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January 12[edit]

A Pascal's triangle question I've never seen before[edit]

Dear Wikipedians:

I'm used to the following type of questions which applies Pascal's triangle, it asks how many ways to form the word "November":

    N
   O O
  V V V
   E E
    M
   B B
  E E E
 R R R R

as you can see each line is always 1 character more or less than the line above it, however, recently my teacher gave the following variation, which totally stumped me:

    N
   O O
  V V V
  E E E
   M M
 B B B B
  E E E
   R R

as you can see there are now lines with same number of characters as previous line, and also line with 2 characters more than the previous line, how do I handle this?

Thanks,

76.65.14.151 (talk) 00:09, 12 January 2009 (UTC)[reply]

I assume the goal is to move from top to bottom without "jumping too far". If no rules are specified then I would guess (and state that as an assumption if possible) that if an M is directly above a B then you can only move directly down to that B, so two of the B's cannot be reached. If the question form allows it then you could also make two answers where the other assumes you can move left-down, straigth down, or right-down, so from each M you can move to 3 B's. PrimeHunter (talk) 00:27, 12 January 2009 (UTC)[reply]

I'm not sure how to interpret the question in the second case - how do you form the word? Are you allowed to move diagonally when letters are directly on top of each other? If not, then the question is easily reduced to one like the first case. If you are, then it depends on precisely what is and isn't allowed. Incidentally, I don't see how Pascal's triangle comes into it - for the first case the answer is just 2ⁿ where n is the number of lines which are longer than the line before (since, when going to those lines, you have a choice of 2 letters to go two, for all the other lines you just have 1 choice).That misses a few options, sorry. Seeing Yanwen's post below, I see where Pascal's triangle comes into it.--Tango (talk) 00:32, 12 January 2009 (UTC)[reply]

The question is understated, can we have some more information please; such as an explicit copy of the said question. In your post you only reference the question you are trying to answer without actually stating what it is that you are trying to do. I too fail to see any relation to pascals triangle. Or the fact that the layers form a word. Or what you mean by 'handle this'. —Preceding unsigned comment added by 92.16.196.156 (talk) 01:41, 12 January 2009 (UTC)[reply]

If you take one letter from each row you get the word "NOVEMBER", the question seems to be how many different routes are there from the top row to the bottom. --Tango (talk) 02:52, 12 January 2009 (UTC)[reply]

One way to do it is to add up the numbers above it. Taking your first example:

    N
   O O
  V V V
   E E
    M
   B B
  E E E
 R R R R

You start with N. There is only 1 way to get to the two O's. Then, from the two O's, there are 2 ways to get to the center V, and 1 way to get to the V's on the side. And you just keep going until you get this:

Then just add up the numbers on the bottom: 6+18+18+6=48

The solution to the variation would then be:

49+49=98

(The two digit numbers throws off the alignment, but hopefully you can still read it alright.)--Yanwen (talk) 03:18, 12 January 2009 (UTC)[reply]

Hi, I'm the original question poster. Thank you for all the input! I think Yanwen's solution is the one that I was looking for. I think the question is really understated and badly formed. But I really appreciate the input from everyone. 70.52.148.139 (talk) 00:53, 13 January 2009 (UTC)[reply]

Leibniz notation[edit]

In calculus, dx, dy, du, etc. seem to be manipulated as if they were variables. This confuses me.

When you're only dealing with first derivatives, this seems pretty straight forward. If you have y = f(x), then ${\frac {dy}{dx}}={\frac {d}{dx}}f(x)=f'(x)$ . You can multiply both sides by dx to get dy = f'(x)dx and then integrate $\int dy=\int f'(x)dx$ --> y = f(x) + C

But when it comes to second derivatives, you get ${\frac {d^{2}y}{dx^{2}}}$ . Does that mean that $d^{2}x$ and $dx^{2}$ are different things? What's the difference. It seems to me like $d^{2}y=d(dy)$ is an infinitesimal change in the infinitesimal change in y. Isn't that just 0, because dy is already infinitesimally small? And is $dx^{2}{\overset {?}{=}}(dx)^{2}$ in the same way that $\sin ^{n}(x)=(\sin(x))^{n}$ ? --Yanwen (talk) 03:01, 12 January 2009 (UTC)[reply]

Maybe this will help.--Shahab (talk) 09:06, 12 January 2009 (UTC)[reply]

Politely speaking, I don't think that link will be of much use to him (most calculus students don't know linear algebra and differential topology). I think the best help would be to note that dy/dx approaches the derivative (if the function is differentiable) as dx approaches 0.

Treating dx, dy etc. as if they were variables is like crossing the road before you see the green man - it's usually okay if you have some experience and take care, but if in doubt, don't do it. For example, "cancelling" the dus appears to work in the chain rule:

{\frac {dy}{dx}}={\frac {dy}{du}}\,{\frac {du}{dx}}

but if you extend this reasoning to the second derivative you would conclude that

{\frac {d^{2}y}{dx^{2}}}={\frac {d^{2}y}{du^{2}}}\,\left({\frac {du}{dx}}\right)^{2}

which is incorrect - the correct extension is

{\frac {d^{2}y}{dx^{2}}}={\frac {d^{2}y}{du^{2}}}\,\left({\frac {du}{dx}}\right)^{2}+{\frac {dy}{du}}\,{\frac {d^{2}u}{dx^{2}}}

A more rigorous approach is to regard d/dx as an operator D which maps each differentiable function y(x) to another function. Then d²y/dx² represents the second iteration of this operator i.e. D(D(y(x))) or D²(y(x)). Gandalf61 (talk) 10:13, 12 January 2009 (UTC)[reply]

One of the articles referenced from differential (mathematics), probably differential (infinitesimal), should really treat this but doesn't. A previous discussion related to this is archived at WP:Reference_desk/Archives/Mathematics/2008_September_10#What_are_the_rules_of_treating_differentials.3F. Dmcq (talk) 12:22, 12 January 2009 (UTC)[reply]

Yes, d ²x and dx ² are different things. d ²x = ddx = d(dx) and dx ² = (dx) ² = (dx) · (dx) = dx · dx . When x is a variable, then so is the differential dx, and the rules of differentiation a sum d(x + y) = (dx) + (dy) = dx + dy and a product d(x · y) = (dx) · y + x · (dy) = dx · y + x · dy lead to correct formulas. So $\scriptstyle dy={\frac {dy}{du}}\,du$ and $\scriptstyle d^{2}y=d(dy)=d({\frac {dy}{du}}\,du)=d({\frac {dy}{du}})du+{\frac {dy}{du}}d^{2}u.$ You also need to know that if x is a constant, then dx = 0, and if x is an independent variable, then dx is constant and consequently d ²x = 0. But there is a problem in interpreting the differentials as infinitesimals which are zero and nonzero at the same time. Bo Jacoby (talk) 17:04, 12 January 2009 (UTC).[reply]

But if we could treat differentials in exactly the same way as other variables then we would have:

{\frac {d^{2}y}{du^{2}}}\,\left({\frac {du}{dx}}\right)^{2}={\frac {d^{2}y}{du^{2}}}\,{\frac {du^{2}}{dx^{2}}}={\frac {d^{2}y}{dx^{2}}}

which we know is incorrect. So to put differentials on a formal basis, we would need some rules to define when we can treat them just like variables and when we can't. Gandalf61 (talk) 09:52, 13 January 2009 (UTC)[reply]

So if treating dy and dx as variables would not yield consistent correct answers, what would be a better way to manipulate differentials? Also, I don't completely understand the page One-form that Shahab referenced. Is it saying that df is a function of x and dx? Wouldn't that just become the difference quotient?--Yanwen (talk) 16:15, 13 January 2009 (UTC)[reply]

The "dy and dx as variables" approach can be made rigorous within the framework of non-standard analysis. But my personal opinion is that a rigorous treatment of non-standard analysis adds more complexities than it removes, whereas a non-rigorous treatment leaves us no better off than before. I prefer the differential operator approach - this means abandoning the idea that dy and dx can be treated as variables, but you still reach the same results as before, with less room for error. Gandalf61 (talk) 11:24, 14 January 2009 (UTC)[reply]

Apart from nonstandard analysis (which I do not understand), the article on differential algebra explains the formal use of dx. I am a little shaken by Gandalf61's example, though. Bo Jacoby (talk) 12:18, 14 January 2009 (UTC).[reply]

Well. The differential is dy = a dx so that dy/dx = a, and da = b dx so that b = da/dx. The second differential is d²y = b dx² + a d²x. Only when x is an independent variable can we set dx = constant and d²x = 0 and d²y = b dx² . Otherwise the second derivative b is a partial derivative, which is not a quotient between differentials. In Gandalf61's example, u is not an independent variable and so d²y / du² is a partial derivative. Bo Jacoby (talk) 21:30, 15 January 2009 (UTC).[reply]

Lebesgue Integrals[edit]

How do you do Lebesgue Integrals? I don't understand it the way our article said it.----~~The Successor of Physics~~ 11:13, 12 January 2009 (UTC)[reply]

I don't know what you mean by 'doing' Lebesgue integrals. But I explain the process to you. Start with a measure space (that is a set together with a sigma algebra (that is, a subcollection of the collection of all subsets of the set). Members of this sigma algebra are called measurable sets. Each measurable set has associated a non-negative real number called its measure (generalizes the notion of 'area' and 'volume'). So in effect, you are defining a function from the sigma algebra to the extended set of non-negative reals. This function is called a measure. A measure space is simply a set, with a sigma algebra and a measure). Call a real valued measurable function on a measure space simple if its range is finite. An indicator function of a measurable set E (as a subset of the measure space) is simply a function that is 1 at E and 0 outside E. Note that every simple function can be expressed as the linear combination of indicator functions. So one can integrate a simple function by factoring out the constants and considering the measure of the sets on which each indicator function is 1. Now the integral of a real-valued non-negative measurable function is simply the supremum of all integrals of non-negative simple functions that are less than or equal to this function.

While reading the article, keep the above in mind. Hope this helps. PST

Try skipping straight to the "Intuitive interpretation" section of Lebesgue integration. If you want to actually calculate the Lebesgue integral of a function f, one way is to find a function g which you can integrate with Riemann integration and which is "sufficiently close" to f; the Lebesgue integral of f is then equal to the Riemann integral of g. And one useful case of "sufficiently close" is if g(x)=f(x) almost everywhere. So, for example, the Dirichlet function is 0 almost everywhere (because the points where it is non-zero, the rationals, are a set of measure 0) so the Lebesgue integral of the Dirichlet function is equal to the Riemann intergal of 0, which is 0. Gandalf61 (talk) 13:06, 12 January 2009 (UTC)[reply]

It's useful to remember that all Riemann-integrable functions are Lebesgue-integrable, and the Lebesgue integral evaluates to the value of the (proper) Riemann integral wherever the latter integral exists. So for all the functions one is taught to integrate in elementary calculus, there's no difference. The power of the Lebesgue integral is mostly a matter of the deeper theory. Ray (talk) 14:25, 12 January 2009 (UTC)[reply]

To summarize, mathematicians don't really care about what the Lebesgue integral of a particular function is; they are worried about the theory (what functions are Lebesgue integrable, certain properties of the integral, how it behaves for uniformly convergent sequences of functions). PST

It's true that certain kinds of mathematicians don't care about calculating the Lebesgue integral (I for one haven't done this since grad school), but what you say above is certainly false for many pure mathematicians, to say nothing of applied folks who are very interested in calculating things. Staecker (talk) 17:05, 12 January 2009 (UTC)[reply]

I could be wrong, but I think the functions those applied folks integrate tend to be at least piecewise continuous, and hence already Riemann/Darboux integrable. Algebraist 17:14, 12 January 2009 (UTC)[reply]

I don't think applied mathematicians study Lebesgue integration that is not Riemann integration (or Riemann-Stieltjes integration). This is because I don't consider probabilists to be applied mathematicians. PST

Yes, but if the OP means how does one compute a Lebesgue integral of a function that is only L¹, I'd answer: using all the machinery of the theory, starting with convergence theorems. The power of Lebesgue integration with respect to the Riemann's is not that it allows to treat functions of a wider class, but rather that it allows to do much more general operations on them, both of algebraic and topological nature; and these operations in some cases also allow the computation of integrals. --PMajer (talk) 14:14, 13 January 2009 (UTC)[reply]

PS: however, Superwj5: just give these people any L¹ function in bones and flesh, and we are going to compute its integral for you as an example (hopefully) --PMajer (talk) 15:21, 13 January 2009 (UTC)[reply]

Thanks a lot!!!----~~The Successor of Physics~~ 13:35, 16 January 2009 (UTC)[reply]

eⁱ[edit]

What is eⁱ? It is used in some of Euler's identities and formulae but I can't solve them unless I get eⁱ!----~~The Successor of Physics~~ 11:31, 12 January 2009 (UTC)[reply]

Hint: $i^{i}$ is $e^{-pi/2}$ .

It's a transcendental number about which (I think) little can usefully be said except that it is eⁱ. What do you want to know for? Algebraist 12:31, 12 January 2009 (UTC)[reply]

You can of course write it as

\cos 1+i\sin 1

, which helps to get an estimate if you have a calculator that can handle trigonometry but not complex numbers. But for any deeper meaning, I don't believe the sine and cosine of 1 radian are very interesting. -- Jao (talk) 12:59, 12 January 2009 (UTC)[reply]

Superwj5, if you want to see eⁱ in the complex plane: it is on the unit circle, first quadrant, on one end-point of an arc of lenght 1, the other end-point being e⁰=1. If you understand how works geometrically the multiplication of complex numbers, you can visualize it as the limit of (1+i/n)ⁿ. For a fixed n draw the piecewise linear arc through the points (1+i/n), (1+i/n)²,..(1+i/n)ⁿ; then you can figure out what happens taking the limit of (1+i/n)ⁿ and why the final arc has lenght 1. But what does "Successor of Physics" mean, and why it's cancelled? .) --78.13.140.37 (talk) 14:34, 12 January 2009 (UTC)[reply]

Maybe it means "no metaphysics". —Tamfang (talk) 05:17, 14 January 2009 (UTC)[reply]

Thanks a lot!!! I want to know it for the equations sin(x)=(e^ix+e^-ix)/2 and cos(x)=(e^ix-e^-ix)/2i. ----~~The Successor of Physics~~ 13:36, 16 January 2009 (UTC)[reply]

what's the problem? these equations just came from e^ix=cos(x)+isin(x) and e^-ix=cos(x)-isin(x), that just tell you what are the real and the imaginary parts of e^ix and e^-ix. --PMajer (talk) 11:28, 17 January 2009 (UTC)[reply]

combination and permutation[edit]

in a maximum group of 1-50 numbers both inclussive,how many times can a sub group of 5 numbers divide it without repetitions of any group?

This is what you mean:

Question: Consider the set of numbers from 1-50; call this X. How many different subsets of X exist, each having the property that every member divides 50.

Answer: This problem is trivial and will be removed. Please state it more clearly. If my interpretation is correct, the answer is 6C5 (since there are six factors of 50, and every subset of five elements must be a subset of this set of factors).

PST Preceding unsigned comment by Point-set topologist (talk) 20:49, 13 January 2009 (UTC)[reply]

Well, of course the statement about the triviality of your problem reflects at most the personal opinion of PLT, who is however not sure about the interpretation of it. So I do not think it will be removed, and if you would restate it more clearly, even by examples, somebody will certainly give you an answer. --PMajer (talk) 13:48, 13 January 2009 (UTC)[reply]

We usually only remove questions if they violate the legal/medical disclaimer, or are cross-postings on other reference desks. PST, if you don't think a question should be answered (e.g., homework, or vague, or trivial), then just don't answer it.

As for the original question, I too cannot understand what you are asking for help with, could you please restate the problem? Eric. 68.18.55.236 (talk) 14:35, 13 January 2009 (UTC)[reply]

I agree with the above - we don't remove questions for being trivial and I also don't understand the question. PST's interpretation is possibly correct, but it's far from certain. --Tango (talk) 14:57, 13 January 2009 (UTC)[reply]

I feel that sometimes people ask these problems (the question is both grammatically wrong and mathematically imprecise) just to waste our times. If we spend our time and put a lot of effort into answering their questions, can't they be just a little considerate about this when they ask a question (just put a tiny bit of effort into making their questions explicit)? That is what I meant. PST

Sure, it would be nice if they did, but how does removing the question help? --Tango (talk) 22:12, 13 January 2009 (UTC)[reply]

Law of excluded middle[edit]

Is the law of excluded middle equivalent to the axiom of choice? Stifle (talk) 16:22, 12 January 2009 (UTC)[reply]

This is a meaningless question without specifying the base system over which you want the equivalence to hold. The axiom of choice implies the law of excluded middle over a fairly weak fragment of intuitionistic set theory (something like extensionality, separation, and pairing suffices, see Diaconescu's theorem). OTOH, I do not know of any nontrivial (i.e., not already including AC) nonartificial theory where the law of excluded middle implies the axiom of choice. — Emil J. 16:37, 12 January 2009 (UTC)[reply]