Taylor expansion of $\sqrt{1+x^4}$ for $x>0$

Question

This is the first question that I am asking here and as the title says, I'm curious about the Taylor expansion of $\sqrt{1+x^4}$ for $x>0$. If anybody can help with this, I'd greatly appreciate your input.

Answer 1

You can't get such a series for all positive values of $x.$ What you may get will work for positive $x$ less than $1.$ You will have to check specially whether it works also when $x=1.$ In any case, we can definitely say that no such series exists for your function for $x>1.$

To explicitly compute this series, write your function as $$(1+x^4)^{1/2},$$ and use the binomial theorem to deduce that this is equal (in the range $|x|<1$) to the series $$1+\sum_{k=1}^{\infty}\frac{1(1-2)(1-4)(1-6)\cdots(1-2k+2)}{2^kk!}(x^4)^k=1+\frac12x^4+\sum_{k=2}^{\infty}\frac{(-1)(-3)(-5)\cdots(3-2k)}{2^kk!}x^{4k}.$$

Answer 2

$$\frac{1}{\sqrt{1-x}}=\sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}x^n $$ is fairly well-known and can be proved in many ways. By applying $\int_{0}^{z}(\ldots)\,dx$ to both sides we get $$ \sqrt{1-z}=\sum_{n\geq 0}\frac{1}{4^n(1-2n)}\binom{2n}{n}z^n $$ then by replacing $z$ with $-x^4$ we have $$ \sqrt{1+x^4} = \sum_{n\geq 0}\frac{(-1)^n}{4^n(1-2n)}\binom{2n}{n}x^{4n}.$$ The radius of convergence is $1$, of course, also because $\frac{1}{4^n}\binom{2n}{n}\sim\frac{1}{\sqrt{\pi\left(n+\frac{1}{4}\right)}}$.