Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $V=C[-\pi,\pi]$ and define $\displaystyle \langle f,g\rangle:=\frac{1}{\pi}\int_{-\pi}^\pi f(x)g(x)\,dx$.

Let $B:=\{1/\sqrt2,\cos(x),\dotsc,\cos(nx),\sin(x),\dotsc,\sin(nx)\}$.

  1. Show $B$ is a orthonormal set
  2. What is the dimension of $W=\operatorname{span}B$?
  3. For case $n=1$, find orthogonal projection of $f(x)=x$ in $W$ and compute $\min\{\|x-p(x)\|:p\in W\}$

For 1, I can do the following:

$$\begin{align} \frac{1}{\pi}\int_{-\pi}^\pi\cos(nx)\sin(nx)\,dx &= \frac{1}{2\pi}\int_{-\pi}^\pi\sin(nx+nx)-\sin(nx-nx)\,dx \\ &= \frac{1}{2\pi}\int_{-\pi}^\pi\sin(2nx)-\sin(0)\,dx \\ &= \frac{1}{2\pi}\int_{-\pi}^\pi\sin(2nx)\,dx=0. \end{align}$$

But I'm a bit confused as to what to do with that lingering $\frac{1}{\sqrt2}$.

For 2, wouldn't the dimension simply be three?

For 3, I think I could get the explicit form of the projection, but to get the '$\min$,' I'm not quite sure.

share|improve this question
    
Please examine the edits I made to your question carefully. I've tried to make them as close to "proper MathJax+Markdown style" as possible. –  kahen Nov 22 '12 at 19:49
add comment

1 Answer 1

up vote 1 down vote accepted

For 1, you are correctly proving that $\sin(nx)$ is orthogonal to $\cos(nx)$. That's part of what you have to prove, but you have to prove that any two guys on this list are orthogonal. You also have to check that the inner product of any guy on this list with itself is 1.

For 2, no. Since $B$ is an orthonormal set, its elements are linearly independent, so the dimension is the number of elements in $B$, which is $2n + 1$.

For 3, to find your min, express a generic element of $W$ as $v = a\frac{1}{\sqrt{2}} + b\sin(x) + c\cos(x)$ and compute $\langle v - x, v - x \rangle$; that's a quadratic function of $a$, $b$, and $c$ which you are trying to minimize (really you're trying to minimize it's square root).

share|improve this answer
    
Thank you, the computation on 3 is rather long. –  Edgar Aroutiounian Nov 12 '12 at 4:16
    
shouldn't 3 be 0? –  Edgar Aroutiounian Nov 12 '12 at 7:49
    
I get $\frac{2}{3}\pi^2 - 3$ (for the square of the distance), achieved when $a = c = 0$ and $b = 1$. It shouldn't be 0, that would say that $x$ is in the span of these other three functions, which is false. –  user29743 Nov 12 '12 at 16:18
    
but I may have made an error. The computation is less involved if you use $\langle v - x, v - x \rangle = a^2 + b^2 + c^2 - 2\langle v, x\rangle + \langle x, x \rangle$ –  user29743 Nov 12 '12 at 16:19
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.