# Homework Help: Maclaurin Series f(x) = (2x)/(1+x2)

1. Dec 10, 2009

### jacoleen

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

Write the Machlaurin series for :

f(x) = (2x)/(1+x2)

2. Relevant equations

3. The attempt at a solution

I tried finding all the derivatives (aka f(x), f'(x), f''(x), etc..) but the equations started getting longer and longer and would always result in 0 when x=0. This was on a final exam last year, and I don't think they would have made students do long computations like that

Also, I tried just finding the derivatives for (1-x)-1, but again, I ended up with a bunch of zeros

Help?

2. Dec 10, 2009

### azure kitsune

Hmmm... I'm not sure, but would it be cheating to use that

$$1 + r + r^2 + r^3 + \cdots = \frac{1}{1-r}$$

for $$|r| < 1$$?

If you do that, you wouldn't need any calculus at all (except for the uniqueness of Maclaurin series).

3. Dec 10, 2009

### jacoleen

I was thinking of doing that in the first place, but on another exam there was a similar question and part c was Find f^2007(0) :(

4. Dec 10, 2009

### azure kitsune

In that case, you can work backwards, because the coefficient of xn in the series gives you $$\frac{f^{n}(0)}{n!}$$

5. Dec 10, 2009

### jacoleen

so my series would just be: Summation of [(-1)n2x2n+1]?

I'm not sure how I would work backwards though :|

6. Dec 10, 2009

### azure kitsune

Usually, you would compute $$\frac{f^{2007}(0)}{2007!}$$ to find the coefficient of x2007 in a Maclaurin series. But in this case, you know the coefficient. What does that tell you about $$\frac{f^{2007}(0)}{2007!}$$ ?

7. Dec 10, 2009

### jacoleen

I'm really not getting it..

do i isolate for f(2007) by equating it to the summation? (without the summation term in front)

8. Dec 10, 2009

### azure kitsune

You're on the right track. What is the coefficient of x2007 in this problem?

9. Dec 10, 2009

### jacoleen

the summation divided by x^-2007?

10. Dec 10, 2009

### jacoleen

* without the negative

11. Dec 10, 2009

### azure kitsune

Actually, you found that

$$\frac{2x}{1+x^2} = \sum_{n=0}^{\infty}(-1)^n2x^{2n+1}$$

When you write out the summation, you get

$$\frac{2x}{1+x^2} = 2 x-2 x^3+2 x^5-2 x^7+2 x^9-2 x^{11} + \cdots$$

From this, can you find the coefficient of x2007?

Last edited: Dec 10, 2009
12. Dec 10, 2009

### jacoleen

is it just -2?

(i'm so sorry I logged off btw, my computer overheat :(

13. Dec 10, 2009

### azure kitsune

Yep!

Now remember that the Maclaurin series of f(x) can be found by:

$$f(x) = \frac{f(0)}{0!} + \frac{f'(0)}{1!} x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 +\cdots$$

So the coefficient of x2007 must be

$$\frac{f^{2007}(0)}{2007!}$$

But you know that the coefficient is -2. So what does this tell you about f2007(0)?

14. Dec 10, 2009

### jacoleen

it's equalled to 2?..and so f(2007) = 2*2007!?

15. Dec 10, 2009

### azure kitsune

Close! It's actually f(2007)(0) = -2 * 2007!, but I think you got the idea. ;)

16. Dec 10, 2009

### jacoleen

OMG..it actually makes sense!!

Thank you so much for your help!! :D

17. Dec 11, 2009

### Staff: Mentor

If all you need to do is get the Maclaurin series for 2x/(1 + x^2), there's something you can do that's much simpler than what I've seen in this thread - just polynomial long division to divide 2x by 1 + x^2. Doing this, I get 2x - 2x^3 + 2x^5 -...