.--Mostargue 06:29, 4 October 2007 (UTC)

complex numbers.

what does a term with "i" represent in an equation.as "i"represent squareroot of -1,a term with it (may be sin or cos )represents what part of physical value when squareroot of -1 dosent exist or is imaginary,n still most of forms uses i,just as in equn of electromagnetic radiations or exponential form of fourier series a term with i is mandatory. —Preceding unsigned comment added by 202.141.149.165 (talk) 13:36, 4 October 2007 (UTC)

Not actually sure what you mean. Lots of letters have implicit meanings and are avoided when it can cause ambiguity (sometimes engineers use j instead of i because i has uses in engineering; sometimes people use p, q and r as indices rather than i, j and k because of vector concerns, and so on).

In polar form, a complex number z with modulus r and argument t has real and complex components in the form r cos t and r sin t. It's just a property of a number, like sign. x42bn6 Talk Mess 14:33, 4 October 2007 (UTC)

I believe the question is asking what is physically represented by the imaginary term of a complex number. As with any number, the answer is "it depends on the application". I can no more say in general terms what "3i" represents than what "3" represents. You may find the applications section of our complex number article useful. Additionally, please consider phrasing your questions more carefully -- inconsistent spelling, spacing, and punctuation does make it more difficult to parse your request (the same holds for below, where I'm really not sure what's being asked). — Lomn 14:52, 4 October 2007 (UTC)

"i" usually represents the square root of minus one as you say, however when using vectors, "i" may represent a vector parallel to the x-axis eg from the origin the vector 4i+5j+10k gives the point (4,5,10).87.102.94.194 15:15, 4 October 2007 (UTC)

fourier series.

a fourier form is used 2 represent a time domain signal to frequency domain signal,n orthogonal signals r used,just as in trignometric form of fourier signal "what does each sine term correspond 2 for say (a simple NRZ signal used for digital communication)". —Preceding unsigned comment added by 202.141.149.165 (talk) 13:49, 4 October 2007 (UTC)

In general fourier analysis can convert one function from a signal in terms of time to a signal as a sum of sinusioudals eg sines and cosines - I assume you are thinking specifically of fourier series.

For this to be possible the function must be periodic ie it must repeat itself over a certain constant time frame

As such a NRZ signal can not be expressed as a fourier series - as it is not a periodic function.

You might want to look at square wave which is a function that can be expressed as a fourier series.

If you want know about methods that make a NRZ signal expressable as a fourier series please say so.87.102.94.194 15:39, 4 October 2007 (UTC)

Actually, the Fourier Transform does not need to operate on a periodic function, there are aperiodic forms of it, but this is mostly only useful for theoretical purposes. More practically, a periodic form of the transform can be applied to chunks of a signal at a time, and can be used for all sorts of things, e.g. filters, convolution, etc. (see Fourier analysis#Applications in signal processing for a few ideas). Specifically for a "non-return-to-zero" signal I don't know if there is much application in terms of the content of the signal; at least, I can't think of any reason (in terms of information) that I would want to alter or analyze that kind of signal in the ways that a FT could. - Rainwarrior 16:49, 4 October 2007 (UTC)

trial method equation solvings

a^3 + 2a+5 how can i solve this within 5 secs —Preceding unsigned comment added by 59.93.243.198 (talk) 17:24, 4 October 2007 (UTC)

Not by asking the question here. It isn't quite clear to me what you mean b y "solving" a³ + 2a + 5; that expression is not an equation. Are you asking for the zeros of the function f(a) = a³ + 2a + 5 (or, equivalently, the solutions of the cubic equation a³ + 2a + 5 = 0)? Our article on cubic equations gives an algebraic method for solving such equations; if you consistently manage to do that within 5 seconds, you should contact the Guinness people.  --Lambiam 17:55, 4 October 2007 (UTC)

Maybe he's asking how to factor it.--Mostargue 12:22, 5 October 2007 (UTC)

It's irreducible over the rationals, so any factorisation is going to be quite ugly. Algebraist 14:28, 5 October 2007 (UTC)

There is one real root, −1.328268856, and one pair of complex conjugate roots, 0.6641344278 ± i·1.822971095 . If you have downloaded the J interpreter you can solve it within 5 seconds by typing this line:

     p. 5 2 0 1

and the result appears as

┌─┬───────────────────────────────────────────────────────────────┐
│1│0.6641344278j1.822971095 0.6641344278j_1.822971095 _1.328268856│
└─┴───────────────────────────────────────────────────────────────┘

showing the high term coefficient, and the 3 roots. Bo Jacoby 19:58, 6 October 2007 (UTC).

Let me warn though that the builtin p. verb for polynomial solving has some precision problems, or at least it did have so last I tried. If you needed to solve polynomials, it might be better to use the lapack bindings (which are not part of the J distribution but has to be downloaded separately). – b_jonas 19:34, 7 October 2007 (UTC)

Or, equivalently, type roots([1,0,2,5]) to an octave session, which uses those lapack routines. – b_jonas 19:37, 7 October 2007 (UTC)

women

there are three types of women in this world, what are they. i begged my proffesor for a clue as to what these types are pertaining to. he said to me, that it deals only with phisical appearence. there are 3 categories of body types that only women fall into, only three. i am desperate for the answer please help. he then said he asked his quantitative research analysis class the question, and only one person was able to find the answer, and he was very impressed. i am the only student he has given this clue to, and it would be very much appreciated to receive the answer... thanks!

---

Etro —Preceding unsigned comment added by 151.196.108.226 (talk) 17:55, 4 October 2007 (UTC)

This has already been discussed on the Miscellaneous RefDesk as well as above on this desk; there is no need to post a third time. — Lomn 20:12, 4 October 2007 (UTC)

Algebra question

x×.8=4032 —Preceding unsigned comment added by 71.225.135.106 (talk) 19:03, 4 October 2007 (UTC)

Are you serious? First off, a × b is the same as b × a, so you can rewrite the left-hand side of your equation as .8 × x. And .8 is another notation for 0.8, which is the same as 8/10. Then, if you have an equation of the form a × x = b in which a is not 0, you can divide both sides by a to arrive at the solution x = b / a.  --Lambiam 20:37, 4 October 2007 (UTC)

Traveling Salesman computation

I expect i'm asking a question which Wikipedia already has an article for, but i'll ask anyway as i can't find one;

(also, should this be in computing or mathematics?)

With Traveling Salesman computation i was thinking about reducing the time of computation by counting the length of an in-construction path as we go (with each added edge) and if the length of path is greater than the current best upper band at that time abandoning the unfinished path - But is it quicker to do it this way (assuming time to calculate length of path increases with increase to the path size) and removing 'over' paths as we go than computing the length only when the path is complete and deciding if the path is shorter - Also at with how many vertices would one system become faster than the other surely as the number of vertices increases so does the calculation time of the first method but so does the possible gain of not having to construct full paths?

Maybe i should start Differentiating?

Thanks!

-Benbread 19:18, 4 October 2007 (UTC)

I am not fully sure I understand your proposed method, but it sounds like branch and bound. If it is possible to answer the more specific questions at all (some techniques improve the time for some inputs but make it worse for other inputs), this requires a more precise description of the proposed algorithms.  --Lambiam 20:24, 4 October 2007 (UTC)

SHA hash websites

Hi. There are probably different links for SHA-1, SHA-2, SHA-512, etc. The problem is, what exactly are those links? The Template:User committed identity gives the link to SHA-1, and the article gives the links to a few, but where are the links to all the SHA websites, say for example where is the link for calculating SHA-512? Can someone list me the links of those websites, and where you got it from? I don't want links to sites other than those used to directly calculate a string into a hash code, so I don't want any sites with info about the SHA websites, I want the actual links themselves. This question was originally from Talk:SHA hash functions. Thanks. ~AH1^(TCU) 20:43, 4 October 2007 (UTC)

The first return of a Googling of SHA-512 hash calculator is a rather fine site that calculates MD5, SHA-1, SHA-256, SHA-384, and SHA-512 hashes; it has worked well for me. Joe 04:06, 5 October 2007 (UTC)

That's the same one listed at SHA hash functions#Online Hash Calculators. Template:User committed identity suggests the other one because being implemented in Javascript, it doesn't involve sending your super secret string to some random person's web server. (The obvious drawback of requiring you to run Javascript is ignored.) The serious answer has to be: forget web forms. Use the sha256sum program (and other similarly named ones) from coreutils. --tcsetattr (talk / contribs) 04:50, 5 October 2007 (UTC)

Hi. Thanks for your replies. However, the first link does calculate various forms of hash, but the problem is, I don't know how to actually scelect an actual form of hash to calculate, say SHA-1, SHA-256, SHA-512, etc. There are links linked from that website, but clicking on one of the links takes me to a website that doens't calculate anything. The problem with SHA-1 is that they say not enough security. Coretils is a link but doesn't take me anywhere. Thanks. ~AH1^(TCU) 12:47, 5 October 2007 (UTC)

What are your security requirements? MD5 is, and always will be, perfectly adequate for many applications, even though it's been broken in others. -- BenRG 16:44, 5 October 2007 (UTC)

In case this is not clear, the http://www.johnmaguire.us/tools/hashcalc/ page will compute all mentioned forms of hashes for you when you click the button labelled "calculate".  --Lambiam 17:52, 5 October 2007 (UTC)

y = x^x

Does the function y = x^x have any significance? Does it have a name? I've always found it a particularly intriguing function in the abstract, especially for –1 ≤ x ≤ 1. (Then there are the related functions y = (–x)^x, y = x^(–x), and y = (–x)^(–x).) Thanks. — Michael J 05:21, 5 October 2007 (UTC)

It doesn't have a name as far as I know. But you'll perhaps be interested in the sophomore's dream. - Fredrik Johansson 05:38, 5 October 2007 (UTC)

Interesting. It's a little over my head, so I'll have to take some time with it, Fredrik. — Michael J 05:52, 5 October 2007 (UTC)

It arises in analyzing Exponentiation#Zero_to_the_zero_power.--Mostargue 11:02, 5 October 2007 (UTC)

It can be written using tetration: y=²x Alpha Omicron 15:21, 5 October 2007 (UTC)

October 5

Logic

I have to prove the following tautology:

$(P\to Q)\land (Q\to R)\to (P\to R)\equiv T$

Here's my logic:

$(\neg P\lor Q)\land (\neg Q\lor R)\to (\neg P\lor R)$

$\neg (\neg P\lor Q)\lor \neg (\neg Q\lor R)\lor (\neg P\lor R)$

$(P\land \neg Q)\lor (Q\land \neg R)\lor (\neg P\lor R)$

Can't see how to get out of that last statement. I'm thinking that maybe I cocked it up at the second line where I flipped the AND to OR. But $\neg (P\land Q)\equiv \neg P\lor \neg Q$, right?

Wrong, $\neg (P\land Q)\equiv \neg P\lor \neg Q$ — see De Morgan's laws.

Woops, typo on my part. That's what I meant. Fixed.

Anyway, open last parentheses, reorder terms and try to reduce $(P\land \neg Q)\lor \neg P$ and $(Q\land \neg R)\lor R$

CiaPan 09:15, 5 October 2007 (UTC)

(contd.)

${\begin{aligned}(P\land \neg Q)\lor \neg P&\equiv (P\land \neg Q)\lor \neg P\lor \neg P\\&\equiv (P\land \neg Q)\lor (\neg P\land 1)\lor (\neg P\land 1)\\&\equiv (P\land \neg Q)\lor (\neg P\land (\neg Q\lor Q))\lor (\neg P\land 1)\\&\equiv (P\land \neg Q)\lor (\neg P\land \neg Q)\lor (\neg P\land Q)\lor (\neg P\land 1)\\&\equiv ((P\lor \neg P)\land \neg Q)\lor (\neg P\land (Q\lor 1))\\&\equiv (1\land \neg Q)\lor (\neg P\land 1)\\&\equiv \neg Q\lor \neg P\end{aligned}}$

Reduce the other part $(Q\land \neg R)\lor R$ in similar way, and you get a logical sum with some $X\dots \lor \neg X\dots$ terms, which will give you, by law of excluded middle, the final result of true, i.e. the tautology proof. --CiaPan 15:35, 8 October 2007 (UTC)

To list all combinations of truth values for P, Q, and R will require a table of only eight entries. If the expression is true for all combinations, that's a proof.

P	Q	R	P→Q	Q→R	P→R	Σ
F	F	F	T	T	T	T
F	F	T	T	T	T	T
F	T	F	T	F	T	T
	⋮			⋮		⋮
T	T	T	T	T	T	T

Or, perhaps it's a proof; that depends on the rules. Formal study of logic introduces many variations, including some in which we must distinguish between P→Q and ¬P∨Q. See Classical logic and its links. --KSmrq^T 11:56, 5 October 2007 (UTC)

Mass-Energy Equivalence

In the famous equation e = mc2, this article says that e measured in Joules; m measured in kilograms, and c measured in meters per second. My question is: does this apply to any kind of matter? Would the energy in a kg of feathers be the same as the energy in a kg of lead?

David A. Johns (e-mail address removed) Doghall 11:35, 5 October 2007 (UTC)

Yes, and it is independent of the specific units of measurement used to express the physical quantities involved. A more appropriate place for this question would have been Wikipedia:Reference desk/Science.  --Lambiam 12:09, 5 October 2007 (UTC)

(edit conflict) Both of Einstein's relativity theories assert that we should detect no difference, as do most theories. However, the universe is not obliged to respect our theories, so this is an experimental question. I believe tests of this sort have been made, and none have been accepted as falsifying the theory. But mathematicians do proofs, not science experiments, so why are you asking us? This question belongs on the science reference desk, not here. --KSmrq^T 12:22, 5 October 2007 (UTC)

MATHEMATICS (Interest homework)

Moses invested a certain sum of money in a housing bank for one year. For the first three months he earned 18% per annum simple interest and for the remaining time he earned 16% per annum compounded interest quaterly.The total interest earned for one year period was 5616.Find the sum invested.Please i beg your help!Georgekalusanga 14:50, 5 October 2007 (UTC)

I can never quite figure out what they mean by "interest", is that the nominal yield not including compounding or effective annual yield ? Let's assume it's the nominal yield not including compounding. Therefore, the quarterly rates based on 18% and 16% are 18/4 and 16/4 or 4.5% and 4%. That gives us:

P × 1.045 × 1.04 × 1.04 × 1.04 = 5616

Or:

P × 1.045 × 1.043 = 5616

On the other hand, if we assume "interest" means effective annual yeild (including compounding), then the quarterly rates are 1.18^1/4 and 1.16^1/4. They might even mean to mix and match the two, with quarterly interest rates of 18/4 and 1.16^1/4. This actually gives us the answer closest to a whole number, so that's my guess. That would give us the formula:

P × 1.045 × 1.161/4 × 1.161/4 × 1.161/4 = 5616

Or:

P × 1.045 × 1.163/4 = 5616

You do the math to finish it at let us know if you have any trouble. StuRat 15:27, 5 October 2007 (UTC)

opposite of disjoint sets

Is there a terms for a pair of distinct sets that share at least one element? 71.29.100.133 17:28, 5 October 2007 (UTC)

I'd call them "overlapping sets", a term that is used, for example. in Distance, Connection form, and Seifert–van Kampen theorem.  --Lambiam 17:38, 5 October 2007 (UTC)

I think the term I've heard most often is just 'sets with nonempty intersection'. Algebraist 18:11, 5 October 2007 (UTC)

Parabola/Catenary

What is the difference between the two? Btw please don't just give the differences in their equations, I would like an explanation about what is physically different. Thanks 172.159.49.31 17:54, 5 October 2007 (UTC)

They're totally different curves? I'm not sure what you want here. If you mean 'how do they look different to the eye', the most obvious difference is apparent on a zoomed-out plot: the catenary is massively steeper for large x (exponential rather than quadratic). Algebraist 18:15, 5 October 2007 (UTC)

Maybe our articles on parabola and catenary can be of help. Both articles mention relations to the physical world and not just equations. PrimeHunter 20:21, 5 October 2007 (UTC)

I would especially recommend the Hanging With Galileo page mentioned in the catenary article. And why are people now asking mathematicians about physics? If we were clever enough to deal with the "real world" we would be engineers, with endless job opportunities and unbounded income potential. ;-) --KSmrq^T 21:24, 5 October 2007 (UTC)

What's wrong with merchant banking and management consultancy? Algebraist 13:21, 6 October 2007 (UTC)

I other words as the link "Kmsqr" supplies says above - the parabola typically arises when a flat heavy thing is supported by a wire (eg humber bridge), whereas the cantenary arises when the majority of the mass is in the wire. eg overhead line

Just out of curiosity can anyone tell the difference between a graph of either - just by looking??87.102.23.214 14:11, 6 October 2007 (UTC)

The difference is that the wire's graph has no bounds for it's derivative (because it's the derivative of a Parabola), versus the Bridge has a bounds of +/-n where is a finite positive constant, because it's the derivative of the upper half of a Hyberbola) em I right?? —Preceding unsigned comment added by 69.54.140.201 (talk) 22:25, 6 October 2007 (UTC)

Did you read the Hanging With Galileo page?  --Lambiam 23:59, 6 October 2007 (UTC)

1 equals to -1

${\frac {1^{n}}{x^{n}}}=\left({\frac {1}{x}}\right)^{n}$

.

Let x == -1

.

${\frac {1^{n}}{(-1)^{n}}}=\left({\frac {1}{-1}}\right)^{n}\,$

.

${\frac {1^{n}}{(-1)^{n}}}=-1^{n}$

.

Let n == 1/2

.

${\frac {\sqrt {1}}{\sqrt {-1}}}={\sqrt {-1}}$

.

${\frac {1}{i}}=i$

.

Multiply both side by i

.

$1=-1\,$

211.28.129.8 22:12, 5 October 2007 (UTC)

See invalid proof.  --Lambiam 22:22, 5 October 2007 (UTC)

Such is the peril of asserting that $i={\sqrt {-1}}$. It's better to say that i is a solution to the equation x² = –1. Strad 23:37, 5 October 2007 (UTC)

Not to mention that when you take non-integer roots in the complex numbers, you really have to consider the impact of multiple solutions. Confusing Manifestation 07:16, 6 October 2007 (UTC)

The flaw in your proof is assuming that a square root and the power of 1/2 are the same thing, I did a proof Sophomore year of high school in Algebra 2 that 4throot(x^2) and (4throot(x))^2 are not equal for negative x. In your case, the fact that square roots have two solutions versus 1/2 power having only one solution has caused you to make an error. If you don't fully understand how this is the case, try looking up De'Moivre's formula. Hope that clarifys things. —Preceding unsigned comment added by 69.54.140.201 (talk) 09:56, 6 October 2007 (UTC)

To be more specific, let's define $\forall x\in \mathbb {C} \quad R(x):=\{y\in \mathbb {C} |y^{2}=x\}$. Then

$\forall x\in \mathbb {C} \setminus \{0\}\quad {\frac {\sqrt {1}}{\sqrt {x}}}={\sqrt {\frac {1}{x}}}$

doesn't make sense since you do not explain how you choose the roots. In particular, as your example demonstrates, it is not true in general that an arbitrary choice for each root would satisfy the equation. What you can say however is:

$\forall x\in \mathbb {C} \setminus \{0\}\quad {\frac {R(1)}{R(x)}}=R\left({\frac {1}{x}}\right)$

In particular:

${\frac {R(1)}{R(-1)}}=R(-1)$

$R(1)={R(-1)}^{2}$

$\{-1;1\}={\{-i;i\}}^{2}$

$\{-1;1\}=\{-1;1\}$

And there is no contradiction. Ceroklis 09:59, 6 October 2007 (UTC)

Hm? $(-i)^{2}=i^{2}=-1$, so it seems that your last set should be the singleton $\{1\}$. Or are you using some non-element-wise definition of exponentiation? Of course, it makes no sense to divide two sets element-wise anyway: sets have no order, so you can permute one of them and change the answer. --Tardis 16:40, 9 October 2007 (UTC)

Denoting $AB=\{ab|a\in A,b\in B\}$ is fairly standard, and likewise for ${\frac {A}{B}}$. $\{-i,i\}^{2}$ was meant as a shorthand for $\{-i,i\}\{-i,i\}$ which is indeed $\{-1,1\}$. -- Meni Rosenfeld (talk) 17:12, 9 October 2007 (UTC)

Ah; my mistake. I didn't realize that a cartesian product (but with the infix operator used, rather than "pair") was the intended understanding. Of course, it's a bit odd to say that $A^{2}\,\not \!:=\{a^{2}|a\in A\}$, but it's comprehensible. Thanks. --Tardis 16:01, 11 October 2007 (UTC)

The product $AB=A*B$ is not at all the same thing as the cartesian product $A\times B$. For instance

$\{1;2\}\times \{2;4\}=\{(1,2);(1,4);(2,2);(2,4)\}$

$\{1;2\}*\{2;4\}=\{2;4;8\}$

And there is in general no bijection between the two. Ceroklis 19:02, 11 October 2007 (UTC)

Tardis did say, with the infix operator used, rather than "pair", the infix operator being in this case multiplication. -- Meni Rosenfeld (talk) 12:38, 12 October 2007 (UTC)

October 6

Solving polynomials by composition of functions

I have been searching on the internet for any information about the use of composition of functions in solving polynomials. I have independently discovered this technique and I doubt I'm the first. It probably goes by another name or notation I am unfamiliar with. Furthermore I can create several formulas depending on the degree of the polynomial that represent factorizations of the highest power of x. It is also possible to achieve linear translations this way, by using 1*n where n is the highest power of x. Quartic equations represent the first good use of this method since 4 is not prime. You can create a formula for all Quartics that can be rewritten as a composition of 2 Quadratics, and then test a specific Quartic to see if it fits. Though this method takes a fair amount of work it can allow you to use basic algebraic techniques to solve problems otherwise requiring advanced knowledge such as the Quartic formula or Chebychev radicals, etc. any help in locating equivalent work would be much appreciated thanks.

I don't recall having heard of such a technique. I am somewhat skeptic as this seems to imply the existence of solutions by radicals to every polynomial equation (which is impossible). Can you give an example of how this technique is used? -- Meni Rosenfeld (talk) 11:19, 6 October 2007 (UTC)

To solve a quartic this way won't I need one of the quadratics first? - which I don't see how to get.14:22, 6 October 2007 (UTC) —Preceding unsigned comment added by 87.102.23.214 (talk)

It can't solve all quartics, not a general solution, but it's benefit is that you only need high school level knowledge to do it. I haven't played around w/ the sandbox enough to give you an example yet, sorry. Does copy and paste from microsoft's equation editor work??? —Preceding unsigned comment added by 69.54.140.201 (talk) 22:12, 6 October 2007 (UTC)

I see where I must have confused you, "You can create a formula for all Quartics that can be rewritten as a composition of 2 Quadratics" I didn't actually mean ALL quartics, rather all of them that fit the form created by a composition of 2 quadratics. —Preceding unsigned comment added by 69.54.140.201 (talk) 22:19, 6 October 2007 (UTC)

Copy+Paste from Microsoft's equation editor wouldn't work. Help:Formula contains information regarding displaying formulae, but you can give your example with plain text if you are uncomfortable with TeX. In any case, if I understand your method correctly, solving for the coefficients of the quadratics would lead back to a quartic equation. -- Meni Rosenfeld (talk) 22:29, 6 October 2007 (UTC)

Well, do you know about Polynomial long division? If you know a binomial factor (e.g. (x-2)), you can divide it out and reduce the degree of your polynomial by one (or if you know a quadratic that is a factor, by two, etc.). The thing is, in order to do this division, you need to know the factor beforehand, which comes back to the same problem you started with. There are a great many polynomials that can be easily factored. However, this is not often useful in practical applications of polynomial-solvers, e.g. in computer simulations the polynomial coming in is not known beforehand and the coefficients may be emperical measurements; a technique that doesn't work for the general case is likely to fail in this situation. - Rainwarrior 23:10, 6 October 2007 (UTC)

I'm assuming here that by "composition" you mean product, and not a composite function. I could be mistaken. - Rainwarrior 23:13, 6 October 2007 (UTC)

You still don't fully understand what I am suggesting here, (I know for sure this method works, I have tested it). I just looked at the help file so I'll try and give you an example.

Consider the function $f(g(x))$, if we let $f(x)=x^{2}+jx+k$ and $g(x)=x^{2}+mx+n$ then $f(g(x))$ is a quartic. Do you recall how to solve for x in a composition of functions?? Well it is done by setting $f(y)=0$ and solving for $y$, where $y=g(x)$. then set $g(x)$ equal to the solutions of $f(y)$ and solve for $x$. So lets calculate the formula that discribe the Quartics the fit this form. Note, I have assumed that you have divided the leading term out first.

$f(g(x))=(x^{2}+mx+n)^{2}+j(x^{2}+mx+n)+k$

$f(g(x))=(x^{4}+2mx^{3}+(m^{2}+2n)x^{2}+2mnx+n^{2})+(jx^{2}+jmx+jn)+(k)$

Drop parentheses and add.

$f(g(x))=x^{4}+2mx^{3}+(m^{2}+2n+j)x^{2}+(2mn+jm)x+(n^{2}+jn+k)$

See now where the quartic comes from?? and that quartic is what we want to solve. So we let $f(g(x))=0$ and apply the method I described above. As long as a Quartic in question can fit the $f(g(x))$ equation above it can be solved this way, not all Quartics can (I believe). So lets find the general solution for the 'fitting' Quartics.

$f(y)=0$

$y^{2}+jy+k=0$

And by the Quadratic formula.

$y={\frac {-j\pm {\sqrt {j^{2}-4k}}}{2}}$

Now before things get confusing let me clarify. Note that $f$ is the last function to act on x, for this reason it is the first equation to solve. And it's solutions are the required outputs of $g(x)$ therefore we must set $g(x)$ not equal to 0, but equal to the solutions of $f$.

$x^{2}+mx+n={\frac {-j\pm {\sqrt {j^{2}-4k}}}{2}}$

Keep in mind that the right side represents two distinct values in most cases, and certainly in the general case. This means that the $\pm$'s are independent and you must account for all possible combinations, (++,+-,-+,--) Which yields (of course) 4 solutions. Also note that the right side is a constant and therefore can be subtracted from both sides and grouped with $n$.

$x^{2}+mx+(n-{\frac {-j\pm {\sqrt {j^{2}-4k}}}{2}})=0$

Once again we will use the quadratic formula.

$x={\frac {-m\pm {\sqrt {m^{2}-4n+{\frac {-j\pm {\sqrt {j^{2}-4k}}}{2}}}}}{2}}$ Preceding As you can see, it can produce radical solutions not unlike the Quartic formula but the lack the presence of any Cube roots indicating that this is clearly not a general solution. It does however allow someone with only high school Algebra knowledge to solve 4th degree polynomials in cases not otherwise possible. And you don't have to have any prior knowledge of any solutions of the Quartic in order to solve it. It should be noted that one must be able to solve a linear system in two varibles in order to use this method. This is because the $x^{2}$ and $x$ terms both have $j$ and $n$ which means to determine both of their values when 'fitting' a Quartic to the formula. The last is handled by $k$. — unsigned comment added by 69.54.140.201 (talk) 23:52, 6 October 2007 (UTC)

There is an error in the coefficient for the linear term in the expanded quartic; the formula should be

f(g(x)) = x⁴ + 2mx³ + (m²+2n+j)x² + (2mn + jm)x + n² + jn + k.

Not all quartic functions can be expressed as the composition of two quadratic ones. To apply this to a quartic of the form x⁴ + ax³ + bx² + cx + d, we have to find unknowns j, k, m and n from the equations resulting from equating the coefficients:

a = 2m

b = m²+2n+j

c = 2mn + jm

d = n² + jn + k

This means that a, b and c have to satisfy the relationship:

a³ − 4ab + 8c = 0.

If so, the unknowns can be found (in fact, we can then always choose n = 0) and we can find the roots of the quartic by repeated application of quadratic root finding as shown above. For example, your method works beautifully on x⁴ + 10x³ + 35x² + 50x + 24. But if that relationship between a, b and c does not hold, the system of equations has no solutions, so a decomposition of the quartic function into quadratic functions is then not possible.  --Lambiam 04:50, 7 October 2007 (UTC)

An addition to my post. The condition on the three coefficients a, b and c can be reformulated as a condition on the multiset of roots of the quartic. For the quartic to be expressible as a composition of quadratic functions, you need that r₁ + r₄ = r₂ + r₃ for some permutation [r₁, r₂, r₃, r₄] of the four roots. This condition is fulfilled, for example, when the quartic function is symmetric about some axis. The example I gave above is actually symmetric about x = −⁵/₂.  --Lambiam 12:45, 7 October 2007 (UTC)

I want to comment that your phrasing gives the impression that you assume some incompetence on our part (e.g. "Do you recall how to solve for x in a composition of functions?", as if we need to be reminded). That was unnecessary; I was only expressing skepticism regarding a method which you have described only vaguely, and Rainwarrior only suggested a different interpretation to what would otherwise be a somewhat unusual method.

Finding the coefficients does turn out to be easier than I had originally suspected (I haven't done the actual calculations then), so as Lambiam has commented, the method works well when it does (which is only rarely). -- Meni Rosenfeld (talk) 12:26, 7 October 2007 (UTC)

By the way, the required condition can be expressed in a neat way in terms of the roots of the polynomial; the roots need to come in pairs which have the same difference (for example, it will work for $(x-1)(x-2)(x-6)(x-7)$, since $2-1=7-6$.

I wonder how well this method will do for a polynomial of degree, say, 6 (too lazy to carry out any calculations though). -- Meni Rosenfeld (talk) 12:36, 7 October 2007 (UTC)

It may be a little unfair to bring up the phrasology the poster used - this style of language is quite polite in some parts of the world - I'm sure no sarcasm was intended.

With regards to the original question - I think the title used "Solving (certain) quartics by composition of functions" would be a name for it - I wonder if the question asker wanted to know if the method had ever been given a name after the original discover of the method - which I imagine is not likely since the method looks (on the whole) technically useless.87.102.115.31 13:59, 7 October 2007 (UTC)

Never said it was impolite. -- Meni Rosenfeld (talk) 16:05, 7 October 2007 (UTC)

Wasn't trying to be insulting, just through in addressing your understandable clarification questions, and thanks for pointing out the error, I have edited in the necessary changes. In the case of a 6th degree polynomial there are two distinct orders of composition of 2nd degree and 3rd degree polynomials, resulting in two formulas. My original question remains open however, has anyone seen similar work to the kind I have now described in detail? If not, this does present the unlikely possibility that this is new work. (One of reasons I'm asking) I think the best feature of this method I have independently discovered is that in general elementry techniques can be used, at least on polynomials who's $n$ is a power of 2. The case of a 6th degree polynomial present the necessity that the Cubic formula be known for most fitting cases to be solvable, especially when the cubic is the inside function (1st to act on x). I should also point out that in the case of an 8th degree polynomial there are three useful substitutions, 4 of 2, 2 of 4, and 2 of 2 of 2. Yes this method can be extended to three or more composits. I think I mentioned the value of my method being that it allows the solving of Quartics and other higher degree polynomials without the need for more advanced methods in 'fitting' cases. In general it seems using composite functions to solve other functions has been rarely attempted or overlooked, at least from my mathematical education so far... I doubt this method is of any real value to applied mathematics but it should be kept in mind when takling theoretical problems, as the form of the solutions is substancially simpler than the Quartic formula. —Preceding unsigned comment added by 69.54.140.201 (talk) 13:54, 7 October 2007 (UTC)

It has occured to me that what we have discussed here is just a generalization of a biquadratic equation, of the form $x^{4}+ax^{2}+b=0$, which is solved using $t=x^{2}$. The method will work (for $n=4$) exactly when the equation has the more general form $(x-p)^{4}+a(x-p)^{2}+b=0$; the substitution $t=(x-p)^{2}$ then solves it. Finding the a, b and p requires steps similar to what we have discussed, but with a slightly different methodology. p is the point where the third derivative is 0; The method will work if the first derivative is also 0 at that point; b is the value of the function at that point, and a is half the second derivative at that point.

In short, a much more restricted version of the method (for $n=4$ and $p=0$) is already well known. The more I think about it, the more it seems that a more general discussion deserves a mention in the literature; I personally have not encountered it. -- Meni Rosenfeld (talk) 16:05, 7 October 2007 (UTC)

Your right it is similar to a Biquadratic function. I would like to point out that if you write your expression above as $((x-p)^{2})^{2}+a((x-p)^{2})+b=0$ it is almost exactly what I have here save that the inside quadratic need not be a perfect square. I agree that it should get a more through mention in the literature as well. By the way, I found and corrected another error in my work, when I subtracted the first solution from both sides of the second equation i solved I forgot to switch the sign. —Preceding unsigned comment added by 69.54.140.201 (talk) 23:29, 7 October 2007 (UTC)

I should note that the use of Composite Functions is much more useful with exponential functions with complicated exponents. A basic example would be

$Y=e^{x^{2}-5x-7}-3$

the appropiate functions being.

$f(x)=e^{x}-3$ and

$g(x)=x^{2}-5x-7$

$f(g(x))=Y$

$f(y)=e^{y}-3$

$e^{y}-3=0$

$e^{y}=3$

$\operatorname {ln} e^{y}=\operatorname {ln} 3$

$y=\operatorname {ln} 3$

$x^{2}-5x-7=\operatorname {ln} 3$

$x^{2}-5x-(7+\operatorname {ln} 3)$

$x={\frac {5\pm {\sqrt {25+28+4\operatorname {ln} 3}}}{2}}$

$x={\frac {5\pm {\sqrt {53+4\operatorname {ln} 3}}}{2}}$

You could also set the original problem equal to zero and move the constant over, then $\operatorname {ln}$ both sides and work from there. —Preceding unsigned comment added by 69.54.140.201 (talk) 05:53, 8 October 2007 (UTC)

Note that once again, it is not necessary to think in terms of composite functions, and I find that it just complicates the matter. If we want to solve $e^{x^{2}-5x-7}-3=0$, we can do so in a straightforward way:

$e^{x^{2}-5x-7}-3=0$

$e^{x^{2}-5x-7}=3$

$x^{2}-5x-7=\ln 3\;\!$

$x^{2}-5x-(7+\ln 3)=0\;\!$

$x={\frac {5\pm {\sqrt {53+4\ln 3}}}{2}}$

By the way, note the calculation error you had. -- Meni Rosenfeld (talk) 13:04, 8 October 2007 (UTC)

See Galois theory. Bo Jacoby 16:55, 10 October 2007 (UTC).

Come to think of it, the composition technique is identical to substitution only, it bothers to more formally define the substitution. The one useful property is that you can work out general substitution cases to create short cut formulas.

using integration to find the lateral surface area of a cone

I'm trying to prepare for SAT II Math level 2, but I find a lot of stuff I haven't studied for 2 years since I'm in calc ... so I figure it would be best to try to rely on calc rather than memorise everything again. I'm stuck on the lateral surface area of a cone, which I know is equal to pi times its slant height times the radius. It's also equal to a Riemann sum of its circumference integrated along its height, right? So if the height is known, we can relate the radius to the height, which can be our dependent variable. So the ratio of the radius to the height can be expressed as a constant k, and the radius is h*k. So the area of each slice is 2*pi*r, or pi*hk. We can let our origin be the tip of the cone, and the height 12 and the radius 5 (one of the example problems I was doing). So we integrate this this from 0 to 12, and end up with 60pi? According to the formula though I should end up with 65pi .... 74.65.173.180 10:23, 6 October 2007 (UTC)

comment removed, was totally wrong. Ceroklis 16:07, 6 October 2007 (UTC)

I'm not sure who is confused more than who. The questioner is entirely correct in stating that the lateral surface area is equal to pi times its slant height times the radius. Slant height is not the same as (just) height, the usual meaning of the symbol h in this context. The radius r, height h and slant height s of a cone are related by the formula of the Pythagorean theorem: r² + h² = s². If r = 5 and h = 12, we find a nice integral value for s.

Integrating the area of the slices along the height will give you the volume of the cone. Perhaps the questioner meant something like the "lateral area" of a slice; however, that is not a directly meaningful or helpful concept. Integrating the circumference of the slices, as is done above, does not result in the correct value for the area of the lateral surface. The problem is that this gives, as it were, the area as seen by an observer at a straight angle to the cone axis, and not at a straight angle to the surface. Therefore the image undergoes foreshortening, in fact by a factor of h/s.

The easiest way of deriving the formula for the lateral surface is to cut the cone open along a straight line to the top and to flatten it, giving a circle segment. This derivation involves a bit of hand waving, because, strictly speaking, it requires a proof that the area does not change as we flatten the shape.  --Lambiam 14:55, 6 October 2007 (UTC)

Finding the surface area of something with integrals involves arc length:

• $\int _{a}^{b}2\pi f(x){\sqrt {1+f'(x)^{2}}}dx$

We let the function start from the point (0,r) and end at (h,0). The distance between these two points is the slant height s. We are given r=5 and h=12. r/h also happens to be the negative of the slope of the line passing through the two points we gave. So we create a function

• $f(x)=-{\frac {r}{h}}x+r$

This function, when graphed, goes through those two points we made. Now, here's where the fun begins.

Suppose we rotate f(x) about the x-axis. What do we get? a cone. Now let's find the surface area of revolution using the formula above:

• $\int _{0}^{h}2\pi (-{\frac {r}{h}}x+r)({\sqrt {1+(-{\frac {r}{h}})^{2}}})dx$

Let's pull out the constants:

• $2\pi ({\sqrt {1+(-{\frac {r}{h}})^{2}}})\int _{0}^{h}(-{\frac {r}{h}}x+r)dx$

Now, to solve the definite integral:

• $\int _{0}^{h}(-{\frac {r}{h}}x+r)dx=-{\frac {r}{2h}}h^{2}+rh-0$

What do we do now? Well, now we need to multiply this by the constant:

• $2\pi ({\sqrt {1+(-{\frac {r}{h}})^{2}}})(-{\frac {r}{2h}}h^{2}+rh)$

Let's simplify a bit:

• $2\pi ({\sqrt {1+{\frac {r^{2}}{h^{2}}}}})(-{\frac {rh}{2}}+rh)$

How to simplify this? Well, first let's make that radical look nicer. Using the identity $1={\frac {h^{2}}{h^{2}}}$, we can rewrite the inside of that radical as:

• ${\sqrt {\frac {h^{2}+r^{2}}{h^{2}}}}$

Now, ${\frac {1}{h^{2}}}$ is a perfect square. We can remove it from inside the radical, leaving us with:

• ${\frac {2\pi }{h}}({\sqrt {h^{2}+r^{2}}})(-{\frac {rh}{2}}+rh)$

But what is that radical? Well, this is where the Pythagorean Theorem comes in. Remember that a^2 + b^2 = c^2. So take the square root of both sides:

• ${\sqrt {a^{2}+b^{2}}}=c$

Now we let a=h and b=r, now we need to find c. Well, looking up, remember the slant height s? Well, c=s. So let's replace it:

• ${\frac {2\pi s}{h}}(-{\frac {rh}{2}}+rh)$

So let's now simplify the second half, rewriting it:

• $(rh-{\frac {1}{2}}rh)={\frac {rh}{2}}$

So now we have:

• ${\frac {2\pi s}{h}}{\frac {rh}{2}}$

Simplifies to:

• $\pi rs\!$

and there you have it.--Mostargue 19:02, 6 October 2007 (UTC)

The formula used above can be found (with a slightly different notation) in Surface of revolution.  --Lambiam 20:40, 6 October 2007 (UTC)

ʃ 1/x dx

So, the power rule states:

${\frac {d}{dx}}x^{n}=nx^{n-1}.$

This can be rewritten as:

$\int x^{n}dx={\frac {x^{n+1}}{n+1}}+C$ for n ≠ -1

If n = -1, then it becomes ln x. So my question is, how come:

$\lim _{n\to -1}\int x^{n}dx\neq {\frac {x^{n+1}}{n+1}}+C$

also, what is the TEX for the inequal sign --Mostargue 13:24, 6 October 2007 (UTC)

You can use \neq or \not= for the unequal sign. The latter, putting \not before an operator, is generic; for example, \not< produces $\not <$.

That the limit above can't equal the right-hand side should be obvious. The variable n is bound on the left hand side, so that we can equivalently replace the whole expression by:

$\lim _{i\to -1}\int x^{i}dx,$

while on the right-hand side the variable n is free.

However, the following equality holds for positive x:

$\lim _{n\to -1}\int _{1}^{x}z^{n}dz=\lim _{n\to -1}{\frac {x^{n+1}-1}{n+1}}.$

 --Lambiam 14:30, 6 October 2007 (UTC)

Right, that's what I meant. There should be a lim on both sides. So $\lim _{n\to -1}{\frac {{|x|}^{n+1}-1}{n+1}}=\ln {|x|}$?--Mostargue 15:01, 6 October 2007 (UTC)

Both numerator and denominator of that expression tend to zero as n tends to -1.

L'Hôpital's rule says that in this case,

$\lim _{x\to c}{\frac {f(x)}{g(x)}}=\lim _{x\to c}{\frac {f\prime (x)}{g\prime (x)}}$.

In this case, we have;

$f(n)=x^{(n+1)}-1\,$ and $g(n)=n+1\,$

We can rewrite f(n) as;

$f(n)=\exp[(n+1)\ln x]=\exp[(n\ln x+\ln x)]\,$

$f'(n)=\exp[(n+1)\ln x].(\ln x)\,$ - and $f'(-1)=\exp(0).\ln x\,$, i.e. $f'(-1)=\ln x\,$

Also, $g(n)=n+1\,$, so $g'(n)=1\,$

So we have the result;

$\lim _{n\to -1}{\frac {f(n)}{g(n)}}=\lim _{n\to -1}{\frac {f\prime (n)}{g\prime (n)}}=\ln x$ Richard B 15:45, 6 October 2007 (UTC)

The point is, yes, $\lim _{n\to -1}{\frac {{|x|}^{n+1}-1}{n+1}}=\ln {|x|}$. Here's an alternate derivation:

Definition of the exponential function: $e^{x}=\lim _{m\to \infty }\left(1+{\frac {x}{m}}\right)^{m}$

Corresponding definition of the natural logarithm: $\ln x=\lim _{m\to \infty }m({\sqrt[{m}]{x}}-1)$.

Substitute $p=1/m\!\,$ to get $\ln x=\lim _{p\to 0}{\frac {x^{p}-1}{p}}$.

Substitute $n=p-1\!\,$ to get $\ln x=\lim _{n\to -1}{\frac {x^{n+1}-1}{n+1}}$.

Jim 18:05, 6 October 2007 (UTC)

While one may choose to define the natural logarithm this way, it is a definition I've never seen before, and I think one should actually first prove that it is equivalent to a more common definition before using it.  --Lambiam 23:36, 6 October 2007 (UTC)

It comes from a fact that $\lim _{h\to 0}{\frac {a^{h}-1}{h}}=\ln a,$ which comes from $(a^{x})'=a^{x}\ln a$, as well as the definition of the derivative. (Igny 23:54, 7 October 2007 (UTC))

If you use that, you can skip most of the alternative derivation:

$\lim _{n\to -1}{\frac {x^{n+1}-1}{n+1}}=\left.{\frac {d}{dn}}x^{n+1}\right|_{n=-1}=\left.x^{n+1}\ln x\right|_{n=-1}=\ln x.$

 --Lambiam 00:53, 8 October 2007 (UTC)

October 7

French Franc 1965

Shall appreciate if anyone can tell me today's value in Euros of 15,000 1965 French Francs. Info needed to illustrate history of a small village.86.197.151.8 08:52, 7 October 2007 (UTC)petitmichel

I have moved this question to the Humanities section, here.  --Lambiam 12:57, 7 October 2007 (UTC)

Thanks. Just for interest the equivalent sum in 2006 (old francs) would have been approx. 115,500 francs! Appreciation of 100,000! Whow! (Yes, this is a crude figure, but sufficient for my needs.86.200.4.62 14:33, 7 October 2007 (UTC)petitmichel

Proof question

Hello. I just did a question that I would like someone to check for me. It would be much appreciated.

The positive integers can be split into five distinct arithmetic progressions, as shown:

A: 1, 6, 11, 16
B: 2, 7, 12, 17
C: 3, 8, 13, 18
D: 4, 9, 14, 19
E: 5, 10, 15, 20

a) Write down an expression for the value of the general term in each progression.
b) Hence prove that the sum of any term in B and any term in C is a term in E.
c) Prove that the square of every term in B is a term in D.
d) State and prove a similar claim about the square of every term in C.

a)

$A=5n-4$

$B=5n-3$

$C=5n-2$

$D=5n-1$

$E=5n$

b)

${\begin{aligned}B+C&=(5m-3)+(5m-2)\\&=(5m+5m)+(-2-3)\\&=10m-5\\&=5(2m-1)\\\end{aligned}}$

Since this number is multiple of 5, it is a term in E.

c)

${\begin{aligned}B^{2}&=(5n-3)^{2}\\&=25n^{2}-30n+9\\\end{aligned}}$

${\begin{aligned}25n^{2}&\equiv 0{\text{ (mod }}5)\\-30n&\equiv 0{\text{ (mod }}5)\\9&\equiv 4{\text{ (mod }}5)\\\implies 25n^{2}-30n+9&\equiv 4{\text{ (mod }}5)\\\end{aligned}}$

${\begin{aligned}D&=5n-1\\5n&\equiv 0{\text{ (mod }}5)\\-1&\equiv 4{\text{ (mod }}5)\\\implies 5n-1&\equiv 4{\text{ (mod }}5)\\\end{aligned}}$

Therefore since both are congruent to 4 (mod 5) it is true.

d)

${\begin{aligned}C^{2}&=(5n-2)^{2}\\&=25n^{2}-20n+4\\\end{aligned}}$

${\begin{aligned}25n^{2}&\equiv 0{\text{ (mod }}5)\\-20n&\equiv 0{\text{ (mod }}5)\\4&\equiv 4{\text{ (mod }}5)\\\implies 25n^{2}-20n+4&\equiv 4{\text{ (mod }}5)\\\end{aligned}}$

So, as proved above, the square of any term in C is also a term in D. —Preceding unsigned comment added by 172.200.78.35 (talk) 15:08, 7 October 2007 (UTC)

a) and b) are good. b) would be even better if you remark that $2m-1$ must be a positive integer (for example, $5(-1)=-5$ is not in E, and you need to mention that this doesn't happen here).

In c) and d), you have the right idea, but the way you have written it doesn't really prove the needed result. You say "every square of an element of B is congruent to 4 modulo 5, and every element of D is congruent to 4 modulo 5, hence every square of an element of B is is an element of D", which is a fallacy (for example, every man breathes, and every woman breathes, but not every man is a woman). What you should have shown is the converse, that every number congruent to 4 modulo 5 is an element of D. However, I think this just complicates matters; it is easier to just say that $(5n-3)^{2}=5(5n^{2}-6n+2)-1$ and show that $5n^{2}-6n+2$ is a positive integer, and hence this is in D. The same goes for part d). -- Meni Rosenfeld (talk) 15:25, 7 October 2007 (UTC)

Thank you Meni, I'll take that on board. 172.200.78.35 15:33, 7 October 2007 (UTC)

Just an after thought, would my problem with b) and your solution to c) and d) have been solved if I said n>o? 172.200.78.35 15:38, 7 October 2007 (UTC)

First, let me correct myself; there is a relatively serious error in your solution to b). You have an element of B and an element of C; You know that the former is some multiple of 5 minus 3, and and latter is some multiple of 5 minus 2. But it doesn't have to be the same multiple. So your numbers are not $5m-3$ and $5m-2$ but rather $5n-3$ and $5m-2$, where n and m are positive integers which may or may not be the same.

Now, stating that n (and m) is positive is just the start. From a) we learn that:

• If n is a positive integer, then $5n-4$ is in A, and so on;
• If some number is in A, then there is some positive integer $n$ such that the number is equal to $5n-4$, and so on.

So a correct solution to b) would be: Let x be in B and y be in C. Then there are some positive integers n and m such that $x=5n-3$ and $y=5m-2$. So $x+y=5(m+n-1)$, and $m+n-1$ is a positive integer. Therefore, $x+y$ is in E. -- Meni Rosenfeld (talk) 16:15, 7 October 2007 (UTC)

October 8

Mathematical Induction

How would one show that 7^k-4^k-3^k is divisible by 12 by math induction?

I know how to do all the pre-requisite stuff... but I am stuck at the point where I have to show that Pk+1 is true (k+1 should be on a lower level).

So now, I sub all the k's with (k+1)'s. What is the next step I should do?

Thanks. Acceptable 02:52, 8 October 2007 (UTC)

If you're trying to prove f(x) = g(x), at that stage you'd be trying to prove that the f(k+1) case is true assuming that case f(k) is true, so, by your assumption:

f(k) = g(k)

So now you're trying to prove:

f(k+1) = g(k+1)

So what you can do now, is try to group terms of f(k+1) into something that is equal to f(k), at which point you can substitute g(k) for f(k), since you've already assumed that. After doing this, it should be easier to prove that whatever's left over is true.

Take a look at the example at Mathematical induction. Ask again if you still have trouble, maybe show your work. (Help:Formula might be useful.) - Rainwarrior 03:19, 8 October 2007 (UTC)

Wouldn't it be much easier to prove that 7^k-4^k-3^k is divisible by 12 for all integers k through modular arithmetic?? —Preceding unsigned comment added by 69.54.140.201 (talk) 06:08, 8 October 2007 (UTC)

That would be easier, but presumably this is an exercise in which the student is required to use induction. In this case it helps to prove something formally stronger (by induction):

for k ≥ 1, 7^k − 4^k − 3^k is divisible by 12 AND 3·4^k + 4·3^k is divisible by 12.

Then, in the induction step, in proving that 7^k+1 − 4^k+1 − 3^k+1 is divisible by 12, you can also use the assumption (part of the induction hypothesis) that 3·4^k + 4·3^k is divisible by 12.  --Lambiam 06:16, 8 October 2007 (UTC)

On second thoughts, taking "3·4^k + 4·3^k is divisible by 12" along in the induction proof is quite unnecessary, since this can be easily proved (for k ≥ 1) in an entirely straightforward algebraic way.  --Lambiam 19:34, 8 October 2007 (UTC)

ASSISTANCE

I THINK THAT I HAVE COME UP WITH ANEW THEORY ABOUT REAL NUMBERS.I NEED SOME HELP TO ASSESS AND FIX THE MATHEMATICAL CONSTRUCTOR OF THIS THEORY.I WENT TO SOME SCHOOLS OF MATH.WHERE I LIVE BUT I DIDNOT GET ANY UNDERSTANDING.MAY I GET ANY HELP HERE?80.123.226.186 12:06, 8 October 2007 (UTC)h.dyeni

I don't know about assistance, but all caps is the equivalent of shouting in these forums, normal English syntax is preferred.—Cronholm¹⁴⁴ 12:26, 8 October 2007 (UTC)

That said, if you describe your theory we can express our opinion of it. -- Meni Rosenfeld (talk) 12:54, 8 October 2007 (UTC)

Your name is not JSH by any chance is it? 211.28.129.8 14:30, 8 October 2007 (UTC)

I don't believe JSH would have said "I think that I have ...". PrimeHunter 22:43, 8 October 2007 (UTC)

I feel like I'm missing something important here... care to spill the beans? Algebraist 23:11, 8 October 2007 (UTC)

From what I can gather fromm google searches, he looks to be a notorious character in Google Groups, pushing "out there" theories. I could be wrong though. --Ybbor^Talk 23:25, 8 October 2007 (UTC)

Right. James S. Harris aka JSH is probably the most notorious crank in the usenet group sci.math (which Google Groups has an interface to). He often claims to have made great discoveries and usually expresses no doubt about his own claims, despite his less than impressive track record. PrimeHunter 00:18, 9 October 2007 (UTC)

And JSH is now a disambig, in case this comes up again. Algebraist 13:30, 9 October 2007 (UTC)

I would be interested in helping, please do show us what you have so far. —Preceding unsigned comment added by 69.54.140.201 (talk) 23:37, 10 October 2007 (UTC)

Variance problem!

i'd like to ask you about the variance..

why cant we sum variance's normally?? without using the covariance??

thax.. —Preceding unsigned comment added by 86.108.109.227 (talk) 16:45, 8 October 2007 (UTC)

Perhaps our article Variance will help. A better question would be "why should we be able to sum variances normally?". There is just no reason why this should be possible.

Also, consider the following 3 scenarios. In all of them, X is a variable with variance 1. Y can be either equal to $X$, equal to $-X$ or an independent variable which also has variance 1. In the first case, $X+Y$ is equal to $2X$, so its variance is 4 (since in calculating variance you take squares). In the second case, $X+Y$ is the constant 0 so its variance is 0. And in the third case you do get the variance to be equal to 2, the sum of variances. -- Meni Rosenfeld (talk) 17:30, 8 October 2007 (UTC)

I fear you have your cases in the wrong order there. Algebraist 19:00, 8 October 2007 (UTC)

Sorry, thanks. Fixed. -- Meni Rosenfeld (talk) 21:33, 8 October 2007 (UTC)

This can be derived as a simple consequence of the definitions of variance and covariance, together with the property of the expected value operator that it distributes over addition. To simplify things, assume we have two random variables X and Y, both with expected value 0 – possibly by taking some other random variables minus their respective expected values, which does not change their variances or covariance. Then

Var(X + Y) = E((X + Y)²) = E(X² + Y² + 2XY) = E(X²) + E(Y²) + 2E(XY) = Var(X) + Var(Y) + 2Cov(X, Y).

The covariance term vanishes precisely when X and Y are statistically independent.  --Lambiam 19:59, 8 October 2007 (UTC)

The covariance term vanishes with independence, yes, but surely 'precisely' is wrong there: events can be uncorrelated without being independent. Algebraist 23:11, 8 October 2007 (UTC)

I stand corrected.  --Lambiam 05:23, 9 October 2007 (UTC)

October 9

Three calculus questions

1. I was asked to differentiate $y=x+cot(x)$, so I proceeded as follows: $y'=1+(d/dx)[1/tan(x)]<br>=1+(1/sec^{2}(x))<br>=1+cos^{2}(x)$, but I was told that the answer is $-cot^{2}(x)$. Is there a trigonometric identity I should know but don't? 2. I was asked to differentiate $h(s)=(1/s)-10csc(s)$, so I came up with $h'(s)=-(1/s^{2})-(10/cos(s))$, but I hear it's really $-(1/s^{2})+10csc(s)cot(s)$. Does this hinge on yet another trig identity unknown to me?

3. I was asked to differentiate $(x(x^{2}-1))/x+3$, and I got the answer $(-x^{3}+2x^{2}+7x)/(x+3)^{2}$, but someone told me it's $(2x^{3}+9x^{2}-3)/(x+3)^{2}$. Could someone please tell me if either of us is right, and why?

All help is appreciated. Thank you very much, anon. —Preceding unsigned comment added by 70.23.86.189 (talk) 02:12, 9 October 2007 (UTC)

In the first two, you seem to have made the following mistake:

${\frac {d}{dx}}{\frac {1}{g(x)}}={\frac {1}{{\frac {d}{dx}}g(x)}}$ (with g(x) = tan(x) in the first one, and sin(x) in the second) where what you should be considering is ${\frac {d}{dx}}{\frac {1}{g(x)}}={\frac {d}{dx}}\left[g(x)\right]^{-1}={\frac {d}{dx}}f\left(g(x)\right)$ where f(x) = 1/x, and using the chain rule. In the third, my very quick differentiation gave me an answer different to both the ones you offerred, but closer to the second. The same chain rule trick applies here (or you can use the quotient rule, which is equivalent to the chain and product rules used together), and just be very careful with your algebra. Confusing Manifestation 03:48, 9 October 2007 (UTC)

Actually, for question 3, the second answer offered is correct. In all cases, I think it is easiest to use the quotient rule ${\frac {d}{dx}}\left({\frac {f(x)}{g(x)}}\right)={\frac {f'(x)g(x)-f(x)g'(x)}{g(x)^{2}}}$. For the third question, it will help to write the function as ${\frac {x^{3}-x}{x+3}}$. -- Meni Rosenfeld (talk) 06:30, 9 October 2007 (UTC)

In regards to the first two there are trigonometric identities that you 'should' know. It would be better if you didn't convert to sine and cosine before you differentiated. would probably make your answer look the way it's supposed to since, aside from a missing - sign or two you have the correct answers. The rules for deriving the other four trigonometric functions should be in your text. —Preceding unsigned comment added by 69.54.140.201 (talk) 09:25, 9 October 2007 (UTC)

Actually, his answer for the first question is dead wrong. Don't mislead the OP. –King Bee ^{(τ • γ)} 13:38, 9 October 2007 (UTC)

Strong topology on distribution space

Simple question: Is the strong topology on distribution space the same as the compact open topology? I think so but I am a little out of touch and hence need confirmation. Thank you in advance. twma 03:51, 9 October 2007 (UTC)

The locally convex inductive topology on a test space is not the same as the compact open topology. I am very sure about this. twma 03:40, 10 October 2007 (UTC)

CAS programming on PC

Hi,
Does anyone know a program I can use to create/edit Ti-89 files (.89p) on my computer? Massive thanks for anyone that can tell me!!! --Fir0002 07:59, 9 October 2007 (UTC)

TI-89 series#Programming mentions some, including TIs official software development kit available here. Algebraist 14:21, 9 October 2007 (UTC)

y= form of the general Quadratic polynomial in 2 variables

I figured out how to graph Conic Sections on graphing utility in Cartesian mode. This is how it's done.

$Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0$

Now since we are solving for y

$Cy^{2}+(Bx+E)y+(Ax^{2}+Dx+F)=0$

Note that since we are solving y as two equations of x, we can treat x like a constant in a one variable Quadratic. This means we can apply the Quadratic formula to get our equation. If your not convince try completing the square and isolating y just you would in solving a one-variable quadratic. Be sure to completely simplify.

$y={\frac {-Bx-E\pm {\sqrt {(Bx+E)^{2}-4C(Ax^{2}+Dx+F)}}}{2C}}$

Simplify

$y={\frac {-(Bx+E)\pm {\sqrt {B^{2}x^{2}+2BEx+E^{2}-4ACx^{2}-4CDx-4CF}}}{2C}}$

$y={\frac {-(Bx+E)\pm {\sqrt {(B^{2}-4AC)x^{2}+(2BE-4CD)x+(E^{2}-4CF)}}}{2C}}$

The $\pm$ is why there are two equations, graphically it is obvious why there must be at least two. Are any of you familiar with this? —Preceding unsigned comment added by 69.54.140.201 (talk) 10:04, 9 October 2007 (UTC)

Yes I'm quite familiar with doing things exactly like that.87.102.87.171 13:12, 9 October 2007 (UTC)

October 10

Supervising kids in groups of 8

class of 32 kids, 1 adult to supervise each group of 8. write algebraic expression to find out how many adults were needed.

well notice that you have 3 numbers, and note that there are 8 students, per adult thus 8x=32, x being the number of adults, needed to supervise. then divide both sides by 8 so that you get x=32/8 and that simplifies (reduces) to x=4

If 'a group' means 'a subset' then there are ${\tbinom {32}{8}}=10518300$ 'groups', so you need over ten million adults. Each kid belongs to ${\tbinom {32-1}{8-1}}=2629575$ groups, so every kid will be supervised by over two-and-half million adults. CiaPan 07:14, 10 October 2007 (UTC)

Possibly the question is about the following generalization (and abstraction): A set of n elements is partitioned into parts, each of which has a size not exceeding k. What is the least number of parts for which this is possible?

Calling the number of parts p, we have n ≤ pk, and the question can be rephrased as: give a closed expression P(n, k) such that the following two statements are equivalent:

n ≤ pk,

P(n, k) ≤ p.

This is solved, of course, by defining P(n, k) = n/k, but presumably we want to remain in the domain of integers. Then the answer is found in the ceiling function:

P(n, k) = ⌈n/k⌉.

This is a Galois connection. For the original question, we find P(32, 8) = ⌈32/8⌉ = ⌈4⌉ = 4. If we increase the class size by one but one adult can still supervise at most 8 kids, we have P(33, 8) = ⌈33/8⌉ = ⌈4.125⌉ = 5.  --Lambiam 13:36, 10 October 2007 (UTC)

Oh God, while this is hilarious, you realize that you've just made some poor high school freshman wet himself with fear of mathematics. OP: the first response is the one that you care about. Donald Hosek 18:11, 10 October 2007 (UTC)

LOL, it actually looks harder than it really is, hope that calms the OP down LOL A math-wiki 05:36, 11 October 2007 (UTC)

What do you call this mathematical concept? Weightage?

If I score 90/100 for my first exam and 70/100 for my second exam, my mean score should be 80/100. However, because my second exam is given a weightage of 75% and my first exam a weightage of 25%, my mean mark is 75/100.

When each value has a weightage in %, to get the mean of all the values, you have to multiply each value by its weightage before taking the mean of all the values.

That would be a weighted mean. Confusing Manifestation 05:10, 10 October 2007 (UTC)

Yes, a weighted mean ... often referred to as a "weighted" average. (Joseph A. Spadaro 22:10, 10 October 2007 (UTC))

Rethinking Infinity

I recently read up on Set Theory, Ordinals, and Cardinal Numbers, and I found the arguments involving different infinite sets to be illogical.

I think there is a much simpler answer, but first I need to point some facts about The Concept of Infinity. First of all, Infinity cannot be treated like a number or a variable, you can't calculate $\infty +1$. You might try and argue that since 1 gets arbitrary as $x\to \infty$, that $\infty +1=\infty$, but now you've inadvertantly violated the most fundamental principle in mathematics, equality, $x\neq x+1$. Instead you should write $\infty \equiv \infty +1$, because the concepts you implied by $\infty$ and $\infty +1$ are the same. The point infinity doesn't have numeric or algebraic value but it does have meaning. The basic idea is that it is equivalent conceptually, but not actually equal numerically.

What does this mean for infinite sets?

It means that the idea of compairing the 'size' of infinite sets is nonsensical. A much better way of handling the problem Cantor tried to address is as follows. Consider the sets $\mathbb {N}$ and $\mathbb {Q}$, the natural numbers, and the rational numbers. The problem as cantor saw it was that $\mathbb {N} \subset \mathbb {Q}$, but yet even $\mathbb {N}$ has an infinite number of elements. There really isn't a problem here because infinity has only conceptual value. A subset is a concept as well, it doesn't have any numeric value, not even an arbitrary (or variable) value, so there is no logical falacy. It is logical to look at the behavior of different infinite sets, and to use conceptual arguments and notation to categorize them, but any attempt to appliy numeric values to (such as aleph null, aleph one, etc.) is illogical. A good case to illustrate where my thinking diverges from Cantor's comes from the work I did above, consider another set $\mathbb {K}$ that is the set also a subset of $\mathbb {Q}$ and is defined by ${\mathbb {Q} :X\notin \mathbb {N} }$. Then we can logically conclude, $\mathbb {N} \subset \mathbb {Q}$, $\mathbb {K} \subset \mathbb {Q}$, $\mathbb {N} \not \cup \mathbb {K}$, and $\mathbb {N} +\mathbb {K} \equiv \mathbb {Q}$. These are also concepts (hence the equivalence). I would like to point out the infinite limits and infinite integrals specifically those that have finite outputs should also use and equivance sign, rather than an equals sign.

That is more or less like saying "Complex numbers are nonsensical, since they violate the most fundamental principle in mathematics, inequality, $x^{2}\geq 0$". First, While I agree that equality is a fundamental concept in mathematics, the existence of an x such that $x=x+1$ says more about addition (and the structure over which it is defined) than about equality. Second, just because for every real number we have $x\neq x+1$, doesn't mean that any mathematical structure that humanity will ever consider must satisfy this as well. The field of real numbers is just a drop in the mathematical ocean. We have plentiful structures where this does not hold, such as the real projective line, extended real number line, and of course, cardinal numbers. In the ordinal numbers, $x\neq x+1$ does hold but $x\neq 1+x$ does not.

In short, dismissing an entire and fundamental theory, only because it exhibits behavior somewhat different from our everyday finite numbers, is just silly. In fact, it would be quite surprising if infinite quantities behaved in exactly the same way as finite ones. -- Meni Rosenfeld (talk) 11:34, 10 October 2007 (UTC)

It would be more productive not to get hung up on concepts of numerical equivalence of infintities here I think.. eg If a set contains infinite subsets each with infinite members - then rather than attempting to 'count them' I would recommend using a method such as venn diagrams to get a simple understanding of how one infinite set can be 'bigger' than others..

Maybe someone else could recommend other methods of visualisation that simplify an overview of sets and set theory.87.102.79.56 15:42, 10 October 2007 (UTC)

Venn diagrams have little value (detrimental value, even) for understanding infinite cardinalities. -- Meni Rosenfeld (talk) 19:42, 10 October 2007 (UTC)

Nice counterarguments, I should mention that what is really confusing me is how cardinal number can compair the size of say $\mathbb {N}$ and $\mathbb {K}$ in my example scenario. Seeing as they have no elements in common how can cardinal numbers justify that $\mathbb {K} >\mathbb {N}$ —Preceding unsigned comment added by 69.54.140.201 (talk) 19:24, 10 October 2007 (UTC)

Actually, your set $\mathbb {K}$ above actually has the same size as $\mathbb {N}$, since $\mathbb {Q}$ is countable. The situation would be different if you took $\mathbb {R}$ instead. Regardless, I don't see any problem - does the fact that the sets $\{1,2,3\}$ and $\{4,5\}$ have no elements in common creates any difficulty in showing that the former is larger? You know how cardinals work (using bijections), right?

It is good that you have read up on set theory, but I can only suggest that you read up some more... -- Meni Rosenfeld (talk) 19:42, 10 October 2007 (UTC)

Uh, no and I'm guessing this is why cardinals seem illogical to me. It's hard to learn math from an encyclopedia LOL. —Preceding unsigned comment added by 69.54.140.201 (talk) 20:05, 10 October 2007 (UTC)

So I'll just say that $\mathbb {R}$ is larger than $\mathbb {N}$ because there is an injection from $\mathbb {N}$ to $\mathbb {R}$ but no injection from $\mathbb {R}$ to $\mathbb {N}$. Finding a good book will be helpful, though - I don't have one to recommend, but hopefully someone else here will. -- Meni Rosenfeld (talk) 20:36, 10 October 2007 (UTC)

Cantor's diagonal argument may or may not help you to understand this, just in case you haven't found it already. 130.88.47.27 13:00, 11 October 2007 (UTC)

Already found that (I have created an account fyi) A math-wiki 07:10, 12 October 2007 (UTC)

Points on a circle

If you have n points evenly distributed on a circle of radius r, what is the average distance between one point and all the other points, i.e. the sum of the distance to from one point to each point in turn, divided by the number of other points. Thanks!

So far I got;

$D_{avg}={\frac {{\sqrt {2}}\sum _{k=1}^{n}(r{\sqrt {1-\cos({\frac {2k\pi }{n}})}})}{n}}$

I think that works, but I really need something in the form of a function without summation series in it. Thank you. ΦΙΛ Κ 10:50, 10 October 2007 (UTC)

If you have points P and Q on a circle of radius r, centre O, and the angle POQ at the centre of the circle is 2kπ/n then I think the length of PQ is 2rsin(kπ/n) - so I don't see where the square root comes from in your sum. Also, if there are n points altogether, you should be summing n-1 terms and dividing by n-1 to find the average - otherwise when n=2 you get the answer r when you should expect to get 2r. Gandalf61 11:22, 10 October 2007 (UTC)

Actually, the expression for the distance is correct, it can be written as either $r{\sqrt {2(1-\cos \theta )}}$ (direct use of the cosine theorem) or $2\sin(\theta /2)r$ (after simplifcation by trigonometric identities). Now, you may or may not include all n terms, depending on whether you want the distance from a point to itself to count. Whatever the case, I doubt the average can be given in closed form (though I'll try to look into it). Will an approximation help? -- Meni Rosenfeld (talk) 11:42, 10 October 2007 (UTC)

A good approximation (including the zero distance) is $r\left({\frac {4}{\pi }}-{\frac {\pi }{3n^{2}}}\right)$, and I suppose this can be improved with higher negative powers of n. -- Meni Rosenfeld (talk) 12:30, 10 October 2007 (UTC)

(OT: This was found empirically, with help from Plouffe's inverter; I was disappointed by its failure to find 3.819718634 being equal to $12/\pi$.) -- Meni Rosenfeld (talk) 12:44, 10 October 2007 (UTC)

Your expression can be simplified. As remarked by Meni,

$1-\cos {\tfrac {2k\pi }{n}}=2\sin ^{2}{\tfrac {k\pi }{n}},$.

so

${\sqrt {1-\cos {\tfrac {2k\pi }{n}}}}={\sqrt {2}}\left|\sin {\tfrac {k\pi }{n}}\right|.$

If 0 ≤ k ≤ n, the value of the sine is nonnegative, so then we can omit taking the absolute value, and the summation becomes:

$\sum _{k=1}^{n}r{\sqrt {1-\cos {\tfrac {2k\pi }{n}}}}=\sum _{k=0}^{n-1}r{\sqrt {2}}\sin {\tfrac {k\pi }{n}}=r{\sqrt {2}}\sum _{k=0}^{n-1}\sin {\tfrac {k\pi }{n}}.$

(Using the fact that the summand vanishes for both k = 0 and k = n, I've shifted the range of summation.) Since sin x = Im(e^ix), the last summation equals the imaginary part of

$\sum _{k=0}^{n-1}e^{{\frac {k}{n}}i\pi }={\frac {2}{1-e^{\frac {i\pi }{n}}}}=1+i{\frac {\sin {\tfrac {\pi }{n}}}{1-\cos {\tfrac {\pi }{n}}}}=1+i\cot {\tfrac {\pi }{2n}}.$

Combining all this gives us that the sum of the distances between one point and the other points is equal to:

$2r\cot {\tfrac {\pi }{2n}}.$

I leave it to you whether you want to divide this by n or by n−1.  --Lambiam 14:42, 10 October 2007 (UTC)

I can't believe I haven't thought of that. I am so embarassed. I will give up on mathematics and start a floor-sweeping career. -- Meni Rosenfeld (talk) 19:46, 10 October 2007 (UTC)

I was more interested in the case where n → ∞. In that case we shift to the integral ${\frac {r}{\pi {\sqrt {2}}}}\int _{0}^{2\pi }{\sqrt {1+\cos \theta }}d\theta$ (the idea being that the average value of f(x) over [a,b] is ${\frac {\int _{a}^{b}f(x)dx}{b-a}}$. From there, it turns into a Calc II problem.

When using the limit from Lambiam's formula, remember that you're talking about ${\frac {2r\cot {\tfrac {\pi }{2n}}}{n}}$ where L'Hôpital's rule applies. You should get the same result either way, if not one of us made a mistake. Donald Hosek 20:32, 10 October 2007 (UTC)

Multivariable calc

So my friend had a multivariable calculus quiz recently, and one of the question was as follows:

“ Let $f(x)$ and $g(x)$ be differentiable functions of one variable. Find $\partial z \over \partial x$ if $f(yz)=g(xz)$. ”

Apparently her TA has no office hours and never goes over anything, so she's not expecting any help from there. I'm not entirely sure how to do this problem myself either. I can get as far as $f'(yz)\cdot y{\partial z \over \partial x}$ on the left, but that's about all (and to be honest, I'm not 100% sure I'm right, either). Perhaps someone could explain how to finish this problem? --M1ss1ontomars2k4 18:50, 10 October 2007 (UTC)

The question makes no sense as presented, since y appears in exactly one place and is not quantified. I suspect it should have been $f(yz)=g(xy)$ (for every y and x, where z is a function of x). In this case, you can take $y={\frac {1}{x}}$ and get $f\left({\frac {z}{x}}\right)=g(1)$, a constant. So, under some smoothness assumptions, as well as the assumption that f is not constant, we have $z=\alpha x$ for some $\alpha$. So we have $g(x)=f(\alpha x)$, so ${\frac {dz}{dx}}=\alpha ={\frac {g'(0)}{f'(0)}}$. -- Meni Rosenfeld (talk) 19:34, 10 October 2007 (UTC)

I'm not entirely sure what was meant by that y either. But my friend was very sure the quiz said $g(zx)$; I asked her multiple times to make sure. I guess you'd have to assume z is a function of x and y since the question asks for the partial derivative of z with respect to x. Either that or her TA is just nuts. Actually, from what she's told me, it's probably both. --M1ss1ontomars2k4 19:41, 10 October 2007 (UTC)

We may or may not need to assume that z is a function of both x and y, but that certainly doesn't fix the problem. The y issue has nothing to do with calculus, multivariate or otherwise. It has to do with much deeper aspects of the role variables play in mathematical statements. A variable must have some sort of quantification, be it either as a dummy ("for all y", "there exists a y") or as an item about which the statement contains information. But this question is not about y and y does not appear anywhere else in it, and any quantification would lead the given $f(yz)=g(xz)$ to be trivial in content (admittedly, this last observation does have something to do with calculus). -- Meni Rosenfeld (talk) 19:53, 10 October 2007 (UTC)

I had to think about this one for a awhile, I never actually studied this, but I did do stuff like this for fun while I was in AP Calculus and I'm pretty sure I know what you need to do. I believe your answer so far is correct. I don't know how to do ${\partial z \over \partial x}$ to $g$ though, forgot. I would guess that once you differentiate both, you can collect all the terms with ${\partial z \over \partial x}$ and then factor that out and divide what's left from that side to get ${\partial z \over \partial x}=$ —Preceding unsigned comment added by 69.54.140.201 (talk) 19:57, 10 October 2007 (UTC)

Ok I remember how to do g now, and I noticed your missing the other half of the product rule on f'

you have ${\partial z \over \partial x}=f'(yz)y{\partial z \over \partial x}$ but it should read

${\partial z \over \partial x}f(yz)=f'(yz)y{\partial z \over \partial x}+f'(yz)z{\frac {dy}{dx}}$

${\partial z \over \partial x}g(xz)=g'(xz){\frac {d}{dx}}[xz]$

From the chain rule. Then use the product rule on

${\frac {d}{dx}}[xz]$

Which gives

$x{\partial z \over \partial x}+z$

So you have

${\partial z \over \partial x}g(xz)=g'(xz)x{\partial z \over \partial x}+g'(xz)z$

Now remember that $f(yz)=g(xz)$, well this implies that $f'(yz)=g'(xz)$ thus

$f'(yz)y{\partial z \over \partial x}+f'(yz)z{\frac {dy}{dx}}=g'(xz)x{\partial z \over \partial x}+g'(xz)z$

and now we isolate ${\partial z \over \partial x}$

${\partial z \over \partial x}={\frac {g'(xz)z+f'(yz)z{\frac {dy}{dx}}}{f'(yz)y-g'(xz)x}}$

You have some things mixed up here. But that's irrelevant, you can't give a good answer to a bad question. -- Meni Rosenfeld (talk) 20:32, 10 October 2007 (UTC)

I think that fixes the mistakes, but more importantly we should consider that the TA in question probably didn't bother to think about the formal implications his lack of variable definitions gives us. He was probably testing whether his students, knew how to do it, not do they understand that this is technically unsound. So as long as dy/dx exists then my answer works, assuming I fixed all the mistakes too. —Preceding unsigned comment added by 69.54.140.201 (talk) 20:42, 10 October 2007 (UTC)

No, you did not fix all the mistakes, and the question cannot be solved, any more than the question "you are given that $a=3$. Find $b^{2}$" can be solved. -- Meni Rosenfeld (talk) 20:51, 10 October 2007 (UTC)

On second thought, if we do assume that y is meant to be a function of x, and that the solution should be given in terms of y, then there might be a correct solution - but I find this an extreme stretch of the question. -- Meni Rosenfeld (talk) 20:58, 10 October 2007 (UTC)

It would make more sense that z is a function of (x,y), and I'm curious what mistakes I've missed, need a good multivar Calculus review, been a year and a half since I did much Calculus. —Preceding unsigned comment added by 69.54.140.201 (talk) 21:11, 10 October 2007 (UTC)

We don't know much about f and g, but let us try some simple example and see where that leads us. Take the functions defined by

f(x) = 0 and g(x) = 0 (for all x).

Using these example function, the question reduces to:

Find ∂z/∂x if 0 = 0.

Something tells me that if we can't solve the problem for this simple example, we may have a problem solving it in general.  --Lambiam 21:15, 10 October 2007 (UTC)

[edit conflict, responding to anon] You can start with the treatment of ${\partial z \over \partial x}g(xz)$ as the derivative of $g(xz)$ with respect to to x, when it is actually the derivative of z with respect to x, multiplied by $g(xz)$.

I do take some of my earlier comments back, though. The question just might, after all, make some sense if we treat z as a function of x and y. I'll think about this some more. Note that in your solution, you do treat y as a function of x. -- Meni Rosenfeld (talk) 21:20, 10 October 2007 (UTC)

Some progress might be possible if we assume that f and g are nonconstant. -- Meni Rosenfeld (talk) 21:23, 10 October 2007 (UTC)

Looking at what I did more closely, I actually derivated both functions with respect to x. So my notation is wrong, not my thinking. Em I right??? Furthermore, if I recall correctly ${d \over dx}xy$ is worked out by the product rule. So i should remove the z on those two, which I have now done. That better? —Preceding unsigned comment added by 69.54.140.201 (talk) 21:41, 10 October 2007 (UTC)

Yes, see below. The improper notation makes it difficult for me to follow your derivation, but your result is almost correct (almost, because there is no ${\frac {dy}{dx}}$) so the thinking is probably right. -- Meni Rosenfeld (talk) 21:48, 10 October 2007 (UTC)

Okay, here goes. We are assuming that z is a function of x and y. Denote $p(x,y)=f(yz(x,y))$, $q(x,y)=g(xz(x,y))$. We are given $f(yz)=g(xz)$ so $p(x,y)=q(x,y)$ for every x and y, so their derivatives with respect to x are also equal:

$f'(yz)y{\frac {\partial z}{\partial x}}={\frac {\partial }{\partial x}}p(x,y)={\frac {\partial }{\partial x}}q(x,y)=g'(xz)\left(z+x{\frac {\partial z}{\partial x}}\right)$

${\frac {\partial z}{\partial x}}(f'(yz)y-xg'(xz))=g'(xz)z$

So wherever $f'(yz)y\neq xg'(xz)$, we have:

${\frac {\partial z}{\partial x}}={\frac {g'(xz)z}{f'(yz)y-xg'(xz)}}$

So other than the notation, anon got most of it right. Was that the intended solution? Only the TA knows. I do give this question the "most poorly worded question of the year" award. -- Meni Rosenfeld (talk) 21:38, 10 October 2007 (UTC)

If anyone is wondering why this question turn out to be solvable despite my "deeper aspects of the role variables play in mathematical statements"... That is because I did not account for notational shortcuts. Keeping in mind that z is shorthand for $z(x,y)$, y actually appears in the question 4 times (under the quantification "for all"). -- Meni Rosenfeld (talk) 21:44, 10 October 2007 (UTC)

Quick question, how is ${\partial \over \partial x}f(yz)\not =f'(yz)y{\partial z \over \partial x}+f'(yz)z{\frac {dy}{dx}}$ —Preceding unsigned comment added by 69.54.140.201 (talk) 21:48, 10 October 2007 (UTC)

Because when taking a partial derivative with respect to x, you treat y as a constant. So instead of using the product rule, you should use the $(cf)'=cf'$ rule. The expression you give is also correct if we agree that ${\frac {dy}{dx}}=0$. -- Meni Rosenfeld (talk) 21:55, 10 October 2007 (UTC)

See, ... I knew I was forgetting something LOL, ya I agree and thus the term goes to 0 and we don't bother writing it. Thanks

It should technically be ${\partial y \over \partial x}$ not ${dy \over dx}$ correct??? —Preceding unsigned comment added by 69.54.140.201 (talk) 23:29, 10 October 2007 (UTC)

It shouldn't be anything, really. If by y we mean then one-variable function which for every value of the argument returns the constant y, then it should be ${\frac {dy}{dx}}$. If by y we mean the function of two variables which returns the second argument, then its partial derivative with respect to the first variable will be ${\frac {\partial y}{\partial x}}$. But neither of these are applicable in our context. -- Meni Rosenfeld (talk) 23:36, 10 October 2007 (UTC)

I would have to disagree, I would say it would be ${\partial y \over \partial x}$ because we began by doing partial differentiation, and because y may equal a function of two variables but could be considered, F(X,Y)=y, where y is a constant. Thus the partial derivative is 0. The regular derivative would also yield zero in the one variable case, but since we started by doing partial differentiation, that's what the operation should stay until it is dropped because it equals zero. —Preceding unsigned comment added by 69.54.140.201 (talk) 23:44, 10 October 2007 (UTC)

And I would have to maintain my position that, as y is a variable rather than a function, we have no business discussing its derivative in the first place. -- Meni Rosenfeld (talk) 23:49, 10 October 2007 (UTC)

Actually, you could argue that y is a constant, since by the definition of a partial derivative, all other variables (not functions like z) are held constant. Thus both of our arguments can coexist, it is fundamentally nonsensical to talk about y's derivative but in practice the notation comes up and it should be written with partials not ordinary derivatives since we started with PARTIAL differentiation not ordinary differentiation. —Preceding unsigned comment added by A math-wiki (talk • contribs) 00:31, 11 October 2007 (UTC)

"Jet Ski Rental Company"

I'm working with a probelm that states a jet ski company suggests a two hour minimum rental fee of $ 125.00 increased by another $ 55 bucks for every aditional hour surpassing the two hours. It says at the last few sentences of the probelm, Macy and Mya's rental was priced at the balance of $ 290.00. Number 20 (the probelm) asks me to write an equation to determine the n number of hours Macy and Mya utilized the jet skis.

As a rule preponderant in the wikipedia guidelines, I did my own homework.

First I wrote the probelm in terms of step 1, knowing the question.

1. The company charges a 120 minute renting of $ 125.00 plus a taxing 55 for use of ski exceeding the two hour minimizer.

2. Well, I thought, 125.00 plus a additional 55 dollars totals 180, one hundred ten less than what the two girls excerised their worth of usage to be.

But where I am stuck is where to ensue for there. We know what half of 150 and five more than half of that 200 dollars is, but where would I go subsequently?

--75.44.155.52 22:37, 10 October 2007 (UTC)

Am I missing something? The charge is 55$ for every hour beyond the first two. So a cost of 290$ means 3 hours more than 2 hours, that is, anywhere between 4 and 5 hours.

The question states you should write an equation; That would be ${\frac {290-125}{55}}+2=5$, that is, up to 5 hours. -- Meni Rosenfeld (talk) 22:56, 10 October 2007 (UTC)

The algebraic expression your looking for is $125+55(n-2)$ for total hours $n$ so set that equal to 290 and solve for n. That's what I think your looking for. —Preceding unsigned comment added by 69.54.140.201 (talk • contribs)

By the way, SineBot is doing such a good job that I haven't bothered telling you this before, but... Please sign your posts by typing ~~~~ (4 tildes). You should also consider creating an account. -- Meni Rosenfeld (talk) 23:44, 10 October 2007 (UTC)

Agreed, I will mention the name here in a few minutes for you. —Preceding unsigned comment added by 69.54.140.201 (talk) 23:46, 10 October 2007 (UTC)

Ok I got an account up now :) A math-wiki 23:55, 10 October 2007 (UTC)

Mazal Tov! -- Meni Rosenfeld (talk) 00:02, 11 October 2007 (UTC)

Trignometry

Hello I wanted to ask that there are many a times the questions in trignometry dont come out with questions why is it so —Preceding unsigned comment added by 122.168.40.46 (talk) 10:29, 11 October 2007 (UTC)

I'm afraid I don't understand your question, but trigonometry might be useful. Algebraist 10:34, 11 October 2007 (UTC)

Perhaps the questioners want to say that many of the identities given at Trigonometry and List of trigonometric identities are presented without any justification as to why they should hold.  --Lambiam 16:05, 11 October 2007 (UTC)

I think the identities should have their proofs, with them. (If they don't currently), it would be nice if the question was clairified a bit more. A math-wiki 06:37, 12 October 2007 (UTC)

Argand diagram

This is, I confess, a homework question, but I'm not looking for an answer, just a route of attack. I've got to draw on an argand diagram, the set of all complex numbers z, such that $\arg {\frac {z}{z-1}}=x$, where x is an angle in radians (I'm keeping this question general). However, I can't see anyway of approaching this problem. Any hints on where to begin? Laïka 21:00, 11 October 2007 (UTC)

Try letting $z=a+bi\;\!$ where $a,b\in \mathbb {R}$, finding the real and imaginary parts of ${\frac {z}{z-1}}$, and writing an equation describing the fact that its argument is x. -- Meni Rosenfeld (talk) 21:26, 11 October 2007 (UTC)

Thanks, but could you be a bit more specific - I get: $\cos {x}+i\sin {x}={\frac {longformula}{otherlongformula}}$. Is there any way to solve this without having to perform the division, which just gives a long string of as and bs with no obvious representation on the Argand? Laïka 22:13, 11 October 2007 (UTC)

Hint: If you divide the imaginary part of ${\frac {z}{z-1}}$ by its real part, you get $\tan x\;\!$ (I understand that in your actual question x is some given number, so $\tan x$ is also a number). This should give you enough information about the relation between a and b to deduce the shape of the requested set. -- Meni Rosenfeld (talk) 22:22, 11 October 2007 (UTC)

Exponential problem

I recently came across the following question:

"Solve for x: 6^x+1 = (2^x+2)(3^2x)"

Upon trying, I found this question to be quite easy and got x=-1/2 as an answer. However, when I plugged it back into the equation, it didn't work. What I did was simply divide both sides by (3^2x) and easily got a common base of 2, then simply went on from there.

Could someone tell me how to solve this?

Thanks =). Acceptable 21:12, 11 October 2007 (UTC)

I assume that you mean $6^{x}+1=(2^{x}+2)3^{2x}\;\!$. The solution method you have described does not seem correct to me, so perhaps you should present it in more detail. I will also admit that I can think of no way of solving this equation, that I doubt there is any elementary way of solving it, and that for its numerical solution, -0.250447..., I can find no symbolic representation. -- Meni Rosenfeld (talk) 21:35, 11 October 2007 (UTC)

Oh no, I meant "6 to the power of x+1 = 2 to the power of x+2 times 3 to the power of 2x" —Preceding unsigned comment added by Acceptable (talk • contribs) 22:08, 11 October 2007 (UTC)

What I did was:

• 6^x+1 = (2^x+2)(3^2x)
• (6^x+1)/(3^2x) = (2^x+2)
• 2^-x+1 = (2^x+2)
• Therefore: -x+1 = x+2
• -2x = 1
• x = -1/2

But somewhere there, I must have made a mistake. Acceptable 22:05, 11 October 2007 (UTC)

You should use parentheses to specify the order of operations; the correct way to write this equation in plain text is 6 ^ (x+1) = 2 ^ (x+2) 3 ^(2x).

Your mistake is claiming that ${\frac {6^{x+1}}{3^{2x}}}=2^{-x+1}$. -- Meni Rosenfeld (talk) 22:12, 11 October 2007 (UTC)

Oh, am I able to simplify (6^(x+1)/(3^2x) any further? If not, what is the proper method that I should approach this problem with? Acceptable 22:27, 11 October 2007 (UTC)

Perhaps the article Exponentiation can help. The most relevant rules here are $(ab)^{n}=a^{n}b^{n}\;\!$ and ${\frac {a^{m}}{a^{n}}}=a^{m-n}$. -- Meni Rosenfeld (talk) 22:42, 11 October 2007 (UTC)

There is, quite possibly, a much easier way of looking at it. Instead of just using brute force (mathematically speaking), if we assume your only interested in real solutions, then we know that $6=2\cdot 3$, thus we con conclude that $x+2=2x$ and thus $x-2=0$ and so $x=2$ there probably exist nonreal solutions to this equation however, but I'm going to infer that your not interested in them (Em I right??). A math-wiki 23:07, 11 October 2007 (UTC)

$x=2\;\!$ is not a solution to our equation, which is $6^{x+1}=2^{x+2}3^{2x}\;\!$. -- Meni Rosenfeld (talk) 23:39, 11 October 2007 (UTC)

How could logarithms be employed to solve this? Acceptable 00:03, 12 October 2007 (UTC)

EDIT: Oh I see, I can split the 6 into (2x3), then distribute the exponent, this will leave with with identical bases on both sides. Thanks for your help. Acceptable 00:12, 12 October 2007 (UTC)

EDIT: Meni, I noticed that while thinking about it afterwards, now that I'm back home, the one problem with my solution is that it IGNORES $6^{x+1}$, which must also agree which must equal the other side when x=2, but the OP seems to have figured out how to use my original thinking appropriately. I assume the OP is thinking...

$2^{x+1}3^{x+1}=2^{x+2}3^{2x}$

Now $x+1=x+2$ has no solution, so the original equation doesn't have any solutions. Thus, $6^{x+1}\not =2^{x+2}3^{2x}$.

I should note that if the x on the left side of the equation was actually a y, then my solution would've existed. A math-wiki 06:15, 12 October 2007 (UTC)

You can't conclude from $2^{x+1}3^{x+1}=2^{x+2}3^{2x}$ that $x+1=x+2$. It looks like you're trying to apply a unique factorization rule but the best you can do with that is rule out integer solutions. In reals, it doesn't work. You need to do some logarithm work:

$2^{x+1}3^{x+1}=2^{x+2}3^{2x}$

$\log(2^{x+1}3^{x+1})=\log(2^{x+2}3^{2x})$

$\log(2^{x+1})+\log(3^{x+1})=\log(2^{x+2})+\log(3^{2x})$

$(x+1)\log 2+(x+1)\log 3=(x+2)\log 2+2x\log 3$

Now it's a linear equation in x. The rest is easy. It does have a solution. I won't spoil the ending, but it's between 0 and 1. --tcsetattr (talk / contribs) 07:57, 12 October 2007 (UTC)

Electrons on a Circle

I asked a similar question this yesterday, though this one seems much more complicated to me. I will go through my calucaltions so far, so if there are any errors, you can help me rectify them.

If you have n electrons distributed evenly around a circle radius r, distance x from another electron, what is the force of repulsion experienced by the electron not on the circle from the circle of electrons. Electrostatic repulsion is a inverse square law, and the coefficent of electrostatic force can be considered as k.

• r = radius (OA, OC)
• x = CY
• y = OB
• a = AY
• α = AOY
• β = BAY

using cosine rule; $a={\sqrt {x^{2}+(x+r)^{2}-2r(r+x)\cos \alpha }}$ so the force, being an inverse square is; $F={\frac {k}{x^{2}+(x+r)^{2}-2r(r+x)\cos \alpha }}$ now, seeing as always in the circle the components in y-direction will cancel, we can consider only those in the x-direction in order to caluclate a resultant force we need to consider only the x-directions components, to find this consider the triangle ABY, the angle β and distance OB are neccesary. using sine rule. $y=-r\cos(\alpha )\,$
therefore using sine rule,
$\sin \beta ={\frac {r-r\cos(\alpha )+x}{\sqrt {x^{2}+(x+r)^{2}-2r(r+x)\cos \alpha }}}$
so the x direction component of F, or Fsinβ is;
${\mathit {F_{x}}}={\frac {k(r-r\cos(\alpha )+x)}{{\big [}x^{2}+(x+r)^{2}-2r(r+x)\cos \alpha {\big ]}^{\frac {3}{2}}}}$

So to consider all the electrons; then $\alpha ={\frac {2\pi {\mathit {k}}}{N}}$ where N number of electrons. and k varies from 1 to N. Therfore a sum of all the forces would be.

${\mathit {F_{total}}}={\frac {k}{N}}\sum _{k=1}^{N}{\bigg [}{\frac {r-r\cos({\frac {2\pi k}{N}})+x}{{\big [}x^{2}+(x+r)^{2}-2r(r+x)\cos({\frac {2\pi k}{N}}){\big ]}^{\frac {3}{2}}}}{\bigg ]}$

Now this is it, I really need some help simplifying this massive mess, and like yesterday I really need the summation series out of the equation if that's possible. Thank you. ΦΙΛ Κ 22:04, 11 October 2007 (UTC)

I looks like to me; one, you have mistakenly assumed that the outside electron is in line with two electrons that are on opposite sides, and two, the question should be able to be answered by finding the relative locations of the circle electrons to eachother and then connecting the circle electrons to the outside one. A math-wiki 06:15, 12 October 2007 (UTC)

Need help

a cow is tied to a circular silo. The rope connecting the cow to the silo is half the circumference of the silo. Upon how much area can the cow graze?

How do I start? We just finished parametric equations in class.--Mostargue 02:15, 12 October 2007 (UTC)

It depends on how a cow is tied to a circular silo. if it is tied to a fixed point or if the rope is free to move around the silo. 202.168.50.40 04:24, 12 October 2007 (UTC)

I think it's safe to assume that the point is completely fixed and is on the circle. Be careful, the region created by the rope is not actually going to be circular, the reason why is as it gets close to the silo, the silo will get in the way of the rope and prevent the cow from having his rope in a straight line. There will be a semi-circular region the cow may graze, plus the region where the rope is effectively shortened by the silo. A math-wiki 06:27, 12 October 2007 (UTC)

It should be fairly easy to write a set of parametric equations for that curve and then use integration to find the area, I am not familiar enough with parametrics, and specifically parametric calculus to be of any real help, but Algebraicly, I may be able to solve this A math-wiki 06:41, 12 October 2007 (UTC)

EDIT: On thinking about it, disregard some of that. I just noticed that, since the rope is half the circumference, your algebraic equation will be a lot simpler. The basic equation your trying to right create a tangent line from a point (x,y) on the silo circle such that the distance from endpoint to silo PLUS the distance still against the silo is always equal to $\pi r$. A math-wiki 06:53, 12 October 2007 (UTC)

I should note that if you think about the shape, it's a well known polar equation (that I can't remember the name of!!) It looks like a circle with one side partially flattened. Hope that helps. A math-wiki 06:55, 12 October 2007 (UTC)

I believe it's a form of the rose curve, specifically one that looks like $r=\alpha sin(k\theta )$ A math-wiki 07:04, 12 October 2007 (UTC)

Rational Number

Consider the following:

${\frac {0}{1}}={\frac {0}{2}}={\frac {0}{3}}$

Now if we have a set of

$\left\{{\frac {0}{1}},{\frac {0}{2}},{\frac {0}{3}}\right\rbrace$.

Do we have a set of 3 rational numbers or a set of 1 rational number? 202.168.50.40 05:11, 12 October 2007 (UTC)

I would say set of 1 rational number, namely, $\{0\}$ --Spoon! 05:51, 12 October 2007 (UTC)

Are you sure? I don't know much set theory, but I think that'd be a pretty sticky technical question. The rational numbers are ordered pairs of integers, right? So that set contains three elements, each an ordered pair. (Each ordered pair (a,b) is then a set of two elements, the set {a} and the set {a,b}.) I'm not getting much out of the intuitive definition, though, that an element of a set is entirely determined by whether the set contains it. Asking the question, "What rational numbers does this set contain", the answer would be "0/1". That's the only yes-answer to the question, something like 5/4 would be a no-answer. But if you asked what ordered pairs the set contains, you'd get three things that claimed they were in it. Black Carrot 06:27, 12 October 2007 (UTC)

I believe if I read the set theory section here correctly, it would actually be three. This is because (at least in naive set theory) it doesn't matter whether or not the elements are equal. A math-wiki 06:30, 12 October 2007 (UTC)

The answer is one rational number. The numbers 0 ⁄ 1, 0 ⁄ 2, 0 ⁄ 3 are all equal, so the set contains only one element (namely the number zero). With respect to the other posters, there is absolutely no legitimate argument that the set contains three elements, and this should not be considered a difficult or technical question. Trust me on this. Jim 07:44, 12 October 2007 (UTC)

What about the set {0, 0, 0, 1}? Does this contain 4 elements or 2? -- JackofOz 07:53, 12 October 2007 (UTC)

a related question. is the set { {}, {}, {} } a subset of {42}? Why or why not? What about simply { {} } (a set containing only the null set), as opposed to just {}?