Question

find a power series representation for the function amd determine the radius of convergence f(x)=(x^2+x)/(1-x)^3

Answer 1

ok have you found the power series representation?

Answer 2

\[\frac{1}{(1-x)^2}=\frac{d}{dx}\frac{1}{(1-x)}\]
\[\frac{1}{(1-x)^3}=\frac{1}{2} \frac{d}{dx} \frac{1}{(1-x)^2}\]

so it looks like we need the power series of 1/(1-x)
then we need the second derivative of that power series
then multiply it by 1/2 to get the power series of 1/(1-x)^3

Answer 3

then when we are done multiply by the (x^2+x)

Answer 4

so what does that series look like? is it just \[x^{2}+x \Sigma x^n\]

Answer 5

well not exactly you need to find derivative of that one series like twice

Answer 6

\[\frac{1}{1-x}=1+x+x^2+x^3+x^4 + \cdots \\ \frac{1}{(1-x)^2}=0+1+2x+3x^2+4x^3 + \cdots \]
now you still need to differentiate one more time

Answer 7

i will let you do that

Answer 8

differentiate both sides

Answer 9

basically the terms in first line is x^n
derivative of that is nx^{n-1}
and derivative of that derivative is n(n-1)x^(n-2)

Answer 10

you can actually use that to write in sigma notation

Answer 11

@n.sorge are you still there?

Answer 12

so would n=2

Answer 13

?

Answer 14

in the sigma notation

Answer 15

n is n

Answer 16

does it start at 0 or 2 i mean

Answer 17

0(0-1)x^(0-2)
+
1(1-1)x^(1-2)
+
2(2-1)x^(2-2)
+
so on...

Answer 18

either

Answer 19

those first two terms are 0

Answer 20

the first two terms of the series right

Answer 21

then 2+6x.....

Answer 22

\[\frac{1}{1-x}=1+x+x^2+x^3+x^4+ \cdots +x^{n}+\cdots \\ \text{ differentiating both sides } \\ \frac{1}{(1-x)^2}=0+1+2x+3x^2+4x^3+\cdots + nx^{n-1}+\cdots \\ \text{ differentiating both sides } \\ \frac{2}{(1-x)^3}=0+0+2+6x+12x^2+ +\cdots n(n-1)x^{n-2}+c\dots \]

Answer 23

but you weren't looking for 2/(1-x)^3
you were looking for 1/(1-x)^3

Answer 24

that c there is a type-o

Answer 25

ignore it

Answer 26

so divide both sides of that last equation by 2

Answer 27

\[\frac{1}{1-x}=1+x+x^2+x^3+x^4+ \cdots +x^{n}+\cdots \\ \text{ differentiating both sides } \\ \frac{1}{(1-x)^2}=0+1+2x+3x^2+4x^3+\cdots + nx^{n-1}+\cdots \\ \text{ differentiating both sides } \\ \frac{2}{(1-x)^3}=0+0+2+6x+12x^2+ \cdots +n(n-1)x^{n-2}+\cdots \]

Answer 28

so 0+0+1+3x+...

Answer 29

\[\frac{1}{(1-x)^3}=\frac{0}{2}+\frac{0}{2}+\frac{6x}{2}+\frac{12x^2}{2}+\cdots +\frac{n(n-1)x^{n-2}}{2}+\cdots \]

then finally you should know that we are looking for (x^2+x)/(1-x)^3
so take the series from above and multiply it by the (x^2+x)

Answer 30

so what would that look like would you have to distribute that

Answer 31

n.sorge i have basically done the whole problem for you

it should be pretty easy to write what I have up there in sigma notation
and then take that sum and multiply it by (x^2+x)

Answer 32

\[\frac{1}{(1-x)^3}=\sum_{n=0}^{\infty}\frac{n(n-1)x^{n-2}}{2}\]

Answer 33

your function is \[\frac{x^2+x}{(1-x)^3}\]
so just mutliply both sides of that equation I just wrote by (x^2+x) to get the power series for (x^2+x)/(1-x)^3