Two routes to the same square-root series · source reference

Printed question

It is given that $y=\sqrt{e^x\cos x}$.

(i) Show that $2y\frac{dy}{dx}=y^2-e^x\sin x$. [2]

(ii) By further differentiation of the result in part (i), find the Maclaurin series for $y$, up to and including the term in $x^2$. [4]

(iii) Using standard series from the List of Formulae (MF26), expand $\sqrt{e^x\cos x}$ as far as the term in $x^2$ and verify that the same result is obtained in part (ii). [3]

[2020/MI/Prelim/I/6]

Printed working

Method 1, implicit differentiation. $y^2=e^x\cos x$; $2yy'=e^x(-\sin x)+\cos x(e^x)=y^2-e^x\sin x$.

Method 2, direct differentiation. $y'=\frac12(e^x\cos x)^{-1/2}[e^x(-\sin x)+\cos x(e^x)]=\frac1{2y}(y^2-e^x\sin x)$; multiply by $2y$.

Differentiate $2yy'=y^2-e^x\sin x$: $2yy''+2(y')^2=2yy'-[e^x\cos x+e^x\sin x]=2yy'-(y^2+e^x\sin x)$.

At 0, $y=1$, $2(1)y'=1\Rightarrow y'=1/2$; $2y''+2(1/2)^2=2(1)(1/2)-1=0\Rightarrow y''=-1/4$.

$y=1+\frac{1/2}{1!}x+\frac{-1/4}{2!}x^2+\cdots=1+x/2-x^2/8+\cdots$.

Standard-series method 1. $e^{x/2}(\cos x)^{1/2}=[1+x/2+(x/2)^2/2!+\cdots][1+\frac12(-x^2/2!)+\cdots]=1+x/2+x^2/8-x^2/4+\cdots$.

Standard-series method 2. $[(1+x+x^2/2+\cdots)(1-x^2/2+\cdots)]^{1/2}=(1+x+\cdots)^{1/2}=1+x/2+\frac{(1/2)(-1/2)}{2!}x^2+\cdots$. Both verify the same result.

Teacher note 1

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

Teacher note 2

Use implicit differentiation and product rule:

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

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:

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

No need to make $y''$ the subject first; substitute the values at zero into their respective equations.

Teacher note 3

From MF27, $\cos x=1-x^2/2!+x^4/4!-\cdots$. For $e^{x/2}$ replace $x$ by $x/2$ in $e^x=1+x+x^2/2!+\cdots$.

Mathematical review

(i) $y^2=e^x\cos x$, hence $2yy'=e^x\cos x-e^x\sin x=y^2-e^x\sin x$.

(ii) Differentiating gives $2(y')^2+2yy''=2yy'-e^x(\sin x+\cos x)$. At zero, $y=1,y'=1/2$; therefore $1/2+2y''=0$, so $y''=-1/4$. Thus $y=1+x/2-x^2/8+O(x^3)$.

(iii) $e^{x/2}=1+x/2+x^2/8+O(x^3)$ and $(\cos x)^{1/2}=1-x^2/4+O(x^4)$. Their product gives the same polynomial. The real square-root branch is valid near zero where $\cos x>0$.

Provenance

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