Question

taylor series of f(x)=x-x^3, a=-2
with radius of convergence and interval of convergence

Answer 1

define f' f'' f''' and such tell it goes to zero

Answer 2

f = x - x^3

f' = -3x^2

f'' = -6x

f''' = -6

f'''' and beyond = 0

Answer 3

yes i got that far

Answer 4

define the f's at=a=-2

Answer 5

6, -11, 12, and ,6

Answer 6

f = -2 - (-2)^3 = 6

f' = -3(-2)^2 = -12

f'' = -6(-2) = 12

f''' = -6 = -6

f'''' and beyond = 0 = 0

Answer 7

i did forget a 1 didnt i

Answer 8

6,-11,12,-6, 000000

Answer 9

i can't find a pattern for the series

Answer 10

set these up into the standard taylor getup as the coeefs

\[x-x^3=6\frac{x^0}{0!}-11\frac{x^1}{1!}+12\frac{x^2}{2!}-6\frac{x^3}{3!}+0000\]

Answer 11

\[x-x^3=6-11(x+2)+6(x+2)^2-(x+2)^3\]

Answer 12

I'm stuck right there!

Answer 13

lol, umm, just expand it out and you should see that it siplifies to x-x^3

Answer 14

the polynomial expansion of a polynomial expansion ... is just the polynomial you started with :)

Answer 15

so I don't need to put it in the form of \[\sum_{n=0}^{\infty} (x-a)^n\] etc etc?

Answer 16

no. polynomials are unique; therefore you cant really have 2 different polynomials of the same curve.

the exercise is meant to get you familiar with the concept of createing a power series using something simple

Answer 17

in other words, it means to solidify techniques

Answer 18

so how would I find out the radius and interval of convergence?

Answer 19

the radius of convergence is the distance from "a" wherein the taylor matches the function

since the taylor matches the function at all point; the radius in infinity

Answer 20

the interval is the range that the radius spans; -R < x < R ; and since R = inf on this .. interval is just (-inf,inf)

Answer 21

dont live life in a formula :) understand the concepts

Answer 22

okay thanks again!!!

Answer 23

yep