Taylor expansion for $\sqrt{x+2}$

Question

I'm enrolled in Coursera's calculus with a single variable and am trying to solve one of the homework problems.

In lecture, it was stated that to expand $\sqrt x$ about $x=a$, you would have:

$$\sqrt{x} = \sqrt{a} + {1 \over 2 \sqrt{a}}(x-a)- {1\over 8 \sqrt{a^3}}(x-a)^2 + H.O.T$$

The homework hint says you can us the Binomial series to find the Taylor series expansion for expressions with non-integer powers.

Wikipedia says the Binomial series expands to $$(x +1)^{ \alpha }= \sum \limits_{k=0}^{\infty} {\alpha \choose k} x^k$$ $${\alpha\choose{k}} = \frac{\alpha \cdot (\alpha - 1) \cdot (\alpha - 2) \cdot \dots \cdot (\alpha - k + 1)}{k!}$$

My first question is where the term $$a^{1/2 - k}$$ comes from, given the Binomial series formula.

My second question is how to properly evaluate the series about a particular value other than zero.

The homework problem asks me to compute the Taylor series for $$f(x) = \sqrt{x+2}$$ about $x=2$. I also tried to use substitution with $h=x+2$, $x=h-2$ and then compute the Taylor series expansion about h=0 using the definition of Taylor series formula with

$$\sum_{n=0} {{f^{(n)}\over n!}(x-a)^n}$$

$$f(h) = \sqrt{h-2}$$

But with $f(h=0)$, I get imaginary numbers.

Answer 1

Hint: I am assuming you want to expand $\sqrt{x}$ about $x=2$.

If you want to use the "formula" for the Taylor expansion, you need the derivatives of $\sqrt{x}$ at $x=2$. These derivatives are well-behaved, and you can find an explicit formula for the $n$-th derivative at $x=2$.

If you are allowed to quote the general Binomial Theorem, note that $$\sqrt{x}=\sqrt{2+(x-2)}=\sqrt{2}\left(1+\frac{x-2}{2}\right)^{1/2}.$$ Then we are looking at $(1+t)^{1/2}$ for $t=\frac{x-2}{2}$.

Answer 2

This is what I have now:

Evaluating derivatives for $\sqrt{x+2}$ at x=2, I get:

$ f(0) = \sqrt{2+ (2)} = 2 $

$ f'(0) = {1\over2} {1 \over \sqrt{2+(2)} }= 1/4$

$ f''(0) = -{1\over 4} { 1 \over (\sqrt{ 2+(2)})^{3} }= -1/32$

The taylor series expansion for $ \sqrt{x+2} \space \space at \space \space x=2 :$

$2 + {1\over 4\cdot 1!} (x-2) - {1 \over 32 \cdot 2! }+...$

Which I think this is correct, but when I try to use the general Binomial theorem, I run into trouble.

$\sqrt{2+(x-2)}=\sqrt{2}\left(1+\frac{x-2}{2}\right)^{1/2}$

$(1+t)^{1/2}$ where $t=\frac{x-2}{2}$

$1 { {1/2} \choose 0} + \sqrt{2}{ 1/ 2 \choose 1}(1+t)+ 2^{3/2}{ 1/ 2 \choose 2}(1+t)^2 = $

$1 + \sqrt{2}\space{ {1\over 2} \over 1!}(1+t)+ 2^{3/2}\space{ {1\over 2} \cdot {-1 \over 2} \over 2!}(1+t)^2 + ...$

$1 + \sqrt{2}\space{ {1\over 2} \over 1!}(1+\frac{x-2}{2})+ 2^{3/2}\space{ {1\over 2} \cdot {-1 \over 2} \over 2!}(1+\frac{x-2}{2})^2 + ...$

I assume that ${ {1/2} \choose 0} = 1$, since there is only one way to choose 0 things, but I'm not sure if that is correct.

Answer 3

Say we're given $f_\alpha(x)=x^\alpha$.

Thus $f_\alpha'(x)=\alpha x^{\alpha-1}$, and $f_\alpha''(x)=\alpha(\alpha-1)x^{\alpha-2}$.

In general, $$f_\alpha^{(n)}(x)=p(\alpha,n)x^{\alpha-n}$$ where $$p(\alpha,n)=\prod_{k=1}^n(\alpha-k+1)=\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-n+1)$$ So, it would make sense to evaluate out Taylor Series around $x=2$: $$f_\alpha(x)=\sum_{n\geq0}\frac{f_\alpha^{(n)}(2)}{n!}(x-2)^n$$ $$f_\alpha(x)=\sum_{n\geq0}\frac{p(\alpha,n)\cdot2^{\alpha-n}}{n!}(x-2)^n$$ $$f_\alpha(x)=\sum_{n\geq0}2^{\alpha-n}\frac{p(\alpha,n)}{n!}(x-2)^n$$ $$f_\alpha(x)=\sum_{n\geq0}2^{\alpha-n}{\alpha\choose n}(x-2)^n$$ $$x^\alpha=\sum_{n\geq0}2^{\alpha-n}{\alpha\choose n}(x-2)^n$$ Plugging in $x+2$ for $x$, and $\alpha=1/2$ gives $$(x+2)^{1/2}=\sum_{n\geq0}2^{1/2-n}{1/2\choose n}(x+2-2)^n$$ $$\sqrt{x+2}=\sum_{n=0}^{\infty}2^{1/2-n}{1/2\choose n}x^n$$ $$\sqrt{x+2}=\sqrt{2}\,\sum_{n=0}^{\infty}{1/2\choose n}\biggr(\frac{x}{2}\biggl)^n$$ If you want to simplify more, recall that ${a\choose b}=\frac{\Gamma(a+1)}{\Gamma(b+1)\Gamma(a-b+1)}$, where $\Gamma(x)$ is the Gamma function. Using this gives $$\sqrt{x+2}=\sqrt{2}\,\sum_{n=0}^{\infty}\frac{\Gamma(1/2+1)}{\Gamma(n+1)\Gamma(1/2-n+1)}\biggr(\frac{x}{2}\biggl)^n$$ $$\sqrt{x+2}=\sqrt{2}\,\sum_{n=0}^{\infty}\frac{\Gamma(3/2)}{n!\Gamma(3/2-n)}\biggr(\frac{x}{2}\biggl)^n$$ $$\sqrt{x+2}=\sqrt{2}\,\Gamma(3/2)\sum_{n=0}^{\infty}\frac{(x/2)^n}{n!\,\Gamma(3/2-n)}$$ And recalling that $\Gamma(3/2)=\frac{\sqrt{\pi}}{2}$ gives $$\sqrt{x+2}=\frac{\sqrt{2\pi}}{2}\sum_{n=0}^{\infty}\frac1{\Gamma(3/2-n)}\frac{(x/2)^n}{n!}$$ Which you really don't need to simplify anymore. :)