Taylor and Fourier analysis

Power series

Power series

of the form:

\(\sum_{n=0}a_n(x-c)^n\)

Smoothness of power series

Power series are all smooth. That is, they are infinitely differentiable.

Taylor series

Taylor series

\(f(x)\) can be estimated at point \(c\) by identifying its repeated differentials at point \(c\).

The coefficients of an infinate number of polynomials at point \(c\) allow this.

\(f(x)=\sum_{i=0}^{\infty }a_i(x-c)^i\)

\(f'(x)=\sum_{i=1}^{\infty }a_i(x-c)^{i-1}i\)

\(f''(x)=\sum_{i=2}^{\infty }a_i(x-c)^{i-2}i(i-1)\)

\(f^j(x)=\sum_{i=j}^{\infty }a_i(x-c)^{i-j}\dfrac{i!}{(i-j)!}\)

For \(x=c\) only the first term in the series is non-zero.

\(f^j(c)=\sum_{i=j}^{\infty }a_i(c-c)^{i-j}\dfrac{i!}{(i-j)!}\)

\(f^j(c)=a_ij!\)

So:

\(a_j=\dfrac{f^j(c)}{j!}\)

So:

\(f(x)=\sum_{i=0}^\infty (x-c)^i \dfrac{f^i(c)}{i!}\)

Convergence

If \(x=c\) then the power series will be equal to \(a_0\).

For other values the power series may not converge.

Cauchy-Hadamard theorem

Radius of convergence:

\(\dfrac{1}{R}={\lim \sup}_{n_\rightarrow \infty} (|a_n|^{\dfrac{1}{n}})\)

Maclaurin series

A Taylor series around \(c=0\).

\(f(x)=\sum_{i=0}^\infty (x-c)^i \dfrac{f^i(c)}{i!}\)

\(f(x)=\sum_{i=0}^\infty (x)^i \dfrac{f^i(0)}{i!}\)

For example, for:

\(f(x)=(1-x)^{-1}\)

\(f^i(0)=i!\)

So, around \(x=0\):

\(f(x)=\sum_{i=0}^\infty (x)^i\)

Fourier transforms

Taylor series of matrices

We can also use Taylor series to evaluate functions of matrices.

Consider \(e^M\)

We can evaluate this as:

\(e^M=\sum_{k=0}^\infty \dfrac{1}{k!}M^k\)

Analytic functions

(root test, direct comparison test, rate of convergence, radius of convergence)

Fourier analysis

Representing wave functions

Wave function are of the form:

\(\cos(ax + b)\)

\(\sin(ax + b)\)

We can use the following identities:

  • \(\cos(x)=\sin(x+\dfrac{\tau }{8})\)

  • \(\sin(-x)=-\sin(x)\)

  • \(\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)\)

So we can write any function as:

Using \(e\)

Harmonics

Fourier series

Fourier series

Motivation: we have a function we want to display as another sort of function.

More specifically, a function can be shown as a combination of sinusoidal waves.

To frame this let’s imagine a sound wave, with values \(f(t)\) for all time values \(t\). We can imagine this as a summation of sinusoidal functions. That is:

\(f(t)=\sum_{n=0}^{\inf } a_ncos(nw_0t)\)

We want to get another function \(F(\xi )\) for all frequencies \(\xi \).

Combinations of wave functions

We can add sinusoidal waves to get new waves.

For example

\(s_N(x)=2\sin(x+3)+\sin(-4x)+\dfrac{1}{2}\cos(x)\)

As a summation of series

We can simplify arbitrary series using the following identities:

\(\cos(x)=\sin(x+\dfrac{\tau }{8})\)

\(\sin(-x)=-\sin(x)\)

So we have:

\(s(x)=2\sin(x+3)-\sin(4x)+\dfrac{1}{2}\sin(x+\dfrac{\tau }{8})\)

We can put this into the following format:

\(s(x)=\sum^m_{i=1}a_i\sin(b_ix+c_i)\)

Where:

\(a=[2,-1,\dfrac{1}{2}]\)

\(b=[1,4,1]\)

\(c=[3,0,\dfrac{\tau}{8}]\)

Ordering by \(b\)

We can move terms around to get:

\(s(x)=\sum^m_{i=1}a_i\sin(b_ix+c_i)\)

Where:

\(a=[2,\dfrac{1}{2},-1]\)

\(b=[1,1,4]\)

\(c=[3,\dfrac{\tau}{8},0]\)

Adding waves with same frequency

We know that:

\(\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)\)

So:

\(\sin(b_ix+c_i)=\sin(b_ix)\cos(c_i)+\sin(c_i)\cos(b_ix)\)

If \(2\) terms have the same value for \(b_i\), then:

\(a_i\sin(b_ix+c_i)+a_j\sin(b_jx+c_j)=a_i\sin(b_ix+c_i)+a_j\sin(b_ix+c_j)\)

\(a_i\sin(b_ix+c_i)+a_j\sin(b_jx+c_j)=a_i\sin(b_ix)\cos(c_i)+a_i\sin(c_i)\cos(b_ix)+a_j\sin(b_ix)\cos(c_j)+a_j\sin(c_j)\cos(b_ix)\)

So we now get for:

\(s(x)=\sum^m_{i=1}a_i\sin(b_ix+c_i)\)

\(a=[,-1]\)

\(b=[,4]\)

\(c=[,0]\)

Fourier transforms

Fourier transform

\(\hat f(\Xi )=\int_{-\infty}^{\infty }f(x)e^{-2\pi ix\Xi }dx\)

Inverse Fourier transform

\(f(x)=\int_{-\infty}^{\infty }\hat f(\Xi )e^{2\pi ix\Xi }d\Xi \)

Fourier inversion theorem