PX153 - J3 - proofs and derivations

Q : given a function, $f (x)$ , how can $a_{0}$ , $a_{n}$ , and $b_{n}$ be calculated?
- orthogonality relations:
  (1) $\int_{- π}^{π} \sin n x \cos m x d x = 0$ where, $n, m \in Z$
  (2)

\int_{- π}^{π} \sin n x \sin m x d x = {\begin{cases} 0 & n \neq m \\ π & n = m \end{cases}

(3)

\int_{- π}^{π} \cos n x \cos m x d x = {\begin{cases} 0 & n \neq m \\ π & n = m \end{cases}

- trigonometric identities:
$(a) : \cos α \cos β = \frac{1}{2} (\cos (α + β) + \cos (α - β))$
$(b) : \sin α \sin β = \frac{1}{2} (\cos (α - β) - \cos (α + β))$
$(c) : \sin α \cos β = \frac{1}{2} (\sin (α + β) + \sin (α - β))$

proofs of relations

(1)

I_{1} = \int_{- π}^{π} \sin n x \cos m x d x

using relation $(c) :$

\begin{aligned} I_{1} & = \frac{1}{2} \int_{- π}^{π} (\sin [(n + m) x] + \sin [(n - m) x]) d x \\ = \frac{1}{2} {[\frac{- \cos [(n + m) x]}{n + m} - \frac{\cos [(n - m) x]}{n - m}]}_{- π}^{π} \\ = 0 \end{aligned}

I_{2} = \int_{- π}^{π} \sin n x \sin m x d x

using relation $(b) :$

I_{2} = \frac{1}{2} \int_{- π}^{π} (\cos [(n - m) x] - \cos [(n + m) x]) d x

when $n \neq m :$

\begin{aligned} I_{2} & = \frac{1}{2} {[\frac{\sin [(n - m) x]}{n - m} - \frac{\sin [(n + m) x]}{n + m}]}_{- π}^{π} \\ = 0 \end{aligned}

when $n = m :$

\begin{aligned} I_{2} & = \frac{1}{2} \int_{- π}^{π} (1 - \cos (2 n x)) d x \\ = \frac{1}{2} {[x - \frac{\sin (2 n x)}{2 n}]}_{- π}^{π} \\ = π \end{aligned}

(3)

I_{3} = \int_{- π}^{π} \cos n x \cos m x d x

using relation $(a) :$

\begin{aligned} I_{3} & = \frac{1}{2} \int_{- π}^{π} (\cos [(n + m) x] + \cos [(n - m) x]) d x \\ = \frac{1}{2} {[\frac{\sin [(n + m) x]}{n + m} + \frac{\sin [(n - m) x]}{n - m}]}_{- π}^{π} \end{aligned}

when $n \neq m :$

I_{3} = 0

when $n = m :$

I_{3} = π

finding constants

f (x) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} [a_{n} \cos (n x) + b_{n} \sin (n x)]

integrating both sides:

\int_{- π}^{π} f (x) d x = {[\frac{a_{0} x}{2} + \sum_{n = 1}^{\infty} [\frac{a_{n} \sin (n x)}{n} - \frac{b_{n} \cos (n x)}{n}]]}_{- π}^{π} = a_{0} π

\frac{a_{0}}{2} = \frac{1}{2 π} \int_{- π}^{π} f (x) d x

now, multiplying by $\cos (m x)$ and integrating over $x$ :

\begin{aligned} \int_{- π}^{π} f (x) \cos (m x) d x & = {[\frac{a_{0} x}{2} \frac{\sin (m x)}{m}]}_{- π}^{π} + \sum_{n = 1}^{\infty} (a_{n} \int_{- π}^{π} \cos (n x) \cos (m x) d x + b_{n} \int_{- π}^{π} \sin (n x) \cos (m x) d x) \\ = a_{m} π \end{aligned}

- because:

{[\frac{a_{0} x}{2} \frac{\sin (m x)}{m}]}_{- π}^{π} = 0

- from relation $(3) :$

\int_{- π}^{π} \cos (n x) \cos (m x) d x = {\begin{cases} 0 & n \neq m \\ π & n = m \end{cases}

- from relation $(1) :$

\int_{- π}^{π} \sin (n x) \cos (m x) d x = 0

∴ a_{n} = \frac{1}{π} \int_{- π}^{π} f (x) \cos (m x) d x

finally, multiplying by $\sin (m x)$ and integrating over $x$ :

\begin{aligned} \int_{- π}^{π} f (x) \sin (m x) d x & = {[\frac{a_{0} x}{2} \frac{\cos (m x)}{m}]}_{- π}^{π} + \sum_{n = 1}^{\infty} (a_{n} \int_{- π}^{π} \cos (n x) \sin (m x) d x \\ + b_{n} \int_{- π}^{π} \sin (n x) \sin (m x) d x) \\ = a_{m} π \end{aligned}

- because:
$$\left[\frac{a_{0}x}{2} \frac{\cos(mx)}{m}\right]{-\pi}^{\pi}=0$$
- from relation $(1) :$
$$\int{-\pi}^{\pi} \cos(nx) \sin(mx) ,dx =0$$
- from relation $(2) :$
$$\int_{-\pi}^{\pi} \sin(nx) \sin(mx),dx = \pi$$
for $m = n$

∴ b_{n} = \frac{1}{π} \int_{- π}^{π} f (x) \sin (n x) d x

summary

\begin{aligned} \frac{a_{0}}{2} & = \frac{1}{2 π} \int_{- π}^{π} f (x) d x \\ a_{n} & = \frac{1}{π} \int_{- π}^{π} f (x) \cos (n x) d x \\ b_{n} & = \frac{1}{π} \int_{- π}^{π} f (x) \sin (n x) d x \end{aligned}

eg: $f (x) = | x |$ in $[- π, π]$

f (x) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} (a_{n} \cos (\frac{n π x}{L}) + b_{n} \sin (\frac{n π x}{L}))

\begin{aligned} \frac{a_{0}}{2} & = \frac{1}{2 π} \int_{- π}^{π} | x | d x \\ = \frac{1}{2 π} (\int_{0}^{π} x d x + \int_{- π}^{0} - x d x) \\ = \frac{1}{2 π} (2 \int_{0}^{π} x d x) \\ = \frac{π}{2} \end{aligned}

\begin{aligned} a_{n} & = \frac{1}{π} [\int_{- π}^{0} - x \cos (n x) d x + \int_{0}^{π} x \cos (n x) d x] \\ l e t x^{'} = - x ⟹ d x^{'} = - d x \\ = \frac{1}{π} [\int_{π}^{0} x^{'} \cos (- n x) (- d x^{'}) + \int_{0}^{π} x \cos (n x) d x] \\ = \frac{1}{π} [\int_{0}^{π} x^{'} \cos (- n x) d x^{'} + \int_{0}^{π} x \cos (n x) d x] \\ = \frac{1}{π} [\int_{0}^{π} x \cos (- n x) d x + \int_{0}^{π} x \cos (n x) d x] \\ = \frac{2}{π} \int_{0}^{π} x \cos (- n x) d x \\ = \frac{2}{π} {[\frac{x \sin n x}{n}]}_{0}^{π} - \int \frac{2}{π} \frac{\sin n x}{n} d x \\ = - \frac{2}{π} {[\frac{- \cos n x}{n^{2}}]}_{0}^{π} \\ = \frac{2}{n^{2} π} ((- 1)^{n} - 1) \\ ∴ a_{n} & = {\begin{cases} - \frac{4}{n^{2} π} & n \in o d d \\ 0 & n \in e v e n \end{cases} \end{aligned}

\begin{aligned} b_{n} & = \frac{1}{π} [\int_{- π}^{0} - x \sin (n x) d x + \int_{0}^{π} x \sin (n x) d x] \\ = \frac{1}{π} [\int_{π}^{0} x^{'} \sin (n x^{'}) d x^{'} + \int_{0}^{π} x \sin (n x) d x] \\ = \frac{1}{π} [\int_{π}^{0} x^{'} \sin (- n x^{'}) (- d x^{'}) + \int_{0}^{π} x \sin (n x) d x] \\ = \frac{1}{π} [- \int_{0}^{π} x^{'} \sin (n x^{'}) d x^{'} + \int_{0}^{π} x \sin (n x) d x] \\ = \frac{1}{π} [- \int_{0}^{π} x \sin (n x) d x + \int_{0}^{π} x \sin (n x) d x] \\ = 0 \end{aligned}

∴ f (x) = | x | = \frac{π}{2} - \frac{4}{π} (\cos x + \frac{\cos 3 x}{9} + \frac{\cos 5 x}{25} + . . .)