Determine a Function by a List of its Nth Derivative for each N. Is there any easy way to take a list of the derivative of a function , 2nd derivative, etc, and determine the original function?  I am wondering because I am trying to turn the Taylor Series 
$$1+(x^4)/4! + (x^8)/8! + (x^{12})/12! + (x^{16})/16! \ldots$$
into the original function.  How might I do this?
 A: New answer after OP has modified his question:
Recall that the exponential function can be written as
$$e^x = \sum_{k=0}^\infty \frac{x^k}{k!}$$
This looks a bit like your function, but it has some additional terms. Let's see if we can get rid of them.
By adding $e^{-x}$ and dividing by two we can eliminate all odd exponents: 
$$\frac{1}{2}(e^x + e^{-x}) = \frac{1}{2} \left (\sum_{k=0}^\infty \frac{x^k}{k!} + \sum_{k=0}^\infty \frac{(-x)^k}{k!}\right ) = \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}\tag{1}$$
This is the power series of $\cosh(x)$, by the way.
The exponents in our series are now $0, 2, 4, 6, 8, \ldots$. In order to arrive at your series, we therefore have to eliminate every other term of $(1)$. 
This can be achieved by adding $\sum_{k=0}^\infty \frac{(-1)^kx^{2k}}{(2k)!}$ to to $(1)$ and dividing by two:
$$\frac{1}{2} \left (\sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}+\sum_{k=0}^\infty \frac{(-1)^kx^{2k}}{(2k)!} \right )  = \sum_{k=0}^\infty \frac{x^{4k}}{(4k)!}$$
As stated before, $\sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}$ is $\cosh(x)$. Now recall that $\sum_{k=0}^\infty \frac{(-1)^kx^{2k}}{(2k)!}$ is simply $\cos(x)$.
Therefore the function you're looking for is $f(x) = \frac{1}{2} (\cosh(x) + \cos(x))$.
A: $1+x^4 + x^8 + x^{12} + x^{16} \ldots$
is a geometric series with ratio $x^4$ and the sum is
$\dfrac{1}{1-x^4},\;x^4<1\to |x|<1$
