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)?
0^anything = 0
expanding the series, we have
f(x) = 3/0! x^0 + 3/1! x^1 + 3/2! x^2 + 3/3! x^3 + ...
f'(x) = 0 + 3/1! + 3*2/2! x + 3*3/3! x^2 + ...
f"(x) = 0 + 0 + 3*2/2! + 3*6/3! x^1 + ...
So, f"(0) = 3
To find the value of f''(0), we need to differentiate the given function twice. Let's start by differentiating the power series term by term:
The given function f(x) = ∑(n=0 to ∞) (3x^n/n!)
First, let's find the first derivative of f(x):
f'(x) = d/dx [∑(n=0 to ∞) (3x^n/n!)]
= ∑(n=0 to ∞) d/dx [(3x^n/n!)]
= ∑(n=0 to ∞) [3(n)x^(n-1)/n!] [Using the power rule of differentiation]
= ∑(n=1 to ∞) [3(n)x^(n-1)/n!] [Shifting the index to start from 1]
= ∑(n=0 to ∞) [3(n+1)x^n/(n+1)!] [Replacing n with n+1]
Now let's differentiate again to find the second derivative of f(x):
f''(x) = d/dx [∑(n=0 to ∞) (3(n+1)x^n/(n+1)!)]
= ∑(n=0 to ∞) d/dx [3(n+1)x^n/(n+1)!]
= ∑(n=0 to ∞) [3(n+1)(n)x^(n-1)/(n+1)!] [Using the power rule of differentiation]
= ∑(n=0 to ∞) [3(n)x^(n-1)/n!]
= ∑(n=1 to ∞) [3(n)x^(n-1)/n!] [Shifting the index to start from 1]
= ∑(n=0 to ∞) [3(n+1)x^n/(n+1)!] [Replacing n with n+1]
Now, let's evaluate f''(0). Plugging in x = 0 into the power series formula, we can see that all terms will become zero, except for the n = 0 term. Therefore, the value of f''(0) is:
f''(0) = 3(0+1)(0)^(0-1)/(0+1)!
= 3(0)(1)/1!
= 0
Hence, f''(0) is equal to 0.