# Homework Help: Equivalent expression

1. Aug 30, 2014

### cosmictide

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

Hi guys, I'm getting confused trying to manipulate an expression. Any help here would be hugely appreciated.

How do I manipulate the expression -4 / (4x + 1)1/2 to become (1/4 + x)-1/2

3. The attempt at a solution

I've tried dividing by -4 to get and used an index law to get (1/2)*(x+1)-1/2 but that's clearly not the expression I want. I'd really appreciate if someone could point me in the right direction.

Last edited by a moderator: Aug 30, 2014
2. Aug 30, 2014

### nrqed

The two expressions you wrote are actually not equal. You can easily check by plugging a value of x, for example x=0.

3. Aug 30, 2014

### cosmictide

I know they're not equivalent, what I'm trying to do is manipulating the first expression to become the second expression if that makes sense.

4. Aug 30, 2014

### nrqed

Since they are not equivalent it is impossible to manipulate one to bring it in the form of the other.

5. Aug 30, 2014

### cosmictide

Okay so I must be missing something else. I'm trying to find the Taylor series about 0 for the function g(x) = (1/4 + x)-1/2 from the series of the function f(x) =

(1/4 + x)-3/2 = 8 - 48x + 240x2 - 1120x3 by using integration.

I've done the integration to get

-4 / (4x + 1)1/2 = c + 8x - 24x2 + 80x3 - ...

Taking x=0 I calculated the constant c = -4 leaving me with

-4 / (4x + 1)1/2 = -4 + 8x - 24x2 + 80x3 - ...

This is where I'm struggling. If I am asked to find the Taylor series about 0 for the function g(x) = (1/4 + x)-1/2 then shouldn't the left-hand side of the equation above be equivalent to this or am I missing something else?

Thanks.

6. Aug 30, 2014

### Staff: Mentor

Since the problem involves Taylor series, I moved it to the Calculus section.

7. Aug 30, 2014

### Ray Vickson

You had a similar problem in the Calculus and Beyond Forum, and made the same error there. The fact is that you cannot get the series for $(1/4 + x)^{-1/2}$ directly by integrating the series for $(1/4+x)^{-3/2}$. You ought to be able to see the reason: if $F'(x) \neq G(x)$ the series for $G(x)$ is not gotten from that of $F(x)$ by differentiating and vice-versa (integrating instead of differentiating). However, if $F'(x) = c G(x)$ for some known constant $c$ then it is not difficult to pass from one to the other. Do you see how?

Note: typo edited out now.

Last edited: Aug 30, 2014
8. Aug 30, 2014

### nrqed

I did not check all the numbers but it does look good to me, so far.
Ok, now you do need to rewrite the left side in the form

$\frac{C}{(1/4 +x)^{1/2}}$

where C is a constant.
Once you have that, just divide both sides of your result by the constant and you will have your final answer.

So what you need to do now is to find the constant required so that

$\frac{C}{(1/4 +x)^{1/2}} = - \frac{4}{(4x +1)^{1/2}}$

9. Aug 30, 2014

### cosmictide

Thanks guys, much appreciated.