of the form:

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

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

\(f(x)\) 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:

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

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

For other values the power series may not converge.

Radius of convergence:

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

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 \(x=0\):

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

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

Consider \(e^M\)

We can evaluate this as:

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

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

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:

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 letâ€™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 \(\xi \).

We can add sinusoidal waves to get new waves.

For example

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

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:

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

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

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

We can move terms around to get:

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

Where:

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

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

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

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:

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

\(a=[,-1]\)

\(b=[,4]\)

\(c=[,0]\)

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

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