How to find taylor series of $ \sqrt{x+1} $ I need to find the infinite taylor series of $ \sqrt{x+1} $.
I tried to just find the deriviatives of $ \sqrt{x+1} $ and search for a pattern, and then prove by induction that indeed the pattern I found was correct, and then prove the Lagrange reminder strive to $ 0 $  as $ n\to \infty $ but it got very complicated and seems not effiecient.
Is there a reasonable way to find this taylor series ? This is a question from an exam, so I guess it shouldnt take too much work (as what I tried) because the time is limited.
Thanks in advance.
Here's what I found:
For any $ 1\leq k $ :
$ f^{\left(k\right)}\left(x\right)=\prod_{n=1}^{k}\left(\frac{1}{2}-n+1\right)\left(x+1\right)^{\left(\frac{1}{2}-k\right)} $
Thus, if we'll name the taylor expansion as $ T_{f,0} $ (taylor expansion of $ f=\sqrt{x+1} $ around $ 0 $ ) we'll get:
$ T_{f,0}=1+\sum_{k=1}^{m}\frac{f^{\left(k\right)}\left(0\right)}{m!}x^{m}=\sum_{k=1}^{m}\frac{1}{k!}\left(\prod_{n=1}^{k}\left(\frac{1}{2}-n+1\right)\right)x^{k} $
Thus, the reminder should be:
$ R_{m}\left(x\right)=\frac{f^{\left(m+1\right)}\left(x_{0}\right)}{\left(m+1\right)!}x^{m+1}=\frac{1}{\left(m+1\right)!}\prod_{n=1}^{m+1}\left(\frac{1}{2}-n+1\right)\left(x_{0}+1\right)^{\frac{1}{2}-\left(m+1\right)}x^{m+1} $
 A: The derivatives of $(1+x)^\alpha$ are relatively easy to find:
$$(1+x)^\alpha\to\alpha(1+x)^{\alpha-1}\to\alpha(\alpha-1)(1+x)^{\alpha-1}\to\alpha(\alpha-1)(\alpha-2)(1+x)^{\alpha-3}\to\cdots$$
and evaluate at $x=0$ as the falling factorials $(\alpha)_k$.
Then the Lagrange remainder reads
$$\frac{(\alpha)_{n+1}}{(n+1)!}(1+x^*)^{\alpha-n-1}x^{n+1}=\frac{(\alpha)_{n+1}}{(n+1)!}(1+x^*)^{\alpha}\left(\frac x{1+x^*}\right)^{n+1},$$ where $0\le|x^*|<|x|.$ Then for $-\frac12<x<1$, the last factor ensures an exponential decay (the others are bounded).
A: I understand that you mean the Taylor series at 0. If you differentiate repeatedly, you see the pattern:
$$
\begin{aligned}
f(x)&=\sqrt{x+1}  \\\\
f'(x)&=\frac{1}{2}\frac{1}{\sqrt{x+1}}  \\\\
f''(x)&=-\frac{1}{2\cdot 2}\frac{1}{(x+1)^{\frac{3}{2}}} \\\\
f'''(x)&=\frac{1\cdot 3}{2\cdot 2\cdot 2}\frac{1}{(x+1)^{\frac{5}{2}}} \\\\
f^{(4)}(x)&=-\frac{1\cdot 3\cdot 5}{2\cdot 2\cdot 2\cdot 2}\frac{1}{(x+1)^{\frac{7}{2}}} \\\\
f^{(n)}(x)&=\dots
\end{aligned}
$$
Observe that:

*

*The signs are alternating.

*The numerators have the products of odd numbers.

*The denominators have powers of two.

*The powers of $\sqrt{x+1}$ are decreasing.

So we can guess the general form:
$$
f^{(n)}(x)=(-1)^{n+1}\,\frac{1\cdot 3 \cdots (2n-3)}{2^n}\frac{1}{(x+1)^{\frac{2n+1}{2}}}
$$
for $n\geq1$ whereas $f(0)=1$.
The proof by induction is simply taking another derivative. If you want a nice formula for the numerator, you can use this:
$$
f^{(n)}(x)=(-1)^{n+1}\,\frac{(2(n-1))!}{2^{n-1}\,(n-1)!\,2^n}\frac{1}{(x+1)^{\frac{2n+1}{2}}}
$$
where again the sign for the case $n=0$ is exceptional, since $f(0)=1$.
Then the series is:
$$
S(x)=1+\sum_{n=1}^\infty (-1)^{n+1}\,\frac{(2(n-1))!}{2^{n-1}\,(n-1)!\,2^n}\,\frac{x^n}{n!}
$$
Finally, you could prove that the remainder converges to zero, but I think that it is easier (yet a bit less rigourous) to argue that the function is analytic in a neighbourhood of 0 (assuming known that $\sqrt{x}$ is analytic at $x=1$) then the series converges to $f(x)$ in the interval of convergence (which can be easily proved to be $(-1,1]$).
As for the requirement of being reasonable, I would put this question in a first-year Calculus exam and give half an hour.
