some maths

CHALLENGE

So here’s a maths problem I’ve had for an oral test that I found really cool and interesting :)

First, let’s suppose that sin(θ) ≠ 0 and cos(θ) ≠ 0 Then you can consider those 2 sums S and T

( cos^k (θ) is the same as (cos(θ))^k and θ is a real number and n is a natural number)

“Find a simple expression for S + iT and from that deduce a simple expression for S” ( i is the imaginary unit ) “What does S equals to if sin(θ) = 0 ?” ( because what you will find above doesn’t work for sin(θ) = 0 so you have to find another way around )

It’s not too hard since it’s meant for an oral test ( 4 hours written tests are much tougher than 1 hour oral tests in France <3 ) so I hope you guys can give it a try :)

You have oral maths tests? That is just cruel.

There are tricks to this, rite? Because I am absolutely not doing those sums. Especially if this is supposed to be an oral test.

Well Tsskyx, there is Euler’s identity : e^it = cos t + i sin t :)

And for PJ, yup we have those orals test also in physics, english, engineering/computer sciences and philosophy :p It usually goes like this: Me: Hello Sir Examinator: Good afernoon young man, show me the proof of Pascal’s formula dealing with binomial coefficients. Me: proceeds to prove Pascal’s formula on the blackboard Examinator: You made a mistake here. Me: proceeds to correct the mistake Examinator: Mhmm, alright. Now do that exercice. Me: omg sums with sin and cos

Oh it sounded like you just had to do the math on your head and say the answer xD

Well, for those tests, you have to write your stuff on the board and then explain what you did to the examinator, easier said than done :p Last time I was like: “so the nth roots of 1 can be expressed in the form e^i2kπ/n” and the guy was like “orly ?!??”

Lol

I got the sums down to (1+itantheta)^k, but I don’t know much about sums, so I stopped there.

Which sum are you talking about? I have no idea what you did xd

Full solution :D

First, ‘need to change S + iT into a compact expression c:

step 1 to step 2: S and T both got n terms so you can bring them into one big sigma sum thing step 2 to step 3: Euler’s identity: e^it = cos t + i sin t (there’s an easy proof for that if you assume that derivation rules work the same in ℂ, but if you don’t want to, there’s another proof that uses Taylor series) step 3 to step 4: the cosine and the exponential have got an exponent in common step 4 to step 5: it’s the sum the terms of a geometric sequence q^0 + q^1 + q^2 + … + q^n = (q^n+1 - 1)/(q - 1) (in France you’re taught the proof for that in highschool) step 5 to step 6: bringing the -1s into the fractions step 6 to step 7: cos^n+1 θ / cos θ = cos^n θ and then changing the whole thing into a single nice fraction step 7 to step 8: Euler’s identity again, but the cos θ from the exponential disappear with the -cos θ, so only i sin θ is left step 8 to step 9: bring the i to the front because it’s cool and 1/i = i^4 /i = i^3 = i^2 i = -i

Then you notice that S = Re(S + iT), because both S and T are sums of real numbers :p

step 2 to step 3: i cos^n+1 θ is purely imaginary so it’s real part is 0 step 3 to step 4: Euler’s identity again xd step 4 to step 5: again, the i cos((n+1)θ) term is purely imaginary, and then there are only real numbers left since i(-1) = 1

Aaaaand for that last question of the test

since cos 0 = 1

Et voilà ! \o/