Question

Taylor series/Lagrange remainder question:
Let f(x) = (1+x)^(1/3). Write a fraction that estimates (1.2)^(1/3)

Answer 1

Also, the previous part of the question asked to find the Taylor polynomial of degree 2, which I found to be T(x) = 1+x/3-(x^2)/9

Answer 2

Would we simply write (1.2)^(1/3) is approximately T(1.2) = 1+1.2/3 -(1.2)^2/9 ??

Answer 3

Oh and it's centered at c=0 (so I guess it's a Maclaurin series but whatever)

Answer 4

Well Maclaurin series is an approximation so that would give you an estimate.

Answer 5

I doubt this is the answer because substitution for T(1.2) gives 1.24, yes (1.2)^(1/3)=1.0626... (and the next question is asking to show that the error in your estimate is at most 1/2025...)

Answer 6

give me a second

Answer 7

Did you calculate your Maclaurin series correctly?

Answer 8

Yes, I double-checked on Wolfram as well: http://www.wolframalpha.com/input/?i=taylor+polynomial+%281%2Bx%29%5E%281%2F3%29

Answer 9

What's \(f''(0)\)?

Answer 10

f''(0) = -2/9

Answer 11

http://www.wolframalpha.com/input/?i=y%3D+1%2Bx%2F3-%28x%5E2%29%2F9%2C+x%3D1.2

Answer 12

Well since your math is right, basically it just so happens to give a high error this time.

Answer 13

I think the fraction they're looking for though is not right? :( I dunno because I'll need it to have an error of at most 1/2025 according to the next question. So I'm off already at the first decimal :S

Answer 14

Ohhh

Answer 15

Hold on ._. I think I know the problem

Answer 16

Well intuitively when I think about a Taylor polynomial I think about a linear approximation

Answer 17

This is only one degree higher of approximation

Answer 18

I know what went wrong.. the original function is (1+x)^(1/3), so I need to use T(0.2) not T(1.2) since there's already a 1...

Answer 19

I feel so dumb loool

Answer 20

Okay, that explains it.

Answer 21

ok now it Thanks for your help anyways though :) Always appreciated