Question

Find a taylor series and radius of convergence for f(x) centered at a = 1 where f(x) = (x^4)-(3x^2)+1

Answer 1

So f(x) = (x^4)-(3x^2)+1
f'(x) = (4x^3)-(6x)
f''(x) = (12x^2)-6
f'''(x) = 24x
f''''(x) = 24

and f(1) = -1
f'(1) = -2
f''(1) = 12-6=6
f'''(1) = 24
f''''(1) = 24

Answer 2

so taylor series is defined as \[\sum_{0}^{\infty} [f ^{n}(a)(x-a)^n]/n!\]

Answer 3

so plugging that in i got -1 - 2(x-1)/1! + 3(x-1)^2 + 4(x-1)^3 + (x-1)^4 as the series

Answer 4

@zepdrix some assistance? :D

Answer 5

@ganeshie8 por favor? :-)

Answer 6

@Kainui

Answer 7

@ILovePuppiesLol

Answer 8

hi

Answer 9

whats that big symbol that looks like an E

Answer 10

thatd be a sigma.

Answer 11

why are n exponents

Answer 12

first ones actually the nth derivative, other ones an exponent

Answer 13

okey i officially am going to bed! i do not know this

Answer 14

k

Answer 15

Taylor series is
\(f(x-a) =\\
\color{red}{ f(a)} +\color{blue}{f'(a)} (x-a)+\color{green}{f''(a)} (x-a)^2+\color{orange}{f^{(3)}}(a) (x-a)^3+\color{purple}{f^{(4)}}(x-a)^4+f^{(5)}(x-a)^5+\cdots\)
Your \(f^{(5)}(x)=0\) hence, the series stops at \(f^{(4)}\)
Just plug the value in, you have
\(f(x-1) = \color{red}{-1}+\color{blue}{(-2)}(x-1) +\color{green}{(6)}(x-1)^2 +\color{orange}{(24)}(x-1)^3+\color{purple}{(24)}(x-1)^4\)
See what is going on??

Answer 16

For the radius, just plug x=1 in, all terms are gone but f(a) , hence radius is |-1|=1 . that means at a =1, inside the circle center 1, radius 1, the series converges.

Answer 17

use polynomial division or manually manipulate it into form as a polynomial in \(x-1\)
let \(x=u+1\) so
$$x^4 -3x^2+1=(u+1)^4-3(u+1)^2+1=u^4+4u^3+3u^2-2u-1
$$
so $$f(x)=-1-2(x-1)+3(x-1)^2+4(x-1)^3+(x-1)^4
$$

Answer 18

ah. so that essentially is the taylor series form? I dont need to represent it as some sort of sum because the derivative stopped after the fifth one? Meaning its finite?

Answer 19

the radius of convergence for a polynomial is clearly infinite as the polynomial expression is a finite series. @Loser66 you forgot to divide by \(n!\) in your expression for the Taylor expansion

Answer 20

yes, exactly @jtug6--the Taylor series gives a limit of successive polynomial approximations to a function using local data (derivatives of all orders); a polynomial itself though is already its best local polynomial approximation

Answer 21

thank you!

Answer 22

:) yes, I did @oldrin.bataku