Why Do Signs Change in Fourier Series Coefficients Calculation?

[teo]

hi, first off just have to say these boards kick serious ass! Too much interesting posts on here, really like the Mechanical eng board since that's what I am doing at uni :)

Right, to the problem,

I do the integral and get the below answer,

Bn = 1/pi [-Cos nt/n] between pi/2 and 0
Bn = 1/pi [[-Cos n(pi/2)/n]-1/n]

but then my notes skip from this to another line,

Bn = 1/npi [1-Cos nt]

The n gets taken out and 1 comes from Cos n0 but why do the signs change?

 

[teo]

LOL just looked at it again, and i forgot that there's a minus infront of the cos nt/n which makes

Bn = 1/npi [(-cos nt)-(-1)]

which answers my question :)

 

[M98Ranger]

Hi there,

Glad to hear you are enjoying the Mechanical Eng board! As for your problem, it looks like you are working with Fourier series and are trying to find the coefficients Bn. The first step is to find the integral of the function you are trying to represent using the Fourier series. In this case, it seems like you have done that and have found the integral to be Bn = 1/pi [-Cos nt/n] between pi/2 and 0.

The next step is to evaluate the integral at the upper and lower limits. In this case, the upper limit is pi/2 and the lower limit is 0. When you plug in these values, you get Bn = 1/pi [-Cos n(pi/2)/n] for the upper limit and Bn = 1/pi [-Cos 0/n] for the lower limit. The reason for the change in signs is due to the negative sign in front of the Cos term and the fact that Cos(pi/2) = 0. So for the upper limit, you get Bn = 1/pi [0/n] = 0 and for the lower limit, you get Bn = 1/pi [-1/n] = -1/n.

Now, to simplify the expression, the -1/n can be rewritten as -1/(n*pi) and the n can be factored out to give Bn = 1/npi [1-Cos nt]. This is the same as the line in your notes, just written in a different order.

I hope this helps clarify the change in signs and how the expression is simplified. Keep up the good work in your studies!