Question

The maximum value of
1cosx + 2cos2x + 3cos3x + ..... + 99cos99x

Answer 1

The problem is that each cos is different

Answer 2

Hint : cos(0)

Answer 3

yeah max is 1

Answer 4

any Fourier series :D
if it goes to infinity it would be much easy xD it give the exact but here is a solution :-
0<=cos theta <=1

Answer 5

but cos x = 1 for cos x
for cos 3x = cos 0
so 3x =0 x =0 , fuk i tried it , it forms an ap yeah yeah

Answer 6

:)

Answer 7

i should have contiued further

Answer 8

lol , but if x=0 then all goes to 1 :P
so when u have x=0 it give u the maximum value :D

Answer 9

yes

Answer 10

all goes 1 ?

Answer 11

so no problem :3

Answer 12

yeb
cos 0 = cos 2(0)= cos n (0) =1 for n in N :)

Answer 13

i guess there was a theorem , which i cant remember or idk if its true
sum_n=0 ^ infinity n cos nx =1 for pi>n>0 hmm but not sure ( dont consider this case )
im only thinking in a comment :)

Answer 14

i see

Answer 15

so now u know
1cosx + 2cos2x + 3cos3x + ..... + 99cos99x <=1+2+3....+99 =n(n+1)/2 for n=99

Answer 16

This is what it looks like when Gauss and Fourier hang out. =P

Answer 17

nice one ;)

Answer 18

yeah lol exactly xD