How would you solve f(t)=1+t with -pi<x<pi?

[hawaiidude]

foueri series

how would you solve f(t)=1+t with -pi<x<pi?

we know that a0/2+sigma n=1 (an cos( nt) + bn(sin( nt) ?

 

[ahrkron]

Hi hawaiidude,

You need to provide some of the work you have done with this. How far have you gone?

Or, if you want, you can ask some general questions about Fourier series (which would not be moved to the homework help forum).

 

[M98Ranger]

we have to find the coefficients an and bn

since f(t)=1+t , we can rewrite it as f(t)=1+0t

so, a0/2=1 and an=0 for all n

now, to find bn

bn= 2/pi integral of f(t) sin(nt) dt from -pi to pi

= 2/pi integral of (1+t) sin(nt) dt from -pi to pi

= 2/pi integral of sin(nt) dt + integral of t sin(nt) dt from -pi to pi

= 0 + 2/pi integral of t sin(nt) dt from -pi to pi

= 2/pi (-cos(nt)) from -pi to pi

= 2/pi (-cos(npi) + cos(-npi))

= 2/pi (-cos(npi) + cos(npi))

= 4/pi cos(npi)

= 0

since bn=0 for all n, the Fourier series for f(t)=1+t is simply a0/2, which is equal to 1.

Therefore, the solution for f(t)=1+t with -pi<x<pi is f(t)=1.