Fourier series of sec

Real Analysis
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Tolaso J Kos
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Fourier series of sec

#1

Post by Tolaso J Kos »

We have seen (somewhere in the past) the Fourier series of $\cos x$ as well as $\sin ax$. How about the Fourier series of $\sec x$?

Let $x \in \left( - \frac{\pi}{4} , \frac{\pi}{4} \right)$. Expand $\sec x$ in a Fourier series.
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mathofusva
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Re: Fourier series of sec

#2

Post by mathofusva »

Since $\sec x$ is even, we have $b_n = 0$ and
$$a_n = \frac{8}{\pi}\,\int_0^{\pi/4}\sec x\cos(4nx)\,dx, \,\,(n = 0, 1, \ldots.)$$
In view of the trigonometric identity:
$$\cos(4(n+1)x) - \cos(4nx) = -2\sin(2x)\sin((4n+2)x) = -4\sin x\cos x\sin((4n+2)x),$$
we find that
\begin{eqnarray*}
a_{n+1} - a_n & = & -\frac{32}{\pi}\,\int_0^{\pi/4} \sin x\sin((4n+2)x)\,dx\\
& = & - \frac{16}{\pi}\, \int_0^{\pi/4} (\sin((4n+3)x)-\sin((4n+1)x))\,dx\\
& = & -\frac{32}{\pi}\,\frac{(-1)^n}{\sqrt{2}(4n+1)(4n+3)}\\
& = & \frac{(-1)^{n+1}16\sqrt{2}}{(4n+1)(4n+3)\pi}.
\end{eqnarray*}
Since
$$\int_0^{\pi/4} \sec x\,dx = \log(1 + \sqrt{2}),$$
it follows that
$$a_0 = \frac{8}{\pi}\,\log(1 + \sqrt{2})$$
and so
\begin{eqnarray*}
a_n & = & a_0 + \sum_{k=0}^{n-1}(a_{k+1} - a_k)\\
& = & a_0 + \sum_{k=0}^{n-1}\frac{(-1)^{k+1}16\sqrt{2}}{(4k+1)(4k+3)\pi}.
\end{eqnarray*}
Thus,
$$\sec x = \frac{8}{\pi}\,\log(1 + \sqrt{2}) + \sum_{n=1}^\infty a_n\cos(4nx),$$
where $a_n$ is given by the formula above.
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