STEP I, II, III 1999 solutions

The number of currants X in a portion x is actually $\sim Po(4x)$

So the integrals are actually $\displaystyle \int (2x)[e^{-4x}\frac{(4x)^{4}}{4!}] \, dx$

Aaah. I see. I will try that now.

In the meantime, these are my solutions for questions 1, 3, 4, 5, and 7 of paper III. Incidentally, one of my interview questions was to prove that there were only five regular polyhedra (last part of question 4), but by considering the sum of angles at each vertex instead... Much less long-winded, I think, even though it would be necessary to assume that all regular polyhedra are convex.

Also, I think this post needs updating... (There's also a thread for 2000 now.)

STEP III - Question 8

i)

$0 \leq x \leq 2\pi \Rightarrow n = 1$.

So the differential equation becomes:

$\frac{d^{2}y}{dx^{2}} + y = 0$, whose general solution is: $y(x) = A\cos(x) + B\sin(x)$.
Now, since when x=0, y=0 and $\frac{dy}{dx} = 1$, the particular solution is $y(x) =\sin(x)$ (*).

ii) Since $2\pi \leq x \leq 4\pi, \Rightarrow n = 2$.

So this time the differential equation is:

$\frac{d^{2}y}{dx^{2}} + 4y = 0$

general solution: $y(x) = C\cos(2x) + D\sin(2x)$ (**).

Now, since y and $\frac{dy}{dx}$ are continuous at $x=2\pi$, (**) has the same value as (*) at $x = 2\pi$ and also their derivatives take the same value. From this we can work out the values of C and D, (C=0 and D= $\frac{1}{2}$). Which gives us: $y(x) = \frac{1}{2}\sin(2x)$.
Using the same method in the previous part, we can deduce that $y(x) = \frac{1}{n}\sin(nx)$.

iii)

$\displaystyle\int^\infty_0 y^{2} \, dx = \displaystyle\int^{2\pi}_{0} y^{2} \, dx + \displaystyle\int^{4\pi}_{2\pi} y^{2} \, dx + \displaystyle\int^{6\pi}_{4\pi} y^{2} \, dx + ...........+ \displaystyle\int^{2r\pi}_{2(r-1)\pi} y^{2} \, dx + ....$




= $\displaystyle\int^{2\pi}_{0} \sin^{2}(x) \, dx + \displaystyle\int^{4\pi}_{2\pi} \frac{1}{2^{2}}\sin^{2}(x) \, dx + \displaystyle\int^{6\pi}_{4\pi} \frac{1}{3^{2}}\sin^{2}(x) \, dx + .....$



= $\displaystyle\int^{2\pi}_{0} (1-\cos(2x)) \, dx + \displaystyle\int^{4\pi}_{2\pi} \frac{1}{2^{2}}(1-\cos(4x)) \, dx + \displaystyle\int^{6\pi}_{4\pi} \frac{1}{3^{2}}(1-\cos(6x))\, dx + .....$

(using $\cos(2x) = 1 - 2\sin^{2}(x)$)

= $\frac{1}{2}[ 2\pi - 0] + \frac{1}{2^{2}}[\frac{1}{2}( 4\pi - 2\pi )] + \frac{1}{3^{2}}[\frac{1}{2}( 6\pi - 4\pi)] + ....$


= $\pi + \frac{\pi}{2^{2}} + \frac{\pi}{3^{2}} + ..... = \pi \displaystyle\sum_{n=1}^\infty \frac{1}{n^{2}}$

STEP III - Question 6

Parametrization: $x=a\cos^{n}(t)$and $y=a\sin^{n}(t)$

When n=3, $x=a\cos^{3}(t)$ and $y=a\sin^{3}(t)$.

Using the chain rule: $\frac{dy}{dx} = \frac{dy}{dt}.\frac{dx}{dt} = \frac{3a\sin^{2}(t)\cos(t)}{-3a\cos^{2}(t)\sin(t)} = -\tan(t).$

So $\frac{dy}{dx} = 0$ when $t= 0, \pi$and $2\pi$ and as $t \longrightarrow \frac{\pi}{2}$and $\frac{3\pi}{2}, \frac{dy}{dx} \longrightarrow \pm \infty.$

And also x=a when y=0, and y=a when x=0.

*graph is attached*

see later in thread for the rest.

1999 Paper 3 numbers 9,11 and 14

Step II, Q.8- Can anyone walk me through Brianeverit's solution, from the part attached below.