# Math20512

### FAQ

The coursework is available here.

### Question 1

I've been asked what shape a cone with a square base is. It is the same as a pyramid. I have now amended the coursework to make this (I hope) clear.

### Final part of Question 2

The period of the solutions to a linear differential equation is easy to find: just solve it (sines and cosines) and see what the period is.

If the equation is not linear (like the pendulum $$\ddot x = - \sin x$$ ), one needs to approximate it by a linear equation.

Method: Do a Taylor series expansion of all the terms, and ignore all nonlinear terms.

Artificial example: Consider $\ddot x + x\dot x +x\cos x - \dot x^2 =0.$ (Not solvable!) To "linearize" expand everything as Taylor series (the only thing that needs expanding here is $$\cos x$$), and then ignore all nonlinear terms. We get $\ddot x + x\dot x + x(1-\frac12 x^2+\frac1{4!}x^4-\dots) - \dot x^2 =0.$ The only linear terms are the first, and the $$x.1$$ from the cos series, to get $\ddot x + x=0.$ The general solution to this equation is $x(t) = A\cos t + B\sin t,$ and this has period $$2\pi$$.

Another method
(Step 1) Let $$\varepsilon$$ be a constant (small) and replace $$x$$ by $$\varepsilon u$$. So put $$x=\varepsilon u$$ and $$\dot x = \varepsilon\dot u$$ etc, to get $\varepsilon\ddot u +\varepsilon^2 u\dot u +\varepsilon u\cos(\varepsilon u) - \varepsilon^2\dot u^2=0.$ (Step 2) Now divide by $$\varepsilon$$ (notice it's a factor here- but you might have to take Taylor series to get it as a factor) to get $\ddot u +\varepsilon u\dot u + u\cos(\varepsilon u) - \varepsilon\dot u^2=0.$ (Step 3) Put $$\varepsilon=0$$. We then get (with $$\cos(0)=1$$) $\ddot u + u=0.$ (Step 4) Solve this as before.

Note: the two methods are essentially the same, the advantage of the second is that it makes systematic what one means by the linear terms