A shifted cotangent is analytic at zero · source reference

Printed question

(i) Given that $y=\cot(2x+\frac\pi4)$, find an equation relating $\frac{dy}{dx}$ and $y$. [2]

Hence show that $\frac{d^3y}{dx^3}=k[y\frac{d^2y}{dx^2}+(\frac{dy}{dx})^2]$, where $k$ is a constant to be found. [2]

(ii) Find the Maclaurin’s series for $y$ up to and including the term in $x^3$ if $x$ is sufficiently small for powers of $x$ higher than $x^3$ to be neglected. [2]

(iii) Using the Maclaurin’s expansion for $y$, estimate the value of $\csc^2(\frac{13\pi}{50})$, giving your answer in the form $a+b\pi+c\pi^2$, where $a,b,c$ are constants to be determined. [4]

(iv) Using your result from part (ii), state the equation of the tangent to the curve $y=\cot(2x+\frac\pi4)$ at $x=0$. [1]

(v) Explain why a Maclaurin series for $y=\cot2x$ cannot be found. [1]

[modified 2009/ACJC/Prelim/I/11]

Printed working

$y'=-2\csc^2(2x+\pi/4)=-2[1+\cot^2(2x+\pi/4)]=-2(1+y^2)$.

$y''=-4yy'$, $y'''=-4(y')^2-4yy''=-4[yy''+(y')^2]$, so $k=-4$.

At 0, $y=1$, $y'=-4$, $y''=16$, $y'''=-128$; $y=1-4x+16x^2/2!-128x^3/3!+\cdots\approx1-4x+8x^2-64x^3/3$.

Differentiate: $-2\csc^2(2x+\pi/4)\approx-4+16x-64x^2$; $\csc^2(2x+\pi/4)\approx2-8x+32x^2$.

$2x+\pi/4=13\pi/50\Rightarrow x=\pi/200$; $\csc^2(13\pi/50)\approx2-\pi/25+\pi^2/1250$. $a=2,b=-1/25,c=1/1250$.

Tangent is $y=1-4x$. At 0, $\cot2x$ is undefined, hence no Maclaurin expansion exists there.

Teacher note 1

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:

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

Teacher note 2

The first 2 terms (constant and term in $x$) in the Maclaurin’s series for $y$ give the tangent equation.

Mathematical review

(i) $y'=-2(1+y^2)$, $y''=-4yy'$, and $y^{(3)}=-4[(y')^2+yy'']$, so $k=-4$.

(ii) At zero, $y=1,y'=-4,y''=16,y^{(3)}=-128$. Thus $y=1-4x+8x^2-64x^3/3+O(x^4)$.

(iii) $\csc^2(2x+\pi/4)=-y'/2\approx2-8x+32x^2$. Set $x=\pi/200$: the estimate is $2-\pi/25+\pi^2/1250$, so $a=2,b=-1/25,c=1/1250$.

(iv) The tangent is $y=1-4x$. (v) $\cot2x$ has a pole at zero, so an ordinary power series there cannot represent it (a Laurent series is possible).

Provenance

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