Math Help - parseval law prove

1. parseval law prove

http://i50.tinypic.com/a9pkpe.gif

1.cant see how the input in *
gives the result
??

there is no mathmatical description

2. Originally Posted by transgalactic
http://i50.tinypic.com/a9pkpe.gif

1.cant see how the input in *
gives the result
??

there is no mathmatical description

1) Define everything that appears in that file: what's f, e_k, etc.

2) Write clearer

3) Write way, waaaay larger: evern with the + is hard to distinguish what's going on there..

4) If possible, much preferable to use LaTeX to write mathematics.

Tonio

3. i cant get this expression
http://i50.tinypic.com/2aq6pu.jpg

why when we input this basis we get this expression

i cant see how its done mathematickly

4. Originally Posted by transgalactic
i cant get this expression
http://i50.tinypic.com/2aq6pu.jpg
The functions $\frac{1}{\sqrt(2\pi)}$, $\frac{cos(nx)}{\sqrt{2\pi}}$, and $\frac{sin(nx)}{\sqrt{2\pi}}$ form an "orthonormal basis". That means specifically, that that the "inner product", defined here as the integral of the product of two functions, is 0 for any two distinct functions of this set and 1 for the product of a function with itself. If f(x) is written as a combination of those functions, multiplying it by itself and then integrating gives the coefficient, squared, for each such function it self- that's the reason for the sum of " $a_n^2+ b_n^2$.

I don't understand your reference to "two sums". $\sum_{n=0}^\infty{a_n cos(nx)}+ \sum_{n=0}^\infty b_n cos(nx)$ can as easily be written as the single sum [tex]\sum_{n=0}^\infty (a_n sin(nx)+ b_n cos(nx)). Similarly $\sum_{n=0}^\infty (a_n^2+ b_n^2)$ can be written as the two sums $\sum_{n=0}^\infty a_n^2+ \sum_{n=0}^\infty b_n^2$.

5. ok why then we have the norm divided by pi
?

6. We don't divide the norm by pi, we divide each sine or cosine by $\sqrt{\pi}$. That way, when we multiply one by itself and integrate, to find the norm, we have the entire product multiplied by $\pi$. Of course the integral of $sin^2(nx)$ is $]int_0^{2\pi} sin^2(nx)dx= \frac{1}{2}\int_0^{2\pi} (1- cos(2nx))dx$ which gives $\frac{1}{2}(x- \frac{1}{2}sin(2nx))$ evaluated between 0 and $2\pi$. Since sin(2nx) is 0 at 0 and at $2\pi$, that integral is $\pi$. A similar calculation gives the same result for cos(nx). Dividing each "basis" function by $\sqrt{\pi}$ divides the integral of the square by $\pi$ and so gives an integral (norm) of 1.

If we start with the "constant" function, 1, the integral is $\int_0^{2\pi} 1 dx= 2\pi$ so we divide that by $\sqrt{2\pi}$.

That way our "basis" is an orthonormal basis.

7. thanks