Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Official TSR Mathematical Society

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Oh I Really Don't Care

DFranklin

If your integral is of the form

\int f(t) \, dt

you are only guaranteed to get the right result with a substitution of form u = g(t) if g' is never zero (in the interval under consideration).

Why is that?

Basically you end up dividing by g'(t).

Show that if

a+b+c=0

and a,b,c lie on the unit circle then a,b,c form the vertices of an equilateral triangle

SimonM

Show that if

a+b+c=0

and a,b,c lie on the unit circle then a,b,c form the vertices of an equilateral triangle

I'm not sure if this would be considered a stupid question, but by a + b + c = 0, do you mean the x-co-ordinates, or the y, or am I just showing how little maths I know?

Is it the sum of the x-coordinates and also (separately) the sum of the y-coordinates?

Aha, thank you.

SimonM

Show that if

a+b+c=0

and a,b,c lie on the unit circle then a,b,c form the vertices of an equilateral triangle

roots of unity?

I think you need to consider a,b,c, as vectors

a,b,c are complex numbers

I can't think of a good hint which doesn't give it all away, I guess "is there something you can do WLOG"

if a+b+c = 0 then the circumcircle of the triangle formed by these three points is centred at the origin the WLOG we can considerer only a,b,c of unit modulus we can also WLOG rotate it as this conserves its shape. If we note that we can rotate it till one of the points, say for instance a, is 1; we now need only show if 1+b+c= 0 then 1,b,c is an equilateral triangle.
letting b = cos(t)+isin(t) and c = cos(s)+isin(s) and spiting into real and imaginary parts, Im: sin(t)+sin(s)=0 => sin(t)=-sin(s) => t=-s and Re: 1+cos(s)+cos(t)=0 => 2cos(t)=1 => cos(t)= -1/2 => t = 2pi/3 , s = -2pi/3, so the points must form an equilateral triangle. thus for any complex numbers a+b+c=0 a,b,c must be the points of an equilateral triangle QED

Oh I Really Don't Care

Not tough but in the MIT integration speed challenge - see how quick it can be done;

\displaystyle \int \frac{x^4}{x^2 + 1} dx

DeanK22

Not tough but in the MIT integration speed challenge - see how quick it can be done;

\displaystyle \int \frac{x^4}{x^2 + 1} dx

\displaystyle \int \frac{x^4-1+1}{x^2+1} = \int \left ( x^2-1+\frac{1}{x^2+1} \right) dx = \frac{x^3}{3} - x + \tan^{-1} x + C

DeanK22

in the MIT integration speed challenge

What's this?

Oh I Really Don't Care

Swayum

What's this?

http://www.youtube.com/watch?v=HHAHiEgtyT8

That was

\int \frac{x^4}{1-x^2}\,dx

, surely?

(Pretty easy, to be honest - I suspect many people here could have done as well).

Daniel Freedman

DFranklin

That was

\int \frac{x^4}{1-x^2}\,dx

, surely?

(Pretty easy, to be honest - I suspect many people here could have done as well).

Yeah, it seems more a test of how fast you can write. The contestants weren't impressively quick.

Oh I Really Don't Care

\displaystyle \int \frac{x}{(1+x)^3(1+x+x^2)^{\frac{1}{2}}} dx

Spoiler

DeanK22

Not tough but in the MIT integration speed challenge - see how quick it can be done;

\displaystyle \int \frac{x^4}{x^2 + 1} dx

I had this integral in a STEP question yesterday.

A problem I came up with and did whilst I was bored in physics:

By making a substitution of the form

y = \frac{dx}{dt} + kx

, prove (without guessing the form of the solution) that the differential equation

\frac{d^2x}{dt^2} + b\frac{dx}{dt} + cx = 0

has general solution

x = Ae^{\lambda t} + Be^{\mu t}, \mu \neq \lambda

or

x = (At + B)e^{\lambda t}, \mu = \lambda

where

\mu

and

\lambda

are the roots of the equation

z^2 + bz + c = 0

.

Eurgh, I prefer the argument using vector spaces tbh

I know nothing about vector spaces :p:

:p:

It was more interesting than whatever pseudo-quantum-physics crap we were supposed to be doing, anyway.

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