2
$\begingroup$

Does this series converge?

$$\left(\frac{f_{i}^2}{2!} +\frac{f_{i}^4}{4!}\left(2\kappa c\right)+\frac{f_{i}^6}{6!}\left(2\kappa c\right)^2 + \frac{f_{i}^8}{8!}\left(2\kappa c\right)^3+ \cdots\right), \tag{1}$$

where $k, c \geq 0$. I am trying to establish a relation with the Maclaurin series of $\cosh x$:

$$ 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \cdots=\sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}. \tag{2}$$

$\endgroup$
2
  • $\begingroup$ Hint: if $2 \kappa c < 1$, then you can upper bound $(2 \kappa c)^n < 1$. If instead $2 \kappa c > 1$, then you can bound $(2 \kappa c)^n < (2 \kappa c)^{2n+2}$ and combine it with the $f_i$s. Either way you can make the desired comparison with $\cosh$ $\endgroup$ Commented Dec 16, 2020 at 3:09
  • 1
    $\begingroup$ Indeed, it converges for all $f_i, \kappa, c$ to $$ \frac{\cosh(f_i\sqrt{2\kappa c})-1}{2\kappa c}. $$ $\endgroup$ Commented Dec 16, 2020 at 3:34

1 Answer 1

2
$\begingroup$

Multiply the series given by Eq. (1) by $\frac{2 \kappa c}{2\kappa c}$

$$\frac{1}{2\kappa c}\left(\frac{f_{i}^2}{2!}\left(2\kappa c \right) +\frac{f_{i}^4}{4!}\left(2\kappa c\right)2+\frac{f_{i}^6}{6!}\left(2\kappa c\right)^3 + \frac{f_{i}^8}{8!}\left(2\kappa c\right)^4+ \cdots\right), $$

Add a zero to the above expression

$$\frac{1}{2\kappa c}\left(1+\frac{f_{i}^2}{2!}\left(2\kappa c \right) +\frac{f_{i}^4}{4!}\left(2\kappa c\right)2+\frac{f_{i}^6}{6!}\left(2\kappa c\right)^3 + \frac{f_{i}^8}{8!}\left(2\kappa c\right)^4+ \cdots -1\right), $$

comparing with Eq. (2) of the question, we can establish the relation with $\cosh x$

$$\frac{1}{2\kappa c}\left(\sum_{n=0}^{\infty} \frac{\left[f_{i}\left(2\kappa c\right)^{\frac{1}{2}}\right]^{2n}}{2n!}- 1\right) =\frac{\cosh \left[f_{i}\left(2\kappa c\right)^{\frac{1}{2}} \right]-1}{2\kappa c}.$$

Which is the result given by @Sangchul Lee.

$\endgroup$

You must log in to answer this question.

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