The function is \(f(x)=2^x\) and center at a=1

The definition of a Taylor series :

\(\sum_{k=0}^\infty c_k(x-a)^k,\text{ where }c_k=\frac{f^k(a)}{k!},\text{ for }k=0,1,2,3,...\)

a)To use the definition of a Taylor series to find the first four nonzero terms of the Taylor series for the given function centered at a

Find derivative of the function .

\(f(x)=2^x\)

\(f'(x)=2^x\ln(2)\)

\(f''(x)=2^x\ln^2(2)\)

\(f'''(x)=2^x\ln^3(2)\)

\(f^4(x)=2^x\ln^4(2)\)

To find derivative f(x) at x=1

\(f(1)=2\)

\(f'(1)=2\ln(2)\)

\(f''(1)=2\ln^2(2)\)

\(f'''(1)=2\ln^3(2)\)

\(f^4(1)=2\ln^4(2)\)

To find the first four nonzero terms of the Taylor series .

\(c_0=f(1)=2,c_1=\frac{f'(1)}{1!}=\frac{2\ln(2)}{1!},c_2=\frac{f''(1)}{2!}=\frac{2\ln^2(2)}{2!},c_3=\frac{f'''(1)}{3!}=\frac{2\ln^3(2)}{3!},c_4=\frac{f^4(1)}{4!}=\frac{2\ln^4(2)}{4!}\)

The in general \(c_k=\frac{2\ln^k(2)}{k!}\)

b) To write the power series using summation notation .

By using definition of a Taylor series and part a.

The series for centered at a=1 is

\(\sum_{k=0}^\infty c_k(x-a)^k=\sum_{k=0}^\infty\frac{2\ln^k(2)}{k!}(x-1)^k\)