If ∑ n=0 to inf of 3x^n/n! is a Taylor series that converges to f(x) for all real x, what is the value of f ''(0)?
To find the value of f''(0), we need to differentiate the given Taylor series twice and then evaluate the result at x = 0.
First, let's differentiate the given series term by term.
The general term of the series is 3x^n / n!.
To differentiate it once, we use the power rule:
d/dx (x^n) = nx^(n-1).
We obtain:
d/dx (3x^n / n!) = 3 * (n * x^(n-1)) / n! = 3 * x^(n-1) / (n-1)!
Now, let's differentiate it again:
d^2/dx^2 (3x^n / n!) = d/dx (3 * x^(n-1) / (n-1)!)
Using the power rule again, we obtain:
= 3 * ((n-1) * x^(n-2)) / (n-1)!
Note that (n - 1)!/(n - 1)! cancels out:
= 3 * x^(n-2)
Now, let's evaluate this expression at x = 0:
f''(0) = 3 * 0^(n-2) = 0^(n-2) = 0
Therefore, f''(0) is equal to 0.