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`f(x)=e^(2x) ,c=0` Use the definition of Taylor series to find the Taylor series, centered at c for the function.

Taylor series is an example of infinite series derived from the expansion of f(x) about a single point. It is represented by infinite sum of `f^n(x)` centered at` x=a` . The general formula for Taylor series is:


`f(x) = sum_(n=0)^oo (f^n(a))/(n!) (x-a)^n`


or


`f(x) =f(a)+f'(a)(x-a) +(f''(a))/(2!)(x-a)^2 +(f^3(a))/(3!)(x-a)^3 +(f'^4(a))/(4!)(x-a)^4 +...`


To apply the definition of Taylor series for the given function `f(x) = e^(2x)` , we list` f^n(x)` using the derivative formula for exponential function: `d/(dx) e^u = e^u * (du)/(dx)` .


Let `u =2x ` then `(du)/(dx)= 2`


Applying the values on the derivative formula for exponential function, we get:


`d/(dx) e^(2x) = e^(2x) *2`


              ` = 2e^(2x)`


Applying `d/(dx) e^(2x)= 2e^(2x)`   for each` f^n(x)` , we get:


`f'(x) = d/(dx) e^(2x)`


           `= 2e^(2x)`


`f^2(x) = 2 *d/(dx) e^(2x)`


           `= 2*2e^(2x)`


           `=4e^(2x)`


`f^3(x) = 4*d/(dx) e^(2x)`


            `= 4*2e^(2x)`


`f^4(x) = 8*d/(dx) e^(2x)`


           `= 8*2e^(2x)` 


           `=16e^(2x)`


Plug-in `x=0` , we get:


`f(0) =e^(2*0) =1`


`f'(0) =2e^(2*0)=2 `


`f^2(0) =4e^(2*0)=4`


`f^3(0) =8e^(2*0)=8`


`f^4(0) =16e^(2*0)=16 `


Note: `e^(2*0)=e^0 =1` .


Plug-in the values on the formula for Maclaurin series. 


`e^(2x)= sum_(n=0)^oo (f^n(0))/(n!) (x-0)^n`


         ` = sum_(n=0)^oo (f^n(0))/(n!) x^n`


          `= 1+2x+4/(2!)x^2+8/(3!)x^3+16/(4!)x^4+...`


          `=1+2x+4/(1*2)x^2+8/(1*2*3)x^3+16/(1*2*3*4)x^4+...`


          `= 1+2x+4/2x^2+8/6x^3+16/24x^4+...`


         `= 1+2x+2x^2+4/3x^3+2/3x^4+...`


The Taylor series for the given function `f(x)=e^(2x)` centered at `a=0` will be:


`e^(2x) =1+2x+2x^2+4/3x^3+2/3x^4+...`

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