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Find the series expansion of 1/cosx from basic series expansions.
I tried to find 1/cosx from the expansion of cosx but was unsure how to continue. When I found 1/cosx from the basic formula for finding series expansions I didn't get the same answer.
$\sec\left(x\right)=\dfrac{1}{\cos \left(x\right)}$ is well-behaved about $x=0$, so assume $$ \frac{1}{\cos \left(x\right)} = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots $$ Also, $\cos \left(x\right)$ is $$ \cos \left(x\right) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \ldots $$ So, on performing multiplication of the two series, $$ 1 \equiv a_0 + a_1 x + \left(a_2 - \frac{1}{2!}\right)x^2 + \left(a_3 - \frac{a_1}{2}\right)x^3 + \left(a_4 - \frac{a_2}{2} + \frac{1}{24}\right)x^4 \ldots $$ which, \left(on equating the coefficients on powers of x on the left and right hand sides\right) gives the first few terms as $a_0=1, a_1=0, a_2=\dfrac{1}{2}, a_3=0, a_4=\dfrac{1}{4}-\dfrac{1}{24} = \dfrac{5}{24}$. You can proceed to get more terms or a general formula for the nth term.
(going to the fifth term for an example purpose)
Using the basic expansions of cos(x) gives us
$$ \frac{1}{\cos(x)} = \frac{1}{1-\frac{x^2}{2}+\frac{x^4}{24} + \cdots} $$
of the form $ \frac{1}{1-X} $ which has a known and easy expansion : $$ 1+X+X^2+X^3+X^4+X^5+\cdots $$
where $ X = \frac{x^2}{2} + \frac{x^4}{24} $ (approching $0$ when $x$ approaches $0$). The smaller $x$ term is $x^2$, so we don't need to take more terms than $X^2$ in the above expansion (otherwise terms would exceed $x^5$ and be negligeable).
Hence, $$ \frac{1}{\cos(x)} = 1+(\frac{x^2}{2}−\frac{x^4}{24})+(\frac{x^2}{2})^2+o_{x\to0}(x^5) $$
$$ = 1+\frac{x^2}{2}+\frac{5x^4}{24} $$
$\displaystyle\dfrac1{\cos x}=\sum_{n=0}^\infty a_{n}\dfrac{x^{2n}}{(2n)!},~$ where $a_n$ form this sequence. See Euler number or zigzag number for
more details.
By the Euler formula $$ \operatorname{e}^{\operatorname{i} x}=\cos x+\operatorname{i} \sin x, $$ we find the relation $$ \cos x=\frac{\operatorname{e}^{\operatorname{i} x}+\operatorname{e}^{-\operatorname{i} x}}{2}. $$ Then \begin{align*} \sec x&=\frac1{\cos x}\\ &=\frac{2}{\operatorname{e}^{\operatorname{i} x}+\operatorname{e}^{-\operatorname{i} x}}\\ &=\frac{2\operatorname{e}^{\operatorname{i} x}}{\operatorname{e}^{2\operatorname{i} x}+1}\\ &=\frac{2\operatorname{e}^{\frac12 (2\operatorname{i} x)}}{\operatorname{e}^{2\operatorname{i} x}+1}\\ &=\sum_{n=0}^\infty E_n\biggl(\frac12\biggr)\frac{(2\operatorname{i}x)^n}{n!}, \end{align*} where we used the generating function \begin{equation} \frac{2\operatorname{e}^{xt}}{\operatorname{e}^t+1} =\sum_{n=0}^\infty E_n(x)\frac{t^n}{n!}, \quad |t|<\pi \end{equation} of the Euler polynomials $E_n(x)$. Since the function $\sec x=\frac1{\cos x}$ is even, we derive \begin{equation} E_{2n+1}\biggl(\frac12\biggr)=0 \Longleftrightarrow E_{2n+1}=0,\quad n\ge0 \end{equation} and \begin{align} \sec x&=\sum_{n=0}^\infty E_{2n}\biggl(\frac12\biggr)\frac{(2\operatorname{i}x)^{2n}}{(2n)!}\\ &=\sum_{n=0}^\infty (-1)^nE_{2n}\biggl(\frac12\biggr)\frac{(2x)^{2n}}{(2n)!}\\ &=\sum_{n=0}^\infty (-1)^n\frac{E_{2n}}{2^{2n}}\frac{(2x)^{2n}}{(2n)!}\\ &=\sum_{n=0}^\infty (-1)^nE_{2n}\frac{x^{2n}}{(2n)!}, \end{align} where $E_{n}$ denotes the Euler numbers.
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