Question

a) Find the Taylor polynomial of order 4 that you would use to approximate the given number, b) give the approximation of the number and c) estimate the error of the approximation d) verify the error bound.

I'll center this around 1 so the IOC would be [1,1.05].

a) \[P_4(x) = \ln 2 + \frac{ 1 }{ 2 } (x-1) - \frac{ 1 }{ 8 }(x-1)^2 + \frac{ 1 }{ 24 }(x-1)^3 - \frac{ 1 }{ 64 }(x-1)^4\]
b) \[\ P_4(1.05) = 0.717840 \]
And then I get lost at part c and d..

Answer 1

Meh, i'll attempt this..
\[R_4 (x) = | \frac{ f^5(c) }{ 5! } (x-1)^5|\]
\[f(x) = \ln (1+x)\]
\[f^5(x) = \frac{ 24 }{(1+x)^5 } \]
\[f^5(c) = \frac{ 24 }{ (1+c)^5 }\]

\[R_4(x) = \left| \frac{ \frac{ 24 }{ (1+c)^5 } }{ 120 } (x-1)^5 \right|\]

Answer 2

basically u want to see, at what orders the terms are contributing to the approximation

Answer 3

For part a, You seem to be off by too much. I think you want to center around x=0 :

\[\ln(1+x) = x-\frac{x^2}{2!}+\frac{x^3}{3!}-\frac{x^4}{4!}+\cdots \]

Answer 4

there are no factorials

Answer 5

I simplified the factorials in the polynomial

Answer 6

Oops corrected :)

\[\ln(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots \]

Answer 7

Wouldn't the error be off by a lot more if i center it around x = 0 ?