Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

up vote 3 down vote accepted

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 $t=\frac{x-2}{2}$.

share|improve this answer

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.

share|improve this answer
1  
In the expansion for $(1+t)^{1/2}$ you get terms with powers $t^k$, not $(1+t)^k$. That's the main point of the formula. –  Marc van Leeuwen Apr 5 '13 at 12:56

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.