Question

How to determine bn of a fourier series and find the 6 non zero terms where :

bn = 1/2 * integral t*sin(npit/2) dt

Where the integral is from -2 to 2

Answer 1

Can one confirm that bn =0

Answer 2

find n , bn=0?

Answer 3

i got this 1/2((-2t/npi*cos(npit/2)) from integral -2 to 2 - 1/2 *integral -2 to 2 * -2npi*cos(npit/2)) dt

Answer 4

it is \[a_{n}\]

Answer 5

No the question states as bn as the funtion was odd... im just finishing of the integration part off, to form a fourier series.

Answer 6

find expansion fourie?

Answer 7

bn can't be 0

Answer 8

what did any one else get ?

Answer 9

bn = \(\large1/2 [\frac{tcos(n \pi t/2)}{(n \pi/2)}]^2_{-2}-1/2\int_{-2}^2\frac{cos(n \pi t/2)}{n \pi/2}\)
u got this, right ?

Answer 10

the 2nd term will only go to 0

Answer 11

because sin n*pi=0

Answer 12

but the 1st term will be still there

Answer 13

f(t)=t is odd function
f~\[\sum_{1}^{\infty}bnsin(npit/2)\]

Answer 14

oh right i see now .. so that means whenever there is sinnpi its always going to give 0!

Answer 15

yeah, sin(npi) = 0

Answer 16

so for the first part i got 1/2(-4/npi*cosnpi) +1/2(4/npicos(npi))) how is the 8 come out and the n^2 ....?

Answer 17

bn=-4cosnpi/npi ?

Answer 18

even i get -4cosnpi/npi

Answer 19

the ans should be 8sin(pin)-8pin(cosspin)/n^2*pi^2 should be the ans but how is it worked out ?

Answer 20

^ reduces to -4 cos n pi/n pi

Answer 21

-4conpi/npi=\[(-1)^{n}(-4)/npi\]

Answer 22

cosnpi=(-1)^n

Answer 23

yes i understand that but the ans should be 8sin(pin)-8pin(cosspin)/n^2*pi^2 not -4cospi/npi ???

Answer 24

any ideas why?

Answer 25

why do you think there should be 8's in the answer. The integration you posted in Wolfram is missing the leading 1/2 that is in the original problem

Answer 26

i understand finally thankyou everyone.