# Differentiation of a Taylor series

1. Aug 29, 2014

### cosmictide

1. The problem statement, all variables and given/known data

Hi guys, any help on this question would be hugely appreciated.

The Taylor series about 0 for the function f(x)=(1/4+x)-3/2 is

f(x)=8 - 48x + 240x^2 - 1120x^3 + ...

used differentiation to find the Taylor series about 0 for the function g(x)=(1/4+x)-5/2

3. The attempt at a solution

I tired differentiating each term to obtain 0-48+480x-3360x^2 but that can't be right since the correct answer seems to be g(x)=(1/4+x)^-5/2 to be 32-320x+2240x^2....

I'd be greatful if someone could shed some light on how I go about approaching this.

2. Aug 29, 2014

### Orodruin

Staff Emeritus
Is the derivative of (1/4 + x)-3/2 equal to (1/4 + x)-5/2?

I also suggest using etc or LaTeX for readability.

3. Aug 29, 2014

### cosmictide

The derivative of (1/4 + x)-3/2 is -48/(4x+1)5/2 so no they're not equal. I have no idea how to put that right though! Any help would be hugely appreciated.

4. Aug 29, 2014

### Orodruin

Staff Emeritus
Where did the 48 come from? It is correct that it should be a factor, but not 48 ...

Once you get the factor right and have done the differentiation of the Taylor series, you will have a factor times the expression you want on one side and a Taylor series on the other. How would you then solve for the expression you want?

5. Aug 29, 2014

### cosmictide

Okay this time using the chain rule I calculated the derivative to be 3 / 2(1/4+x)5/2.....is that any better?

I think once I get the factor right I just use it on the left-hand side to make the equation equal but not sure what it is yet.

6. Aug 29, 2014

### nrqed

Almost correct. The overall sign is not correct.

Yes, set this equal to the derivative of the expansion given at the beginning and then isolate $(1/4+x)^{-5/2}$

7. Aug 29, 2014

### cosmictide

So I'm left with:

-3 / 2(1/4+x)5/2= -48x + 240x2 - 1120x3...

How do I isolate it? I've tried multiplying and dividing both sides by 3 and 2 but that's clearly wrong! Any help here would be greatly appreciated.

8. Aug 29, 2014

### nrqed

You obtained the left side by differentiating the left side of the initial expression, right? So you must also differentiate the right side. You have not done that yet. Do this and then isolate$(1/4+x)^{-5/2}$. That will work.

And you are welcome :-)

9. Aug 29, 2014

### cosmictide

Thank you so much.

10. Aug 29, 2014

### nrqed

Good work!
You are very welcome.

11. Aug 29, 2014

### Ray Vickson

You wrote\$$f(x) = \left(\frac{1}{4}+x\right) - \frac{3}{2}$$
Is that what you meant, or did you want
$$f(x) = \left(\frac{1}{4}+x\right) ^{-3/2} \, ?$$
If so, use parentheses, like this: f(x) = (1/4+ x)^(-3/2), or better still, ((1/4)+x)^(-3/2) or maybe [(1/4)+x]^(-3/2).

Anyway, assuming you want the second form above, we have
$$f'(x) = -\frac{3}{2} \left(\frac{1}{4}+x\right)^{-5/2} \neq \left(\frac{1}{4}+x\right)^{-5/2}$$