Equation

**lehder** · August 17th 2009, 03:30 PM

HI everybody,

I must show that the following equation: cosx+cos2x+...+cosnx=0 $(n\in \mathbb{N}*)$ accepts at least one solution in $[0,\pi]$ .

1)-We know that f(x)=cosx+cos2x+...+cosnx is continious in $[0,\pi]$ .

2)-So i must show that $f(0)*f(\pi)<0$ , BUT HOW?

CAN YOU HELP ME PLEASE?

AND THANKS ANYWAY.

**pickslides** · August 17th 2009, 04:38 PM

For $f(0)\times f(\pi)$ consider

$f(\pi)= \cos(\pi)+\cos(2\pi)+\cos(3\pi)+\cos(4\pi)+\dots$

$f(\pi)= -1+1-1+1+\dots$

and

$f(0) = \cos(0)+\cos(2\times 0 )+\cos(3\times 0)+\cos(4\times 0)+\dots$

$f(0) = 0+0+0+0+\dots$

Can you see a pattern? What are the implications for n being either even or odd for $f(\pi)$ ?

Thread: Equation

LinkBack

Thread Tools

Display

Equation

Similar Math Help Forum Discussions

problem on matrix ricaati equation for nonlinear singular differential equation

Partial differential equation-wave equation - dimensional analysis

General solution to a trig equation by turning it into a polynomial equation

Represent inverse trigonometric equation as rational integral equation?

vector equation for line perpendicular to cartesion equation through point

Search Tags