Taylor series of $\sqrt{1+x}$

Question

I have to give the Taylor series of $\sqrt{1+x}$ with development point $0$. But I am not quite sure what to do.

Normally you have to give the Taylor series up to a certain order.

Well $T_f(x,0)=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n$

I calculated the derivative of $f(x)=\sqrt{1+x}$ up to order 5 and locked for a regularity.

$f^{(n)}(0)=(-1)^n\cdot\frac{\prod_{k\leq n\,\text{with}\, k\,\text{odd}}k}{2^{n}}$

Then the Taylor series is given by:

$T_{f}(x,0)=\sum_{n=0}^\infty \frac{(-1)^n}{\prod_{1\leq k\leq n\,\text{with}\, k\,\text{even}}k\cdot 2^n}x^n$

Which is pretty ugly to work with.

Can you confirm this solution?

Thanks in advance.

Answer 1

Almost correct:

$$\begin{align*} f^{(0)}(x) & = (1+x)^{\frac{1}{2}} \\[1ex] f^{(1)}(x) & = \frac{1}{2} \cdot (1+x)^{-\frac{1}{2}} \\[1ex] f^{(2)}(x) & = \frac{1}{2} \cdot \left(-\frac{1}{2} \right) \cdot (1+x)^{-\frac{3}{2}} \\[1ex] \vdots \\ f^{(n)}(x) & = \frac{1}{2} \cdot \left(-\frac{1}{2} \right) \cdot \ldots \cdot \left( - \frac{2n-3}{2} \right) \cdot (1+x)^{-\frac{2n-1}{2}} \\[1ex] f^{(n)}(0) & = \frac{(-1)^{\color{red}{n-1}}}{2^n} \cdot \prod_{\substack{k \leqslant \color{red}{2n-3} \\ k \text{ odd}}} k \cdot (1+x)^{-\frac{2n-1}{2}} \end{align*}$$

Usually a convenient notation is used: for $\mu \in \mathbb{R}$ and $n \in \mathbb{N}$ let

$$\binom{\mu}{n} = \frac{\mu \cdot (\mu-1) \cdot \ldots \cdot (\mu - n + 1)}{n!}.$$

Then the Taylor series expresses nicely as

$$T_f(x, 0) = \sum_{n=0}^{\infty} \binom{1/2}{n} x^n.$$

Answer 2

Yes, $f(0)=1, f'(0)=\frac{1}{2}, f''(0)-\frac{1}{4}$ We can continue this all the way to see

$f^{(n)}(x)=1/2\cdot-1/2\cdot-3/2\cdot...\cdot\frac{2n-3}{2}(1+x)^{\frac{-2n+1}{2}}$ So, $$f^{(n)}(0)=(-1)^{n-1}\frac{1*3*5*...*(2n-3)}{2^n}$$

Now you can systematically write down the Taylor expansion for $f(x)=\sqrt{1+x}$

Answer 3

I have only a small problem with the sign, $$ (1+x)^a=\sum_{n\ge 0}\binom an x^n =1 +ax +\frac{a(a-1)}{2\cdot 1}x ^2 +\frac{a(a-1)(a-2)}{3\cdot 2\cdot 1}x ^3+\dots $$ so for $a=\frac 12$, the first two signs are plus, then signs alternate. Explicitly: $$ (1+x)^{1/2} = 1 + \frac{1}{2} \, x - \frac{1}{8} \, x^{2} + \frac{1}{16} \, x^{3} - \frac{5}{128} \, x^{4} + \frac{7}{256} \, x^{5} - \frac{21}{1024} \, x^{6} +\dots\ . $$