Answer:
To find the Taylor series of a function f(x) about a point a, we can use the following formula:
f(x) = f(a) + f'(a)(x-a)/1! + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ...
where f'(a), f''(a), f'''(a), ... denote the first, second, third, ... derivatives of f evaluated at a.
In this case, we have:
f(x) = 1 + 3e^(x+1)
To find the Taylor series about x=0, we need to evaluate the function and its derivatives at x=0.
f(0) = 1 + 3e^(0+1) = 1 + 3e
f'(x) = 3e^(x+1)
f'(0) = 3e^(0+1) = 3e
f''(x) = 3e^(x+1)
f''(0) = 3e^(0+1) = 3e
f'''(x) = 3e^(x+1)
f'''(0) = 3e^(0+1) = 3e
and so on.
Substituting these values into the formula for the Taylor series, we get:
f(x) = f(0) + f'(0)x + f''(0)x^2/2! + f'''(0)x^3/3! + ...
= (1 + 3e) + 3ex + 3ex^2/2! + 3ex^3/3! + ...
= 1 + (3 + 3e)x + (3/2)e x^2 + (1/2)e x^3 + ...
Therefore, the Taylor series for 1+3e^x+1 about x=0 is:
1 + (3 + 3e)x + (3/2)e x^2 + (1/2)e x^3 + ...