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I am having a problem from an old exam saying the following

a) Prove that $$ \langle f,g\rangle = \frac{2}{\pi} \int f(t)g(t)\,\mathrm{d}t $$ gives a inner product on $C[0,\pi]$

b) Find real constants $a,b$ to minimize integral $$ \int_0^\pi \left( 1 - a \sin(t) - b \sin(3t) \right)^2 \, \mathrm{d}t $$ Hint: The functions $\sin(t),\sin(3)t$ form a ortonormal set

I have been able to prove a), but I am usnure what to do about b). I know I have to use some properites of the inner product, to minimize the integral but I am unsure how to proceed.

I know the basic calculus, and some basic linear algebra, but I am terrible at combining them.

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up vote 3 down vote accepted

It is my experience that if you're given a "hint," you still have to prove it. However, assuming you can do this, we can proceed:

$(1-a\sin(t)-b\sin(3t))^2 =$

$1 + a^2\sin^2(t) + b^2\sin^2(t) - 2a\sin(t)- 2b\sin(3t) - 2ab\sin(t)\sin(3t)$.

Thus by linearity of integration (all the integrals have $dt$): $\int_0^\pi(1-a\sin(t)-b\sin(3t))^2 =$

$\int_0^\pi1 + a^2\int_0^\pi\sin^2(t) + b^2\int_0^\pi\sin^2(t) - 2a\int_0^\pi\sin(t)- 2b\int_0^\pi\sin(3t) - 2ab\int_0^\pi\sin(t)\sin(3t)$.

By the 'normal' part of orthonormality, we know $\langle u_i,u_i\rangle =\int_0^\pi\sin^2(t) = \int_0^\pi\sin^2(3t) = \frac{\pi}{2}$, so the above expressions becomes:

$\pi + \frac{\pi}{2}a^2 + \frac{\pi}{2}b^2 - 2a\int_0^\pi\sin(t)- 2b\int_0^\pi\sin(3t) - 2ab\int_0^\pi\sin(t)\sin(3t)$.

By the 'ortho' part, we have $\langle u_i, u_j\rangle = \int_0^\pi\sin(t)\sin(3t)=0$ so it becomes

$\pi + \frac{\pi}{2}a^2 + \frac{\pi}{2}b^2 - 2a\int_0^\pi\sin(t)- 2b\int_0^\pi\sin(3t)$.

Can you take it from here?

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Perfect, and beat me to it. +1 – mixedmath May 28 '12 at 12:11
I got a = 2 and b = 2/3, does this seems correct? Thanks a bunch for your answer! Could you perhaps add a small note on how to show that sin(3t) and sin(t) form a orthonormal set? I will gladly accept this answer! – N3buchadnezzar May 28 '12 at 12:19
You made a small mistake here! $ \int_0^\pi \sin(t) dt = \int_0^\pi \sin(3t) dt \neq 1 $ Since $$ \langle f,g\rangle =\int_0^\pi\frac{2}{\pi} f(t)g(t) dt $$ then $\langle \sin t,\sin t\rangle = 1 \Rightarrow \int_0^\pi \sin(t)^2 dt = \frac{\pi}{2}\langle \sin t , \sin t \rangle = \frac{\pi}{2}$, then I finally get the correct answer! =) – N3buchadnezzar May 28 '12 at 12:45
Ah, yes, I did not read carefully enough. Unfortunately, I don't have enough background to know why these functions are orthonormal, but I do know it is a result which you might see in your textbook as an introduction to Fourier series in the more general form that $\{\sin(nt),\cos(nt) : n\in\mathbb{Z}^+\}$ is an orthonormal set. I imagine it can be proven without much effort from the definition? But that seems to defeat the purpose... – Eric Stucky May 28 '12 at 21:51
Also: yes, $(a,b) = (2,\frac{2}{3})$ looks correct. – Eric Stucky May 28 '12 at 22:45

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