Question

Problem for you:
Find the period of the function in the simplest way:

Answer 1

\[f(x+3) + f(x-3) = \sqrt{3} f(x)\]

Answer 2

Yeah a bit nasty

Answer 3

Haha.
It has an elegant solution, like always.
Best of luck.

Answer 4

Well I think you will excuse me (and may be even ask one of your pals to mdl me, since I want to leave the full soln to others)
\[ (e^{-i3\nu} + e^{i3\nu} ) \cal{F} (\nu) = \sqrt{3} \cal{F} (\nu) \]

Answer 5

\[ \frac{2}{\sqrt3}Cos(3\nu)\cal{F} (\nu) = \cal{F} (\nu)\]

Answer 6

(1)I do not understand how you made that conversion.
(2) This is elementary high school maths. Keep it simple.

Answer 7

let x=x+3\[f(x+6)+f(x)=\sqrt{3} \ f(x-3)\]let x=x-3\[f(x)+f(x-6)=\sqrt{3} \ f(x+3)\]add them\[f(x+6)+f(x-6)+2f(x)=\sqrt{3} \ (f(x-3)+f(x+3))=\sqrt{3} \ (\sqrt{3} \ f(x))=3f(x)\]

Answer 8

so\[f(x+6)+f(x-6)=f(x)\]

Answer 9

let x=x+3\[f(x+9)+f(x-3)=f(x+3)\]let x=x-3\[f(x+3)+f(x-9)=f(x-3)\]add them\[f(x+9)+f(x-9)=0\]period is 18

Answer 10

Are you sure?

Answer 11

emm...maybe ??? check my work...

Answer 12

Noo. Its all correct.
Just the conclusion doesnt seem right.

Answer 13

wait a min

Answer 14

Period is T when:
\[f(x) = f(x +T)\]

Answer 15

\[f(x+18)=-f(x)\]let x=x+9\[f(x+27)=-f(x+9)=f(x-9)\]period is 36

Answer 16

i think we are done

Answer 17

Precisely. Well done!!

Answer 18

@Mikael See. No Fourier's involved. :)

Answer 19

Still - please prove or DISprove the equation above. I would be mcuh obliged in either case. BTW u all should be intrigued by it - since it says f =0 identically
@mukushla @siddhantsharan

Answer 20

Yes, but as is well known, the undeclared but common interpretation of functional equations like that is that they hold FOR ALL real x/domain of definition