Wikipedia:Reference desk/Mathematics

Wikipedia:Reference desk/headercfg

September 13

Adding up numbers with margins of error

How would I solve this expression? Sorry, not sure how to work with LaTeX yet.

(b ± Δ b) + (k ± Δ k) - (r ± Δ r)= b + k - r ± ????

Thanks alot
Señor Purple 01:00, 13 September 2007 (UTC)

± (Δ b + Δ k + Δ r). Geesh, I sure hope this isn't homework. - hydnjo talk 01:09, 13 September 2007 (UTC)

I believe this should be ± (Δ b + Δ k - Δ r). Tesseran 02:16, 13 September 2007 (UTC)

It depends on what is meant by ±. If it means that they are either all pluses or all minuses, Tesseran is correct. If (b ± Δ b) means any number between b - Δ b and b + Δ b, hydnjo is correct. If it means there is a certain probability that the latter is correct i.e. Δ b is the margin of error (which is what you said in the title), they're both incorrect. It's less than what hydnjo said, but the amount depends on what kind of distribution it is. I'd tell you what it is with normal distribution, but I don't know. — Daniel 02:25, 13 September 2007 (UTC)

For independent normal distributions it's $\pm {\sqrt {\Delta b^{2}+\Delta k^{2}+\Delta r^{2}}}$. -- BenRG 02:35, 13 September 2007 (UTC)

Yes, that's the answer I was going to give. If they're measurement uncertainties (and you have no particular reason to assume the distribution is "weird", i.e., not Gaussian), then they "add in quadrature". The square of the total uncertainty is equal to the sum of the squares of the individual uncertainties. —Keenan Pepper 20:39, 13 September 2007 (UTC)

Thanks Daniel -- I wasn't thinking. My interpretation is almost certainly wrong, given the mention of "margins of error". Tesseran 00:27, 14 September 2007 (UTC)

sudoku help

Could someone help me w/this sudoku I can't get any more numbers and I've spent over 15 mins on it!

File:Stuckonsudoku.PNG

Portal111 01:10, 13 September 2007 (UTC)

Start guessing. 202.168.50.40 01:11, 13 September 2007 (UTC)

I shouldn't have to guess on a sudoku, every move should be logical.--Portal111 01:13, 13 September 2007 (UTC)

A case could be made that that assertion is equivalent to P=NP. —Blotwell 23:44, 14 September 2007 (UTC)

The questioner means: a properly constructed sudoku problem is such that a competent solver can, at each stage of solution, using deductive reasoning, find the assignment for some empty cell, so that the whole puzzle can be solved by repeating this, without need to resort to backtracking. This requirement is then part of the definition of "properly constructed". As a criterium it is not sharp: one solver's guess-and-check is another solver's deductive step. After all, from a (purely!) theoretical point of view, sudoku puzzles can be solved by precomputing a table indexed by the solvable puzzles and giving the complete solution by an O(1) lookup step.  --Lambiam 00:37, 15 September 2007 (UTC)

Unbelievable, 15 whole minutes!!! It's perfectly fine to guess and then to eliminate that guess as a result of your subsequent options. A guy named Thomas Edison spoke of this strategy. - hydnjo talk 01:15, 13 September 2007 (UTC)

I usually do them in 6 mins.--Portal111 01:17, 13 September 2007 (UTC)

For a starter, provided the other numbers are correct, you must have a 4 in the top left corner of the top-middle box. i.e. a 4 in the 4th column and top row. Look at the other boxes in the 4th column to see why you can't have a 4 in any of these. You'll probably find the rest of the numbers will slot into place now. Richard B 01:41, 13 September 2007 (UTC)

That's logical. - hydnjo talk 01:47, 13 September 2007 (UTC)

Thanks Richard B that one # let me solve it. Odd that we have the same first name and first letter in our last name too.. —Preceding unsigned comment added by Portal111 (talk • contribs) 02:14, 13 September 2007 (UTC)

Prove identity law

If $ax=a\,$ for some number where $a\,$ is nonzero then $x=1\,$

How would I go about proving this? It's *so* fundamental I can't think where to start.

Depends on how you start. If you're given a mathematical system that contains the identity, such as the real numbers or integers, it's usually already an axiom, and there's no need to prove it because it's already true. If you're trying to show that some mathematical system contains an identity, it's slightly harder. What is this for? I can help you better if you explain what you're doing this for. Gscshoyru 15:46, 13 September 2007 (UTC)

I don't think this particular statement is commonly an axiom (though it's quite close). More likely axioms are that 1 is a multiplicative identity, that if $a\neq 0$ then there is a number $a^{-1}$ such that $a^{-1}a=1$, and that multiplication is associative. If you have all of those, then $1=a^{-1}a=a^{-1}(ax)=(a^{-1}a)x=1x=x\;\!$. -- Meni Rosenfeld (talk) 15:51, 13 September 2007 (UTC)

Oh -- I read the title and didn't read the equation. Oops. Meni's proof works as long as your system has those axioms, and most do. Gscshoyru 15:54, 13 September 2007 (UTC)

It needs to be proved because it is not always true - it depends on certain properties of the algebraic structure in which you are working. In the ring of integers modulo 4, 2x3=2, but 3≠1. This is because 2 has no multiplicative inverse in this ring. Gandalf61 16:04, 13 September 2007 (UTC)

(edit conflict) The nice thing about this question is that it connects with some serious mathematics. The difficult thing is … the same. :-)

We must study the details of the request. Two important questions are: (1) what is "some number", and (2) is x = 1 a unique solution?

Multiplication is evidently required, but that means we may be talking about natural numbers (0,1,2,…), integers (…,−2,−1,0,1,2,…), rational numbers (fractions), real numbers (like √2 and π), or something else. For natural numbers and integers the equation 2x = 2 cannot be solved by multiplying both sides by ¹⁄₂, because it is not available among our numbers; yet we can consult a definition for multiplication and prove that x = 1 is a solution. Do we know that some other value for x is not lurking undiscovered; can we prove it? This is more awkward, but we can do it.

If we have multiplicative inverses (fractions, reciprocals), we can follow the proposed strategy of multiplying both sides by ¹⁄_a. Again we can prove uniqueness, but our method will probably be different.

Number theory and abstract algebra study examples where our arithmetic often requires careful attention to detail. We encounter kinds of "multiplication" where xy = 0 can occur even when neither x nor y is 0. Or we might have xy = −yx, so that the order of multiplication matters. And so on. This teaches us to be very careful about the details of our assumptions, of what properties we use for our proofs.

Uniqueness is something we cannot take for granted. Consider x²+1 = 0. Using real numbers we have no solutions; using complex numbers we have two solutions; and using quaternions we have an infinite number of solutions.

So mathematicians have learned to ask questions that may at first seem peculiar.

1. Does a solution exist?
2. Is the solution unique?
3. Can we find a solution?

The last question is perhaps the most unsettling, because sometimes we can prove that solutions exists, yet be forever denied any way to exhibit one. --KSmrq^T 17:03, 13 September 2007 (UTC)

This looks like a chapter one section one abstract algebra question to me. Meni's approach is the way to go in that case. We're probably assuming a group. Donald Hosek 18:40, 13 September 2007 (UTC)

I doubt the OP had abstract algebra in mind. He is probably just trying to understand better things he has long known about the real numbers. -- Meni Rosenfeld (talk) 18:49, 13 September 2007 (UTC)

Sorry for the late reply. I forgot to mention the universe of discourse for all variables is $\mathbb {R}$. This isn't an intentional study into abstract algebra but obviously questions of this sort demand it. I was just starting to think about proofs on a purely personal, recreational basis and I thought something this simple might be easy to prove. Thanks Meni.

At this point, I think you should be asking yourself less "how do I prove this property of the real numbers?" and more "what are the real numbers, anyway?". There are many approaches for defining the real numbers, and articles like Real number and Construction of real numbers highlight some of them. Each approach comes with its own set of proofs for their most basic properties. One you have those basic properties, you can essentially "forget" what the real numbers are and how did you construct them, and continue to explore based solely on those basic properties. The 3 "axioms" I mentioned are an example; once you have established those for your favored construction, you can move on to using them, in an argument like I presented above, to prove more "advanced" statements like the one in your question.

Of course, the ordered field of real numbers is just one algebraic structure, and you will, no doubt, also want to explore others, like those suggested by KSmrq and Gandalf. -- Meni Rosenfeld (talk) 21:22, 13 September 2007 (UTC)

Working with real numbers, the most important facts we need are that 1 is an identity (two-sided, in fact) for multiplication, and that every nonzero real has a unique multiplicative inverse (also two-sided). Then we deduce

${\begin{aligned}ax&=a&\qquad &({\text{given}})\\a^{-1}(ax)&=a^{-1}a&&(a^{-1}{\text{ exists, multiplication preserves equality}})\\(a^{-1}a)x&=a^{-1}a&&({\text{multiplication is associative}})\\1x&=1&&({\text{inverse property}})\\x&=1&&({\text{identity property}})\end{aligned}}$

By carefully observing the properties we used, we find that the same proof applies without change to other algebras. For example, it works with rational numbers, and with complex numbers, and with integers modulo 17 (and other Galois fields), and with invertible matrices, and so on.

For example, let's use a specific 2×2 matrix for a.

${\begin{aligned}{\begin{bmatrix}4&5\\7&9\end{bmatrix}}X&={\begin{bmatrix}4&5\\7&9\end{bmatrix}}\\{\begin{bmatrix}9&-5\\-7&4\end{bmatrix}}\left({\begin{bmatrix}4&5\\7&9\end{bmatrix}}X\right)&={\begin{bmatrix}9&-5\\-7&4\end{bmatrix}}{\begin{bmatrix}4&5\\7&9\end{bmatrix}}\\\left({\begin{bmatrix}9&-5\\-7&4\end{bmatrix}}{\begin{bmatrix}4&5\\7&9\end{bmatrix}}\right)X&={\begin{bmatrix}9&-5\\-7&4\end{bmatrix}}{\begin{bmatrix}4&5\\7&9\end{bmatrix}}\\{\begin{bmatrix}1&0\\0&1\end{bmatrix}}X&={\begin{bmatrix}1&0\\0&1\end{bmatrix}}\\X&={\begin{bmatrix}1&0\\0&1\end{bmatrix}}\end{aligned}}$

This works in spite of the fact that, for matrices, AB ≠ BA. In fact, we need merely demand that the algebra be a group under multiplication for the proof to go through. Thus we have a powerful motivation to study abstract algebra. --KSmrq^T 16:30, 14 September 2007 (UTC)

You don't need a group, just the cancellation property. Thus the usual generality in which to state this result would be an integral domain. —Blotwell 23:42, 14 September 2007 (UTC)

The proof just above (essentially that proposed by Meni Rosenfeld) requires a multiplicative identity, a guarantee that a multiplicative inverse exists, and associativity (but not commutativity); hence, it requires a multiplicative group.

Perhaps you are thinking of the different method of proof I referred to earlier, in my first post, where we might want to prove this for, say, the natural numbers. But, as I said at the time, if a multiplicative identity, 1, exists, then it clearly satisfies the equation without needing any other properties. We don't need inverses, nor cancellation, nor even associativity. That leaves the question of uniqueness, as I said; but it is not clear that uniqueness is being asked.

Permit me a fanciful example. Suppose a is a guy, x is a gal, ax is a kiss, and the equation ax = a says the kiss leaves the guy unmoved. Certainly a gal whose kisses have no effect on any guy would satisfy the equation (if not the guys). But it could also happen that some guy other than a will respond strongly to x even though a does not; she's not his type, apparently. In fact, a might only respond to a unique "soul mate", so most kissing partners would satisfy the equation. And in these modern times, we need not assume that a is always a guy nor that x is always a gal (so we don't need a two-sorted algebra). Well, I did say it would be a fanciful example; but I could have illustrated the same mathematical points with, say, matrices (albeit less memorably). --KSmrq^T 06:11, 15 September 2007 (UTC)

Solve

solve 25+56+67+89100/678*345-1000 —Preceding unsigned comment added by 59.180.11.224 (talk) 15:40, 13 September 2007 (UTC)

That's what calculators are for. If you don't have one, theree's usually one built into your operating system -- start-programs-accessories on windows. But see the box at the top of this page, we're not here to do your homework for you. Gscshoyru 15:48, 13 September 2007 (UTC)

If you can edit this page, you can access your friend Google. -- Meni Rosenfeld (talk) 15:53, 13 September 2007 (UTC)

Although I imagine that this is a homework problem and you need to show your work. You should take a look at the article on the order of operations as well as your class notes. The key thing to keep in mind for this problem is that you do multiplication and division together, from left to right (so you'll do the division before the multiplication in this case), then, once you have all the multiplications and divisions done, you do the additions and subtractions, again left to right. One other nit: You can solve an equation (no equals sign, then no equation), but an expression like this you can only evaluate. Donald Hosek 16:41, 13 September 2007 (UTC)

Approximately 44,486.496 I think... YeoungBraxx 17:52, 13 September 2007 (UTC)YeoungBraxx

And as an aside for our original poster, if you were my student and gave as your (entire) answer "Approximately 44,486.496", I would give you no points. Or maybe 1/3 if I were feeling generous, but I seldom am. Donald Hosek 18:28, 13 September 2007 (UTC)

Almost [1] - 44 486.4956. Unless of course your teacher wants it rounded down to three decimal places.martianlostinspace ^{email me} 20:15, 13 September 2007 (UTC)

If you like, I will even give you permission to cite my link (to Google) as your working out (that is, how you arrived at your answer. Something I worked very hard for :-)martianlostinspace ^{email me} 20:18, 13 September 2007 (UTC)

You realise, of course, that this value is also rounded? -- Meni Rosenfeld (talk) 20:20, 13 September 2007 (UTC)

So it would appear. Although maybe less so than the other above answer. Maybe Google can only handle nine digits. [2].martianlostinspace ^{email me} 20:58, 13 September 2007 (UTC)

Assuming you are on a Windows machine. You can built your own calculator.

http://windowstipoftheday.blogspot.com/2005/03/ms-dos-calculator.html

Of course, if you are on a windows machine why don't you use the windows calculator. Start - All Programs - Accessories - Calculator

Make sure you select the scientific calculator by selecting : View - Scientific

202.168.50.40 00:47, 14 September 2007 (UTC)

September 14

How does anyone feel about...

The set of boolean values being represented as $\mathbb {B} =\lbrace 0,1\rbrace$? Is $\mathbb {B}$ being used for anything else?

Then, $\mathbb {B} ^{n}$ can be the set of all n-tuples, e.g. $\mathbb {B} ^{3}=\lbrace (x_{1},x_{2},x_{3})|x_{i}\in \mathbb {B} \rbrace$

I think the standard notation is $\mathbb {Z} /2\mathbb {Z}$ or $\mathbb {Z} _{2}$. Gandalf61 14:14, 14 September 2007 (UTC)

But these notations imply that we are referring to the field, where 1+1=0, while I believe the OP is looking for a notation that will emphasize that we have 1+1=1. -- Meni Rosenfeld (talk) 15:17, 14 September 2007 (UTC)

Usually AND is implemented by the multiplication operation, so 1*1=1, whereas addition implements XOR, the "exclusive or", so 1+1=0. $\mathbb {Z} _{2}$ is the simplest example of a Boolean ring. Gandalf61 15:28, 14 September 2007 (UTC)

I've seen $\mathbb {B} ^{n}$ used for the n-dimensional unit ball. The set {0,1} is sometimes written 2 or 2, which makes sense if you think of it as the ordinal 2. -- BenRG 16:56, 14 September 2007 (UTC)

I've seen the notation $\mathbb {B} =\lbrace 0,1\rbrace$ used in the context of computing science, as in this course text (see page 46) on Funmath.  --Lambiam 23:25, 14 September 2007 (UTC)

I guess I wouldn't object, but the notation {0,1}ⁿ itself is quite widely used. Is there really that much gained by shortening {0,1} to $\mathbb {B}$? — PaulTanenbaum 02:22, 16 September 2007 (UTC)

sum

Sum (from n=0 to n=p) of (-1)ⁿ/(p-n)!n! (using 0!=1)

The sum = 0 except when p=0 in which case the sum=1. (from x^p terms of polynomial expansion of: exp(p) x exp (-p) )

Can anyone give an alternative proof of this just out of curiousity (I've already got sum=0 when p=odd, but couldn't go further)83.100.255.59 15:04, 14 September 2007 (UTC)

Have you tried using the Binomial theorem? -- Meni Rosenfeld (talk) 15:16, 14 September 2007 (UTC)

erm, I used the Multinomial theorem (for a power of 2) to get the original result - not sure if there is an alternative method you meant?ignore - not the multinomial (a bit like it but not the same)83.100.255.59 15:40, 14 September 2007 (UTC)

Can't get that to work - would need to find a way to get the n! top term in the binomial coefficent to dissapear83.100.255.59 15:51, 14 September 2007 (UTC)

Assuming that I understand correctly that you are trying to prove that

$\sum _{n=0}^{p}{\frac {(-1)^{n}}{n!(p-n)!}}=0$

The equality does not change if we multiply by $p!$, so we have to show that

$0=\sum _{n=0}^{p}(-1)^{n}{\frac {p!}{n!(p-n)!}}=\sum _{n=0}^{p}(-1)^{n}{p \choose n}$

Now we can use the binomial theorem.

An alternative method is to prove it directly using induction. -- Meni Rosenfeld (talk) 15:52, 14 September 2007 (UTC)

Yes, thanks.83.100.255.59 16:32, 14 September 2007 (UTC)

question

A and B can run 200metres in 22 and 25 seconds respectively.

• How far is B from the finishing line when A reaches it? —Preceding unsigned comment added by 122.164.236.110 (talk) 16:29, 14 September 2007 (UTC)

Do your own homework. Hint: Distance is speed times time, and the time when A crosses the finishing line to the time when B is behind A at this time is, well, the same. x42bn6 Talk Mess 16:32, 14 September 2007 (UTC)

It might help to think that B will be 3 seconds behind when A crosses the line. So, B will have 3/25 of the 200 meter distance yet to go. You take it from there. StuRat 04:45, 19 September 2007 (UTC)

Fourier analysis of resonance

In college I took a semester-long class in the phyics of resonance phenomena; the fundamental idea was that a resonant system has a preferred vibrational frequency, and application of a periodic driving force will drive the sistem to an amplitude that depends on how closely the frequency of the driving force corresponds to the preferred frequency of the system.

As far as I can recall, we considered only sinusoidal driving forces. I was thinking about this recently, and it seemed to me that for nonsinusoidal drivers, one could use Fourier analysis to decompose the driving force into a sum of sinusoids, and then deal with their effects on the system independently. Is that right?

My real question is, where can I read more about this?

Thanks. -- Dominus 16:38, 14 September 2007 (UTC)

This is indeed a standard technique, as long as the driving force is periodic and the differential equation representing the system is linear. As our article on the uses of trigonometry puts it: In almost any scientific context in which the words spectrum, harmonic, or resonance are encountered, Fourier transforms or Fourier series are nearby. You can find some theory and examples at Linear differential equation, although the link with resonance is not made explicit. An easy introduction is the treatment by example found here.  --Lambiam 23:12, 14 September 2007 (UTC)

Seems like you're modeling the system as an LTI system, in which case you can get the response to arbitrary input by convolving with the impulse response in the time domain, or equivalently, multiplying by the transfer function in the frequency domain. —Keenan Pepper 03:11, 15 September 2007 (UTC)

September 15

Blackboard Bold

Does anyone know of a source (freeware or cheap anyway) of a good quality truetype Blackboard bold font? -- SGBailey 08:32, 15 September 2007 (UTC)

You can download it and other math-related TrueType fonts for free at this site from the math department at Union College:

http://www.math.union.edu/~dpvc/jsmath/download/extra-fonts/welcome.html

— Michael J 21:15, 15 September 2007 (UTC)

rainfall

1 inch of rain equals how many gallons of water per acre? jodyjer0801 —Preceding unsigned comment added by Jodyjer0801 (talk • contribs) 13:19, 15 September 2007 (UTC)

http://www.google.com/search?q=1+inch+in+gallons+per+acre --Spoon! 13:38, 15 September 2007 (UTC)

(potentially unnecessary) warning: there are various gallons in the world, so make sure you get the right one. In answer to the question, why take the Google way out? An acre is a furlong by a chain; a furlong is ten chains; a chain is four rods, poles or perch; a rod, pole or perch is five and a half yards; a yard is three feet; a foot is twelve inches; a (US liquid) gallon is 231 cubic inches. Where's the problem? --Algebraist 17:24, 15 September 2007 (UTC)

That way of calculating sounds really funny, because I live in a metric country. I'd just say that an inch is 0.0254 meters, an acre is 4.047*10^3 square meters so an inch times an acre is 102.8 cubic meters; a US gallon is 3.785*10^(-3) cubic meters, so the volume is 27160 gallons. – b_jonas 21:28, 17 September 2007 (UTC)

I suspect that Algebraist's post might be intended to subtly mock the imperial system. -- Meni Rosenfeld (talk) 21:37, 17 September 2007 (UTC)

You do not need to mock the Imperial System, it's quite capable of mocking itself. 210.49.155.132 13:13, 19 September 2007 (UTC)

Catenary

With reference to a catenary, what calculations can be made if you know its length and the horizontal distance between the points at each end that secure it in place? asyndeton 15:57, 15 September 2007 (UTC)

Are the points at the end at the same height? If so, this information completely determines the catenary, as far as I know. Tesseran 18:13, 15 September 2007 (UTC)

Yes, they are at the same height. What do you mean by 'determines the catenary'? asyndeton 18:26, 15 September 2007 (UTC)

It means there is only one catenary with those parameters. --Taejo|대조 12:07, 18 September 2007 (UTC)

See Catenary - you can get the shape of the curve it hangs in, and the 'force/tension' at a given point along, as well as related things such as the amount of droop/angle at a given point.87.102.43.253 18:38, 15 September 2007 (UTC)

The equation of a hanging catenary in (x,y) terms is usually expressed wrt an origin a certain distance directly below the lowest point, this distance (c, say) being a parameter of a particular curve. If the suspension points are 2a apart, then the distance of the lowest point below them (b, say) is b=c[cosh(a/c)-1]. If the total length of the curve is 2s, then s=c[sinh(a/c)]. c cannot be determined explicitly, but is the solution of the equation 2c[sinh(a/2c)]=(s^2-b^2)^0.5 As far as I can see, there is no straightforward answer to such questions as "A heavy flexible inextensible string hangs from two points 10 units apart, with its lowest point 5 units below them. What is its length?"…86.132.239.70 19:19, 15 September 2007 (UTC)

The equation of a catenary is $y=a\cosh \left({\frac {x}{a}}\right)$ You want to find a, which will tell you the exact equation of the catenary, and everything. Suppose that you know the horizontal distance between the ends is w, and you know the length of the catenary is s. The length is given by (from arc length) $s=\int _{-w/2}^{w/2}{\sqrt {1+[f'(x)]^{2}}}\,dx$.

$f'(x)=\sinh \left({\frac {x}{a}}\right)$, so ${\sqrt {1+[f'(x)]^{2}}}=\cosh \left({\frac {x}{a}}\right)$

And the length is $s=\int _{-w/2}^{w/2}\cosh \left({\frac {x}{a}}\right)dx=a[\sinh \left({\frac {x}{a}}\right)]_{-w/2}^{w/2}=2a\sinh \left({\frac {w}{2a}}\right)=s$

You know s and w, and you need to numerically solve for a (with the restriction that a is positive, of course, so that gravity points downward). a will have units of length.

As for the vertical position of the catenary, the height of the place where the ends are hung is $a\cosh \left({\frac {w}{2a}}\right)$, and everything else is relative to that. You can add a constant to adjust the height to be relative to some "ground" if necessary.

--Spoon! 20:35, 15 September 2007 (UTC)

need some help.

If you started with a population of 2 wolves and ended up with a population of 150,000 wolves after 2000 years (with a generation every 5 years), considering the growth rate was relativly the same how many cumulative individuals would have lived in the population? Can someone solve this for me or at least tell me the equation is a way some one who has only finished algebra II would be able understand and solve with a TI-89. Thanks for any help I get! -Icewedge 19:54, 15 September 2007 (UTC)

Depends on what you mean by "generation"... Under some simplified assumptions, we can model this as a geometric progression; Let $a_{0}=2$ be the starting population, and $a_{n}$ the population after n generations. We assume that no wolf survives more than a generation, and that $a_{n}=a_{0}q^{n}$ for some q. Then $150000=2q^{400}$ so $q=75000^{1/400}$. Now we are looking for $\sum _{k=0}^{400}a_{k}={\frac {a_{401}-a_{0}}{q-1}}\approx 5\cdot 10^{6}$ (give or take a million or two, due to sensitivity to starting population and general model inaccuracies). -- Meni Rosenfeld (talk) 21:44, 15 September 2007 (UTC)

(ec: I wrote this before Meni's answer above, but forgot to submit it. :/) I'm assuming that by a constant growth rate you mean that

${\frac {dp}{dt}}=cp\,$,

where $t$ is time, $p$ is the population size and $c$ is the annual growth rate the population. The solution to that differential equation, giving the population size as a function of time, is

$p(t)=p_{0}\,e^{ct}\,$,

where $p_{0}=p(0)$ is the initial size of the population and $e$ is the base of the natural logarithm. To obtain the total number of "wolf-years" elapsing over the 2000 years, $P$, you need to integrate this over time to get

$\int _{0}^{2000}p(t)\,dt=p_{0}\int _{0}^{2000}e^{ct}\,dt=p_{0}{\frac {e^{2000c}-e^{0}}{c}}={\frac {p(2000)-p(0)}{c}}$.

From the constraints $p(0)=2$ and $p(2000)=150,000$ you should be able to solve for $c$ and plug it into the expression above. Note that, to get the (approximate) total number of wolves that have ever lived, you still need to divide the result by the average lifespan of a wolf — which I'd suppose to be the 5 years you've given above, though your terminology is rather unclear. (Bonus exercise: Why is the result only approximate? Consider in particular the limit as the lifespan of a wolf tends to infinity.)

In any case, though my method is slightly different from Meni's, I also get a result a little over 5×10⁶ or so. Which is about as much as can be expected from such a simplistic model. —Ilmari Karonen (talk) 22:37, 15 September 2007 (UTC)

September 16

What does this mean?

I came across this in an (non-wikipedia) article. I have no idea what it means.

$\Sigma \,t^{3}={\frac {t^{4}}{4}}-{\frac {t^{3}}{2}}+{\frac {t^{2}}{4}}$

Can someone explains what it actually means. 220.237.181.98 01:57, 16 September 2007 (UTC)

Which article? I don't know a meaning. Perhaps it is a failed attempt to write the summation

$\sum _{x=1}^{t}x^{3}=\left({\frac {t(t+1)}{2}}\right)^{2}={\frac {t^{4}}{4}}+{\frac {t^{3}}{2}}+{\frac {t^{2}}{4}}$

PrimeHunter 03:37, 16 September 2007 (UTC)

No, it's definately minus t cube on two which is $-{\frac {t^{3}}{2}}$ . There is no typo. 220.237.181.98 04:45, 16 September 2007 (UTC)

If you shift the index by 1, then you get what you want. So maybe it is a failed attempt to write this

$\sum _{x=1}^{t-1}x^{3}=\left({\frac {t(t-1)}{2}}\right)^{2}={\frac {t^{4}}{4}}-{\frac {t^{3}}{2}}+{\frac {t^{2}}{4}}$

--Spoon! 05:52, 16 September 2007 (UTC)

Can you give us some more context? What was the article about? Maelin (Talk | Contribs) 05:02, 16 September 2007 (UTC)

The article is entitled "The Difference Calculus". It's about a collection of mathematical tools for solving difference equations. At least that is what it says. 211.28.126.201 09:19, 16 September 2007 (UTC)

Could you provide a link or other info on where to find it? —Bromskloss 11:47, 16 September 2007 (UTC)

Perhaps they are showing how to solve the recurrence relation $a_{t+1}=a_{t}+t^{3}$ --Spoon! 15:20, 16 September 2007 (UTC)

I don't know the article, but in this context Σ and Δ are possibly operators turning functions on integers into other functions on integers, defined as follows. If F is defined on the integers, then the forward difference ΔF is the function f such that f(t) = F(t+1) – F(t). For example, if F(t) = t⁴ – 2t³ + t², then (ΔF)(t) = 4t³. The operator Σ is such that if Σf = F, then ΔF = f. This allows an indefinite summation constant for Σ, which can be fixed by agreeing that (Σf)(0) = 0. Other definitions are possible; in particular the "backward difference" (∇F)(t) = F(t) – F(t–1). See also Difference operator.  --Lambiam 18:47, 16 September 2007 (UTC)

I agree with Lambiam, it must be the summation operator as defined in Knuth's Concrete Mathematics. – b_jonas 21:10, 17 September 2007 (UTC)

September 17

parabolica

Is the word parabolica a real word i.e. a derivative of parabola (geometry)? —Preceding unsigned comment added by 210.79.26.193 (talk) 05:42, 17 September 2007 (UTC)

The adjective parabolic is parabolica in some languages; it can be transliterated Classical Greek, Latin, and Italian (see, e.g., Traiettoria parabolica). With an accent like this: parabólica, it is Spanish (Castilian), while parabòlica is Catalan. It is not a common English word.  --Lambiam 06:20, 17 September 2007 (UTC)

Anothere example of Italian usage is the "Parabolica" bend at the Monza Grand Prix circuit - bend 9 on this diagram. Gandalf61 09:00, 17 September 2007 (UTC)

Sounds like an artistic use - such as a abstract quadratic sculptor's exhibition of 'parabolica' cf eclectica etc87.102.79.48 11:23, 17 September 2007 (UTC)

logathrims - i dont get it (basic)

for example simplify ${\dfrac {\log 8}{\log 0.25}}$

so far what ive done is log 4^1.5 / log 4^-1 which is (1.5 log 4 / -1 log 4) which gives me log -1.5 and the answer is meant to be log -3/2. Any ideas?? thanks Testeretset 13:58, 17 September 2007 (UTC)

Call me crazy, but I think that -1.5 = -3/2. However, it should be just -1.5 (or -3/2), not log -1.5 or log -3/2. -- Meni Rosenfeld (talk) 14:20, 17 September 2007 (UTC)

Agree with Meni, and for your original question, can you rephrase both numerator and denominator as log(2^x) ? Capuchin 14:23, 17 September 2007 (UTC)

makes sense, yeah the answer was just -3/2 my mistake. thanks... Testeretset 14:34, 17 September 2007 (UTC)

I suggest that the problem lies NOT with logarithms but with faulty mis-understanding of the relationship between rational numbers and floating point numbers. One wonders why the questioner did not take the further step of dividing -3 by 2 on their scientific calculator. 202.168.50.40 00:42, 18 September 2007 (UTC)

1.5 is not a floating-point number. Floating point is the name of a method for storing numbers on a computer in two parts, a mantissa and an exponent. The location of the decimal point can be easily changed by changing the exponent part (hence the name, "floating point").

On the other hand, both 3/2 and 1.5 are rational; you might call the former a "fraction" and the latter a "decimal number" (though I wouldn't expect to encounter such terms in any serious mathematics). -- Meni Rosenfeld (talk) 02:28, 18 September 2007 (UTC)

September 18

Factorization question

I need to simplify the expression $(x^{5}-32)/(x-2)$. I'm guessing $x-2$ is a factor of the numerator, so it will cancel out with the denominator. However, I have forgotten binomial expansion and therefore do not know how to determine the other factor of the numerator. One clue I can use to check any proposed answer is that I have deduced, by means of a graph and a table, that the limit of this expression as $x$ approaches 2 is 80. Can anyone remind me how to arrive at the other factor of the numerator? Thanks, anon. —Preceding unsigned comment added by 70.23.83.108 (talk) 00:12, 18 September 2007 (UTC)

(x^5-32) is a difference of two 5-cubes. If you don't know the formula for that, just use long division.--Mostargue 00:41, 18 September 2007 (UTC)

Having studied computer science, I know the magic sequence in computing which is {1,2,4,8,16,32,64,128,256,512,1024,...} 202.168.50.40 00:45, 18 September 2007 (UTC)

Hopefully write x⁵-32=(x-2)(ax⁴+bx³+cx²+dx+e), then solve for the coefficients; I recommend starting at one end, either a or e, and work inwards. (Of course, this is just polynomial long division.) Tesseran 01:21, 18 September 2007 (UTC)

It should be pointed out that the binomial expansion doesn't seem relevant here; Geometric progression, on the other hand, does. By the way, the fact that $2^{5}-32=0$ guarantees that $x-2$ is a factor of $x^{5}-32$. -- Meni Rosenfeld (talk) 02:24, 18 September 2007 (UTC)

It is true that x⁵−32 can be factored as (x−2)q(x), where q(x) is a polynomial of degree 4. However, it is not true that q(x) has only two terms.

Curiously, it is trivial to show that x−2 is a factor, yet without revealing the other factor. The reasoning is that if we substitute 2 for x we get zero, so 2 is a root and x−2 is a factor.

Even before finding the other factor we can expect it to show a symmetry between powers of x and powers of 2, a property inherited from the original, x⁵−2⁵. Others have suggested polynomial long division, however we are dividing by a monic linear polynomial so we can take advantage of the quick and convenient method called synthetic division. Essentially, we evaluate the polynomial at 2 using Horner's rule, accumulating the intermediate values as coefficients of the new polynomial. Here's an illustration with p(x) = x³−27 factoring out x−3.

p(x)	1		0		0		−27
	↓		+		+		+
	↓		3		9		27
x=3	↓	↗×3	↓	↗×3	↓	↗×3	↓
q(x)	1		3		9		0

This shows that the quotient is q(x) = x²+3x+9. In fact, we can readily surmise the quotient for xⁿ−aⁿ divided by x−a from the regular pattern seen in this example. When a is 1 and n is prime, the quotients are cyclotomic polynomials. --KSmrq^T 05:11, 18 September 2007 (UTC)

Here's my attempt at synthetic division:

             x^4 + 2x^3 + 4x^2 +  8x + 16
x-2 / x^5                            - 32
    -(x^5 - 2x^4)
            2x^4
          -(2x^4 - 4x^3)
                   4x^3
                 -(4x^3 - 8x^2)
                          8x^2
                        -(8x^2 - 16x)
                                 16x - 32
                               -(16x - 32)
                                        0

Note that I wouldn't generally solve a problem for you like this, but you did seem to be genuinely trying (with the graph and all), and seemed like you needed some help learning how to do synthetic division. StuRat 04:37, 19 September 2007 (UTC)

INTEGRALS

Solve:

• $\int \cos {e^{x}}dx$
• $\int {\frac {dx}{\ln {x}}}$
• $\int \sin {x^{2}}dx$
• $\int {\frac {e^{x}}{x}}dx$
• $\int {\sqrt {x^{3}+1}}dx$
• $\int {\frac {\sin {x}}{x}}dx$

This is not homework, for obvious reasons.--Mostargue 00:55, 18 September 2007 (UTC)

have you tried the Wolfram Integrator? —Tamfang 01:28, 18 September 2007 (UTC)

Also, (a) have you already tried you hand at any of these? (b) Is there a specific context for them, or did they just pop into your head? Depending on the context, it may be possible to evaluate some of these for specific limits, but not produce a general closed form solution to the indefinite integral. Confusing Manifestation 01:58, 18 September 2007 (UTC)

My math book says the integrals cannot be expressed in elementary functions. I'd like to see what they CAN be expressed as.--Mostargue 02:02, 18 September 2007 (UTC)

Not much. You can try the Wolfram integrator, which is based on Mathematica. It will express such integrals using non-elementary functions like the exponential integral, Fresnel integral and elliptic integrals. But if you check the definitions of those, you might find some to be defined exactly as those integrals. This is, again, because those integrals are as simple as they are going to get - you just can't simplify them any further.

Of course, you can try to represent them as a power series; for example, $\int {\frac {\sin x}{x}}\ dx=\sum _{k=0}^{\infty }{\frac {x^{2k+1}}{(2k+1)!(2k+1)}}$. -- Meni Rosenfeld (talk) 02:19, 18 September 2007 (UTC)

Thanks Meni =). Is there any algorithm for determining whether or not a function has an elementary antiderivative? --Mostargue 05:24, 18 September 2007 (UTC)

Also, $\int {\frac {e^{x}}{x}}dx$ and $\int {\frac {\sin x}{x}}dx$ are related, especially if you think of these as complex analytic functions. I don't know if any other pairs integrals you've given are related. – b_jonas 06:00, 18 September 2007 (UTC)

I think the Risch algorithm might be capable of that, but I don't understand it well enough to tell for sure. -- Meni Rosenfeld (talk) 09:07, 18 September 2007 (UTC)

Don't know about algorithms, but the area you want is differential Galois theory. Algebraist 12:49, 19 September 2007 (UTC)

What do you call this shape?

A rectangle attached to a semicircle. —Preceding unsigned comment added by 166.121.36.232 (talk) 03:00, 18 September 2007 (UTC)

I've heard a rectangle attached to two semicircles called a slot, so maybe you could call your case a half-slot ?:

 ____
(____) 2 semicircles = "slot"

 __
(__| 1 semicircle = "half-slot" ?

StuRat 04:19, 19 September 2007 (UTC)

Depending on context, it could be called a round arch or Roman arch. —David Eppstein 04:25, 19 September 2007 (UTC)

Recursive integral

Hello, I've been trying to calculate something in the case where, in a game, two groups of units are attacking each other, to see how the damage per second changes according to the two groups.

I've managed to find a simple solution when one groups attacks another but the other group doesn't attack something else, but I can't really see how to find the solution now when both groups attack each other.

The equation I'd like a solution for is

$x_{\Sigma A}(t)=x_{A}\left(n_{A}(0)-\left\lfloor {\frac {x_{\Sigma D}(0)\cdot t_{1}}{y_{A}}}\right\rfloor -\left\lfloor {\frac {\int _{t_{1}}^{t}x_{D}\left(n_{D}(0)-\left\lfloor {\frac {\int _{t_{1}}^{t}x_{\Sigma A}(\tau )\;d\tau }{y_{A}}}\right\rfloor \right)d\tau }{y_{A}}}\right\rfloor \right)$

I know the formula is quite messy, but that's what I ended up with. (I'd really prefer if the solution was analytic and not numerical, but well...)

So, I know everything except $x_{\Sigma A}(t)$ (which represents the damage per second of the group of A units with respect to time.).

I tried to put it in different programs but they didn't help me (I also noticed that they couldn't even figure out $\int _{a}^{b}\lfloor cx\rfloor \;dx$, I'd like to understand why, I've been using $\int _{a}^{b}\lfloor cx\rfloor \;dx=\left[{\frac {1}{2c}}\lfloor cx\rfloor (2cx-\lfloor cx\rfloor -1)\right]_{a}^{b}$ and came to the correct solution for my first case, what is wrong with that ?)

Thanks. --Xedi 04:23, 18 September 2007 (UTC)

Do you really need instantaneous "attacking each other"? Why not use approximate integration?--Mostargue 05:28, 18 September 2007 (UTC)

I have done such a calculation myself in the past (for Heroes II), but I think you won't be getting far if you express it using the floor function - being discontinuous, it's not very closed-form friendly. Even when a closed form exists, most integrators just weren't meant to deal with such nonsmooth, let alone discontinuous, functions (and you probably won't have much luck manually for anything nontrivial). What I have done is assume that the number of units is continuous (which isn't too bad if the number is above, say, 10). Then if N represents the number of units at a given time, a represents the attack power (damage dealt per attacking unit per unit time) and d represents the defense power (hit points per unit), you have the differential equations:

${\frac {dN_{1}}{dt}}={\frac {-N_{2}a_{2}}{d_{1}}}$

${\frac {dN_{2}}{dt}}={\frac {-N_{1}a_{1}}{d_{2}}}$

To which the solution is (denoting $r={\sqrt {\frac {a_{1}d_{1}}{a_{2}d_{2}}}}$ and $q={\sqrt {\frac {a_{1}a_{2}}{d_{1}d_{2}}}}$:

$N_{1}(t)=n_{1}\mathrm {Cosh} (qt)-{\frac {n_{2}}{r}}\mathrm {Sinh} (Qt)$

$N_{2}(t)=n_{2}\mathrm {Cosh} (qt)-n_{1}r\mathrm {Sinh} (Qt)\;\!$

See hyperbolic functions. For armies too small for this approximation to be acceptable, you might have to resort to simulating the battle (by calculating at each step how much time it will take for one unit to be eliminated). -- Meni Rosenfeld (talk) 09:24, 18 September 2007 (UTC)

You might want to look at this patent http://patimg2.uspto.gov/.piw?Docid=06729954&homeurl=http%3A%2F%2Fpatft.uspto.gov%2Fnetacgi%2Fnph-Parser%3FSect1%3DPTO1%2526Sect2%3DHITOFF%2526d%3DPALL%2526p%3D1%2526u%3D%25252Fnetahtml%25252FPTO%25252Fsrchnum.htm%2526r%3D1%2526f%3DG%2526l%3D50%2526s1%3D6,729,954.PN.%2526OS%3DPN%2F6,729,954%2526RS%3DPN%2F6,729,954&PageNum=&Rtype=&SectionNum=&idkey=NONE&Input=View+first+page which calculates damages on groops of soldiers off screen based on various parameters, including formations, supporting units etc . Like all patents there's a lot of useless stuff to read, but there are some good ideas in there if you can be bothered. It's just a model though if you want a continuous solution the use the method given above.87.102.7.192 10:14, 18 September 2007 (UTC)

By the way could you tell us what each symbol in your expression meant - it wasn't immediately apparent.87.102.7.192 10:15, 18 September 2007 (UTC)

(Your differential equation aside.) If you want to use this for a computer simulation I'd really recommend calculating individual combats and using BSP trees or similar to kill the comutation complexity stone dead. (Quadtree is a better link than BSP tree.)87.102.7.192 11:11, 18 September 2007 (UTC)

Thanks a lot, this has been very helpful. For a quick explanation of my equation, x generally means damage per second, y means hitpoints and n the number of units.

This equation came from $x_{\Sigma A}(t)=x_{A}\left(n_{A}(0)-\left\lfloor {\frac {x_{\Sigma D}(0)\cdot t_{1}}{y_{A}}}\right\rfloor -\left\lfloor {\frac {\int _{t_{1}}^{t}x_{\Sigma D}(\tau )d\tau }{y_{A}}}\right\rfloor \right)$ and $x_{\Sigma D}(t)=x_{D}\left(n_{D}(0)-\left\lfloor {\frac {\int _{t_{1}}^{t}x_{\Sigma A}(\tau )d\tau }{y_{D}}}\right\rfloor \right)$

In this case, ΣA starts getting attacked by ΣD at t=0, and ΣA starts attacking ΣD at t=t₁

Basically, the total damage per second of the group A is the damage per second at t=0 (equal to the number of units time the DPS of each unit), minus the damage per second lost between t=0 and t=t₁ (that is, the DPS of one unit times the number of units lost) minus the damage per second lost between t=t₁ and t, that is the integral on the third part. For the total damage of group B, it's the same except for not losing any DPS between t=0 and t=t₁

All this was from some basic assumptions, like every shot hits instantly, no damage is lost (no overkills), and the full group starts attacking at the same time.

I like your method, Meni Rosenfeld, but as you pointed out, it isn't very accurate for small numbers of units (I would of like to compare how well one single unit with DPS 1000 and 1000 HP would have done compared to 20 units with DPS 50 and HP 50 (obviously, with no overkill, it should win, but it would be nice to have the full solution.)

And again, why is the floor function so bad to integrate ? What is wrong with $\int _{a}^{b}\lfloor cx\rfloor \;dx=\left[{\frac {1}{2c}}\lfloor cx\rfloor (2cx-\lfloor cx\rfloor -1)\right]_{a}^{b}$ ? (Well, I suppose this equation wouldn't suffice to resolve the equation I gave...)

Checked it - I'm 90% certain you are right (touch wood) note - you don't need the "-1" at the end - it cancels.87.102.7.192 15:28, 18 September 2007 (UTC)

Finally, how can I obtain numerical solutions here, with the integral in an integral ? Would it be better to start with the two separate equations for ΣA and ΣD ?

Thanks a lot. --Xedi 14:14, 18 September 2007 (UTC)

Here's what I'd do to simplify. Split the equation into before and after t1 - since there's a boundary in the behaviour here. First make an equation for the damage up to t1 - this should be simple. If the unit survives these 'free shots' from the 'defender', then use a second equation with the start values coming from the remaining hit points after time t1 - the differential equations given above should be right for this part.

So for your first part up to t1 the damage seems to be "'Defender number' x time x 'defence factor'" .. if this value exceeds the 'hit points' for A(ttacker) then terminate. If it doesn't just subtract from the original attacker 'hit points' and finish the battle (without the difficult boundary condition) as per Meni Rosenfeld or another model of your choice.87.102.7.192 14:49, 18 September 2007 (UTC)

The floor function doesn't (isn't) look difficult to integrate at all - I get f(b+1)²/2 + f(a)²/2 - f(b)² -f(b) +b f(b) -a f(a) f(a)²/2 - f(b)²/2 +b f(b) -a f(a) (correcting - take with pinch of salt if you haven't checked..) for the integral between b and a of f(x) = floor(x) .. I haven't checked your result but I'd bet it's right (it's easy to verify anyway).87.102.7.192 15:04, 18 September 2007 (UTC)

It's not that $\int _{a}^{b}\lfloor cx\rfloor \;dx=\left[{\frac {1}{2c}}\lfloor cx\rfloor (2cx-\lfloor cx\rfloor -1)\right]_{a}^{b}$ is wrong (though I haven't checked it), but rather that I don't think it can be generalized to more complicated integrands, and that the algorithms used by typical integration software only know how to deal with smooth functions.

The best way to obtain accurate numerical results is the simulation I alluded to above, which I will explain in more detail. You start at time t=0. You calculate the time it will take for a unit in group 1 to be eliminated, and the time it will take for a unit in group 2. You take the smaller value; you jump to that time, remove 1 unit from the appropriate group and do the appropriate amount of damage to the other group. You calculate again the times for elimination, and so on. The time it will take is linear in the number of units, so you can easily handle myriads of units. You can also use all sorts of hybrid algorithms - If both groups have many units, you can use the continuous version until you get to 1000's of units, and then switch to the discrete version. If group 1 consists of a few big units and group 2 consists of many little units, you can have each time step be whenever a unit from group 1 is destroyed, and treat group 2 as a continuous variable. The equations for calculating the time step will be similar to my differential equations above, with the difference that damage to group 2 is constant in each step.

For the simple case of 1 big unit versus many little ones, you can do something similar - try to solve the equations and see what you get. A useful approximation is that the power of a group of n units with attack a and defense d is $adn(n+1)$; the difference in power between two combating groups will always be roughly constant. See where that gets you. This easily gives that in your example of $n_{1}=1,a_{1}=1000,d_{1}=1000,n_{2}=20,a_{2}=50,d_{2}=50$, the 2nd group will be defeated after dealing roughly 500 damage to the first (this makes sense - they will survive one second, in which they do an average damage of roughly 500 per second).

In short, I suggest you take these ideas, try to implement them and experiment with the results. You will get a better understanding by doing so. -- Meni Rosenfeld (talk) 15:14, 18 September 2007 (UTC)

Thanks a lot. I just finished using this method, it seems to be working great. Surely a lot easier than those integrals, but a bit less charming as it is only numeric (my fault for liking analytic expressions too much.). It really helped. --Xedi 17:01, 18 September 2007 (UTC)

Physics momentum question

If I have two masses m/2 connected by a rigid rod of length 2r (at angle A to the horizontal) that is struct by a mass of M, velocity V₀ moving horizontally (assuming elastic collisions) at an offset x from the centre of mass of the mass connected rod..

1.Must the impulse be normal to the rod?

2.Will the collision result in any rotation of the mass connected rod?

3.I'm having diffucultly factoring in any resultant rotation of the rod - though it looks like the rod should rotate if x<>0 because the impulse (change of momentum) occurs at a point off the centre of mass

4.Please explain where and why my reasoning is going wrong in (3.) - and please could someone either explain what I should be doing and maybe give a link.. Thanks.87.102.7.192 12:35, 18 September 2007 (UTC)

5. I could refactor the question to ask what if an impulse H acts normally at distance x from the centre of mass of a body of rotational moment inertia I.. the answer seems to be that no rotational movement is given because of the conservation of angular momentum.. Is this right? If so why am I totally unable to reproduce this result eg pushing a brick off axis with a pen - the brick always starts to rotate..?87.102.7.192 13:49, 18 September 2007 (UTC)

I feel like there's a more clever way to solve this problem, but there's certainly a very straightforward way that works. Write ${\vec {v}}_{0},{\vec {v}}',{\vec {v}},{\vec {v}}_{1},{\vec {v}}_{2}$ for the initial incoming velocity (with no y component), the outgoing velocity, and the velocities immediately after impact of the bar and the masses that are away from and close to the impact, respectively. Write ${\vec {x}}$ for the vector from the center of the bar to the impact point, ${\hat {n}}$ for the normal to the bar (that points up if $A=0$), and $\omega$ for the angular velocity of the bar. Assuming a pure, instantaneous elastic collision with no friction, the force on the bar must be perpendicular to it, so ${\vec {v}}\|{\hat {n}}\Rightarrow v_{x}=v_{y}\tan A,\,v=v_{y}\sec A$. From the rigidity of the bar we have ${\vec {v}}_{12}={\vec {v}}\mp r\omega {\hat {n}}$. Now conservation of linear momentum gives us $M{\vec {v}}_{0}=M{\vec {v}}'+{\frac {m}{2}}({\vec {v}}_{1}+{\vec {v}}_{2})=M{\vec {v}}'+m{\vec {v}}$; conservation of kinetic energy gives us $Mv_{0}^{2}=Mv'^{2}+{\frac {m}{2}}(v_{1}^{2}+v_{2}^{2})=M(v_{x}'^{2}+v_{y}'^{2})+{\frac {m}{2}}(v^{2}-2r\omega {\vec {v}}\cdot {\hat {n}}+r^{2}\omega ^{2}+v^{2}+2r\omega {\vec {v}}\cdot {\hat {n}}+r^{2}\omega ^{2})=M(v_{x}'^{2}+v_{y}'^{2})+m(v^{2}+r^{2}\omega ^{2})$. Conservation of angular momentum about ${\vec {x}}$ (chosen because the projectile then never has any) gives us ${\vec {v}}_{1}\times (-(x+r){\hat {x}})+{\vec {v}}_{2}\times ((r-x){\hat {x}})=0$ (where ${\hat {x}}\neq {\hat {\imath }}$), or (noting that ${\vec {v}},{\hat {n}}\perp {\hat {x}}$ and $x\omega <0$) $-2vx+2r^{2}\omega =0$ or $vx=r^{2}\omega$, so our quantity $r^{2}\omega ^{2}={\frac {v^{2}x^{2}}{r^{2}}}$. From the linear momentum equation we have $v_{y}={\frac {M}{m}}v_{y}'$ and then (using ${\vec {v}}\|{\hat {n}}$) $v_{x}'=v_{0}-v_{y}'\tan A$, so combining everything into the kinetic energy equation gives us (using $v=v_{y}\sec A$) $Mv_{0}^{2}=M(v_{0}^{2}-2v_{0}v_{y}'\tan A+v_{y}'^{2}\tan ^{2}A+v_{y}'^{2})+mv_{y}^{2}\sec ^{2}A\left(1+{\frac {x^{2}}{r^{2}}}\right)=Mv_{0}^{2}-2Mv_{0}v_{y}'\tan A+Mv_{y}'^{2}\sec ^{2}A+{\frac {M^{2}v_{y}'^{2}}{m}}\sec ^{2}A\left(1+{\frac {x^{2}}{r^{2}}}\right)$. This is a quadratic in $v_{y}'$: collecting terms gives $\left(1+{\frac {M}{m}}\left(1+{\frac {x^{2}}{r^{2}}}\right)\right)\sec ^{2}Av_{y}'^{2}-2v_{0}\tan Av_{y}'=0$, with the obvious solution $v_{y}'=0$ that corresponds to no collision occuring. Throwing it away, we have $v_{y}'={\frac {2v_{0}\tan A}{\left(1+{\frac {M}{m}}\left(1+{\frac {x^{2}}{r^{2}}}\right)\right)\sec ^{2}A}}$; the other quantities of interest follow immediately (and $\omega \neq 0$ unless $0\in \{{\frac {M}{m}},{\frac {x}{r}},v_{0},A\}$). Note that I have assumed $A\neq {\frac {\pi }{2}}$ here, although I imagine that continuity guarantees that the results equal their limits in that case. --Tardis 16:49, 18 September 2007 (UTC)

The bit I always get confused about is the 'conservation of angular momentum' which initially total angular momentum is zero. But if ω<>0 (which occurs when x<>0 etc according to the above) - then the system after collision would have angular momentum (mωr).. So that seems contradictory to me.. I'm not sure where the error lies here - in my reasoning or what?87.102.7.192 18:34, 18 September 2007 (UTC)

The law of conservation of angular momentum only applies to a closed system. Whatever delivers the impulse isn't part of the system, so it isn't closed. — Daniel 22:14, 18 September 2007 (UTC)

Of course, if you include the impacting particle in your system then the angular momentum of the whole system is conserved. Note that the solution provided by Tardis implicitly assumes this, but he cleverly calculates angular momentum about the point of impact. As the paths of the impacting particle both before and after impact pass through this point, the angular momentum of the impacting particle about this point is 0, both before and after impact. So calculating angular momentum about this point allows Tardis to ignore the angular momentum of the impacting particle. Gandalf61 12:24, 19 September 2007 (UTC)

Inverse Trig Equations

I have a homework question that says "simplify the following : tan(arcsin(x))"

how do i even go about trying to solve that? —Preceding unsigned comment added by 76.78.16.35 (talk) 20:57, 18 September 2007 (UTC)

You can write tangent in terms of sine and cosine. Cosine, in turn, can be written in terms of sine. The rest should be pretty simple remembering that $f(f^{-1}(x))=x$ for x in the range of f. Donald Hosek 22:14, 18 September 2007 (UTC)

• $\theta =\arcsin {x}$
• $x=\sin {\theta }$

• 
• Proofs_of_trigonometric_identities

Thus, $\sin {\theta }={\frac {a}{h}}$

• $a=x,h=1$
• Therefore:
• $b={\sqrt {1-x^{2}}}$
• Finally,
• $\tan {\arcsin {x}}=\tan {\theta }={\frac {a}{b}}={\frac {x}{\sqrt {1-x^{2}}}}$

--Mostargue 22:32, 18 September 2007 (UTC)

Mostargue, you're new here so I'll fill you in that we don't generally provide complete answers to homework problems here. Providing a good hint is harder (for us!), but in the long run is much more helpful to the questioner. -- Meni Rosenfeld (talk) 00:01, 19 September 2007 (UTC)

oh... lol sorry I got a little carried away. --Mostargue 07:02, 19 September 2007 (UTC)

September 19

Plotting software

Can someone reccomend the most user-friendly, down to earth peice of free software for plotting simple complex functions such as $f(z)=z\times (2+3i)\,$? The less command line syntax crap the better!