By the method used to obtain (12.5)[which is the series(13.1)below], verify each of the other series (13.2)to (13.5)below.

The series are $\cos x,(1+x{)}^p$

The power series is the infinite series cantered at c and have an interval of convergence

The series are $\cos x,(1+x{)}^p$

$\cos x=a_0+a_{1}x+a_2{x}^2+a_3{x}^2+a_4{x}^4+\dots$

For$x=0$

$\begin{array}{r}\cos x=1\\ a_0=1\end{array}$

Differentiate with respect to x.

$\begin{array}{r}d/dx(\cos x)=-\sin x\\ 0+a_1+2a_{2}x+3a_3{x}^2+4a_4{x}^3+\dots .x=0\\ -\sin x=0\\ a_1=0\end{array}$

Solve further

$\begin{array}{r}d/dx(-\sin x)=-\cos x=2a_2+6a_{3}x+12a_4{x}^2+\dots \\ x=0\\ -\cos x=-1\\ a_2=\frac{-1}{2!}\end{array}$

Solve further.

$\begin{array}{r}d/dx(-\cos )=\sin x\\ 6a_3+24a_{4}x+\dots x=0\\ \sin x=0\\ a_3=0\end{array}$

Solve further.

$\begin{array}{r}d/dx(\sin x)=\cos x\\ 24a_4+\dots x=0\\ \cos x=1\\ a_4=\frac{1}{4!}\end{array}$

Hence $\cos x=1-\frac{1}{2!}+\frac{1}{4!}+\dots$

The power series is $\sum _0^{\mathrm{\infty }} \frac{(-1{)}^n{x}^{2n}}{\left(2n\right)!}$

verify $(1+x{)}^p$

$(1+x{)}^p=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4+\dots$

For$x=0$

$\begin{array}{r}(1+x{)}^p=1\\ a_0=1\end{array}$

Differentiate with respect to x.

$\begin{array}{r}d/dx(1+x{)}^p=p(1+x{)}^{p-1}\\ 0+a_1+2a_{2}x+3a_3{x}^2+4a_4{x}^3+\dots x=0\\ p(1+x{)}^{p-1}=p\\ a_1=p\end{array}$

Solve further.

$\begin{array}{r}d/dx\left(p(1+x{)}^{p-1}\right)=p(p-1)p(1+x{)}^{p-2}\\ 2a_2+6a_{3}x+12a_4{x}^2+\dots x=0\\ p(p-1)(1+x{)}^{p-2}=p(p-1)\\ a_2=\frac{p(p-1)}{2!}\end{array}$

Solve further.

$\begin{array}{r}d/dx\left(p(p-1)p(1+x{)}^{p-2}\right)=p(p-1)(p-2)p(1+x{)}^{p-3}\\ 6a_3+24a_{4}x+\dots =p(p-1)(p-2)p(1+x{)}^{p-3}\\ =p(p-1)(p-2)\\ a_3=\frac{p(p-1)(p-2)}{3!}\end{array}$

Hence ,$(1+x{)}^p=1+px+\frac{p(p-1)}{2!}{x}^2+\frac{p(p-1)(p-2)}{3!}{x}^3+\dots$

The binomial expression is$\sum _0^{\mathrm{\infty }} \left(\begin{array}{l}p\\ n\end{array}\right)x^n$

The series has been verified.