Find-the-fourier-series-of-f-x-x-from-0-lt-x-lt-pi-

Question Number 17625 by tawa tawa last updated on 08/Jul/17

Find the fourier series of : f (x) = x, from 0 < x < π

Commented by tawa tawa last updated on 08/Jul/17

please help with this .

Answered by alex041103 last updated on 09/Jul/17

\frac{π}{2} - \underset{n = 1}{\sum^{\infty}} \frac{s i n (2 n x)}{n} = f (x) = x w h e n x \in (0, π)

Commented by tawa tawa last updated on 09/Jul/17

i have a question sir ?

Is the period π or 2 π ?

how can i identify if it is π or 2 π ?

if the period is π \dots how can i reperesent a_{0,} a_{n}, and b_{n}

if it the period is 2 π \dots how can i reperesent a_{0,} a_{n}, and b_{n}

Commented by alex041103 last updated on 09/Jul/17

{I t}^{'} s e a s y t o p r o v e t h a t i f

f (x) = c + \underset{n = 1}{\sum^{\infty}} [a_{n} c o s (\frac{2 π n x}{T}) + b_{n} s i n (\frac{2 π n x}{T})]

t h e n

c = \frac{1}{T} \underset{0}{\int^{T}} f (x) d x

a_{n} = \frac{2}{T} \underset{0}{\int^{T}} f (x) c o s (\frac{2 π n x}{T}) d x

b_{b} = \frac{2}{T} \underset{0}{\int^{T}} f (x) s i n (\frac{2 π n x}{T}) d x

I n t h e c a s e o f f (x) = x w e w i l l s e t t h e p e r i o d

T = π

\Rightarrow c = \frac{π}{2}, a_{n} = 0, b_{n} = - \frac{1}{n}

\Rightarrow x \in (0, π), f (x) = \frac{π}{2} - \underset{n = 1}{\sum^{\infty}} \frac{s i n (2 n x)}{n}

Commented by tawa tawa last updated on 09/Jul/17

That means for period 2 π

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

a_{n} = \frac{1}{π} \int_{0}^{2 π} f (x) \cos (nx)

b_{n} = \frac{1}{π} \int_{0}^{2 π} f (x) \sin (nx)

Again,

That means for period π

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

a_{n} = \frac{2}{π} \int_{0}^{π} f (x) \cos (nx)

b_{n} = \frac{2}{π} \int_{0}^{π} f (x) \sin (nx)

? ? ? ? ? ? ? ? ? ? ? ? ? ?

Commented by alex041103 last updated on 09/Jul/17

N o .

F o r p e r i o d o f π w e h a v e :

b_{n} = \frac{2}{π} \int_{0}^{π} f (x) \sin (\frac{2 π}{π} nx) d x = \frac{2}{π} \underset{0}{\int^{π}} f (x) s i n (2 n x) d x

a_{n} = \frac{2}{π} \int_{0}^{π} f (x) \cos (\frac{2 π}{π} nx) d x = \frac{2}{π} \underset{0}{\int^{π}} f (x) c o s (2 n x) d x

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

A n d i n f a c t i f w e u s e T = 2 π (t h e p e r i o d i s 2 π)

w e a r e g o i n g t o g e t f o u r i e r s e r i e s f o r

f (x) = x f o r x \in [0, 2 π]

Commented by tawa tawa last updated on 09/Jul/17

God bless you sir .

Commented by alex041103 last updated on 10/Jul/17

Commented by alex041103 last updated on 10/Jul/17

Blue \to T = 2 π, i . e . for 0 < x < 2 π

x = π - 2 \underset{n = 1}{\sum^{\infty}} \frac{\sin (nx)}{n}

Red \to T = π, i . e . for 0 < x < π

x = \frac{π}{2} - \underset{n = 1}{\sum^{\infty}} \frac{\sin (2 nx)}{n}

Commented by tawa tawa last updated on 10/Jul/17

i really appreciate your effort sir . God bless you .