Use one series to recover its companion · source reference

Printed question

Let $y=\cos[\ln(1+x)]$ for $|x|<1$.

(a) Show that $(1+x)^2\frac{d^2y}{dx^2}+(1+x)\frac{dy}{dx}+y=0$. By further differentiation of this result, show that the Maclaurin series for $y$ is given by $1-\frac12x^2+\frac12x^3-\frac5{12}x^4+\cdots$. [6]

(b) By using the result in part (a), deduce the Maclaurin series for $\sin[\ln(1+x)]$ in ascending powers of $x$ up to and including $x^3$. [2]

(c) Hence state the equation of tangent to the curve $y=\sin[\ln(1+x)]$ at $x=0$. [1]

[2022/DHS/Prelim/I/6]

Printed working

$y'=-\sin[\ln(1+x)]/(1+x)$, so $(1+x)y'=-\sin[\ln(1+x)]$.

$(1+x)y''+y'=-\cos[\ln(1+x)]/(1+x)=-y/(1+x)$; multiplying by $1+x$ gives the required identity.

Differentiate: $(1+x)^2y'''+2(1+x)y''+(1+x)y''+y'+y'=0$, hence $(1+x)^2y'''+3(1+x)y''+2y'=0$.

Differentiate again: $(1+x)^2y''''+2(1+x)y'''+3(1+x)y'''+3y''+2y''=0$, hence $(1+x)^2y''''+5(1+x)y'''+5y''=0$.

At 0: $y=1$, $y'=0$, $y''=-1$, $y'''=3$, $y''''=-10$. Thus $y=1-x^2/2!+3x^3/3!-10x^4/4!+\cdots=1-x^2/2+x^3/2-5x^4/12+\cdots$.

$\sin[\ln(1+x)]=-(1+x)y'=-(1+x)(-x+3x^2/2-5x^3/3+\cdots)=x-x^2/2+x^3/6+\cdots$.

Tangent at 0 is $y=x$.

Teacher note 1

Let $u=1+x$, $v=\ln(1+x)=\ln u$. Use chain rule, differentiate from outside to inside: $\frac d{dx}\cos[\ln(1+x)]=\frac d{dv}\cos v\frac{dv}{du}\frac{du}{dx}=-\sin v\times\frac1u\times1=-\sin[\ln(1+x)]/(1+x)$.

Teacher note 2

Use implicit differentiation and product rule: $\frac d{dx}[(1+x)^2y'']=(1+x)^2y'''+2(1+x)y''$.

Teacher note 3

For $y=\sin[\ln(1+x)]$, when $x=0$, $y=0$, $y'=1$. Tangent: $y-0=1(x-0)$, so $y=x$ (constant and term in $x$ in the series).

Mathematical review

$y'=-\sin[\ln(1+x)]/(1+x)$. Differentiating gives the stated differential identity. Differentiating the identity yields $(1+x)^2y^{(3)}+3(1+x)y\prime\prime+2y\prime=0$ and $(1+x)^2y^{(4)}+5(1+x)y^{(3)}+5y\prime\prime=0$. At zero, successive derivative values are $y=1,y'=0,y''=-1,y^{(3)}=3,y^{(4)}=-10$, obtained by differentiating the identity successively. Thus $y=1-x^2/2+x^3/2-5x^4/12+O(x^5)$.

Then $-(1+x)y'=-(1+x)(-x+3x^2/2-5x^3/3+\cdots)=x-x^2/2+x^3/6+O(x^4)$. The sine-composite curve has point $(0,0)$ and slope 1, so its tangent is $y=x$.

Provenance

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