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October 16[edit]

cbse mathematics question papers[edit]

i want cbse 11th maths chapterwise question papers. Please help me....... —Preceding unsigned comment added by 59.92.241.69 (talk) 12:45, 16 October 2009 (UTC)[reply]

I think "cbse" means Central Board of Secondary Education. I do not understand the rest of your request. Do you want study questions? –RHolton ≡– 13:44, 16 October 2009 (UTC)[reply]

It sounds like OP's asking for past papers? Vimescarrot (talk) 22:38, 16 October 2009 (UTC)[reply]

The Limiting Function of a Fourier Series[edit]

So given a Fourier series, we can say a lot about the function from it was generated. Things like convergence (in various norms), if the function is even, odd, or neither, if it is continuous, differentiable, how many times is it continuously differentiable, even things like the period of the function. My question is, given a Fourier series, is there some sort of an "inverse" operation we can do to retrieve the original function exactly? Is there any way to tell what function was used to generate the given Fourier series? My questions in particular is about this
$g(x)={\sqrt {5}}-\sum _{n=1}^{\infty }\left({\frac {\sin(n{\sqrt {5}})}{n^{4}}}\cos(nx)+{\frac {1}{n^{4}}}\sin(nx)\right).$
What is this converging to? Thanks! -Looking for Wisdom and Insight! (talk) 23:25, 16 October 2009 (UTC)[reply]

I'm not exactly sure what you're asking. I mean, there exist various manipulations you can do on series to rewrite them as other series, but the series you've given seems to be a perfectly good closed-form expression for a function to me. Ray^Talk 02:03, 17 October 2009 (UTC)[reply]

The answer is no. For example, look at two periodic functions each with one jump discontinuity, differing from each other ONLY in that one is continuous from the right and the other from the left at that jump. They both have the SAME Fourier series. Michael Hardy (talk) 02:17, 17 October 2009 (UTC)[reply]

...but I should add that there is an inverse operation if, instead of pointwise defined functions, you look at Fourier series of certain sorts of generalized functions, or Fourier series of well-behaved measures. Michael Hardy (talk) 02:19, 17 October 2009 (UTC)[reply]

This seems like a circular question. The function that generates the series above is g(x). In other words series converges to a function that generates it. If your asking if there is some way to generate a closed form expression for the function given the series then the answer is probably no.--RDBury (talk) 14:22, 17 October 2009 (UTC)[reply]

"series converges to a function that generates it" is well known to be false. Michael Hardy (talk) 14:40, 17 October 2009 (UTC)[reply]

We have an article on this: Convergence of Fourier series. Michael Hardy (talk) 14:47, 17 October 2009 (UTC)[reply]

I should have added "with assumptions about absolute convergence etc." The terms in the series given are all O(1/n⁴) so I kind of assumed that would be implied by the context.--RDBury (talk) 15:03, 17 October 2009 (UTC)[reply]

The fact remains (as I pointed out above) that more than one function generates the same series. Isn't that what the question asked? Michael Hardy (talk) 15:10, 17 October 2009 (UTC)[reply]

My interpretation of the question is he wants an expression for the function that generates the series. What I should of said is that g(x) is one such function. But I think what the original poster was really looking for is a closed form expression and I doubt it exists. Not really sure why this is turning into an argument, you clearly know a lot more about the subject than I do and I've already agreed that I should have been more careful about the statement I made.--RDBury (talk) 15:30, 17 October 2009 (UTC)[reply]

OK, maybe I'm getting distracted by the language that was used. The poster used the word "inverse". The way I am accustomed to think of it, summing the series is the inverse operation. So if the question is whether the inverse operation can tell you the original function, the answer is that there are certainly limits on how far you can take that, and the question of what the limits are is complicated.

But if the poster meant "How do you sum the series?", that's another matter. In this case n&sqrt;5 looks as if it might not be easy to deal with, but I'm not sure. Michael Hardy (talk) 16:59, 17 October 2009 (UTC)[reply]

Rewrite the sines and cosines as complex exponentials; $g(x)$ can then be written in closed form quite easily in terms of polylogarithms. It's doubtful that it can be reduced to elementary functions, but I'm not sure. Fredrik Johansson 17:22, 17 October 2009 (UTC)[reply]

I see your point. I was thrown by the n root 5 as well but you can expand the product into functions of

x\pm {\sqrt {5}}

. If it was n³ in the denominator then it looks like you could get a piecewise polynomial function by repeated integration of the sawtooth function, but the parity is wrong for that in this case.--RDBury (talk) 18:27, 17 October 2009 (UTC)[reply]

Okay so I understand, in general, two functions may be distinct (on a set of measure zero? or just countably many points? or just a finite number of points?) and yet may still generate the same exact Fourier series. But for this specific example, the convergence is uniform so the limiting function must be continuous for all real numbers. In fact, it is continuously differentiable 3 times (and the Fourier series can also be differentiated term by term 3 times). So the generating function must be unique. The problem is that I was given this Fourier series WITHOUT the function which generated it so I am just curiously investigating if there is any way to retrieve the original generating function. -Looking for Wisdom and Insight! (talk) 04:59, 18 October 2009 (UTC)[reply]

Two L¹ functions have the same Fourier coefficients if and only if they disagree on a set of measure zero. In particular, the generating function is never unique. — Emil J. 11:58, 19 October 2009 (UTC)[reply]