STEP III 1990 question 5 solution - The Student Room
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STEP III 1990 question 5 solution

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

Which is the proposition for n=k+1. Therefore, by induction, the proposition is true for all non-negative integer n.

Solution by Speleo.

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