Taylor Polynomials: Order 4 for ln(1+x), Derivative Patterns, and Error Analysis

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SUMMARY

The discussion focuses on deriving the Taylor polynomial of order 4 for the function ln(1+x) centered at 0, resulting in T4(x) = x - x²/2 + x³/3 - x⁴/4. Participants identified a pattern in the derivatives, leading to a general form for Tn(x) as (-1)^(n+1) x^n/n. The remainder term was established as Rn(x) = (-1)^n (x^(n+1)/(1+c)^(n+1)(n+1)), where c is between x and 0. To ensure that the Taylor polynomial provides an approximation of ln(1.3) with an error less than 0.0002, it was determined that n must be at least 4999.

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Homework Statement


(a) Give Taylor Polynomal of order 4 for ln(1+x) about 0.
(b) Write down Tn(x) of order n by looking at patterns in derivatives in part (a), where n is a positive integer.
(c) Write down the remainder term for the poly. in (b)
(d) How large must n be to ensure Tn gives a value of ln(1.3) which has an error less than 0.0002

Question 2 part d

Homework Equations


The Attempt at a Solution


Okay so we have that the taylor poly is Tn(x) = ∑[f'(a)(x-a)^n]/n! where f' is the nth derivative
And the remainder is Rn(x) = f'(c)(x-a)^(n+1)/(n+1)! where f' is the n+1 th derivative and c lies between x and a
For part a I differentiated ln(1+x) a few times and got a pattern... the values I got at x=0 were 0, 1, -1, 2, -6, 25, -120.
and I got T4 = x - x^2 /2 + x^3 /3 -x^4 /4
b. I got the highest term as (-1)^(n+1) x^n /n
c. I got the remainder term to be (-1)^n (x^(n+1)/(1+c)^(n+1)*(n+1))
d. I got an answer of n=4999 using the approximation that abs(Rn) is always less than abs(1/(n+1))
 
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I am afraid I don't have a log in for the university of Sydney.
 
Sorry about that
 
You want to be taken through the entire exercise, which is fine. However it is not fine to request that without showing your work. So let's start with that, show your work for a b c, write down the formulae you think are relevant. Like Taylor's theorem the definition of the remainder etc.
 
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