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I understand it is essentially a maclaurin series, however, is their an easy method of calculating this? Such as working out the Maclaurin for $\cosh(x)$ then inversing the series? (First few terms)

Botond
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Mathematica
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  • Are you seeking the entire series? Or are you looking for a few terms only? – Mark Viola Dec 01 '17 at 22:03
  • First few terms is sufficient, just wish to understand an easier means of calculating it. Thanks. – Mathematica Dec 01 '17 at 22:04
  • The coefficients of the series can be expressed in terms of Bernoulli numbers. If you want the first few terms only, proceed by computing the first few derivatives evaluated at $0$. – Mark Viola Dec 01 '17 at 22:09
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    Certainly, you could write out $\cosh x = 1 + \frac{x^2}{2} + \frac{x^4}{24} + \cdots$, so $\frac{1}{\cosh x} = a_0 + a_2 x^2 + a_4 x^4 + \cdots$. Then multiply the two series to get $a_0 = 1$, $a_2 + \frac{1}{2} a_0 = 0$, $a_4 + \frac{1}{2} a_2 + \frac{1}{24} a_0 = 0$, etc. Then solve iteratively for $a_0$, then $a_2$, then $a_4$, etc. – Daniel Schepler Dec 01 '17 at 22:19
  • @Mathematica Please, if you are ok, you can accept the answer and set it as solved. Thanks! – user Jan 11 '18 at 23:27

4 Answers4

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$\frac{1}{\cosh x}$ is a meromorphic function with simple poles at $\frac{\pi i}{2}+\pi i\mathbb{Z}$.
By computing the residues we have $$\frac{1}{\cosh x}=\sum_{n\geq 0}\frac{4\pi(2n+1)(-1)^n}{4x^2+\pi^2(2n+1)^2} $$ and by expanding the terms of the last series as geometric series we have: $$\begin{eqnarray*}\frac{1}{\cosh x}&=&\sum_{n\geq 0}(-1)^n\sum_{m\geq 0}(-1)^m\left(\frac{4}{\pi(2n+1)}\right)^{2m+1}x^{2m}\\&=&\sum_{m\geq 0}\frac{(-1)^m 4^m x^{2m}}{\pi^{2m+1}}\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^{2m+1}}\\&\stackrel{(*)}{=}&\sum_{m\geq 0}\frac{(-1)^m x^{2m}}{4\cdot(2m)! }\,E_{2m}\end{eqnarray*}$$ where $(*)$ is a consequence of the relation between $\zeta\left(2m+1,\frac{1}{4}\right)-\zeta\left(2m+1,\frac{3}{4}\right)$, Euler numbers and alternating permutations. Euler numbers can be computed from the recursion $$ \color{#C00000}{\mathrm{E}_{2n}=-\sum_{k=0}^{n-1}\binom{2n}{2k}\mathrm{E}_{2k}} $$ proved here by robjohn, which is just a consequence of $1=\frac{1}{\cosh x}\cosh(x)=\frac{1}{\cosh x}\sum_{m\geq 0}\frac{x^{2m}}{(2m)!}$.

Jack D'Aurizio
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If you fix an order of expansion, the simplest way is division by increasing powers of $1$ by the expansion of $\cosh x$ up to this order. At order $4$, you should obtain $$\frac1{\cosh x}=1-\frac{x^2}2+\frac{5x^4}{24}+O(x^6).$$

Bernard
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you can consider for example that

$$\cosh x=1+\frac{x^2}{2}+\frac{x^4}{24}+o(x^6)$$

then

$$\frac{1}{\cosh x}=\left(1+\frac{x^2}{2}+\frac{x^4}{24}+o(x^6)\right)^{-1}$$

an then use the series for

$$(1+x)^a$$ with $$a=-1$$

user
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Well, to find the Taylor expansion about $0$, we can just compute the derivatives. $$f(0) = \frac{1}{\cosh(0)} = 1$$ $$f'(0) = -\sinh(0)\frac{1}{\cosh^2(0)} = 0$$ $$f^{(2)} = \sinh(0)\frac{2}{\cosh^3(0)} -\cosh(0)\frac{1}{\cosh^2(0)} = -1$$ And in general, we can split up $f'$ into $g(x)=-\sinh(x)$ and $h(x)=\frac{1}{\cosh^2(x)}$ and use the generalized product rule to obtain $$f^{(n+1)}(0)=\sum_{i=1}^{n} \binom{n}{i} g^{(i)}(0)f^{(n-i)}(0) =\sum_{i=1}^{\lfloor n/2 \rfloor}-\binom{n}{2i}\cosh(0)f^{(n-2i)}(0)$$ Using $h^{(k)}(0) = (-1)^k\frac{(k+1)!}{\cosh^{k+2}(0)} = (-1)^k (k+1)!$ we have $$f^{(n+1)}(0) = \sum_{i=1}^{\lfloor n/2 \rfloor}(-1)^{n-2i+1} \binom{n}{n-2i}(n-2i+1)!$$ So we have for our Taylor expansion $$a_{n}\cdot n! = f^{(n)}(0) = \sum_{i=1}^{\lfloor (n-1)/2 \rfloor} (-1)^{n-2i}\frac{(n-1)!}{(2i)!}(n-2i)$$ And our Taylor expansion is thus $$a_0 + a_1 + \frac{a_2}{2!} + \frac{a_3}{3!}...$$ There probably exists a simplified formula for $a_n$. I will come back to this later and see if I can find one. This should, however, let you calculate the first few terms of the expansion.

Isaac Browne
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