An even function has no linear term · source reference

Printed question

It is given that $y=(\cos x)^{1/2}$.

Show that $2y\frac{d^2y}{dx^2}+2(\frac{dy}{dx})^2+y^2=0$. [2]

Find Maclaurin’s series for $y$ in ascending powers of $x$, up to and including the term in $x^2$. [3]

By choosing a suitable value of $x$, deduce the approximate relation $\frac1{\sqrt[4]2}\approx1+k\pi^2$, where $k$ is a constant to be determined. [2]

[2008/AJC/Prelim/I/4]

Printed working

$y^2=\cos x$ (1); $2yy'=-\sin x$; $2yy''+2(y')^2=-\cos x=-y^2$, so $2yy''+2(y')^2+y^2=0$.

At 0: $y=1$, $2y'=0\Rightarrow y'=0$, $2y''+1=0\Rightarrow y''=-1/2$.

$y=1+(-1/2)x^2/2!+\cdots\approx1-x^2/4$.

$(\cos x)^{1/2}=1/\sqrt[4]2\Rightarrow\cos x=1/\sqrt2\Rightarrow x=\pi/4$.

$1/\sqrt[4]2\approx1-\frac14(\pi/4)^2=1-\pi^2/64$; $k=-1/64$.

Teacher note 1

Square the equation so the resulting equation (1) is easier to handle for repeated differentiation using implicit differentiation and product rule.

Use implicit differentiation and product rule:

$$\frac d{dx}(yy')=y\frac d{dx}(y')+y'\frac d{dx}(y)=yy''+(y')^2.$$

Teacher note 2

Let $y=f(x)$. To find the Maclaurin’s series for $y$, up to and including the term in $x^2$, differentiate up to the 2nd derivative. Substitute the values of $y,y',y''$ 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

Differentiate $y^2=\cos x$ twice: $2(y')^2+2yy''=-\cos x=-y^2$. At zero $y=1,y'=0$, so $y''=-1/2$. Hence $y=1-x^2/4+O(x^4)$.

At $x=\pi/4$, $(\cos x)^{1/2}=2^{-1/4}$, giving $1/\sqrt[4]{2}\approx1-\pi^2/64$ and $k=-1/64$. This is a truncated estimate, with no claimed rigorous error bound.

Provenance

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