STEP III 1997 question 6 solution

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STEP III 1997 question 6 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP III 1997 question 6 solution

Solution 1

If $x = \cos \theta$ then $\displaystyle \frac{d}{d\theta} = \frac{dx}{d\theta}\frac{d}{dx} = -\sin \theta \frac{d}{dx}$ .

So $\displaystyle \frac{dy_n}{d\theta} = - \sin \theta \frac{dy_n}{dx}, \frac{d^2y_n}{d\theta^2} = \sin^2 \theta \frac{d^2y_n}{dx^2} - \cos\theta \frac{dy_n}{dx} = (1-x^2)\frac{d^2y_n}{dx^2} - x \frac{dy_n}{dx}$

So $\displaystyle \frac{d^2y_n}{d\theta^2} + n^2 y_n = 0$ as desired.

This has general solution $y_n = A_n \cos n\theta + B_n \sin n\theta$ .

$y_n(1) = 1 \implies y_n(\cos 0) = 1 \implies A_n = 1$ .

To use the 2nd boundary condition, observe $-\cos \theta = \cos(\pi - \theta)$ and so we have:

$\cos n\theta + B_n \sin n\theta = (-1)^n (\cos n(\pi-\theta) + \sin n(\pi-\theta))$

$RHS = (-1)^n(\cos n\pi \cos n\theta -\cos n\pi \sin n\theta) = \cos n\theta - B_n \sin n\theta$ and so B_n = 0 .

So $y_n = \cos n \theta$ (which is also $\cos(n \arccos x)$ ). Finally, we need to show:

$\cos(n+1)\theta - 2\cos \theta \cos n\theta + \cos(n-1)\theta = 0$ .

But $\cos(n+1)\theta = \cos n\theta \cos \theta - \sin n \theta \sin \theta, \cos(n-1)\theta = \cos n\theta \cos \theta + \sin n \theta \sin \theta$ , hence result.

Solution by DFranklin

Solution 2

(1) $\\ \displaystyle \frac{dy_{n}}{dx} = \frac{dy_{n}}{d\theta}\times \frac{d\theta}{dx}\\ =\frac{dy_{n}}{d\theta}\times\frac{1}{-\sin\theta}$

Differentiate (1) to get (2):

$\displaystyle \frac{d^{2}y_{n}}{dx^2}=\frac{d^{2}y_{n}}{d\theta^2}\times\frac{1}{\sin^2\theta}-\frac{dy_{n}}{d\theta}\times\frac{\cos\theta}{\sin^3\theta}$

Substitute (1), (2) and $\displaystyle x=\cos\theta$ into $\displaystyle (1-x^2)\frac{d^{2}y_{n}}{dx^2}-x\frac{dy_{n}}{dx}+n^2y_{n}=0$ to get $\displaystyle \frac{d^2y}{d\theta^2}+n^2y_{n}=0\text{ as required }$

The general solution of this linear homogeneous second order differential equation is $\displaystyle y_{n}(\theta)=A\cos n\theta+B\sin n\theta$ We are given y(1)=1, hence after the tranformation of $\displaystyle x=\cos\theta$ above, we have the boundary condition y(0)=1 which results in A=1. Furthermore, we are given $\displaystyle y_{n}(x)=(-1)^ny_{n}(-x)$ which really means that $\displaystyle y_{n}(x)$ is an even function when n is either 0 or even. And that it's an odd function when n is odd. But after the substitution of $\displaystyle x=\cos\theta$ , $\displaystyle y_{n}(\theta)$ becomes an even function for all n since cosine is an even function.

We have $\displaystyle \\y_{n}(\theta)=A\cos n\theta+B\sin n\theta\\y_{n}(-\theta)=A\cos n\theta-B\sin n\theta$ which can only be even for all n when B=0

$\displaystyle \therefore y_{n}(\theta)=\cos n\theta$ and $\displaystyle y_{n}(x)=\cos (n\arccos x)$ for $\displaystyle |x|\leq1$

It follows that $\displaystyle y_{0}(x)=\cos 0 =1$ and $\displaystyle y_{1}(x)=\cos(\arccos x)=x$ as required.

$\displaystyle \\y_{n}(x)=\cos (n\arccos x)\\y_{n-1}(x)=\cos\{(n\arccos x)-(\arccos x)\}\\y_{n+1}(x)=\cos\{(n\arccos x)+(\arccos x)\}$

Upon expanding with Trigonometric Addition Formulas and adding $\displaystyle y_{n+1}(x)$ and $\displaystyle y_{n-1}(x)$ :

$\displaystyle \\y_{n+1}(x)+y_{n-1}(x)=2x\cos (n\arccos x)\\y_{n+1}(x)+y_{n-1}(x)=2xy_{n}(x)\\y_{n+1}(x)-2xy_{n}(x)+y_{n-1}(x)=0$ as shown.

Solution by khaixiang

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