For the function defined by f(t)=2-t, 0≤t< 1, sketch 3 periods and find:
a) a half-range sine series expansion
b) a half-range cosine series expansion
c) a full range fourier series expansion

i have no idea how to solve a or b. i've tried a) several times but cannot seem to get the value for when n is odd.

will be appreciated.

The half-range sine series is the expansion for with the assumption that is considered to be an odd function over its full range, . So for (a), you're essentially finding the full range expansion of the function



with period 2 so that for and integers .

Now, since is odd, there is no cosine series (you find the cosine series coefficients would vanish), leaving you with



where



In this case, , so





The half-range sine series expansion for is then



which can be further simplified by considering the even/odd cases of , but there's no need for that here.

The half-range cosine series is computed similarly, this time assuming is even/symmetric across its full range. In other words, you are finding the full range series expansion for



Now the sine series expansion vanishes, leaving you with



where



for . Again, . You should find that







Here, splitting into even/odd cases actually reduces this further. Notice that when is even, the expression above simplifies to



while for odd , you have



So the half-range cosine series expansion would be





Attached are plots of the first few terms of each series overlaid onto plots of . In the half-range sine series (right), I use terms, and in the half-range cosine series (left), I use or terms. (It's a bit more difficult to distinguish from the latter because the cosine series converges so much faster.)

Multiple choice or not?