Creating a degree $n$ Taylor polynomial for $\sqrt{1+x}$

Question

I have been asked to produce a general formula for the degree $n$ Taylor polynomial for $\sqrt{1+x}$ using a=0 as the point of approximation.

Given that $p_n(x)=f(a)+(x-a)f^\prime(a)+\frac{(x-a)^2}{2!}f^{(2)}(a)+...+\frac{(x-a)^n}{n!}f^{(n)}(a)=\sum_{j=0}^{n}\frac{(x-a)^j}{j!}f^{(j)}(a)$

$f(x)=\sqrt{1+x}=(1+x)^{1/2} \\ f^\prime (x)=\left ( \frac{1}{2} \right )(1+x)^{-1/2} \\ f^{(2)}(x)=\left ( \frac{1}{2}\right )\left ( \frac{-1}{2} \right )(1+x)^{-3/2} \\ f^{(3)}(x)=\left ( \frac{1}{2}\right )\left ( \frac{-1}{2} \right )\left ( \frac{-3}{2} \right )(1+x)^{-5/2} \\ f^{(4)}(x)=\left ( \frac{1}{2}\right )\left ( \frac{-1}{2} \right )\left ( \frac{-3}{2} \right )\left ( \frac{-5}{2} \right )(1+x)^{-7/2} \\ f^{(j)}(x)=(-1)^{j+1}\left ( \frac{1}{2^{j}} \right )(1\cdot 1\cdot 3\cdot 5\cdot ... \cdot 2j-3)(1+x)^{(2j-1)/2} \\ f^{(j)}(0)=(-1)^{j+1}\left ( \frac{1}{2^{j}} \right )(1\cdot 1\cdot 3\cdot 5\cdot ... \cdot 2j-3)$

I'm thinking I have calculated the j-th derivative correctly, however I can't seem to figure out how to simplify it and the rest of the steps into a compact summation.

Added: It's been pointed out to me that the $(1\cdot 1\cdot 3\cdot 5\cdot ... \cdot 2j-3)$ above should be $\prod_{i=0}^{j-1}2i-1$

Thus $f^{(j)}(0)=(-1)^{j+1}\left ( \frac{1}{2^{j}} \right )\prod_{i=0}^{j-1}2i-1$.

If anyone can give me a hint on how to simplify this further it would be appreciated.

Answer 1

Note that $$1\cdot 3\cdot 5 \cdots (2j - 3) = \dfrac{(2j-3)!}{2\cdot 4 \cdot 6 \cdots (2j -2)} = \dfrac{(2j-3)!}{2^{j-1}\cdot 1\cdot 2 \cdot 3 \cdots (j -1)} = 2^{1-j}\dfrac{(2j-3)!}{(j-1)!}.$$

Answer 2

Useful relations:

$$ \left(1 + x\right)^{m} = \sum_{\ell = 0}^{m}{m \choose \ell}\,x^{\ell}\,, \qquad m = 0, 1, 2, \ldots $$

In general $$ \left(1 + x\right)^{m} = \sum_{\ell = -\infty}^{\infty} {\Gamma\left(m + 1\right) \over \Gamma\left(\ell + 1\right)\Gamma\left(m - \ell + 1\right)}\,x^{\ell} = \sum_{\ell = 0}^{\infty} {\Gamma\left(m + 1\right) \over \ell!\,\Gamma\left(m - \ell + 1\right)}\,x^{\ell}\,, \qquad m\ \in\ {\mathbb R} $$

When $m = 0, 1, 2, \ldots$, the $\Gamma\left(m - \ell + 1\right)$ factor cut the sum terms with $\ell > m$ as it happens to be in the above case.

When $m < 0$, it is convenient to use the $\Gamma\,$'s reflection formula:

\begin{align} {\Gamma\left(m + 1\right) \over \Gamma\left(m - \ell + 1\right)} &= {\pi \over \Gamma\left(-m\right)\sin\left(\pi\left\lbrack m + 1\right\rbrack\right)}\, {\Gamma\left(\ell -m\right) \sin\left(\pi\left\lbrack m - \ell + 1\right\rbrack\right) \over \pi} \\[3mm]&= {\left(-1\right)^{-\ell + 1}\sin\left(\pi m\right) \over -\sin\left(\pi m\right)}\, {\Gamma\left(\ell -m\right) \over \Gamma\left(-m\right)} = \left(-1\right)^{\ell}\ {\Gamma\left(\ell -m\right) \over \Gamma\left(-m\right)} \\[5mm]& \end{align}

Then, $$ \left(1 + x\right)^{m} = \sum_{\ell = 0}^{\infty}\left(-1\right)^{\ell}\, {\Gamma\left(\ell - m\right) \over \ell!\,\Gamma\left(-m\right)}\,x^{\ell}\,, \qquad m\ \in\ {\mathbb R} $$ is a useful formula when $m \in {\mathbb R},\ m < 0$.

In your case:

$$ \left\lbrace% \begin{array}{rcl} \sqrt{1 + x\,} & = & \left(1 + x\right)^{1/2} = \sum_{\ell = 0}^{\infty} {\Gamma\left(3/2\right) \over \ell!\,\Gamma\left(3/2 - \ell\right)}\,x^{\ell} = {1 \over 2}\,\sqrt{\pi\,}\sum_{\ell = 0}^{\infty} {x^{\ell} \over \ell!\,\Gamma\left(3/2 - \ell\right)} \\[3mm] \sqrt{1 + x\,} & = & \left(1 + x\right)^{1/2} = 1 + {1 \over 2\sqrt{\pi\,}}\sum_{\ell = 0}^{\infty} \left(-1\right)^{\ell}\,{\Gamma\left(1/2 + \ell\right) \over \left(\ell + 1\right)!}\,x^{\ell + 1} \\[3mm] && \mbox{Notice that}\quad \Gamma\left({1 \over 2} + \ell\right) = {\sqrt{\pi\,} \over 2^{\ell}}\,\left(2\ell - 1\right)!! \end{array}\right. $$

Answer 3

Hint: Using the Binomial Theorem, prove by induction that $$ (1+x)^{-1/2}=\sum_{k=0}^\infty\frac{(2k-1)!!}{(2k)!!}(-x)^k $$ Truncate as needed. Note that $n!!$ is the Double Factorial: $$ \begin{array}{rll} (2k-1)!!&=1\cdot3\cdot5\cdots(2k-1)&=\frac{(2k)!}{2^kk!}\\[6pt] (2k)!!&=2\cdot4\cdot6\cdot8\cdots(2k)&={2^kk!} \end{array} $$