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Proof of the Orthogonality of the Set of Trigonometric Functions 📂Fourier Analysis

Proof of the Orthogonality of the Set of Trigonometric Functions

Theorem

The set {1, cosπxL, cos2πxL,, sinπxL, sin2πxL, }\left\{ 1,\ \cos \dfrac{\pi x}{L},\ \cos \dfrac{2\pi x}{L}, \cdots ,\ \sin\dfrac{\pi x}{L},\ \sin\dfrac{2\pi x}{L},\ \cdots \right\} of functions 2L2L that are periodic functions is an orthogonal set in the interval [L, L)[-L,\ L). In other words, for m,n=1,2,3,m,n = 1, 2, 3, \dots, the following holds.

1LLLcosmπxLcosnπxLdx=δmn1LLLsinmπxLsinnπxLdx=δmnLLcosmπxLsinnπxLdx=01LLLcosnπxLdx=01LLLsinnπxLdx=0 \begin{align} \dfrac{1}{L} \int _{-L}^{L} \cos\dfrac{m\pi x}{L} \cos\dfrac{n\pi x}{L} dx &= \delta_{mn} \label{eq1} \\ \dfrac{1}{L} \int _{-L}^{L} \sin \dfrac{m\pi x}{L}\sin \dfrac{n\pi x}{L} dx &= \delta_{mn} \label{eq2} \\ \int _{-L}^{L} \cos \dfrac{m\pi x}{L} \sin \dfrac{n\pi x}{L} dx \quad &= 0 \label{eq3} \\ \dfrac{1}{L} \int_{-L}^{L} \cos \dfrac{n\pi x}{L} dx &= 0 \label{eq4} \\ \dfrac{1}{L} \int_{-L}^{L} \sin \dfrac{n\pi x}{L} dx &= 0 \label{eq5} \end{align}

Here, δ\delta is the Kronecker delta.

Corollary

According to (4),(5)(4), (5), the average of one period of cosine and sine is 00.

Explanation

Due to the Euler’s formula, the set of exponential functions also possesses orthogonality. This fact is significant in Fourier analysis as it enables the representation of periodic functions as series of periodic functions, namely the Fourier series.

Proof

(1)(1)

LLcosmπxLcos nπxLdx(1) \int_{-L}^{L} \cos \frac{m\pi x}{L} \cos \ \frac{n\pi x}{L} dx \tag{1}

  • case 1.1 mnm \ne n

LLcosmπxLcosnπxLdx(m,n=1,2,mn)=12LL[cos(m+n)πxL+cos(mn)πxL]dx=12[L(m+n)πsin(m+n)πxL]LL+12[L(mn)πsin(mn)πxL]LL=12[L(m+n)πsin((m+n)π)+L(m+n)πsin((m+n)π)]+12[L(mn)πsin((mn)π)+L(mn)πsin((mn)π)]=0 \begin{align*} & \int_{-L}^{L} \cos \dfrac{m\pi x}{L} \cos \dfrac{n\pi x}{L} dx \quad (m,n=1, 2,\dots\quad m\ne n) \\ &= \frac{1}{2} \int_{-L}^{L} \left[ \cos \frac{(m+n)\pi x}{L}+\cos \frac{(m-n)\pi x}{L} \right] dx \\ &= \frac{1}{2} \left[ \dfrac{L}{(m+n)\pi}\sin \dfrac{(m+n)\pi x}{L} \right]_{-L}^{L} + \frac{1}{2} \left[ \dfrac{L}{(m-n)\pi }\sin \dfrac{(m-n)\pi x}{L} \right]_{-L}^{L} \\ &= \frac{1}{2} \left[ \dfrac{L}{(m+n)\pi}\sin \big( (m+n)\pi \big) + \dfrac{L}{(m+n)\pi}\sin \big( (m+n)\pi \big) \right] \\ &\quad+ \frac{1}{2} \left[ \dfrac{L}{(m-n)\pi}\sin \big( (m-n)\pi \big) + \dfrac{L}{(m-n)\pi}\sin\big( (m-n)\pi \big) \right] \\ &= 0 \end{align*}

The first equality holds due to the product-to-sum identities of trigonometric functions. The last equality holds because m+n, mnm+n,\ m-n is an integer not equal to 00, so all terms are 00.

  • case 1.2 m=nm = n

    LL(cosmπxL)2dx=12LL(1+cos2mπxL)dx=12[x+L2mπsin2mπxL]LL=12(2L)=L    1LLL(cosmπxL)2dx=1 \begin{align*} \int _{-L}^{L} \left( \cos \dfrac{m\pi x}{L} \right)^2 dx &=\dfrac{1}{2} \int_{-L}^{L} \left( 1+ \cos \dfrac{2m\pi x}{L} \right) dx \\ &= \frac{1}{2}\left[ x+\frac{L}{2m\pi}\sin \frac{2m\pi x}{L} \right]_{-L}^{L} \\ &= \frac{1}{2}(2L) \\ &= L \end{align*} \\ \implies \dfrac{1}{L}\int _{-L}^{L} \left( \cos \dfrac{m\pi x}{L} \right)^2dx = 1

    The first equality holds due to the half-angle identities of trigonometric functions.

(2)(2)

LLsinmπxLsinnπxLdx(2) \int_{-L}^{L} \sin \dfrac{m\pi x}{L} \sin \dfrac{n\pi x}{L } dx \tag{2}

  • case 2.1 mnm \ne n

    LLsinmπxLsinnπxLdx (m,n=1,2,,mn)=12LL[cos(mn)πxLcos(m+n)πxL]dx=12[L(mn)πsin(mn)πxL]LL12[L(m+n)πsin(m+n)πxL]LL=12[L(mn)πsin((mn)π)+L(mn)πsin((m+n)π)]12[L(m+n)πsin((m+n)π)+L(m+n)πsin((m+n)π)]=0 \begin{align*} &\int_{-L}^{L} \sin \dfrac{m\pi x}{L} \sin \dfrac{n\pi x}{L} dx \ \quad (m,n= 1,2,\cdots,\quad m\ne n) \\ &= \frac{1}{2} \int_{-L}^{L} \left[ \cos \frac{(m-n)\pi x}{L} -\cos \frac{(m+n)\pi x}{L} \right] dx \\ &= \frac{1}{2} \left[ \dfrac{L}{(m-n)\pi}\sin \dfrac{(m-n)\pi x}{L} \right]_{-L}^{L} - \frac{1}{2} \left[ \dfrac{L}{(m+n)\pi}\sin \dfrac{(m+n)\pi x}{L} \right]_{-L}^{L} \\ &= \frac{1}{2} \left[ \dfrac{L}{(m-n)\pi}\sin \big( (m-n)\pi \big) + \dfrac{L}{(m-n)\pi}\sin \big( (m+n)\pi \big) \right] \\ &\quad- \frac{1}{2} \left[ \dfrac{L}{(m+n)\pi}\sin \big( (m+n)\pi \big) +\dfrac{L}{(m+n)\pi}\sin \big( (m+n)\pi \big) \right] \\ &= 0 \end{align*}

    The first equality holds due to the product-to-sum identities of trigonometric functions. The last equality holds for the same reason as in case 1.1, since all terms are 00.

  • case 2.2 m=nm = n

    LL(sinmπxL)2dx=12LL(1cos2mπxL)dx=12[xL2mπsin2mπxL]LL=12(2L)=L    1LLL(sinmπxL)2dx=1 \begin{align*} && \int _{-L}^{L} \left( \sin \dfrac{m\pi x}{L} \right)^2 dx &= \dfrac{1}{2} \int_{-L}^{L} \left( 1- \cos \dfrac{2m\pi x}{L} \right)dx \\ && &=\frac{1}{2}\left[ x-\frac{L}{2m\pi}\sin \frac{2m\pi x}{L} \right]_{-L}^{L} \\ && &= \dfrac{1}{2}(2L) \\ && &= L \end{align*} \\ \implies \dfrac{1}{L}\int _{-L}^{L} \left( \sin \dfrac{m\pi x}{L} \right) ^2 dx=1

    The first equality holds due to the half-angle identities of trigonometric functions.

(3)(3)

LLcosmπxLsinnπxLdx(3) \int _{-L}^{L} \cos \dfrac{m\pi x}{L} \sin \dfrac{n\pi x}{L} dx \tag{3}

  • case 3.1 mnm \ne n

    LLcosmπxLsinnπxLdx=12LL[sin(m+n)πxLsin(mn)πxL]dx=12[L(m+n)πcos(m+n)πxL]LL12[L(mn)πcos(mn)πxL]LL=12[L(m+n)πcos((m+n)π)+L(m+n)πcos((m+n)π)]12[L(mn)πcos((mn)π)+L(mn)πcos((mn)π)]=0 \begin{align*} & \int _{-L}^{L} \cos \dfrac{m\pi x}{L} \sin \dfrac{n\pi x}{L} dx \\ &= \dfrac{1}{2} \int_{-L}^{L} \left[ \sin \dfrac{ (m+n) \pi x}{L} - \sin \dfrac{ (m-n) \pi x}{L} \right] dx \\ &= \dfrac{1}{2} \left[ - \dfrac{L}{(m+n)\pi} \cos \dfrac{ (m+n)\pi x}{L} \right] _{-L}^{L} -\dfrac{1}{2}\left[ - \dfrac{L}{(m-n)\pi} \cos \dfrac{ (m-n)\pi x}{L} \right] _{-L}^{L} \\ &= \dfrac{1}{2} \left[ - \dfrac{L}{(m+n)\pi} \cos \big( (m+n)\pi \big) + \dfrac{L}{(m+n)\pi} \cos \big( (m+n)\pi \big) \right] \\ &\quad- \dfrac{1}{2}\left[ - \dfrac{L}{(m-n)\pi} \cos \big( (m-n)\pi\big) + \dfrac{L}{(m-n)\pi} \cos \big((m-n)\pi\big)\right] \\ &= 0 \end{align*}

    The first equality holds due to the product-to-sum identities of trigonometric functions.

  • case 3.2 m=nm = n

    LLcosmπxLsinnπxLdx=LLcosmπxLsinmπxLdx=12LL2cosmπxLsinmπxLdx=12LLsin2mπxLdx=12(L2mπcos2mπ+L2mπcos2mπ)=0 \begin{align*} \int _{-L}^{L} \cos \dfrac{m\pi x}{L} \sin \dfrac{n\pi x}{L} dx &= \int_{-L}^{L} \cos \dfrac{m\pi x}{L} \sin \dfrac{m\pi x}{L} dx \\ &= \dfrac{1}{2} \int _{-L}^{L} 2\cos \dfrac{m\pi x}{L} \sin \dfrac{m\pi x}{L} dx \\ &= \dfrac{1}{2} \int _{-L}^{L} \sin \dfrac{2m\pi x}{L} dx \\ &= \dfrac{1}{2} \left( -\dfrac{L}{2m\pi}\cos 2m\pi + \dfrac{L}{2m\pi} \cos 2m\pi\right) \\ &= 0 \end{align*}

(4),(5)(4), (5)

LLcosnπxLdx=[LnπsinnπxL]LL=Lnπsinnπ+Lnπsinnπ=0 \begin{align*} \int_{-L}^{L} \cos\dfrac{n \pi x}{L} dx &=\left[ \dfrac{L}{n\pi}\sin \dfrac{n \pi x}{L} \right]_{-L}^{L} \\ &=\dfrac{L}{n\pi}\sin n\pi + \dfrac{L}{n\pi}\sin n\pi \\ &=0 \end{align*}

The last equality holds because nn is an integer. The sine function also follows for the same reason

LLsinnπxLdx=[LnπcosnπxL]LL=Lnπcosnπ+Lnπcosnπ=0 \begin{align*} \int_{-L}^{L} \sin\dfrac{n \pi x}{L} dx &=\left[ -\dfrac{L}{n\pi}\cos \dfrac{n \pi x}{L} \right]_{-L}^{L} \\ &=-\dfrac{L}{n\pi}\cos n\pi + \dfrac{L}{n\pi}\cos n\pi \\ &=0 \end{align*}