# Taylor series a. 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. b. Write the power series using summation notation. f(x)=2^x,a=1

Taylor series
a. 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.
b. Write the power series using summation notation.
$$f(x)=2^x,a=1$$

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