STEP III 1990 question 5 solution

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STEP III 1990 question 5 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP III 1990 question 5 solution

$\begin{pmatrix} n \\ r \end{pmatrix} + \begin{pmatrix} n \\ r-1 \end{pmatrix} = \frac{n!}{r!(n-r)!} + \frac{n!}{(r-1)!(n-r+1)!} = \frac{n!(n-r+1)}{r!(n-r+1)!} + \frac{n!r}{r!(n-r+1)!}$ $= \frac{n!(n-r+1) + n!r}{r!(n-r+1)!} = \frac{n!(n+1)}{r!(n-r+1)!} = \frac{(n+1)!}{r!(n+1-r)!} = \begin{pmatrix} n+1 \\ r \end{pmatrix}$

By induction. Case when n=1 is clearly just the product rule: $(uv)^{(1)} = u^{(1)}v + uv^{(1)}$

Assume true for n=k, then differentiate again to get: $(uv)^{(k+1)} = u^{(k+1)}v + u^{(k)}v^{(1)} + \begin{pmatrix} k \\ 1 \end{pmatrix}u^{(k)}v^{(1)} + \begin{pmatrix} k \\ 1 \end{pmatrix}u^{(k-1)}v^{(2)} + \begin{pmatrix} k \\ 2 \end{pmatrix}u^{(k-1)}v^{(2)} + ... + uv^{(k+1)}$ $= u^{(k+1)}v + (\begin{pmatrix} k \\ 0 \end{pmatrix}+\begin{pmatrix} k \\ 1 \end{pmatrix}) u^{(k)}v^{(1)} + (\begin{pmatrix} k \\ 1 \end{pmatrix}+ \begin{pmatrix} k \\ 2 \end{pmatrix})u^{(k-1)}v^{(2)} + ... + uv^{(k+1)}$ $= u^{(k+1)}v + \begin{pmatrix} k+1 \\ 1 \end{pmatrix}u^{(k)}v^{(1)} + \begin{pmatrix} k+1 \\ 2 \end{pmatrix}u^{(k-1)}v^{(2)} + ... + uv^{(k+1)}$

Which is the proposition with n=(k+1), therefore by induction it is true for all natural n.

$y=\arcsin x => \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} => \frac{d^2y}{dx^2} = \frac{x}{(1-x^2)^\frac{3}{2}}$ Therefore, the proposition for n=0 reduces to: $\frac{x}{\sqrt{1-x^2}} - \frac{x}{\sqrt{1-x^2}} = 0$ which is obviously true.

Now assume that the proposition is true for n=k and differentiate:

$(1-x^2)y^{(k+3)} - 2xy^{(k+2)} - (2k+1)xy^{(k+2)} - (2k+1)y^{(k+1)} - k^2y^{(k+1)} = 0$ $(1-x^2)y^{(k+3)} - (2(k+1)+1)xy^{(k+2)} - (k+1)^2y^{(k+1)} = 0$