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Derivation of Fourier Series 📂Fourier Analysis

Derivation of Fourier Series

Definition

The series for 2L2L-periodic function ff is defined as the Fourier series of ff as follows:

lim⁡N→∞SNf(t)=lim⁡N→∞[a02+∑n=1N(ancos⁡nπtL+bnsin⁡nπtL)]=a02+∑n=1∞(ancos⁡nπtL+bnsin⁡nπtL) \begin{align*} \lim \limits_{N \rightarrow \infty} S^{f}_{N}(t) &= \lim \limits_{N \to \infty}\left[ \dfrac{a_{0}}{2}+\sum \limits_{n=1}^{N} \left( a_{n} \cos \dfrac{n\pi t}{L} + b_{n}\sin\dfrac{n\pi t}{L} \right) \right] \\ &= \dfrac{a_{0}}{2}+\sum \limits_{n=1}^{\infty} \left( a_{n} \cos \dfrac{n\pi t}{L} + b_{n}\sin\dfrac{n\pi t}{L} \right) \end{align*}

Here, each coefficient a0,an,bna_{0}, a_{n}, b_{n} is called the Fourier coefficient, and its value is as follows:

a0=1L∫−LLf(t)dtan=1L∫−LLf(t)cos⁡nπtLdtbn=1L∫−LLf(t)sin⁡nπtLdt \begin{align*} \\ a_{0} &=\dfrac{1}{L}\int_{-L}^{L}f(t)dt \\ a_{n} &= \dfrac{1}{L}\int_{-L}^{L} f(t)\cos\dfrac{n\pi t}{L} dt \\ b_{n} &=\dfrac{1}{L}\int_{-L}^{L}f(t)\sin \dfrac{n\pi t}{L}dt \end{align*}

Description

The Fourier series represents any function as a series expansion of trigonometric functions, famously developed by the French mathematician Joseph Fourier for solving heat equations. The term “any function” is used because if there is a function defined on a certain interval (a,b)(a,b), it can be replicated (Ctrl+C, Ctrl+V) to produce a (b−a)(b-a)-periodic function.

The core principle is to express it as a linear combination of orthogonal trigonometric functions, analogous to decomposing (4,−1,7)(4,-1,7) as follows in three-dimensional vectors:

(4,−1,7)=a1e^1+a2e^1+a3e^1 (4,-1,7) = a_{1}\hat{\mathbf{e}}_{1} + a_{2}\hat{\mathbf{e}}_{1} + a_{3}\hat{\mathbf{e}}_{1}

Indeed, the Fourier series of ff not only has minimal error with ff but also converges pointwise under well-defined conditions to ff.

f(t)=a02+∑n=1∞(ancos⁡nπtLt+bnsin⁡nπtL) f(t) = \dfrac{a_{0}}{2}+\sum \limits_{n=1}^{\infty} \left( a_{n} \cos \dfrac{n\pi t}{L}t + b_{n}\sin\dfrac{n\pi t}{L} \right)

Derivation

Regression Analysis1

  • Part 1

    The goal is to express the function f(t)f(t) as a linear combination of 1,cos⁡πtL,cos⁡2πtL,⋯ ,sin⁡πtL,sin⁡2πtL,⋯1, \cos \dfrac{\pi t}{L}, \cos\dfrac{2\pi t}{L}, \cdots, \sin \dfrac{\pi t}{L}, \sin \dfrac{2\pi t}{L}, \cdots s. Thus, assuming SNf(t)=12α0+∑n=1N(αncos⁡nπtL+βnsin⁡nπtL)S^{f}_{N}(t)=\dfrac{1}{2}{\alpha_{0}}+\sum \limits_{n=1}^{N} \left( \alpha_{n} \cos \dfrac{n \pi t}{L}+\beta_{n}\sin\dfrac{n\pi t}{L} \right), f(t)f(t) can be represented as follows:

    f(t)=SNf(t)+eN(t) f(t)=S^{f}_{N}(t)+e_{N}(t)

    eN(t)e_{N}(t) is the difference between f(t)f(t) and the approximation SNf(t)S_{N}^{f} (t). The smallest difference SNf(t)S_{N}^{f}(t) leads to the closest series expansion to f(t)f(t). Let’s define eNe_{N} as the mean square error2.

    eN=12L∫−LL[eN(t)]2dt=12L∫−LL[f(t)−SNf(t)]2dt e_{N}=\dfrac{1}{2L}\int_{-L}^{L} [e_{N}(t) ]^{2}dt=\dfrac{1}{2L}\int_{-L}^{L} \left[ f(t)-S^{f}_{N} (t) \right]^{2} dt

  • Part 2

    eN=12L∫−LL[f(t)−SNf(t)]2dt=12L∫−LL[f(t)−12α0−∑n=1N(αncos⁡nπtL+βnsin⁡nπtL)]2dt \begin{align*} e_{N} &= \dfrac{1}{2L}\int_{-L}^{L} \left[ f(t)-S^{f}_{N}(t) \right]^{2} dt \\ &= \dfrac{1}{2L}\int_{-L}^{L} \left[ f(t)-\dfrac{1}{2}{\alpha_{0}}-\sum \limits_{n=1}^{N} \left( \alpha_{n} \cos \dfrac{n \pi t}{L}+\beta_{n}\sin\dfrac{n\pi t}{L} \right) \right]^{2} dt \end{align*}

    Let the coefficients that minimize the mean square error eNe_{N} be α0, αn, βn\alpha_{0},\ \alpha_{n},\ \beta_{n}, a0a_{0}, ana_{n}, respectively. The conditions that minimize eNe_{N} are called the normal equations.

    ∂eN∂α0=0,  ∂eN∂αn=0,  ∂eN∂βn=0(m=1, 2, ⋯ , N) \dfrac{\partial e_{N}}{\partial \alpha_{0}}=0,\ \ \dfrac{\partial e_{N}}{\partial \alpha_{n}}=0,\ \ \dfrac{\partial e_{N}}{\partial \beta_{n}}=0\quad (m=1,\ 2,\ \cdots,\ N)

    Then, a0a_{0}, ana_{n}, bnb_{n} can be calculated as follows:

    • Part 2.1 a0a_{0}

      ∂eN∂α0=12L∫−LL∂∂α0[f(t)−12α0−∑n=1N(αncos⁡nπtL+βnsin⁡nπtL)]2dt=2⋅−12⋅12L∫−LL[f(t)−12α0−∑n=1N(αNcos⁡nπtL+βnsin⁡nπtL)]dt=−12L∫−LLf(t)dt+12L∫−LL12α0dt+12L∫−LL∑n=1N(αncos⁡nπtL+βnsin⁡nπtL)dt=−12L∫−LLf(t)dt+12L∫−LL12α0dt=−12L∫−LLf(t)dt+12α0=0 \begin{align*} \dfrac{\partial e_{N}}{\partial \alpha_{0}} &= \dfrac{1}{2L}\int_{-L}^{L} \dfrac{\partial}{\partial \alpha_{0}} \left[ f(t)-\dfrac{1}{2} {\alpha_{0}}-\sum \limits_{n=1}^{N} \left( \alpha_{n} \cos \dfrac{n \pi t}{L}+\beta_{n}\sin\dfrac{n\pi t}{L} \right) \right]^{2} dt \\ &= 2\cdot \dfrac{-1}{2} \cdot \dfrac{1}{2L} \int_{-L}^{L} \left[ f(t)-\dfrac{1}{2}{\alpha_{0}}-\sum \limits_{n=1}^{N} \left( \alpha_ {N} \cos \dfrac{n \pi t}{L}+\beta_{n}\sin\dfrac {n\pi t}{L} \right) \right] dt \\ &= \dfrac{-1}{2L}\int_{-L}^{L}f(t) dt + \dfrac{1}{2L}\int_{-L}^{L}\dfrac{1}{2}\alpha_{0} dt +\dfrac{1}{2L}\int_{-L}^{L} \sum \limits_ {n=1}^{N}\left( \alpha_{n}\cos \dfrac{n\pi t}{L}+\beta_{n} \sin \dfrac{n \pi t}{L} \right) dt \\ &= \dfrac{-1}{2L}\int_{-L}^{L}f(t) dt + \dfrac{1}{2L}\int_{-L}^{L}\dfrac{1}{2}\alpha_{0} dt \\ &= \dfrac{-1}{2L}\int_{-L}^{L}f(t) dt +\dfrac{1}{2}\alpha_{0} \\ &= 0 \end{align*}

      The fourth equality holds because the integral of a trigonometric function over one period is 00.

      a0=1L∫−LLf(t)dt a_{0} = \dfrac{1}{L} \int_{-L}^{L}f(t)dt

    • Part 2.2 ana_{n}

      Choose any m∈{1,2,…,N}m \in \left\{ 1,2,\dots,N \right\}.

      ∂eN∂αm=12L∫−LL∂∂αm[f(t)−12α0−∑n=1N(αncos⁡nπtL+βnsin⁡nπtL)]2dt=2⋅12L∫−LL(−cos⁡mπtL)[f(t)−12α0−∑n=1N(αncos⁡nπtL+βnsin⁡nπtL)]dt=−1L∫−LLf(t)cos⁡mπtLdt+1L∫−LL12α0cos⁡mπtLdt+1L∫−LL∑n=1N(αncos⁡nπtL+βnsin⁡nπtL)cos⁡mπtLdt=−1L∫−LLf(t)cos⁡mπtLdt+1Lαm∫−LLcos⁡mπtLcos⁡mπtLdt=−1L∫−LLf(t)cos⁡mπtLdt+αm=0 \begin{align*} \dfrac{\partial e_{N}}{\partial \alpha_{m}} &= \dfrac{1}{2L}\int_{-L}^{L} \dfrac{\partial}{\partial \alpha_{m}} \left[ f(t)-\dfrac{1}{2} {\alpha_{0}}-\sum \limits_{n=1}^{N} \left( \alpha_{n} \cos \dfrac{n \pi t}{L}+\beta_{n}\sin\dfrac{n\pi t}{L} \right) \right]^{2} dt \\ &= 2\cdot \dfrac{1}{2L} \int_{-L}^{L} \left( - \cos \dfrac{m\pi t}{L} \right)\left[ f(t)-\dfrac{1}{2}{\alpha_{0}}-\sum \limits_{n=1}^ {N} \left( \alpha_{n} \cos \dfrac{n \pi t}{L} +\beta_{n}\sin\dfrac{n\pi t}{L} \right) \right]dt \\ &= -\dfrac{1}{L} \int_{-L}^{L} f(t)\cos\dfrac{m\pi t}{L} dt +\dfrac{1}{L}\int_{-L}^{L} \dfrac{1}{2}\alpha_{0}\cos\dfrac{m\pi t}{L} dt \\ &\quad + \dfrac{1}{L} \int_{-L}^{L} \sum \limits_{n=1}^{N} \left( \alpha_{n} \cos \dfrac{n\pi t}{L} + \beta_{n}\sin\dfrac{n\pi t}{L} \right) \cos\dfrac{m\pi t}{L}dt \\ &= -\dfrac{1}{L}\int_{-L}^{L} f(t)\cos\dfrac{m\pi t}{L} dt + \dfrac{1}{L}\alpha_{m} \int_{-L}^{L}\cos\dfrac{m\pi t}{L}\cos\dfrac{m\pi t} {L} dt\\ &= -\dfrac{1}{L}\int_{-L}^{L} f(t)\cos\dfrac{m\pi t}{L} dt + \alpha_{m} \\ &= 0 \end{align*}

      The fourth and fifth equalities hold due to the orthogonality of trigonometric functions.

      an=1L∫−LLf(t)cos⁡nπtLdt(n=1,2,⋯ ,N) a_{n}= \dfrac{1}{L}\int_{-L}^{L} f(t)\cos\dfrac{n\pi t}{L} dt \quad (n=1, 2, \cdots, N)

    • Part 2.3 bnb_{n}

      Choose any m∈{1,2,…,N}m \in \left\{ 1,2,\dots,N \right\}.

      ∂eN∂βm=12L∫−LL∂∂βm[f(t)−12α0−∑n=1N(αncos⁡nπtL+βnsin⁡nπtL)]2dt=2⋅12L∫−LL(−sin⁡mπtL)[f(t)−12α0−∑n=1N(αncos⁡nπtL+βnsin⁡nπtL)]dt=−1L∫−LLf(t)sin⁡mπtLdt+1L∫−LL12α0sin⁡mπtLdt+1L∫−LL∑n=1N(αncos⁡nπtL+βnsin⁡nπtL)sin⁡mπtLdt=−1L∫−LLf(t)sin⁡mπtLdt+1Lβm∫−LLsin⁡mπtLsin⁡mπtLdt=−1L∫−LLf(t)sin⁡mπtLdt+βm=0 \begin{align*} \dfrac{\partial e_{N}}{\partial \beta_{m}} &= \dfrac{1}{2L}\int_{-L}^{L} \dfrac{\partial}{\partial \beta_{m}} \left[ f(t)-\dfrac{1}{2} {\alpha_{0}}-\sum \limits_{n=1}^{N} \left( \alpha_{n} \cos \dfrac{n \pi t}{L}+\beta_{n}\sin\dfrac{n\pi t}{L} \right) \right]^{2} dt \\ &= 2\cdot \dfrac{1}{2L} \int_{-L}^{L} \left( - \sin \dfrac{m\pi t}{L} \right)\left[ f(t)-\dfrac{1}{2}{\alpha_{0}}-\sum \limits_{n=1}^ {N} \left( \alpha_{n} \cos \dfrac{n \pi t}{L} +\beta_{n}\sin\dfrac{n\pi t}{L} \right) \right]dt \\ &= -\dfrac{1}{L} \int_{-L}^{L} f(t)\sin\dfrac{m\pi t}{L} dt +\dfrac{1}{L}\int_{-L}^{L} \dfrac{1}{2}\alpha_{0}\sin\dfrac{m\pi t}{L} dt \\ &\quad +\dfrac{1}{L} \int_{-L}^{L} \sum \limits_{n=1}^{N} \left( \alpha_{n} \cos \dfrac{n\pi t}{L} + \beta_{n}\sin\dfrac{n\pi t}{L} \right) \sin\dfrac{m\pi t}{L}dt \\ &= -\dfrac{1}{L}\int_{-L}^{L} f(t)\sin\dfrac{m\pi t}{L} dt + \dfrac{1}{L}\beta_{m} \int_{-L}^{L}\sin\dfrac{m\pi t}{L}\sin\dfrac{m\pi t} {L} dt \\ &= -\dfrac{1}{L}\int_{-L}^{L} f(t)\sin\dfrac{m\pi t}{L} dt + \beta_{m} \\ &=0 \end{align*}

      The fourth and fifth equalities hold due to the orthogonality of trigonometric functions.

      bn=1L∫−LLf(t)sin⁡nπtLdt(n=1,2,⋯ ,N) b_{n}=\dfrac{1}{L}\int_{-L}^{L}f(t)\sin\dfrac{n\pi t}{L}dt \quad (n=1, 2, \cdots, N)

  • Part 3 Using the obtained a0a_{0}, ana_{n}, bnb_{n} to express f(t)f(t) results in the same.

    f(t)=SNf(t)+eN(t)where SNf(t)=a02+∑n=1N(ancos⁡nπtL+bnsin⁡nπtL)a0=1L∫−LLf(t)dtan=1L∫−LLf(t)cos⁡nπtLdtbn=1L∫−LLf(t)sin⁡nπtLdt \begin{align*} f(t) &= S^{f}_{N}(t)+e_{N}(t) \\[1em] \text{where } S^{f}_{N}(t) &= \dfrac{a_{0}}{2}+\sum \limits_{n=1}^{N} \left( a_{n} \cos \dfrac{n\pi t}{L} + b_{n}\sin\dfrac{n\pi t} {L} \right) \\ a_{0} &= \dfrac{1}{L}\int_{-L}^{L}f(t)dt \\ a_{n} &= \dfrac{1}{L}\int_{-L}^{L} f(t)\cos\dfrac{n\pi t}{L} dt \\ b_{n} &= \dfrac{1}{L}\int_{-L}^{L}f(t)\sin\dfrac{n\pi t}{L}dt \end{align*}

    Taking the limit for NN yields

    lim⁡N→∞SNf(t)=a02+∑n=1∞(ancos⁡nπtL+bnsin⁡nπtL) \lim \limits_{N \rightarrow \infty} S_{N}^{f} (t)=\dfrac{a_{0}}{2}+\sum \limits_{n=1}^{\infty} \left( a_{n} \cos \dfrac{n\pi t}{L} + b_{n} \sin\dfrac{n\pi t}{L} \right)

    The above series is called the Fourier series of ff, and a0a_{0}, ana_{n}, bnb_{n} are called the Fourier coefficients of ff.

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  1. Byung Sun Choi, Introduction to Fourier Analysis (2002), pp. 51-53 ↩︎

  2. RSS is the mean square error. ↩︎