An inverse tangent generates a reciprocal series · source reference

Printed question

If $y=\sqrt2\tan^{-1}(\frac{x\sqrt2}{x+3})$, show that $(x^2+2x+3)\frac{dy}{dx}=2$. [1]

By further differentiation of the above result, find the Maclaurin’s series expansion for $y$ in ascending powers of $x$ up to and including the term in $x^3$.

Hence find the first three non-zero terms in the expansion of $\frac1{x^2+2x+3}$. [7]

[2008/HCI/Prelim P2/1]

Printed working

$y'=\sqrt2\left[\frac1{1+2x^2/(x+3)^2}\right]\frac{\sqrt2(x+3)-x\sqrt2}{(x+3)^2}=\sqrt2\frac{3\sqrt2}{(x+3)^2+2x^2}=2/(x^2+2x+3)$.

$(x^2+2x+3)y'=2$; $(x^2+2x+3)y''+2(x+1)y'=0$.

$(x^2+2x+3)y'''+2(x+1)y''+2(x+1)y''+2y'=0$; simplify to $(x^2+2x+3)y'''+4(x+1)y''+2y'=0$.

At 0, $y=0$, $y'=2/3$, $y''=-4/9$, $y'''=4/27$. Thus $y\approx2x/3-2x^2/9+2x^3/81$.

$1/(x^2+2x+3)=y'/2\approx\frac12[2/3-4x/9+2x^2/27]=1/3-2x/9+x^2/27$.

Teacher note 1

$\frac d{dx}\tan^{-1}f(x)=\frac1{1+f(x)^2}f'(x)$, where $f(x)=x\sqrt2/(x+3)$. Use quotient rule to find $f'(x)$.

Teacher note 2

Let $y=f(x)$. To find the Maclaurin’s series for $y$, up to and including the term in $x^3$, differentiate up to the 3rd derivative. Substitute the values of $y,y',y'',y'''$ when $x=0$ into Maclaurin’s theorem:

$$f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+\cdots.$$

Mathematical review

Differentiation simplifies to $y'=2/(x^2+2x+3)$. Differentiate $Py'=2$, $P=x^2+2x+3$: $Py''+2(x+1)y'=0$ and $Py^{(3)}+4(x+1)y''+2y'=0$.

At zero, $y=0,y'=2/3,y''=-4/9,y^{(3)}=4/27$. Hence $y=2x/3-2x^2/9+2x^3/81+O(x^4)$. Since the reciprocal equals $y'/2$, it is $1/3-2x/9+x^2/27+O(x^3)$. Use the branch of the inverse tangent continuous near zero; the original quotient is undefined at $x=-3$.

Provenance

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