By successively differentiating $\ln(3-x)$, find the Maclaurin's series for $\ln(3-x)$, up

Question

By successively differentiating $\ln(3-x)$, find the Maclaurin's series for $\ln(3-x)$, up to and including the term in $x^3$.[3] Using appropriate expansion from the List of Formulae (MF27), verify the correctness of the series found in (i). [2] Find the Maclaurin's series for $\sqrt{1-2x}[\ln(3-x)]$ up to and include the term in $x^3$.[3] Find the set of values of $x$ for which the expansion in part (iii) is valid. [2]

Accepted Answer

(i) $\ln(3-x)=\ln3-\frac{x}{3}-\frac{x^2}{18}-\frac{x^3}{81}+\cdots$ (ii) Verified using the MF27 series for $\ln(1+u)$. (iii) $\sqrt{1-2x}\ln(3-x)=\ln3-(\ln3+\tfrac13)x+(\tfrac5{18}-\tfrac12\ln3)x^2+(\tfrac{17}{81}-\tfrac12\ln3)x^3+\cdots$ (iv) $-\tfrac12<x<\tfrac12$

2022 TGM P1 Q3

2022 TGM P1 Q3

Video Solution:

Solution:

Solution locked