# Maclaurin Series concept help

1. Aug 4, 2007

### kj13529

Im really having trouble wrapping my mind around the topic of maclaurin series, my problem requires me to find the maclaurin series of f(x) = 1/(1+x^2), then use that to find the maclaurin series of g(x)= arctan(x)...i don't even know where to begin.

2. Aug 5, 2007

### Gib Z

Well you could do it either one of 2 ways: Use the definition of the Maclaurin series for a function f(x), $$f(x) = f(0) + f'(0)x + f''(0)\frac{x^2}{2!} + f'''(0) \frac{x^3}{3!}...$$.

Or you could divide 1 by 1+x^2 via polynomial division, and it gets the same series.

After you get that series, it wants you to realize that since that series is the same as 1/(1+x^2), the integral of the series is equal to the integral of 1/(1+x^2), which is arctan(x).

3. Aug 5, 2007

### kj13529

So basically I plug into the general formula, then what, how do I use that information to find the series of arctan?

4. Aug 5, 2007

### Gib Z

you plug into that formula, then you must realise that series is equal to 1/(1+x^2). Integrate both sides, since you know the integral of 1/(1+x^2) is arctan(x), you know the series for arctan x.

5. Aug 5, 2007

### HallsofIvy

Staff Emeritus
A power series is a power series! If you have two power series that converge to the same function, then they must be exactly the same power series- in other words, it doesn't matter how you actually find the series.

You could also do 1/(1+ x2) as a geometric series. You know, of course, that
$$\sum_{n=0}^\infty ar^n= \frac{a}{1- r}$$
For this problem, you must have
$$\frac{a}{1- r}= \frac{1}{1+ x^2}$$
so a= 1 and r= -x2: the power series is
$$\sum_{n=0}^\infty (-x^2)^n= \sum_{n=0}^\infty (-1)^n x^{2n}$$

As Gib Z said, since
[tex]arctan(x)= \int \frac{1}{1+ x^2} dx[/itex]
You can integrate the series term by term to get the series for arctan(x).

6. Jul 8, 2010

### Jibe

I'm working on the same problem, and the trouble I'm having is that when you take the first, second, third etc. derivatives of 1/(1+x^2) I'm not seeing a clear pattern in the derivatives. The only problems I've seen with the Maclaurin series are problems with some function such as sin(x) that will repeat after a few derivations. So this ones giving me lots of trouble It could be that I'm deriving it wrong, but I don't think so. Can you guys help me figure out the pattern?

7. Jul 8, 2010

### hunt_mat

Maclaurin series is simple. Let's see have an example, f(x)=sin(x). Then suppose I can write the function as sin (x)=a+bx+cx^{2}+dx^{3}+...
Our task is to compute a,b,c,d,... To find a, set x=0 to find a=sin(0)=0.
To find b, differentiate w.r.t x to find cos(x)=b+2cx+3dx^{2}. To find b, set x=0 to see that:
b=cos(0)=1. So the series is sin(x)=x+cx^{2}+dx^{3}+...
To find c, differentiate twice to find:
-sin(x)=2c+6dx+...
Set x=0 to see that c=0.
Again apply the same methodology to find d. sin(x)=x+dx^{3}+...
Differentiate three times to get -cos(x)=6d+...
set x=0 to find d=-1/6=-1/3!
Then sin(x)=x-x^{3}/3!+...
That is the idea of Maclaurin's series.