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The series are $\cos 鈦�x,(1+x{)}^p$

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

The series are $\cos 鈦�x,(1+x{)}^p$

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

For$x=0$

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

Differentiate with respect to x.

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

Solve further

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

Solve further.

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

Solve further.

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

Hence $\cos 鈦�x=1鈭�\frac{1}{2!}+\frac{1}{4!}+鈥�$

The power series is $\underset{0}{\overset{\mathrm{鈭�}}{鈭�}}鈥�\frac{(鈭�1{)}^n{x}^{2n}}{\left(2n\right)!}$

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

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

For$x=0$

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

Differentiate with respect to x.

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

Solve further.

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

Solve further.

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

Hence ,$(1+x{)}^p=1+px+\frac{p(p鈭�1)}{2!}{x}^2+\frac{p(p鈭�1)(p鈭�2)}{3!}{x}^3+鈥�$

The binomial expression is$\underset{0}{\overset{\mathrm{鈭�}}{鈭�}}鈥�\left(\begin{array}{l}p\\ n\end{array}\right)x^n$

The series has been verified.