You can make (1+D+D2+...) work out, if you are prepared to make your mathematics even more deranged.
So lets look at (1+D+D2+...)0
Think of the 0 not as 0 but as some infinitesimal ϵ times some unknown function g.
If that function is g=e0.5x then we get (1+12+14+...)e0.5x which is finite, so multiplied by ϵ it becomes infinitesimal.
If g=e2x then we get (1+2+4...)e2x and as we know 1+2+4+...=−1 because (1+2+4+...)×2=2+4+8...=(1+2+4...)−1
So this case is the same as before.
But for g=ex we get (1+1+1...)ex 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+D2...)0=Cex
Great. Now lets apply the same reasoning to
(1+D+D2...)ex. First note that this is infinite, it’s ∞ex, so undefined. Can we make this finite. Well think of ex as actually being ex+ϵg and in this case, take g=xex
Dn(ex+ϵg)=ex+ϵnex+ϵxex
For the final term, the smallness of epsilon counteracts having to sum to infinity. For the first and middle term, the sum is (1ϵ+(1ϵ+1)+(1ϵ+2)+...)ϵex
Which is (1+2+3+...)ϵex−(1+2+...+(1ϵ−1)+1ϵ)ϵex
Now 1+2+3+...=−112
So we have (−112)ϵex−121ϵ(1ϵ−1)ϵex
The first term is negligible. So −12(1ϵ−1)ex
Note that the 12ex can be ignored, because we have Cex for arbitrary (finite) C as before.
Now −12ϵ is big, but it’s probably less infinite than ∞ somehow. Let’s just group it into the C and hope for the best.
You can make (1+D+D2+...) work out, if you are prepared to make your mathematics even more deranged.
So lets look at (1+D+D2+...)0
Think of the 0 not as 0 but as some infinitesimal ϵ times some unknown function g.
If that function is g=e0.5x then we get (1+12+14+...)e0.5x which is finite, so multiplied by ϵ it becomes infinitesimal.
If g=e2x then we get (1+2+4...)e2x and as we know 1+2+4+...=−1 because (1+2+4+...)×2=2+4+8...=(1+2+4...)−1
So this case is the same as before.
But for g=ex we get (1+1+1...)ex 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+D2...)0=Cex
Great. Now lets apply the same reasoning to
(1+D+D2...)ex. First note that this is infinite, it’s ∞ex, so undefined. Can we make this finite. Well think of ex as actually being ex+ϵg and in this case, take g=xex
Dn(ex+ϵg)=ex+ϵnex+ϵxex
For the final term, the smallness of epsilon counteracts having to sum to infinity. For the first and middle term, the sum is (1ϵ+(1ϵ+1)+(1ϵ+2)+...)ϵex
Which is (1+2+3+...)ϵex−(1+2+...+(1ϵ−1)+1ϵ)ϵex
Now 1+2+3+...=−112
So we have (−112)ϵex−121ϵ(1ϵ−1)ϵex
The first term is negligible. So −12(1ϵ−1)ex
Note that the 12ex can be ignored, because we have Cex for arbitrary (finite) C as before.
Now −12ϵ is big, but it’s probably less infinite than ∞ somehow. Let’s just group it into the C and hope for the best.