Finding the Correct Function that fits the Scenario

Question

i have been trying to find a function that fits the following scenario: $$ f'(c) = 1^0 $$ $$ f''(c) = 2^1 $$ $$ f^{(3)}(c) = 3^2 $$ $$ f^{(4)}(c) = 4^3 $$ and so on, the purpose is to derive a way to find: $$ B_{n,k}(1,2,3^2,4^3,\ldots) = B_{n,k}(f'(c),f''(c),f^{(3)}(c),f^{(4)}(c),\ldots) $$ In terms of a function. I have derived a new way to find identities using discrete convolution of terms with this method. I am very interested in finding this function. The $c$ values must all be the same.

Answer 1

I barely understand this, but that won't stop me from giving my understanding of the problem. I'll assume that you mean for the derivative of this theoretical function at the single value $c$ to be equal to the above.

In general you want.

$${{d^n f} \over {d t^n}}=n^{n-1}$$

Your solution will be a function, but without knowing initial conditions, the solution given here will be incomplete.

Take the Laplace transform of the equation. You'll get this...

$$p^n \cdot L={{n^{n-1}} \over p}$$

Solve for $L$...

$$L={{n^{n-1}} \over {p^{n+1}}}$$

Take the inverse Laplace transform. (look it up in a table if you don't do complex analysis)

$$f(t,n)=n^{n-1} \cdot {{t^n} \over {n!}}$$

Take the nth derivative of this function with respect to t, and you'll see that it satisfies your conditions.

Answer 2

The function you are looking for (for $c=0$) is $\dfrac{-W(-x)}{1 + W(-x)}$ where $W$ is the Lambert W function.

Answer 3

The power series given by Robert Israel converges when the absolute value of x is less than 1/e, as n! grows about as fast as the nth power of n/e (Stirling's Formula). This does give a solution, if you replace x by x-c in the series.