Recent content by bbq pizza

g(t)=½( f(t)+f(-t) ) h(t)=½( f(t)-f(-t) ) show its 2π periodic so: g(t+2π) = ½( f(t+2π)+f(t-2π) ) why does -t become t-2π ? ½( f(t)+f(-t) ) = g(t) h(t+2π)=½( f(t+2π)-f(t-2π) ) ½( f(t)-f(-t) ) = h(t) is this correct? can...

i got π/4 + ( ( -1)n-1 ) cos(nx) ) /π*n2+( (-1)nsin(nx) ) /n with the an=(-1)n-1) / (πn2) because during the an calculation i get an=( cos(nπ)-cos(n0) ) / (πn2) and cos1π=-1 and cos(0)=cos2π=1 so -1n=+1 for n=even and =-1 for n=odd and bn= - ( (-1)n ) / n from ( -nπcos(nπ) )/πn2 I'm still...

thanks BvU. I looked over it and noticed some mistakes, so.. an=( (-1)^n -1) / (π * n^2) and bn= - ( (-1)^n ) / n i see the π/2 getting rid of cosine but how do i achieve the π/4 = 1 - 1/3 + 1/5 ... using suitable values of x. do i set x=π/4 then get 1 for ??

http://imageshack.com/a/img911/5796/jhLmo0.jpg http://imageshack.com/a/img673/1972/xQeDHn.jpg x=pi gives too high so i must have made a mistake on calculation of ceoffiscients but i don't see it

hi, if there is a initial value problem with a ζ in it with specified values what do you do with it when taking the laplace transform?

< Mentor Note -- thread moved to HH from the technical forums, so no HH Template is shown >[/color] hi I've got a problem that I've partially worked but don't understand the next part/have made a mistake? f(x)=0 for -π<x<0 and f(x)=x for 0≤x≤π i got a0=π/4 and an=0 and bn=0 if n is even and...