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fourier series short intro

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Fourier series short intro: Proofs of some integral identities.

int sin

∫−ππsin⁡mx dx=[−1mcos⁡mx]−ππ=−1m[cos⁡mπ−cos⁡m(−π)]=−1m(cos⁡mπ−cos⁡mπ)=0.(21)\tag{21} \begin{array}{rcl} \displaystyle \int_{-\pi}^\pi \sin mx \ dx & = & \displaystyle \left[ -\frac{1}{m} \cos mx \right]_{-\pi}^\pi \newline \newline & = & \displaystyle -\frac{1}{m} \left[ \cos m \pi - \cos m(-\pi) \right] \newline \newline & = & \displaystyle -\frac{1}{m} ( \cos m \pi - \cos m \pi ) \newline \newline & = & 0. \end{array}

int cos

∫−ππcos⁡mx dx=[−1msin⁡mx]−ππ=−1m[sin⁡mπ−sin⁡m(−π)]=−1m(sin⁡mπ+sin⁡mπ)=−1m(0+0)  =  0.(22)\tag{22} \begin{array}{rcl} \displaystyle \int_{-\pi}^\pi \cos mx \ dx & = & \displaystyle \left[ -\frac{1}{m} \sin mx \right]_{-\pi}^\pi \newline \newline & = & \displaystyle -\frac{1}{m} \left[ \sin m \pi - \sin m(-\pi) \right] \newline \newline & = & \displaystyle -\frac{1}{m} ( \sin m \pi + \sin m \pi ) \newline \newline & = & \displaystyle -\frac{1}{m} ( 0 + 0 ) \ \ = \ \ 0. \newline \newline \end{array}

int sin · sin, a≠ba \ne b

sin⁡axsin⁡bx=12cos⁡(ax−bx)−12cos⁡(ax+bx)=12cos⁡(a−b)x−12cos⁡(a+b)x(23a)\tag{23a} \begin{array}{rcl} \sin ax \sin bx & = & \frac12 \cos (ax - bx) - \frac12 \cos (ax + bx) \newline \newline & = & \frac12 \cos (a - b)x - \frac12 \cos (a + b)x \end{array}

∫−ππsin⁡axsin⁡bx dx=12∫−ππcos⁡(a−b)x dx−12∫−ππcos⁡(a+b)x dx=12∫−ππcos⁡nx dx−12∫−ππcos⁡mx dx=0−0  =  0.(23b)\tag{23b} \begin{array}{rcl} \displaystyle \int_{-\pi}^\pi \sin ax \sin bx \ dx & = & \displaystyle \tfrac12 \int_{-\pi}^\pi \cos (a - b)x \ dx \newline \newline & & \displaystyle - \tfrac12 \int_{-\pi}^\pi \cos (a + b)x \ dx \newline \newline & = & \displaystyle \tfrac12 \int_{-\pi}^\pi \cos nx \ dx - \tfrac12 \int_{-\pi}^\pi \cos mx \ dx \newline \newline & = & 0 - 0 \ \ = \ \ 0. \end{array}

int cos · cos; sin, a≠ba \ne b

cos⁡axcos⁡bx=12cos⁡(ax+bx)+12cos⁡(ax−bx)=12cos⁡(a+b)x+12cos⁡(a−b)x(24a)\tag{24a} \begin{array}{rcl} \cos ax \cos bx & = & \frac12 \cos (ax + bx) + \frac12 \cos (ax - bx) \newline \newline & = & \frac12 \cos (a + b)x + \frac12 \cos (a - b)x \end{array}

∫−ππcos⁡axcos⁡bx dx=12∫−ππcos⁡(a+b)x dx+12∫−ππcos⁡(a−b)x dx=12∫−ππcos⁡mx dx+12∫−ππcos⁡nx dx=0+0  =  0.(24b)\tag{24b} \begin{array}{rcl} \displaystyle \int_{-\pi}^\pi \cos ax \cos bx \ dx & = & \displaystyle \tfrac12 \int_{-\pi}^\pi \cos (a + b)x \ dx \newline \newline & & \displaystyle + \tfrac12 \int_{-\pi}^\pi \cos (a - b)x \ dx \newline \newline & = & \displaystyle \tfrac12 \int_{-\pi}^\pi \cos mx \ dx + \tfrac12 \int_{-\pi}^\pi \cos nx \ dx \newline \newline & = & 0 + 0 \ \ = \ \ 0. \end{array}

int sin · cos; sin, a≠ba \ne b

sin⁡axcos⁡bx=12sin⁡(ax+bx)+12sin⁡(ax−bx)=12sin⁡(a+b)x+12sin⁡(a−b)x(25a)\tag{25a} \begin{array}{rcl} \sin ax \cos bx & = & \frac12 \sin (ax + bx) + \frac12 \sin (ax - bx) \newline \newline & = & \frac12 \sin (a + b)x + \frac12 \sin (a - b)x \end{array}

∫−ππsin⁡axcos⁡bx dx=12∫−ππsin⁡(a+b)x dx+12∫−ππsin⁡(a−b)x dx=12∫−ππsin⁡mx dx+12∫−ππsin⁡nx dx=0+0  =  0.(25b)\tag{25b} \begin{array}{rcl} \displaystyle \int_{-\pi}^\pi \sin ax \cos bx \ dx & = & \displaystyle \tfrac12 \int_{-\pi}^\pi \sin (a + b)x \ dx \newline \newline & & \displaystyle + \tfrac12 \int_{-\pi}^\pi \sin (a - b)x \ dx \newline \newline & = & \displaystyle \tfrac12 \int_{-\pi}^\pi \sin mx \ dx + \tfrac12 \int_{-\pi}^\pi \sin nx \ dx \newline \newline & = & 0 + 0 \ \ = \ \ 0. \end{array}

int sin · sin, a=ba = b

sin⁡axsin⁡ax=12cos⁡(ax−ax)−12cos⁡(ax+ax)=12−12cos⁡2ax(26a)\tag{26a} \begin{array}{rcl} \sin ax \sin ax & = & \frac12 \cos (ax - ax) - \frac12 \cos (ax + ax) \newline \newline & = & \frac12 - \frac12 \cos 2ax \end{array}

∫−ππsin⁡axsin⁡ax dx=12∫−ππdx−12∫−ππcos⁡2ax dx=12∫−ππdx−12∫−ππcos⁡mx dx=π−0  =  π.(26b)\tag{26b} \begin{array}{rcl} \displaystyle \int_{-\pi}^\pi \sin ax \sin ax \ dx & = & \displaystyle \tfrac12 \int_{-\pi}^\pi dx - \tfrac12 \int_{-\pi}^\pi \cos 2ax \ dx \newline \newline & = & \displaystyle \tfrac12 \int_{-\pi}^\pi dx - \tfrac12 \int_{-\pi}^\pi \cos mx \ dx \newline \newline & = & \pi - 0 \ \ = \ \ \pi. \end{array}

int cos · cos; sin, a=ba = b

cos⁡axcos⁡ax=12cos⁡(ax+ax)+12cos⁡(ax−ax)=12cos⁡2ax+12(27a)\tag{27a} \begin{array}{rcl} \cos ax \cos ax & = & \frac12 \cos (ax + ax) + \frac12 \cos (ax - ax) \newline \newline & = & \frac12 \cos 2ax + \frac12 \end{array}

∫−ππcos⁡axcos⁡bx dx=12∫−ππcos⁡2ax dx+12∫−ππdx=12∫−ππcos⁡mx dx+12∫−ππdx=0+π  =  π.(27b)\tag{27b} \begin{array}{rcl} \displaystyle \int_{-\pi}^\pi \cos ax \cos bx \ dx & = & \displaystyle \tfrac12 \int_{-\pi}^\pi \cos 2ax \ dx + \tfrac12 \int_{-\pi}^\pi dx \newline \newline & = & \displaystyle \tfrac12 \int_{-\pi}^\pi \cos mx \ dx + \tfrac12 \int_{-\pi}^\pi dx \newline \newline & = & 0 + \pi \ \ = \ \ \pi. \end{array}

int sin · cos; sin, a=ba = b

sin⁡axcos⁡ax=12sin⁡(ax+ax)+12sin⁡(ax−ax)=12sin⁡2ax+0(28a)\tag{28a} \begin{array}{rcl} \sin ax \cos ax & = & \frac12 \sin (ax + ax) + \frac12 \sin (ax - ax) \newline \newline & = & \frac12 \sin 2ax + 0 \end{array}

∫−ππsin⁡axcos⁡bx dx=12∫−ππsin⁡2ax dx=12∫−ππsin⁡mx dx=0.(28b)\tag{28b} \begin{array}{rcl} \displaystyle \int_{-\pi}^\pi \sin ax \cos bx \ dx & = & \displaystyle \tfrac12 \int_{-\pi}^\pi \sin 2ax \ dx \newline \newline & = & \displaystyle \tfrac12 \int_{-\pi}^\pi \sin mx \ dx \newline \newline & = & 0. \end{array}

int sin · sin, cos · cos, sin · cos

∫−ππsin⁡axsin⁡bx dx={0,a≠bπ,a=b.(29)\tag{29} \int_{-\pi}^\pi \sin ax \sin bx \ dx = \left\{ \begin{array}{rcl} 0, & a \ne b \newline \pi, & a = b. \end{array} \right.

 

∫−ππcos⁡axcos⁡bx dx={0,a≠bπ,a=b.(30)\tag{30} \int_{-\pi}^\pi \cos ax \cos bx \ dx = \left\{ \begin{array}{rcl} 0, & a \ne b \newline \pi, & a = b. \end{array} \right.

 

∫−ππcos⁡axsin⁡bx dx=0.(31)\tag{31} \int_{-\pi}^\pi \cos ax \sin bx \ dx = 0.

interval changes

∫−ππsin⁡mx dx=∫−12λ12λsin⁡(m2πλx) dx−[1mcos⁡mx]−ππ=−[λm2πcos⁡(m2πλx)]−12λ12λ1m[cos⁡mπ−cos⁡m(−π)]=λm2π[cos⁡(m2πλ)(12λ)−cos⁡(m2πλ)(−12λ)]=λm2π[cos⁡mπ−cos⁡m(−π)]1m(cos⁡mπ−cos⁡mπ)=λm2π(cos⁡mπ−cos⁡mπ)0=0.(32)\tag{32} \begin{array}{rcl} \displaystyle \int_{-\pi}^\pi \sin mx \ dx & = & \displaystyle \int_{-\tfrac12 \lambda}^{\tfrac12 \lambda} \sin \left( m\frac{2\pi}{\lambda} x \right) \ dx \newline \newline \displaystyle -\left[ \frac{1}{m} \cos mx \right]^\pi_{-\pi} & = & \displaystyle -\left[ \frac{\lambda}{m2\pi} \cos \left( m \frac{2\pi}{\lambda} x \right) \right]_{-\tfrac12 \lambda}^{\tfrac12 \lambda} \newline \newline \displaystyle \frac{1}{m} [ \cos m \pi - \cos m(-\pi) ] & = & \displaystyle \frac{\lambda}{m2\pi} \left[ \cos \left( m \frac{2\pi}{\lambda} \right) (\tfrac12 \lambda) - \cos \left( m \frac{2\pi}{\lambda} \right) (-\tfrac12 \lambda) \right] \newline \newline & = & \displaystyle \frac{\lambda}{m2\pi} [ \cos m\pi - \cos m(-\pi) ] \newline \newline \displaystyle \frac{1}{m} (\cos m\pi - \cos m\pi) & = & \displaystyle \frac{\lambda}{m2\pi} (\cos m\pi - \cos m\pi) \newline \newline 0 & = & 0. \end{array}

 

∫−ππsin⁡mt dt=∫−12T12Tsin⁡(m2πTt) dt−[1mcos⁡mt]−ππ=−[Tm2πcos⁡(m2πTt)]−12T12T1m[cos⁡mπ−cos⁡m(−π)]=Tm2π[cos⁡(m2πT)(12T)−cos⁡(m2πT)(−12T)]=λm2π[cos⁡mπ−cos⁡m(−π)]1m(cos⁡mπ−cos⁡mπ)=λm2π(cos⁡mπ−cos⁡mπ)0=0.(33)\tag{33} \begin{array}{rcl} \displaystyle \int_{-\pi}^\pi \sin mt \ dt & = & \displaystyle \int_{-\tfrac12 T}^{\tfrac12 T} \sin \left( m\frac{2\pi}{T} t \right) \ dt \newline \newline \displaystyle -\left[ \frac{1}{m} \cos mt \right]^\pi_{-\pi} & = & \displaystyle -\left[ \frac{T}{m2\pi} \cos \left( m \frac{2\pi}{T} t \right) \right]_{-\tfrac12 T}^{\tfrac12 T} \newline \newline \displaystyle \frac{1}{m} [ \cos m \pi - \cos m(-\pi) ] & = & \displaystyle \frac{T}{m2\pi} \left[ \cos \left( m \frac{2\pi}{T} \right) (\tfrac12 T) - \cos \left( m \frac{2\pi}{T} \right) (-\tfrac12 T) \right] \newline \newline & = & \displaystyle \frac{\lambda}{m2\pi} [ \cos m\pi - \cos m(-\pi) ] \newline \newline \displaystyle \frac{1}{m} (\cos m\pi - \cos m\pi) & = & \displaystyle \frac{\lambda}{m2\pi} (\cos m\pi - \cos m\pi) \newline \newline 0 & = & 0. \end{array}

period

∫−ππsin⁡mx dx=∫−12λ12λsin⁡(m2πλx) dx=∫0λsin⁡(m2πλx) dx=∫0λsin⁡(mkx) dx  =  ∫0λsin⁡(kmx) dx(34)\tag{34} \begin{array}{rcl} \displaystyle \int_{-\pi}^\pi \sin mx \ dx & = & \displaystyle \int_{-\tfrac12 \lambda}^{\tfrac12 \lambda} \sin \left( m\frac{2\pi}{\lambda} x \right) \ dx \newline \newline & = & \displaystyle \int_0^{\lambda} \sin \left( m\frac{2\pi}{\lambda} x \right) \ dx \newline \newline & = & \displaystyle \int_0^{\lambda} \sin ( m k x ) \ dx \ \ = \ \ \int_0^{\lambda} \sin ( k_m x ) \ dx \end{array}

 

∫−ππsin⁡mt dt=∫−12T12Tsin⁡(m2πTt) dt=∫0Tsin⁡(m2πTt) dt=∫0Tsin⁡(mωt) dt  =  ∫0Tsin⁡(ωmt) dt(35)\tag{35} \begin{array}{rcl} \displaystyle \int_{-\pi}^\pi \sin mt \ dt & = & \displaystyle \int_{-\tfrac12 T}^{\tfrac12 T} \sin \left( m\frac{2\pi}{T} t \right) \ dt \newline \newline & = & \displaystyle \int_0^T \sin \left( m\frac{2\pi}{T} t \right) \ dt \newline \newline & = & \displaystyle \int_0^T \sin ( m \omega t ) \ dt \ \ = \ \ \int_0^T \sin ( \omega_m t) \ dt \end{array}

wavenumber and angular requency

km=mk1,    m=1,2,3,..(36)\tag{36} k_m = m k_1, \ \ \ \ m = 1, 2, 3, .. λm=2πkm \lambda_m = \frac{2\pi}{k_m}

 

ωm=mω1,    m=1,2,3,..(37)\tag{37} \omega_m = m \omega_1, \ \ \ \ m = 1, 2, 3, .. Tm=2πωm T_m = \frac{2\pi}{\omega_m}

other forms

f(x)=∑m=1∞amcos⁡[m(2πL)x]+∑m=1∞bmsin⁡[m(2πL)x]+c0(38)\tag{38} f(x) = \sum_{m=1}^\infty a_m \cos \left[ m \left( \frac{2\pi}{L} \right) x \right] + \sum_{m=1}^\infty b_m \sin \left[ m \left( \frac{2\pi}{L} \right) x \right] + c_0

am=2L∫0Lf(x) cos⁡[m(2πL)x] dx(39)\tag{39} a_m = \frac{2}{L}\int_0^L f(x) \ \cos \left[ m \left( \frac{2\pi}{L} \right) x \right] \ dx

bm=2L∫0Lf(x) cos⁡[m(2πL)x] dx(40)\tag{40} b_m = \frac{2}{L}\int_0^L f(x) \ \cos \left[ m \left( \frac{2\pi}{L} \right) x \right] \ dx

c0=1L∫0Lf(x) dx.(41)\tag{41} c_0 = \frac{1}{L} \int_0^L f(x) \ dx.

f(t)=∑m=1∞amcos⁡[m(2πL)t]+∑m=1∞bmsin⁡[m(2πL)t]+c0(42)\tag{42} f(t) = \sum_{m=1}^\infty a_m \cos \left[ m \left( \frac{2\pi}{L} \right) t \right] + \sum_{m=1}^\infty b_m \sin \left[ m \left( \frac{2\pi}{L} \right) t \right] + c_0

am=2T∫0Tf(t) cos⁡[m(2πL)t] dt(43)\tag{43} a_m = \frac{2}{T}\int_0^T f(t) \ \cos \left[ m \left( \frac{2\pi}{L} \right) t \right] \ dt

bm=2T∫0Tf(t) cos⁡[m(2πL)t] dt(44)\tag{44} b_m = \frac{2}{T}\int_0^T f(t) \ \cos \left[ m \left( \frac{2\pi}{L} \right) t \right] \ dt

c0=1T∫0Tf(t) dt.(45)\tag{45} c_0 = \frac{1}{T} \int_0^T f(t) \ dt.

 

km=m2πL,    m=1,2,3,..(46)\tag{46} k_m = m \frac{2\pi}{L}, \ \ \ \ m = 1, 2, 3, ..

ωm=m2πT,    m=1,2,3,..(47)\tag{47} \omega_m = m \frac{2\pi}{T}, \ \ \ \ m = 1, 2, 3, ..

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