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Problem statement: Suppose f(x) = cosx in the interval $\left[-\frac\pi2, \frac\pi2\right]$. Determine $a_0$ of the Fourier expansion of f(x).

What I think: Fourier expansion should be for a periodic function of $2\pi$. Fourier expansion for any function $f$ integrable between $\left(-\pi, \pi\right)$:

$$()=_0 + \sum_{n=1}^\infty a_n\cos(nx) + \sum_{n=1}^\infty b_n\sin(nx) $$

So, integrating both sides from $\left[-\frac\pi2, \frac\pi2\right]$. should give:

$$\int_{-\pi}^\pi f()dx=_0\int_{-\pi}^\pi dx + \sum_{n=1}^\infty \left(\int_{-\pi}^\pi a_n\cos(nx)\,dx + \int_{-\pi}^\pi b_n\sin(nx)\,dx\right) $$ Now, $$\int_{-\pi}^\pi a_n\cos(nx)\,dx = 0$$ $$\int_{-\pi}^\pi b_n\sin(nx)\,dx = 0$$ which gives: $$a_0 = \frac1 {2\pi} \int_{-\pi}^\pi f()\,dx$$

My doubt: What will change due to the given range $\left[-\frac\pi2, \frac\pi2\right]$? The above expansion is valid for $\left[-\pi, \pi\right]$, will the answer obtained from it hold true for the range $\left[-\frac\pi2, \frac\pi2\right]$ too?

Bernard
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Alpha001
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  • Your $f(x)$ is not periodic as given. You can extend it to an even periodic function of period $\pi$ by simply repeating $f(x)$ every $\pi$ units. Then $a_0$ will be given as an integral from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$. Alternatively, simply use $cos(x)$ with period $2\pi$ as your extension of $f(x)$ so that $a_0 = 0$. Other periodic extensions are possible too, each with their own fourier series. – Paul Jul 18 '21 at 14:52

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