Find the Taylors series for f (x)=e2x about 0

To find the Taylor series representation of a function f(x) about a particular point, we can use the formula:

f(x) = f(a) + f'(a)(x-a)/1! + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ...

In this case, we need to find the Taylor series for f(x) = e^(2x) about x = 0.

Step 1: Find the values of f(a), f'(a), f''(a), ... at x = a = 0.
- f(0) = e^(2*0) = e^0 = 1
- f'(x) = d(e^(2x))/dx = 2e^(2x), so f'(0) = 2e^0 = 2
- f''(x) = d(2e^(2x))/dx = 4e^(2x), so f''(0) = 4e^0 = 4
- f'''(x) = d(4e^(2x))/dx = 8e^(2x), so f'''(0) = 8e^0 = 8
- and so on...

Step 2: Plug the values into the formula.
f(x) = 1 + 2(x-0)/1! + 4(x-0)^2/2! + 8(x-0)^3/3! + ...

Simplifying, we get:

f(x) = 1 + 2x + 2(x^2)/2 + 8(x^3)/6 + ...

Simplifying further, we have:

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

So, the Taylor series representation of f(x) = e^(2x) about x = 0 is:

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