Factor a logarithm argument · source reference

Printed question

(i) Given that $f(x)=\ln(2x+1)$, find $f(0),f'(0),f''(0),f'''(0)$. Hence obtain the first three non-zero terms in the Maclaurin’s series for $f(x)$. [5]

(ii) Hence, or otherwise, show that the first three non-zero terms in the expansion of $\ln(2x^2+3x+1)$ are $ax+bx^2+ax^3$, where $a$ and $b$ are constants to be found. [3]

[2010/MI/Prelim/I/7]

Printed working

$f(x)=\ln(2x+1)$; $f'=2/(2x+1)=2(2x+1)^{-1}$; $f''=-2(2x+1)^{-2}(2)=-4(2x+1)^{-2}$; $f'''=-4(-2)(2x+1)^{-3}(2)=16(2x+1)^{-3}$.

$f(0)=0$, $f'(0)=2$, $f''(0)=-4$, $f'''(0)=16$.

$f(x)=0+2x+(-4)x^2/2!+16x^3/3!+\cdots=2x-2x^2+8x^3/3+\cdots$.

$\ln(2x^2+3x+1)=\ln[(2x+1)(x+1)]=\ln(2x+1)+\ln(x+1)=2x-2x^2+8x^3/3+x-x^2/2+x^3/3+\cdots=3x-5x^2/2+3x^3+\cdots$.

$a=3$, $b=-5/2$.

Teacher note 1

$\frac d{dx}\ln g(x)=g'(x)/g(x)$, where $g(x)=2x+1$; $\frac d{dx}[g(x)]^n=n[g(x)]^{n-1}g'(x)$.

Teacher note 2

Since the question states “hence or …”, use the result for $\ln(2x+1)$. Factorise $2x^2+3x+1$ and re-express its logarithm in terms of $\ln(2x+1)$ using $\ln(AB)=\ln A+\ln B$.

Mathematical review

(i) $f'=2/(1+2x)$, $f''=-4/(1+2x)^2$, $f^{(3)}=16/(1+2x)^3$. Values at zero are $0,2,-4,16$. Therefore $f=2x-2x^2+8x^3/3+O(x^4)$.

(ii) $\ln(2x^2+3x+1)=\ln(1+2x)+\ln(1+x)$ near zero. Add $2x-2x^2+8x^3/3$ and $x-x^2/2+x^3/3$ to obtain $3x-5x^2/2+3x^3+O(x^4)$. Thus $a=3,b=-5/2,c=3$.

Provenance

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