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I have an even function with the Fourier series as below:

$f(t) = \frac{a_0}2 + \sum_{n=1}^\infty a_n \cos nt$

with $a_n = \frac{1}\pi $$\int_{-\pi}^{\pi} f(x)\cos nx\ dx$

And I have to demonstrate the Fourier series for $f(\pi (t-2))$ in the interval {1,3} using $a_n$

Here what's I am thinking: Let $f(\pi (t-2)) = \frac{a'_0}2 + \sum_{n=1}^\infty a'_n \cos(n\pi (t-2))$

So $a'_0 = \frac{1}\pi $$\int_{-\pi}^{\pi} f(\pi (x-2))\ dx$

Let $ u = \pi (x-2)$, then $du = \pi dx$, and x: 1 to 3 give u: $-\pi$ to $\pi$

Then $a'_0 = \frac{1}{\pi^2} $$\int_{-\pi}^{\pi} f(u)\ du$ = $\frac{a_0}\pi$

But when I calculate $a_n$ with the same method, it gave $a_n = \frac{1}{\pi^2} $$\int_{-\pi}^{\pi} f(u)\cos n\pi u du$

and I stuck here.

nghiahandai
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1 Answers1

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You don't have to evaluate any coefficients. Just change $t$ to $\pi (t-2)$ in the given series and use the identity $\cos (n\pi (t-2))=\cos (n \pi t) \cos 2n\pi) -\sin (n \pi t) \sin 2n\pi)=\cos (n \pi t)$ The Fourier series on $[1,3]$ is $f=\frac {a_0} 2 +\sum_{n=1}^{\infty} a_n \cos (n\pi t)$. Note that $\{\cos (n \pi t)\}$ is orthonormal on $[1,3]$ (and $\{\cos (n t)\}$ is not!). Hence all you have to do is replace $\{\cos (n t)\}$ by $\{\cos (n \pi t)\}$.

Kavi Rama Murthy
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  • Thank you. But sorry, I don't understand the word "orthonormal" here. I just know "orthonormal vector" but seem like it's unlikely. So could you explain more? And what is the meaning of the given interval [1,3] in this problem? Cause if the solution like what you did, seem like it's irrelevant. – nghiahandai Jun 20 '18 at 09:59
  • I am unable guess how much you know about Fourier series. I suppose you can assume that the series we get by changing $t$ to $\pi (t-2)$ is the Fourier series on $[1,3]$ because the transformation $t \to \pi (t-2)$ takes the original interval $[-\pi, \pi]$ to $[1,3]$. For future study let me inform you that Fourier series of a function is its expansion w.r.t. an orthonormal sequence of functions. Orthonormal means $\int_a^{b} f_n (x)f_m(x), dx=1$ if $n=m$ and $0$ if $n \neq m$. – Kavi Rama Murthy Jun 20 '18 at 10:24
  • Ah, I see. Thank you so much! – nghiahandai Jun 20 '18 at 10:26