#import "@preview/physica:0.9.6": *
#import "lib.typ": ccases

= problem 4

== a)

$
  f(x) = cases(
    -pi - x quad & "if" -pi < x < -pi/2,
    x quad & "if" -pi/2 < x < pi/2,
    pi - x quad & "if" pi/2 < x <= pi
  )
$

$f(x)$ is odd, so $a_0 = a_n = 0$.

$
  b_n & = 2/pi integral_0^pi f(x) sin(n x) dd(x) \
      & = 2/pi [integral_0^(pi/2) x sin(n x) dd(x)
          + integral_(pi/2)^pi (pi - x) sin(n x) dd(x)]
$

$
  integral_0^(pi/2) x sin(n x) dd(x)
  & = [-x/n cos(n x) + 1/n^2 sin(n x)]_0^(pi/2)\
  & = -pi/(2n) cos(n pi/2) + 1/n^2 sin(n pi/2)
$

$
  integral_(pi/2)^pi (pi - x) sin(n x) dd(x)
  & = (-1)^n integral_0^(pi/2) u sin(n u) dd(u)\
  & = (-1)^n [-pi/(2n) cos(n pi/2) + 1/n^2 sin(n pi/2)]
$

thus
$
  b_n = 2/pi (1 + (-1)^n)[-pi/(2n) cos(n pi/2) + 1/n^2 sin(n pi/2)]
$

for even $n$: $cos(n pi/2) in {-1,0,1}$ and $sin(n pi/2) = 0$, so $b_n = 0$

for odd $n$: $cos(n pi/2) = 0$ and $sin(n pi/2) = (-1)^((n-1)/2)$
$
  b_n = 4/(pi n^2) sin(n pi/2)
$

for $n = 2k+1$:
$
  b_(2k+1) = 4/(pi(2k+1)^2) (-1)^k
$

applying parseval's identity:
$
  1/pi integral_(-pi)^pi f^2(x) dd(x) = sum_(n=1)^oo b_n^2
$

$
  2/pi [integral_0^(pi/2) x^2 dd(x) + integral_(pi/2)^pi (pi-x)^2 dd(x)]
  = 2/pi [pi^3/24 + pi^3/24] = pi^2/12
$

$
  sum_(k=0)^oo 16/(pi^2(2k+1)^4) = pi^2/12
$

so
$
  sum_(k=0)^oo 1/(2k+1)^4 = pi^4/96
$

== b)

even extension of $f(x) = x$ on $[0,1]$ is $f(x) = |x|$ with period 2.

$
  a_0 & = 2 integral_0^1 x dd(x) = 1 \
  a_n & = 4 integral_0^1 x cos(n pi x) dd(x) \
      & = 4[x/(n pi) sin(n pi x) + 1/(n^2 pi^2) cos(n pi x)]_0^1 \
      & = 4/(n^2 pi^2) ((-1)^n - 1)
$

$a_n = 0$ for even $n$

$a_n = -8/(n^2 pi^2)$ for odd $n$

the series is
$
  f(x) = 1/2 - 8/pi^2 sum_(k=0)^oo 1/(2k+1)^2 cos((2k+1) pi x)
$

evaluating at $x = 0$ gives $f(0) = 0$:
$
  0 = 1/2 - 8/pi^2 sum_(k=0)^oo 1/(2k+1)^2
$

therefore
$
  sum_(k=0)^oo 1/(2k+1)^2 = pi^2/16
$

== c)

odd extension of $f(x) = x$ on $[0,1]$ is $f(x) = x$ with period 2.

$
  b_n & = 4 integral_0^1 x sin(n pi x) dd(x) \
      & = 4[-x/(n pi) cos(n pi x) + 1/(n^2 pi^2) sin(n pi x)]_0^1 \
      & = -4/(n pi) (-1)^n = 4/(n pi) (-1)^(n+1)
$

the can be written as
$
  f(x) = 4/pi sum_(n=1)^oo (-1)^(n+1)/n sin(n pi x)
$

evaluating at $x = 1/2$:
$
  1/2 = 4/pi sum_(n=1)^oo (-1)^(n+1)/n sin(n pi/2)
$

since $sin(n pi/2) = 0$ for even $n$ and $sin((2k+1)pi/2) = (-1)^k$ for odd $n$:
$
  1/2 = 4/pi sum_(k=0)^oo (-1)^(k+1)/(2k+1) (-1)^k = 4/pi sum_(k=0)^oo (-1)^(2k+1)/(2k+1)
$

therefore
$
  sum_(k=0)^oo (-1)^k/(2k+1) = pi/4
$