# Maclaurin series for f(x)= ((1-x^2)/(1+x^2))

1. Jan 15, 2014

### christian0710

Hi I'm studying for an upcoming exam and I have to find the Maclaurin series for f(x)= ((1-x^2)/(1+x^2))
And I gotta admit i feel stuck.

I know i need to find the terms f(0) +f'(0) +f''(0)/2 etc.

Frist of all I can't find the first derivative f´(x) because my TI89 calculater comes up with a Dimension error.
Sexond: As an alternative I really can't see which other known function/series i could use to substitute the function into.

2. Jan 15, 2014

### Office_Shredder

Staff Emeritus
Are you not expected to be able to calculate the derivative of that function by hand? Because it's a simple application of the quotient rule.

As far as calculating the Maclaurin series, can you calculate the series for
$$\frac{1}{1+x^2} ?$$
Because this is a good place to start.

3. Jan 15, 2014

### Staff: Mentor

Are you saying that you don't know how to determine the derivative of this function by hand? Have you learned how to take the derivative of a quotient?

4. Jan 15, 2014

### christian0710

#2 That is like the geometric series 1/(1-x) which is SUM(1*x^n)
So the sereis for 1/(1+x^2) would be an alternating series and if you substitue x^2 the series would be:
SUM(-1)^n*X^(n +1) Right'?

It's the top part of the fraction which confuses me :(

5. Jan 15, 2014

### christian0710

Yes It's (f/g`= ((1-x^2)'(1+x^2) -(1+x^2)'(a-x^2))/(1+x^2)^2 But i doubt i'd have to do all that in my head and differentiate it multiple times like that in an exam situation? seems to take a long time.

6. Jan 15, 2014

### Office_Shredder

Staff Emeritus
You have the right idea, but you substituted x2 incorrectly.

Once you have that result correctly, notice that you are multiplying a Taylor series that you know by a polynomial, 1-x2. If someone asked you to calculate the Taylor series for xcos(x) you should be able to do it very quickly from the Taylor series of cos(x), the same principle applies here.

As for the derivative, you aren't expected to calculate it for this problem but it seemed odd that you felt you were stuck because you couldn't do it on your calculator.

7. Jan 15, 2014

### christian0710

Okay i'll try again :)
The geometric series is 1+x+(1/2)x^2 +(1/3!)*x^3 so if we substitute in x^2 we get 1+x^2 + (1/2)x^4 +(1/3!)*x^6 Right?

The sum should be SUM; M(-1)^n*X^(n +1) Right'?
Except i did the signs wrong :)

8. Jan 15, 2014

### Staff: Mentor

What's the big deal? Do it (on paper) and collect terms. See what you get. Then do it for f''. See if a pattern develops. Maybe you have to do only a few terms before you see the pattern. The suggestion of office shredder is also very good. His approach is related to the sum of a certain geometric progression. Can you guess the geometric progression involved?

Chet

9. Jan 15, 2014

### ehild

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

1/(1+x2) is the sum of a geometric series. What is the quotient ?
ehild

Last edited: Jan 15, 2014
10. Jan 15, 2014

### christian0710

So we get 1-2*SUM(0-infinity)(1*x^n+2) but if we subtract the series from 1 then the first term 1-1= 0 so instead of going from 0 to infinity it goes n goes from 1 to infinity?
Gosh, I don't know how I'm going to do those mathmatical manipulations as well as you for the exam.

I think i might be a bit unsure of what you mean by "what's the quotient"?

11. Jan 15, 2014

### ehild

The sum of an infinite series with quotient q is

$$\sum _0 ^{\infty}{q^n}=\frac{1}{1-q}$$

when |q<1|.

Compare $\frac{1}{1-q}$ to $\frac{1}{1+x^2}$. It can be written as $\frac{1}{1-(-x^2)}$. So what corresponds to q?

ehild

12. Jan 16, 2014

### jk22

You can also make a polynomial division :

1-x^2/1+x^2 gives 1 remains
-2x^2/1+x^2 gives -2x^2 remains
2x^4/1+x^2 gives 2x^4 remains
-2x^6 aso...

hence 1-x^2/1+x^2=1-2x^2+2x^4-2x^6+....

13. Jan 16, 2014

### ehild

Use parentheses. What you wrote 1-x^2/1 + x^2 = 1-x^2+x^2=1.

ehild