Faster way to find the first four non-zero terms of the Maclaurin series for $\frac{1-x}{1+x}\cosh\sqrt{x}$

Question

I want to find the first 4 non-zero terms for :

$$\frac{1-x}{1+x}\cosh\sqrt{x}$$ Before expanding, I rewrite this as $$(1-x)\left(\frac{1}{1+x}\right)\cosh\sqrt{x}$$

Then I expand to get $$(1-x)\left(1-x+x^2-x^3+\cdots\right)\left(1+\frac x {2!}+\frac {x^2} {4!}+\frac{x^3}{6!}+\cdots\right)$$

Now multiplying these brackets and simplifying takes a while (pretty long for my exam time), so is there a faster method to do this or to multiply these brackets?

Answer 1

It's unnecessary to break up $1-x$ and $\frac{1}{1+x}$ into two separate series. You can just differentiate $\frac{1-x}{1+x}$ directly. As J.G. pointed out in his comment, it may be easier to rewrite the expression as $\frac{2}{1+x} - 1$ or $2(x + 1)^{-1} - 1$. This gives you $\frac{1-x}{1+x} = 1 - 2x + 2x^2 - 2x^3 + 2x^4 - 2x^5 + 2x^6 - 2x^7 + ...$.

Then you just have two series to multiply together.

Answer 2

The general expression is

$$ \left(\sum_{n} a_{n}\right)\left(\sum_{n} b_{n}\right)\left(\sum_{n} c_{n}\right)=\sum_{j} d_{j} $$ where $$ d_{k}=\sum_{i+j+l=k} a_{i} b_{j} c_{l} $$

Maybe it will help you.

Answer 3

A slightly faster way might be to first compute the first 4 nonzero terms of $$\frac{1}{1+x}\cosh\sqrt{x}$$ and then multiply the result by $x$ to get the first 4 nonzero terms of $$\frac{x}{1+x}\cosh\sqrt{x}$$ and then subtract.