Donald Hobson comments on Explaining a Math Magic Trick

[Donald Hobson]

You can make $(1+D+D^2+...)$ work out, if you are prepared to make your mathematics even more deranged.

So lets look at $(1+D+D^2+...)0$

Think of the $0$ not as $0$ but as some infinitesimal $ϵ$ times some unknown function $g$.

If that function is $g=e^{0.5x}$ then we get $(1+\frac{1}{2}+\frac{1}{4}+...)e^{0.5x}$ which is finite, so multiplied by $ϵ$ it becomes infinitesimal.

If $g=e^{2x}$ then we get $(1+2+\mathrm{4...})e^{2x}$ and as we know $1+2+4+...=-1$ because $(1+2+4+...)\times 2=2+4+\mathrm{8...}=(1+2+\mathrm{4...})-1$

So this case is the same as before.

But for $g=e^x$ we get $(1+1+\mathrm{1...})e^x$ which doesn’t converge. The infinite largeness of this sum cancels with the infinitesimally small size of $ϵ$ (Up to an arbitrary finite constant).

So $(1+D+D^2...)0=C{e}^x$

Great. Now lets apply the same reasoning to

$(1+D+D^2...)e^x$. First note that this is infinite, it’s $\infty e^x$, so undefined. Can we make this finite. Well think of $e^x$ as actually being $e^x+ϵg$ and in this case, take $g=x{e}^x$

$D^n(e^x+ϵg)=e^x+ϵn{e}^x+ϵx{e}^x$

For the final term, the smallness of epsilon counteracts having to sum to infinity. For the first and middle term, the sum is $(\frac{1}{ϵ}+(\frac{1}{ϵ}+1)+(\frac{1}{ϵ}+2)+...)ϵe^x$

Which is $(1+2+3+...)ϵe^x-(1+2+...+(\frac{1}{ϵ}-1)+\frac{1}{ϵ})ϵe^x$

Now $1+2+3+...=-\frac{1}{12}$

So we have $(-\frac{1}{12})ϵe^x-\frac{1}{2}\frac{1}{ϵ}(\frac{1}{ϵ}-1)ϵe^x$

The first term is negligible. So $-\frac{1}{2}(\frac{1}{ϵ}-1)e^x$

Note that the $\frac{1}{2}{e}^x$ can be ignored, because we have $C{e}^x$ for arbitrary (finite) C as before.

Now $-\frac{1}{2ϵ}$ is big, but it’s probably less infinite than $\infty$ somehow. Let’s just group it into the $C$ and hope for the best.