STEP Tricks and Hacks Thread

Not a STEP-er but I'm joining this thread to pick up on useful techniques. Here's an easy one that might be helpful:

Factorising $x^{4} + 1$ and polynomials of the like -

It's fairly easy to write out $\displaystyle x^{4} + 1 \equiv \left ( x^{2} + \alpha x +1 \right )\left( x^{2}+\beta x +1 \right )$ and then compare coefficients using C3 techniques, but in case you're feeling adventurous,



This could come in handy in certain integrals. Try the classic $\displaystyle \int \sqrt{\tan x} \ \mathrm{d}x$ using this.

I gave a fair amount of effort to do this, so please hope it's correct!!!

this post is more useful i think, and im surprised it hasnt been posted about yet; (maybe everyone knows it already)

$\displaystyle \left(a^n + b^n \right) = \left(a-b\right)\left(a^{n-1} - a^{n-2}b + a^{n-3}b^2 - a^{n-4}b^3 + ... - ab^{n-2} + b^{n-1}\right)$

^^ this is only valid when n is even. Notice how you alternate signs the whole way down the right hand side bracket, starting and ending with plus signs

$\displaystyle \left(a^n - b^n\right) = \left(a-b\right)\left(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + ... + ab^{n-2} + b^{n-1}\right)$

valid for all integer n, and notice how the right hand bracket is all plus signs

very useful factorisation, but one of your signs is wrong on the first one, and it works for odd n, not even

The answer looks correct at a first glance. If you're interested in a slightly neater way of doing this, check here.

I gave a fair amount of effort to do this, so please hope it's correct!!!
P.s sorry for my messy writing, I've been ill for the past few days so don't have the steadiest hand ever
(also final answer is at the top)

The answer is spot on with this method. But please do check out the elegant solution Zacken posted!

Is it worth mentioning T substitutions?

As in the tangent sub? I think it's already been mentioned.

hmm I was thinking:
$[br]\displaystyle t = \tan \frac{ \theta }{2}[br][br]\displaystyle \cos \theta = \frac{ 1 - t^2 }{ 1 + t^2}[br][br]\displaystyle \sin \theta = \frac{ 2t }{ 1 + t^2}[br][br]\displaystyle \frac{d\theta}{dt} = \frac{2}{1 + t^2}[br][br]$
couldn't see that anywhere earlier

p.s. anyone know how to space lines in latex?

A link had been posted by shamika, but you can definitely expand on it.

I just do

...

hmm I was thinking:
$[br]\displaystyle t = \tan \frac{ \theta }{2}[br][br]\displaystyle \cos \theta = \frac{ 1 - t^2 }{ 1 + t^2}[br][br]\displaystyle \sin \theta = \frac{ 2t }{ 1 + t^2}[br][br]\displaystyle \frac{d\theta}{dt} = \frac{2}{1 + t^2}[br][br]$
couldn't see that anywhere earlier

p.s. anyone know how to space lines in latex?

p.s. anyone know how to space lines in latex?

Roots of unity

If
$[br]\omega[br]$
is the nth root of unity,

we can write

$[br](\omega)^{n} - 1 = 0[br](\omega - 1)(\omega^{n-1} + \omega^{n-2} ... + 1) = 0[br]$

Since (normally) the first bracket isn't zero, you have the somewhat useful result that the latter is zero. The geometrical implication of this (by considering an n-sided polygon in the Argand diagram, centred at the origin) is that the sum of the vectors from the centre of a polygon to its individual vertices is zero.

See III/97/3, I/90/2.

on the verge of anal leakage

There is a very clever way of doing this, but I do not quite remember it.