Question

For the following sets of functions: i) show that each is orthogonal
on the given interval, and ii) determine the corresponding orthonormal
set.

Answer 1

\[{\sin2nx},n=1,2,3,..., 0\le x \le \pi\]
\[cosn \pi x, n=0,1,2,... , 0 \le x \le 2\]
\[\sin((n \pi x)/L), n=1,2,3,... , x \in [-L,L]\]

Answer 2

Right, these are the key results you'll need to prove your formula for the Fourier coefficients. And to show this result, just evaluate the integrals.

That is, two of these functions from the first list say, are orthogonal if there inner product is zero. That is, by definition of that inner product
\[ \int_0^\pi \sin(2nx)\sin(2mx) \ dx = 0 \]
when \( m \neq n ). Hence to show that, just evaluate the integral.

Repeat for each these these sets.

Answer 3

* \( m \neq n \)

Answer 4

That will prove orthogonalility? But how do I find the orthnormal set? Or am I overlooking something?

Answer 5

As with any collection of vectors { v1, v2 , ... , vn }, once you have established they are orthogonal, to make the set orthnormal, you'll need to normalize each vector. Here, as in every inner product space, the norm you use for that normalization is the square root of the inner product of the vector with itself: i.e.,
\[ ||v|| = \sqrt{ < v, v> } \]
and
\[ < v, v > = \int_a^b v^2(x) dx \]

Answer 6

For example,
\[ || \sin(2x) || = \sqrt{ \int_0^\pi \sin^2(2x) dx } = \sqrt{ \pi/2} \]
Thus the function \( \sin(2x) \) normalized is
\[ \frac{2/pi} \sin(2x) \]

Answer 7

**correction:
\[ \sqrt{2/\pi} \sin(2x) \]