# `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+...`