Fourier Series in Complex Form

[JamesEllison]

Hi,
Just wanted to verify my process to complete the Fourier series in complex form.
With the Function
f(t) = 8, 0≤t≤2
-8, 2≤t≤4

f(t) = $\sum$ Cn e^-int
n=1 to ∞

Where

Cn = $\frac{1}{T}$ ∫ f(t) e^{-int$\frac{2∏}{T}$}

Cn = $\frac{1}{T}$ 0∫T f(t) e^-int2∏/T dt

Cn = $\frac{1}{4}$ 0∫4 f(t) e^-int2∏/T dt ]

Cn = $\frac{1}{4}$ 0∫2 8e^-int2∏/T dt + 2∫4 -8 e^-int2∏/T dt + -∞∫0 0 e^-in2∏/t dt + 4∫∞ 0 e^-in2∏/t dt

Cn = $\frac{8}{4}$0∫2 e^-int∏/2 dt + 2∫4 e^-int∏/2 dt

Cn = 2 ( [itex]\frac{2e^{$\frac{-int∏}{2}$}}{-in∏}[/itex] 0 ->2 + [itex]\frac{2e^-int∏/2}{-in∏}[/itex] ) 2 -> 4

Cn = 2 ( [itex]\frac{2e^-in∏}{-in∏}[/itex] + [itex]\frac{2e⁰}{in∏}[/itex] ] + [ [itex]\frac{2e^-in2∏}{in∏}[/itex] - [itex]\frac{2e^-in∏/}{in∏}[/itex] ] }


Cn = ( $\frac{4}{in∏}$ ) { e^-in∏ + 1 + e^-in2∏ - e^-in∏ }

Cn = $\frac{4}{in∏}$ ( 2(e^-in∏ + e^-in2∏ + 1 ))

Use e^-iθ = Cosθ - iSinθ and let θ = n∏ and θ = n2∏

Cn = $\frac{4}{in∏}$ ( 2 (Cosθ - iSinθ) + (Cosθ - iSinθ) + 1 )

Cn = $\frac{4}{in∏}$ ( 2Cos(n∏) - 2iSin(n∏) + Cos (2∏n) - iSin(2∏n) )

Cn = $\frac{4}{in∏}$ ( 2 Cos (n∏) + 1)



Something along those lines??

PS sorry about my terrible notation within the question, i tried using as many symbols with the code as I could. Mainly for Tiny Tim :D

 

[gtoguy97]

Yes, your process looks correct. However, you made a minor mistake in the last step. The answer should be Cn = \frac{4}{in∏} ( 2 Cos (n∏) + 1 - Cos(2nπ) ).