Register to reply 
Taylor polynomial 1/(1x^2) 
Share this thread: 
#1
Apr310, 08:48 PM

P: 253

1. The problem statement, all variables and given/known data
The question asks me to write out a taylor polynomial for 1/(1x^2) of degree 2n+1 at 0. 3. The attempt at a solution My answer was 1 + x^2 + x^4 + x^6 + ..... + (x^4)/(1x^2) which I just got from using hte geometric series formula. The textbook answer however is this: 1  x^2 + x^4  ............. + (1)^n x^2n Am I wrong? I thought so, but I took out a graphing calculator, and the larger I make my polynomial degree, the closer it looks to 1/(1x^2). So my answer must be correct? thanks 


#2
Apr310, 08:59 PM

P: 253

omg, silly me. The question was 1/(1 + x^2) , not 1 / ( 1 x^2 )



#3
Apr310, 09:05 PM

P: 253

so should I just keep differentiating ? :S



#4
Apr310, 09:07 PM

Emeritus
Sci Advisor
PF Gold
P: 4,500

Taylor polynomial 1/(1x^2)
If you know what the taylor series polynomial is for 1/(1x) it's just a simple substitution



#5
Apr310, 09:13 PM

P: 253




Register to reply 
Related Discussions  
Taylor Polynomial  Calculus & Beyond Homework  21  
Taylor Polynomial  Calculus & Beyond Homework  3  
Taylor Polynomial  Calculus & Beyond Homework  2  
Taylor Polynomial  Introductory Physics Homework  2  
Taylor Polynomial  Introductory Physics Homework  3 