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.


4 Answers 4


$\sec\left(x\right)=\dfrac{1}{\cos \left(x\right)}$ 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.

  • $\begingroup$ Do you have any idea how to derive it from basic series expansions like cosx? $\endgroup$
    – user183782
    Oct 22, 2014 at 16:41
  • $\begingroup$ @user183782: It's simply the Taylor series expansion of $\dfrac1{\cos x}$ for x around $0$. $\endgroup$
    – Lucian
    Oct 22, 2014 at 16:46
  • $\begingroup$ I know, but I'm trying to figure out how to derive it from the 5 basic series expansions, without using the Taylor series formula or just plugging in 0 to the derivatives. $\endgroup$
    – user183782
    Oct 22, 2014 at 16:49
  • $\begingroup$ @user183782: Perhaps by writing $1=\cos^2x+\sin^2x$, and using the Cauchy product ? $\endgroup$
    – Lucian
    Oct 22, 2014 at 16:55

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.


  1. M. Abramowitz and I. A. Stegun (Eds), Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, National Bureau of Standards, Applied Mathematics Series 55, Reprint of the 1972 edition, Dover Publications, Inc., New York, 1992.
  2. Xue-Yan Chen, Lan Wu, Dongkyu Lim, and Feng Qi, Two identities and closed-form formulas for the Bernoulli numbers in terms of central factorial numbers of the second kind, Demonstratio Mathematica 55 (2022), no. 1, 822--830; available online at https://doi.org/10.1515/dema-2022-0166.

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .