Math Is Fun Forum

  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 2007-06-13 16:47:41

dukebdx12
Member
Registered: 2007-03-21
Posts: 9

Power Series and Interval of Convergence

Find Power Series and determine the Interval of Convergence

g(x)=3x/(x^2+x-2) , c=0
- after doing partial fractions and getting the form into a/1-r and finding endpoints, then combining both and I got the Interval of Convergence to be (-1,1) is that right?

f(x)=1/2-x ,c=5
- I tried this 1 way but the book did it another way and got a diff. answer so can someone work this ?

Use the Power Series 1/1+x = E (-1)^n*x^n to determine a power series , c=0. Identify Interval of Converg.

f(x)=1/(x+1)^2 = d/dx [1/x+1]
- I dont get how to do this

f(x)=ln(x+1) = S 1/x+1 dx
- this one either

Offline

#2 2007-06-20 05:20:46

HallsofIvy
Guest

Re: Power Series and Interval of Convergence

dukebdx12 wrote:

Find Power Series and determine the Interval of Convergence

g(x)=3x/(x^2+x-2) , c=0
- after doing partial fractions and getting the form into a/1-r and finding endpoints, then combining both and I got the Interval of Convergence to be (-1,1) is that right?

Yes.  Generally speaking a power series converges as long as nothing "bad" happens (in fact, the convergence of the power series is the definition of "analytic"). For a fraction like that, the only "bad" thing that can happen is that the denominator becomes 0. x^2+ x- 2= (x-1)(x+2) so nothing "bad" happens until you hit either 1 or -2.  Since the the "radius" of convergence really is a radius, and the first problem occurs at 1, the radius of convergence around 0 is 1 and the interval of convergence is (1, -1).

f(x)=1/2-x ,c=5
- I tried this 1 way but the book did it another way and got a diff. answer so can someone work this ?

Can I assume you mean 1/(2- x)?  Obviously, that goes "bad" at x= 2 which has distance |2-5|= 3 from c= 5 so the radius of convergence is 3.  You could also write this as 1/(2- (x-5)-5)= 1/((-3)-(x-5)= (-1/3)/(1- (x-5)/(-3)).  That would be the sum of a geoemetric sequence with r= (x-5)/(-3) which converges as long as |r|< 1: i.e.  |x-5|< 3.

Use the Power Series 1/1+x = E (-1)^n*x^n to determine a power series , c=0. Identify Interval of Converg.

f(x)=1/(x+1)^2 = d/dx [1/x+1]
- I dont get how to do this

Well, for power series, it is legal to differentiate "term by "term" so if 1/(x+1)= sum((-1)^n x^n) then d/dx [1/(x+1)]= sum (-1)^n d(x^n)dx.  What is that?  Again, since the power series of a fraction will converge as long as the denominator of the fraction is not 0, the interval of convergence should be obvious.

f(x)=ln(x+1) = S 1/x+1 dx
- this one either

Integrate "term by term": This is sum (-1)^n integral(x^n).

Board footer

Powered by FluxBB