What Is the Power Series and Interval of Convergence for f(x) = x/(x^2 + 1)?

[vande060]

Homework Statement

find the power series for the function and determine the interval of convergence

$$f(x) = {{x} \over {x^2 +1}}$$

im trying latex for the first time so here it is if it doesn't show well: f(x) = x/(x^2 + 1)

Homework Equations

The Attempt at a Solution

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

f(x) = x/(x^2 + 1)f(x) = x/(2x^2 ( 1/2x^2 + 1) )
f(x) = x/(2x^2 ( 1 - -1/2x^2 ) ) $$= \frac{x}{2x^2} \sum_{n=1}^{\infty}$$ (-1/(2x^2))^n

$$\sum_{n=1}^{\infty}$$ ((-1)^n x)/(2x^2)^(n+1)

once I take the limit though by using the ratio test i get 1/16x^8

 

[Apphysicist]

I would find the power series for 1/(x^2+1) then multiply each term by x, then look at it and write the summation notation.

f(x)= f(0)+f '(0)*x +f ''(0)*x^2/2 + f '''(0)*x^3/6 +...

f(x) = (x^2+1)^-1
f '(x) = - 2x*(x^2+1)^-2
f'' (x) = -2x*-2*2x*(x^2+1)^-3 + (x^2+1)^-2*-2

f(0) = (0+1)^-1 = 1
f'(0) = 0
f''(0) = 0 -2*(1^-2) = -2

f(x), centered at 0 for 1/(x^2+1) = 1-x^2+...(I'll leave a couple more terms for you so you can see the pattern), then multiply through by x.

OR, if you've got some intuition, you could note that a/(1-r) is the sum of a geometric series for abs(r)<1, then you'd write: 1-x^2+...(still going to leave those terms for you), the multiply through by x.

You can probably write the summation by looking at the pattern of the terms.

Then you can apply the ratio test, or note from the geometric series that abs(r)<1...to get the radius of convergence. Don't forget to also check the =1 case.

 

[vande060]

I would find the power series for 1/(x^2+1) then multiply each term by x, then look at it and write the summation notation.

f(x)= f(0)+f '(0)*x +f ''(0)*x^2/2 + f '''(0)*x^3/6 +...

f(x) = (x^2+1)^-1
f '(x) = - 2x*(x^2+1)^-2
f'' (x) = -2x*-2*2x*(x^2+1)^-3 + (x^2+1)^-2*-2

f(0) = (0+1)^-1 = 1
f'(0) = 0
f''(0) = 0 -2*(1^-2) = -2

f(x), centered at 0 for 1/(x^2+1) = 1-x^2+...(I'll leave a couple more terms for you so you can see the pattern), then multiply through by x.

OR, if you've got some intuition, you could note that a/(1-r) is the sum of a geometric series for abs(r)<1, then you'd write: 1-x^2+...(still going to leave those terms for you), the multiply through by x.

You can probably write the summation by looking at the pattern of the terms.

Then you can apply the ratio test, or note from the geometric series that abs(r)<1...to get the radius of convergence. Don't forget to also check the =1 case.

okay I understand what I have to do now thank you