184 lines
5.0 KiB
Markdown
184 lines
5.0 KiB
Markdown
#Math #Calculus
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# Starting the Proof Off
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The Taylor Series uses $x^n$ as building blocks for a function:
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[[Taylor Series Proof]]
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However, we can use $\sin (nx)$ and $\cos(nx)$ as well. This will be our starting point to derive the Fourier Series:
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$$
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f(x) = a_0\cos (0x) + b_0\sin(0x) + a_1\cos (x) + b_1\sin(x) + a_2\cos (2x) + b_2\sin(2x)... \\
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f(x) = a_0 + \sum _{n = 1}^\infty (a_n\sin(nx) + b_n\cos(nx))
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$$
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This will be the basic equation we will use.
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# Finding $a_0$
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Let’s integrate the equation on both sides, and bound by $[-\pi, \pi]$:
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$$
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\int _{-\pi}^\pi f(x) dx = \int _{-\pi}^\pi a_0 dx + \sum _{n = 1}^\infty \int _{-\pi}^\pi a_n\cos(nx) dx + \sum _{n = 1}^\infty \int _{-\pi}^\pi b_n\sin(nx) dx
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$$
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The first integral evaluates to $2\pi a_0$. Since the third integral is an odd function, it evaluates to $0$. The second integral can be expressed as:
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$$
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a_n \int _{-\pi}^\pi \cos(nx) dx \\
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= \frac {a_n} n (\sin(n\pi) - \sin(-n\pi)) \\
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= 0
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$$
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So now we have:
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$$
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2\pi a_0 = \int _{-\pi}^\pi f(x) dx \\
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a_0 = \frac 1 {2\pi} \int _{-\pi}^\pi f(x) dx
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$$
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# Finding $a_n$
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Let’s multiply the entire equation by $\cos(mx)$, where $m \in \mathbb{Z}^+$ ($m$ is a positive integer):
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$$
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f(x)\cos(mx) = a_0\cos(mx) + \sum _{n = 1}^\infty a_n\cos(nx)\cos(mx) + b_n\sin(nx)\cos(mx)
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$$
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Now integrate on both sides, and bound by $[-\pi, \pi]$:
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$$
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\int _{-\pi}^\pi f(x)\cos(mx) dx = \int _{-\pi}^\pi a_0\cos(mx) dx + \sum _{n = 1}^\infty \int _{-\pi}^\pi a_n\cos(nx)\cos(mx) dx + \sum _{n = 1}^\infty \int _{-\pi}^\pi b_n\sin(nx)\cos(mx) dx
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$$
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We have three integrals on the right hand side to evaluate:
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## First Integral
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$$
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\int _{-\pi}^\pi a_0 \cos(mx) dx \\
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= \frac{a_0} m \sin(m\pi)- \frac{a_0} m \sin(-m\pi)
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$$
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Since $m\pi$ is always a multiple of $\pi$:
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$$
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=0
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$$
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## Second Integral
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$$
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\int _{-\pi}^\pi a_n\cos(nx)\cos(mx) dx
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$$
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Using $\cos$ addition formula:
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$$
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= \frac {a_0} 2 \int _{-\pi}^\pi \cos(nx + mx) + \cos(nx - mx) dx \\
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= \frac {a_0} 2 (\int _{-\pi}^\pi \cos(nx + mx) dx + \int _{-\pi}^\pi \cos(nx - mx) dx) \\
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= [\frac {a_0} 2 (\frac {\sin(nx + mx)} {n + m} + \frac {\sin(nx - mx)} {n - m})]_{-\pi}^{\pi} \\
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$$
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Here you will notice that this integral doesn’t work for $n = m$. We’ll circle back to that later. For now, this is two odd functions being added together. Since the bounds are the negatives of one another:
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$$
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= 0
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$$
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Now, circling back to the extra case, where $n = m$:
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$$
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a_m\int _{-\pi}^\pi \cos^2(nx)dx \\
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= a_m\int _{-\pi}^\pi \frac {1 + \cos(2x)} 2 dx \\
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= a_m[\frac x 2 + \frac {\sin 2x} 4 ]_{-\pi}^\pi \\
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= a_m[(\frac {\pi} 2 + \frac {\sin 2\pi} 4 ) - (\frac {-\pi} 2 + \frac {\sin -2\pi} 4 )] \\
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= a_m\pi
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$$
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So, the second term in the right hand side evaluates to $a_m\pi$.
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## Third Integral
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$$
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\int _{-\pi}^{\pi} \sin(nx)\cos(mx) dx \\
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= \frac 1 2 \int _{-\pi}^{\pi} \sin(nx + mx) dx + \frac 1 2 \int _{-\pi}^\pi \sin(nx - mx) dx \\
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= [-\frac 1 2(\frac {\cos(nx + mx)} {n + m} + \frac {\cos(nx - mx)} {n - m})]_{-\pi}^\pi \\
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$$
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Remember that $\cos x = -cos(x + \pi)$:
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$$
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= 0
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$$
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## Putting it Together
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Now we have:
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$$
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\int _{-\pi}^\pi f(x)\cos(mx) dx = a_m\pi \\
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\frac 1 \pi \int _{-\pi}^\pi f(x)\cos(mx) dx = a_m
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$$
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Note in this case $m$ and $n$ both represent any positive integer, and are therefore interchangeable:
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$$
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a_n = \frac 1 \pi \int _{-\pi}^\pi f(x)\cos(nx) dx \\
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$$
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# Finding $b_n$
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Multiply the equation by $\sin mx$, where $m \in \mathbb{Z}^+$,integrate, and bound between $[-\pi, \pi]$:
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$$
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\int _{-\pi}^\pi f(x)\sin(mx) dx = \int _{-\pi}^\pi a_0\sin(mx) dx + \sum _{n = 1}^\infty \int _{-\pi}^\pi a_n\cos(nx)\sin(mx) dx + \sum _{n = 1}^\infty \int _{-\pi}^\pi b_n\sin(nx)\sin(mx) dx
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$$
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The first two terms are already covered, so let’s focus on the final term.
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## Last Integral
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$$
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\int _{-\pi}^\pi b_n\sin(nx)\sin(mx) dx \\
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= b_n\int _{-\pi}^\pi \cos(nx - mx) - \cos(nx + mx) dx \\
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= b_n [\frac {\sin(nx - mx)} {n - m} - \frac {\sin(nx + mx)} {n + m}]_{-\pi}^\pi
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$$
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Again, there is a special case where $n = m$. Remember $\sin \pi = 0$, so:
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$$
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= 0
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$$
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With the special case:
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$$
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b_m\int _{-\pi}^\pi \sin^2(mx) dx \\
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= b_m\int _{-\pi}^\pi \frac {-\cos(2mx) + 1} 2 dx \\
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= b_m[\frac 1 2 (x - \frac {\sin(2mx)} {2m})]_{-\pi}^\pi \\
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= b_m\pi
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$$
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## Putting it Together
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$$
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b_m\pi = \int _{-\pi}^\pi f(x)\sin(mx) dx \\
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b_m = \frac 1 \pi \int _{-\pi}^\pi f(x)\sin(mx) dx \\
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b_n = \frac 1 \pi \int _{-\pi}^\pi f(x)\sin(nx) dx
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$$
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# Fourier Series
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Using the above, let’s express $f(x)$ as a Fourier Series:
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$$
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f(x) = \frac 1 {2\pi} \int _{-\pi}^\pi f(x) dx + \sum _{n = 1}^\infty \frac {\cos (nx)} \pi \int _{-\pi}^\pi f(x)\cos(nx) dx + \sum _{n = 1}^\infty \frac {\sin (nx)} \pi \int _{-\pi}^\pi f(x)\sin(nx) dx
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$$
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Note that this representation only works for when the function repeats from $[0, 2\pi]$. Using a similar proof, we can get:
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$$
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f(x) = \frac 1 P \int _{-\frac P 2}^{\frac P 2} f(x) dx + \sum _{n = 1}^\infty \frac {2 \cos (\frac {2\pi nx} P)} P \int _{-\frac P 2}^{\frac P 2} f(x)\cos(\frac {2\pi nx} P) dx + \sum _{n = 1}^\infty \frac {2 \sin (\frac {2\pi nx} P)} P \int _{-\frac P 2}^{\frac P 2} f(x)\sin(\frac {2\pi nx} P) dx
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$$ |