# series representation

• April 11th 2011, 05:15 PM
smray7
series representation
Represent the function f(x) = x/(1+4x)^2 as a power series and determine the radius of convergence.

so the correct answer is (-1)^n * (n+1) *4^n*x^(n+1) but this is what I got:

x/(1+4x)^2

x/[1-(-4x)^2

x series sign (-4x)^n

series sign (-1)^n * 4^n x^(n+1)

Where did the (n+1) in front of the 4^n come from?
• April 11th 2011, 07:01 PM
Sambit
Sorry, but I can't understand what you mean. What is the question exactly?
Quote:

so the correct is what I have multiplied by (n+1)
And what does this mean? What have you multiplied by (n+1) and where?
• April 12th 2011, 04:46 AM
Jester
Are you starting with the power series

$\dfrac{1}{1+x} = 1 - x + x^2 - x^3 + \cdots$

and if so, what are you doing to this series to result in your series?
• April 12th 2011, 09:51 AM
smray7
OK so I fixed the question. What I get from this section is that you want to make your function look like 1/1-x and go from there according to the definition.
• April 12th 2011, 10:08 AM
Jester
Right, so differentiate my series

$\dfrac{1}{1+x} = 1 - x + x^2 - x^3 \cdots + (-1)^{n}x^{n}$

so

$\dfrac{-1}{(1+x)^2} = -1 + 2x - 3x^2 + 4x^3 - \cdots + (-1)^{(n+1)} (n+1)x^n$

or

$\dfrac{1}{(1+x)^2} = 1 - 2x + 3x^2 - 4x^3 + \cdots - (-1)^{(n+1)} (n+1)x^n$

or

$\dfrac{1}{(1+x)^2} = 1 - 2x + 3x^2 - 4x^3 + \cdots + (-1)^n (n+1)x^n$

Now replace $x$ with $4x$ and multiply the whole thing by $x$.
• April 12th 2011, 10:33 AM
smray7
So once you have the power series representation of a function, you the take the derivative?
• April 12th 2011, 11:08 AM
ForumUserX
Quote:

Originally Posted by smray7
So once you have the power series representation of a function, you the take the derivative?

You showed in the OP that you already know the series for 1/(1 - x). The derivative of that series is the series of that derivative, -1/(1 - x)^2. This more closely resembles the original problem which you are working towards.

This is just one tool you can use to find a series for a function you didn't know.
• April 13th 2011, 08:21 AM
TheCoffeeMachine
You could show that, if $\displaystyle f(x) = \frac{x}{(1+4x)^2}$, then:

$\displaystyle f^{(k)}(x) = \frac{4^{k-1}(-1)^k(k)!}{(1+4x)^{k+1}}-\frac{4^{k-1}(-1)^k(k+1)!}{(1+4x)^{k+2}}$

and use the general formula for Taylor/Maclaurin series.