http://i50.tinypic.com/a9pkpe.gif
1.cant see how the input in *
gives the result
??
there is no mathmatical description
http://i50.tinypic.com/a9pkpe.gif
1.cant see how the input in *
gives the result
??
there is no mathmatical description
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
The functions $\displaystyle \frac{1}{\sqrt(2\pi)}$, $\displaystyle \frac{cos(nx)}{\sqrt{2\pi}}$, and $\displaystyle \frac{sin(nx)}{\sqrt{2\pi}}$ 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 "$\displaystyle a_n^2+ b_n^2$.
I don't understand your reference to "two sums". $\displaystyle \sum_{n=0}^\infty{a_n cos(nx)}+ \sum_{n=0}^\infty b_n cos(nx)$ can as easily be written as the single sum [tex]\sum_{n=0}^\infty (a_n sin(nx)+ b_n cos(nx)). Similarly $\displaystyle \sum_{n=0}^\infty (a_n^2+ b_n^2)$ can be written as the two sums $\displaystyle \sum_{n=0}^\infty a_n^2+ \sum_{n=0}^\infty b_n^2$.
We don't divide the norm by pi, we divide each sine or cosine by $\displaystyle \sqrt{\pi}$. That way, when we multiply one by itself and integrate, to find the norm, we have the entire product multiplied by $\displaystyle \pi$. Of course the integral of $\displaystyle sin^2(nx)$ is $\displaystyle ]int_0^{2\pi} sin^2(nx)dx= \frac{1}{2}\int_0^{2\pi} (1- cos(2nx))dx$ which gives $\displaystyle \frac{1}{2}(x- \frac{1}{2}sin(2nx))$ evaluated between 0 and $\displaystyle 2\pi$. Since sin(2nx) is 0 at 0 and at $\displaystyle 2\pi$, that integral is $\displaystyle \pi$. A similar calculation gives the same result for cos(nx). Dividing each "basis" function by $\displaystyle \sqrt{\pi}$ divides the integral of the square by $\displaystyle \pi$ and so gives an integral (norm) of 1.
If we start with the "constant" function, 1, the integral is $\displaystyle \int_0^{2\pi} 1 dx= 2\pi$ so we divide that by $\displaystyle \sqrt{2\pi}$.
That way our "basis" is an orthonormal basis.