Week 9: First few proofs successfully tweaked!

Jun 02, 2022

Hello everyone and welcome back to my blog!

This week I managed to find a way to tweak a large amount of proofs by generating functions of equations involving n!, such as the equation 0*0!+1*1!+2*2!+…+n*n!=(n+1)!-1. While this particular equation and others can be proven much more directly using induction or in this case writing the left hand side as a telescoping sum, I wanted to see if this could be done by generating functions. The only problem with this was that we would need to consider the generating function of n!, which would be 0!+1!x+2!x^2+3!x^3+…, which is divergent for all nonzero x, meaning we can’t really use or even consider this series. However, the Borel sum of this series does exist (for negative x, it’s equal to the integral from 0 to infinity of e^(-t)/(1-xt) dt), and by changing precisely how we use generating functions in proofs, this can be used as a substitute for the generating function of n!.

In most proofs by generating functions, the power series being considered is convergent to some function for at least some values of x, so a statement about these functions (for example the statement e^(ax)e^(bx)=e((a+b)x) can be translated into a statement about power series, and to obtain the desired identity look at the coefficient of x^n in these power series.

If instead we say that we are looking at the nth derivative of the functions evaluated at 0 divided by n!, for functions with convergent power series, this will indeed be the same as looking at the coefficient of x^n, but this operation can be applied to any (analytic) function, not just ones with a convergent power series.

We can now prove, for example, the identity from before. The structure of these revised proofs are instead:

  1. Prove 2 functions are equal using algebra (for example e^(ax)e^(bx)=e((a+b)x))
  2. Perform the operation f^(n)(0)/n! to both sides of the equation, and obtain an identity

At no point do any power series even need to be considered!

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