Taylor series of $f(x)=\sqrt{1+x^4}$

Question

what is the taylor series of $f(x)=\sqrt{1+x^4}$ at $x=0$ I know by the binomial theorem $(1+x^4)^\frac{1}{2}= 1+\sum_{n=1}^{\infty}\frac{(1/2)(1/2-1)(1/2-2)...(1/2-n+1)}{n!}x^{4n}$ but is there any way to simply this?

Answer 1

Let's ignore the first $1/2$ and the $1/n!$ for now and focus in the interesting part:$$(0.5-1)(0.5-2)\cdots(0.5-(n-1))$$

Notice that every element in this product is negative, so this changes from positive to negative every element, we will now "factor" out $-1$ from every element:$$(0.5-1)(0.5-2)\cdots(0.5-(n-1))=((-1)(1-0.5))((-1)(2-0.5))\cdots((-1)((n-1)-0.5))=(-1)^{n-1}\cdot (1-0.5)(2-0.5)\cdots(n-1.5)=(-1)^{n-1}\cdot 0.5\cdot 1.5\cdots \frac{2n-3}2=(-1)^{n-1}\frac12\cdot\frac32\cdots\frac{2n-3}2$$

In a similar way we did with $-1$ we will multiply all the $1/2$ and move them aside:$$(-1)^{n-1}\frac12\cdot\frac32\cdots\frac{2n-3}2=(-1)^{n-1}0.5^{n-1}\cdot1\cdot3\cdot(2n-3)=(-0.5)^{n-1}\cdot1\cdot3\cdot(2(n-1)-1)$$

Now we have a product of the first $n-1$ odd numbers, this is called double factorial and for odd numbers we have $(2k-1)!!=\frac{(2k)!}{2^k(k!)}$, for us $k$ is $n-1$ so we get:$$(-0.5)^{n-1}\cdot1\cdot3\cdot(2(n-1)-1)=(-0.5)^{n-1}\cdot(2(n-1)-1)!!=(-0.5)^{n-1}\cdot\frac{(2n-2)!}{2^{n-1}\cdot(n-1)!}=(-1)^{n-1}\cdot \left(\frac12\right)^{2n-2}\cdot\frac{(2n-2)!}{(n-1)!}$$

Now we will add the previous terms(the one half and and $1/n!$ to get to:$$(-1)^{n-1}\cdot \left(\frac12\right)^{2n-1}\cdot\frac{(2n-2)!}{\underbrace{(n-1)!\cdot n!}_{\text{also equal to: }((n-1)!)^2n}}$$

This is the simplest I could get, so in my comments I had a mistake somewhere.

Answer 2

$$\begin{align*} \prod_{k=0}^{n-1} \left(\frac{1}{2} - k\right) &= \prod_{k=0}^{n-1} \frac{1 - 2k}{2} \\ &= \frac{1 - 2(0)}{2} \prod_{k=1}^{n-1} \frac{(-1)(2k-1)(2k)}{2(2k)} \\ &= \frac{1}{2} (-1)^{n-1} \frac{1}{2^{2(n-1)}} \prod_{k=1}^{n-1} \frac{(2k-1)(2k)}{k} \\ &= (-1)^{n-1} 2^{1-2n} \frac{\prod_{k=1}^{n-1} (2k-1)(2k)}{\prod_{k=1}^{n-1} k} \\ &= (-1)^{n-1} 2^{1-2n} \frac{(2n-2)!}{(n-1)!}, \end{align*}$$ consequently $$\frac{1}{n!} \prod_{k=0}^{n-1} \left(\frac{1}{2} - k\right) = \frac{(-1)^{n-1}}{2^{2n-1} n} \binom{2n-2}{n-1}.$$