Answer:
Explanation:
To write the Taylor series for the function f(x) = e^(-x) about x = x0 = -1, we first need to find the k-th derivative of f(x) and evaluate it at x = -1.
f(x) = e^(-x)
f'(x) = -e^(-x)
f''(x) = e^(-x)
f'''(x) = -e^(-x)
f''''(x) = e^(-x)
...
The k-th derivative alternates between e^(-x) and -e^(-x), depending on the parity of k.
To evaluate the k-th derivative at x = -1, we substitute -1 into the k-th derivative expression and simplify:
f^k(-1) = (-1)^k * e^1
Now we can write the Taylor series using the formula:
f(x) = f(x0) + (x - x0) f'(x0) / 1! + (x - x0)^2 f''(x0) / 2! + (x - x0)^3 f'''(x0) / 3! + ...
Plugging in the values we obtained earlier, we have:
f(-x0) = e
f'(-x0) = -e
f''(-x0) = e
f'''(-x0) = -e
f''''(-x0) = e
...
Substituting these values into the formula, we get:
f(x) = e - (x + 1)e + (x + 1)^2 e / 2! - (x + 1)^3 e / 3! + (x + 1)^4 e / 4! - ...
Using sigma notation, we can write this as:
f(x) = ∑_{k=0}^∞ (-1)^k (x + 1)^k e / k!
where ∑_{k=0}^∞ represents the sum from k = 0 to infinity.