I took the course which covered this material (complex analysis), but I retained very little of it. Roger Penrose gives a brief summary for the layperson in few chapters of a book of his, “The Road to Reality”, which is reasonably clear. I’m going to provide the example from his book, the sum
1 + 4 + 16 + 64 + 256 …
because I don’t have a solid enough grasp of the material to consider the sum in question
1 + 2 + 3 + 4 + 5 + …
Let’s consider several sums.
S1 = 1 + 1/4 + 1/16 + 1/64 + 1/256 + …
S2 = 1 + 1 + 1 + 1 + 1 + 1 + …
S3 = 1 + 4 + 16 + 64 + 256 + …
Most of you would say that S1 converges to 4/3, while S2 and S3 are divergent. These are all in fact special cases of the sum
S(x) = 1 + x^2 + x^4 + x^6 + …
where
in S1, x = 1/2
in S2, x = 1
in S3, x = 2.
Instead of as an infinite sum, we can write S as a simple expression. (Penrose gives this as an exercise to the reader. I’ll solve it here.)
Consider the summation T(n)
T(n) = 1 + n^1 + n^2 + n^3 + …
Multiply by n to give
T(n)*n = n^1 + n^2 + n^3 + n^4 + …
Subtract T*x from T to give
T(n) - T(n)*n = 1
Factor and rearrange
T(n) = 1/(1-n)
Now make the substitution n –> x^2
1 + (x^2)^1 + (x^4)^2 + (x^6)^3 + … = 1/(1-x^2)
1 + x^2 + x^4 + x^6 + … = 1/(1-x^2)
Call this a new sum, S, as a function of x:
S(x) = 1 + x^2 + x^4 + x^6 + … = 1/(1-x^2)
Let’s return to our three initial series, S1, S2, and S3, and use our new formula, S = 1/(1-x^2), to reconsider them
For S1, when we plug 1/2 in for x, we get that S1 = 4/3. No problems here.
For S2, when we plug 1 in for x, we get that S2 = 1/0 = ∞ (or S2 is undefined). OK, this agrees with our notion that S2 diverges. (This is a pathological value of x, which someone else can expand on.)
For S3, when we plug in 2 for x, we get that S3 = -1/3, so we have that
1 + 4 + 16 + 64 + 256 + … = -1/3
This is peculiar, since we expect this to diverge as well. Let’s plot the first few terms of the summation S(x) to see if we can get a more intuitive feel for what’s happening. Specifically, let’s plot:
1
1 + x^2
1 + x^2 + x^4
1 + x^2 + x^4 + x^6
1 + x^2 + x^4 + x^6 + x^8
plot1
As we add more and more terms, our plot becomes a sort of infinitely high bowl bounded at x = -1 and x =1. So, this plot would suggest that our summation has a value for x such that -1 < x <1, and is undefined otherwise.
But let’s now plot 1/(1-x^2)
plot2
y = 1/(1-x^2) is defined for all real x except for x = -1 and x = 1, and those values “outside” the bowl are negative! At x = -1 and x = 1 are “poles” (also at x = i and x = -i). So x = 1 is really just one of four little blips that needs further mathematical treatment to understand. Again, I can’t really delve more into the topic because I’ve forgotten everything. It seems that there are many people here, though, who can go over this in much greater detail. I do, however, hear controls and systems engineers (who practice a very practical discipline – think feedback, automation, robotics) talk about poles a lot.