Maclaurin series for $f(x)=\cosh(3x^4)$

Question

is there a faster way to calculate the first $4$ non-zero terms of the Maclaurin series for $f(x)=\cosh(3x^4)$ rather than differentiating the whole function over and over?

ie. does the Maclaurin series of $\cosh(x)$ help reach the answer faster?

Answer 1

To answer your question, yes, the Maclaurin series of $\cosh(x)$ does help find the answer faster. See that if we have

$$\cosh(x)=\sum_{n=0}^\infty a_nx^n$$

Then it follows that we have

$$\cosh(3x^4)=\sum_{n=0}^\infty a_n(3x^4)^n=\sum_{n=0}^\infty a_n3^nx^{4n}$$

Answer 2

Use the fact that

$$ {\rm cosh}t = \frac{1}{2}(e^{t} + e^{-t}) \tag{1} $$

and

$$ e^t = \sum_{n = 0}^{+\infty}\frac{t^n}{n!} \tag{2} $$

Replacing (2) in (1):

$$ {\rm cosh}t = \frac{1}{2}\sum_{n = 0}^{+\infty}\frac{t^n + (-t)^n}{n!} = \sum_{n = 0}^{+\infty}\frac{t^{2n}}{(2n)!} $$

This last step is because all odd terms cancel out in the numerator while the even terms add up. Now, setting $t = 3x^4$ you get

$$ {\rm cosh}(3x^4) = \sum_{n = 0}^{+\infty}\frac{(3x^4)^{2n}}{(2n)!} = \sum_{n = 0}^{+\infty}\frac{9^n x^{8n}}{(2n)!} $$