1 Answer

Ask a Question

Dmlittle · Answer 1 · 2018-09-30T10:00:49+0000

The half-range sine series is the expansion for
f(t)

with the assumption that
f(t)

is considered to be an odd function over its full range,
-1<t<1

. So for (a), you're essentially finding the full range expansion of the function

$f(t)=\begin{cases}2-t&\text{for }0\le t<1\\-2-t&\text{for }-1<t<0\end{cases}$

with period 2 so that
f(t)=f(t+2n)

for

and integers

.

Now, since
f(t)

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

$f(t)=\displaystyle\sum_(n\ge1)b_n\sin\frac{n\pi t}L$

where

$b_n=\displaystyle\frac2L\int_0^Lf(t)\sin\frac{n\pi t}L\,\mathrm dt$

In this case,
L=1

, so

$b_n=\displaystyle2\int_0^1(2-t)\sin n\pi t\,\mathrm dt$

$b_n=\frac4{n\pi}-(2\cos n\pi)/(n\pi)-(2\sin n\pi)/(n^2\pi^2)$

$b_n=(4-2(-1)^n)/(n\pi)$

The half-range sine series expansion for
f(t)

is then

$f(t)\sim\displaystyle\sum_(n\ge1)(4-2(-1)^n)/(n\pi)\sin n\pi t$

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
f(t)

is even/symmetric across its full range. In other words, you are finding the full range series expansion for

$f(t)=\begin{cases}2-t&\text{for }0\le t<1\\2+t&\text{for }-1<t<0\end{cases}$

Now the sine series expansion vanishes, leaving you with

$f(t)\sim\frac{a_0}2+\displaystyle\sum_(n\ge1)a_n\cos\frac{n\pi t}L$

where

$a_n=\displaystyle\frac2L\int_0^Lf(t)\cos\frac{n\pi t}L\,\mathrm dt$

for
$n\ge0$ . Again,
L=1

. You should find that

$a_0=\displaystyle2\int_0^1(2-t)\,\mathrm dt=3$

$a_n=\displaystyle2\int_0^1(2-t)\cos n\pi t\,\mathrm dt$

$a_n=\frac2{n^2\pi^2}-(2\cos n\pi)/(n^2\pi^2)+(2\sin n\pi)/(n\pi)$

$a_n=(2-2(-1)^n)/(n^2\pi^2)$

Here, splitting into even/odd cases actually reduces this further. Notice that when

is even, the expression above simplifies to

$a_(n=2k)=(2-2(-1)^(2k))/((2k)^2\pi^2)=0$

while for odd

, you have

$a_(n=2k-1)=(2-2(-1)^(2k-1))/((2k-1)^2\pi^2)=\frac4{(2k-1)^2\pi^2}$

So the half-range cosine series expansion would be

$f(t)\sim\frac32+\displaystyle\sum_(n\ge1)a_n\cos n\pi t$

$f(t)\sim\frac32+\displaystyle\sum_(k\ge1)a_(2k-1)\cos(2k-1)\pi t$

$f(t)\sim\frac32+\displaystyle\sum_(k\ge1)\frac4{(2k-1)^2\pi^2}\cos(2k-1)\pi t$

Attached are plots of the first few terms of each series overlaid onto plots of
f(t)