#import "@preview/physica:0.9.6": *

= problem 3

$
  a_0(f') = 0, quad a_n (f') = n b_n (f), quad b_n(f') = -n a_n(f)
$

if $f(-pi) != f(pi)$ then the formulas don't work anymore, since there will be
a discreet jump every time the period is restarted. to fix the formulas we need
to account for this. let $[f] = f(pi) - f(-pi)$
$
  cases(
    a_0 (f') & = 1/(2 pi) integral_(-pi)^pi f'(x) dd(x)
               = 1/(2 pi) [f(x)]_(-pi)^pi = [f]/(2 pi), ,
    a_n (f') & = 1/pi integral_(-pi)^pi f'(x) cos(n x) dd(x) \
             & = 1/pi ([f(x) cos(n x)]_(-pi)^pi
                 + n integral_(-pi)^pi f(x) sin(n x) dd(x)) \
             & = 1/pi ((-1)^n [f] + n pi b_n (f)) \
             & = n b_n (f) + 1/pi (-1)^n [f], ,
    b_n (f') & = - n a_n (f)
  )
$

recall the fourier series for a $p$-periodic function $f$
$
  f(x) approx f_N (x) = a_0 + sum_(n=1)^N
  (a_n cos((2 pi)/p x) + b_n sin((2 pi)/p x))
$
where
$
  cases(
    a_0 = 1/p integral_p f(x) dd(x),
    a_n = 2/p integral_p f(x) cos((2 pi)/p n x) dd(x),
    b_n = 2/p integral_p f(x) sin((2 pi)/p n x) dd(x)
  )
$

let $f(x) = sin^2(x) + 3 x^2 - 4 x + 5$ be periodically continued based on
$[-pi, pi]$.

notice that this series has a jump $[f] > 0$ so we need to use the modified
formulas we found above.

also remark that $sin^2(x) = (1 - cos(2x))/2$, which is already its fourier
series.

since fourier series are linear, we can compute the fourier series of each
term and we can use the provided properties to compute the coefficients using
the derivative of $f$.
$
  f'(x) = dots.c + 6 x - 4
$
we then need to compute the fourier series for $x^2$, $x$ and $1$ using
$
    "fourier"(1) & = cases(
                     a_0 (1) = 1/(2 pi) integral_(-pi)^pi 1 dd(x) = 1,
                     a_n (1) = 1/pi integral_(-pi)^pi cos(x) dd(x) = 0,
                     b_n (1) = 0
                   ) \
    "fourier"(x) & = cases(
                     a_0 (x) = 1/(2 pi) integral_(-pi)^pi x dd(x) = 0,
                     a_n (x) = 1/n (b_n (1)) = 0,
                     b_n (x) = 1/n (a_n (1) - 1/pi (-1)^n [x])
                     = 2/n (-1)^(n + 1)
                   ) \
  "fourier"(x^2) & = cases(
                     a_0 (x^2) = 1/(2 pi) integral_(-pi)^pi x^2 dd(x)
                     = 1/(6 pi) [x^3]_(-pi)^pi = 1/3 pi^2,
                     a_n (x^2) = -1/n b_n (2 x)
                     = -2/n b_n (x) = 4/n^2 (-1)^n,
                     b_n (x^2) = 1/n (a_n (2 x) - 1/pi (-1)^n [x^2]) = 0
                   )
$

combining these we get
$
  "fourier"(f(x)) & = "fourier"(sin^2(x)) \
                  & + "fourier"(3 x^2) - "fourier"(4 x) + "fourier"(5) \
                  & = 1/2 (1 - cos(2 x)) + 3 "fourier"(x^2) \
                  & - 4 "fourier"(x) + 5 "fourier"(1) \
                  & = 1/2 (1 - cos(2 x)) x + cases(
                      a_0 (f(x)) & = 3 a_0 (x^2) - 4 a_0 (x) + 5 a_0 (1) \
                                 & = pi^2 + 5,
                      a_n (f(x)) & = 12/n^2 (-1)^n,
                      b_n (f(x)) & = -8/n (-1)^(n+1) = 8/n (-1)^n
                    ) \
                  & = pi^2 + 11/2 - 1/2 cos(2 x) \
                  & + sum_(n=1)^oo (12/n cos(n x)
                      + 8 sin(n x))(-1)^n / n
$