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

• jacoleen
In summary, to find the Maclaurin series for f(x) = (2x)/(1+x2), you can simply use polynomial long division to divide 2x by 1 + x^2. This gives you the series 2x - 2x^3 + 2x^5 -... and so on.
jacoleen

## Homework Statement

Write the Machlaurin series for :

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

## 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?

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).

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) :(

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

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

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

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

I'm really not getting it..

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

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

the summation divided by x^-2007?

* without the negative

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:
is it just -2?

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

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)?

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

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

OMG..it actually makes sense!

Thank you so much for your help! :D

jacoleen said:

## Homework Statement

Write the Machlaurin series for :

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

## 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?

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 -...

## 1. What is a Maclaurin series?

A Maclaurin series is a mathematical representation of a function using a sum of terms that are powers of x. It is a type of Taylor series, which is used to approximate a function at a specific point by using its derivatives.

## 2. How do you find the Maclaurin series for a function?

To find the Maclaurin series for a function, you must first find the derivatives of the function at x=0 and then substitute them into the general formula for a Maclaurin series, which is f(x) = f(0) + f'(0)x + (f''(0)x^2)/2! + (f'''(0)x^3)/3! + ...

## 3. What is the Maclaurin series for f(x) = (2x)/(1+x^2)?

The Maclaurin series for f(x) = (2x)/(1+x^2) is 2x - 2x^3 + 2x^5 - 2x^7 + ...

## 4. How many terms are needed to approximate f(x) = (2x)/(1+x^2) to within a certain accuracy?

The number of terms needed to approximate f(x) = (2x)/(1+x^2) to within a certain accuracy depends on the degree of accuracy desired. Generally, the more terms included in the series, the more accurate the approximation will be.

## 5. What is the radius of convergence for the Maclaurin series of f(x) = (2x)/(1+x^2)?

The radius of convergence for the Maclaurin series of f(x) = (2x)/(1+x^2) is 1, meaning that the series will converge for all values of x within a radius of 1 from the center, which is x=0.

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