Matt Springer is a graduate student of physics at Texas A&M university. He is also an occasional writer and tinkerer, and he is probably too curious for his own good.
when you order from Amazon, and a fraction of the purchase price will be sent to me at zero cost to you. Much obliged, and thanks for your patronage.A tricky one:
At x = 2, f(x) = 1 + 2 + 4 + 8 + ...
Which, uh, is clearly not going to end anywhere finite. The series fails the single most basic necessary (but not sufficient) condition for convergence of infinite sums - that the terms of the series approach zero as you go farther and farther up the chain of addition.
On the other hand, f(x) is equally clearly not divergent for all x.
For instance, at x = 0, f(x) = 1 + 0 + 0 + ... = 1
So that's a start. Maybe there's other numbers x can equal where the series has a finite value. What are they? This is one of those problems where it's easier if you already know something extra about the question. Take a look at this function:
If you find the series expansion of that function around x = 0, you'll get precisely the f(x) we posed earlier. But there's a complication. While our neat fraction is perfectly well defined at x = 2, we already saw that our series entirely failed to converge there. What gives?
The answer lies in the one place our fraction isn't defined - where x = 1, the denominator is zero and it's impossible to divide by zero. And here's the catch. If you create a series expansion of a function around a particular point (here we chose x = 0) that expansion will converge as it should right up until you hit the place where the function isn't analytic. Ours has such a point at x = 1, and so the series will fail to converge there or anywhere past it. Further, that distance between our series start and the bad point serves to define a radius of convergence. Not only will our series not work for x equal to or greater than 1, it won't work for x equal to or less than -1. In the complex plane, it won't work outside the circle centered at the origin.
We can if we want create another series expansion on the other side of the discontinuity, but of course the same problem would still exist. It wouldn't work outside the radius defined by the distance of our expansion point from the point x = 1.
Here, for comparison, are the graphs of the fractional version of our function and the series version of our function graphed on a domain running from -1.5 to 1.5. The fractional version of our function is in purple, the series (the first eleven terms) is in blue.
This particular kind of series is called the geometric series, and find application not just in physics but also in finance, engineering, and pure mathematics.
Physical Science
Comments
f(0)=1+0+0+...=1
Posted by: Anonymous | March 1, 2009 12:34 PM
Indeed. Typo fixed.
Posted by: Matt Springer | March 1, 2009 1:45 PM
Ah, but what if you were to move your expansion point off the real line...
Posted by: Michael F. Martin | March 1, 2009 2:28 PM
This is also a very good example to give people an idea of what happens with analytic continuation. In essence, we have a function f(x)=1+x+x^2+x^3... that only makes sense for |x|
Posted by: Joshua Zelinsky | March 1, 2009 3:46 PM
Ray Kurzweil, Vernor Vinge, et al. wax eloquet and voluminous about the Singularity! - social and technological convergence beyond conceptualization by past civilizations (including this one),
http://en.wikipedia.org/wiki/File:ParadigmShiftsFrr15Events.svg
First derivatives sell books, second derivatives sell guns. Know whether extrema are peaks (Manhattan Project) or troughs (NINJA loans). As your plot shows, an epsilon skew makes all the difference in the world.
Posted by: Uncle Al | March 1, 2009 3:48 PM
ok, don't know why the rest of my post got cut off. Trying that again. Our function only makes sense |x| less than 1 (this is true even if we look at complex valued x). But we have this other function 1/(1-x) that agrees with f everywhere they are both defined. This second function is defined and infinitely differentiable everywhere by x=1. Thus, we have a continuation of x to the entire complex plane. This is very similar to the process that allows us to get the analytic continuations of the Riemann zeta function even though its series is only defined for x with the real part of x greater than 1.
Posted by: Joshua Zelinsky | March 1, 2009 3:57 PM
And all hell breaks loose when 1 + 2 + 4 + ... goes to -1
http://en.wikipedia.org/wiki/1_%2B_2_%2B_4_%2B_8_%2B_%C2%B7_%C2%B7_%C2%B7
(I remember reading that 1 + 2 + 4 + ... is considered the representation of -1 in a machine with infinite storage space).
Posted by: tcmJOE | March 1, 2009 8:45 PM
Just to be difficult, 1+2+...=-1 is no problem if you use the right norm. Specfically, if you define a "size" of rational numbers so that |2^a b| is 2^-a when b is prime to 2, you get a perfectly reasonable metric on the rational numbers. In this metric that series converges nicely, and to -1. Of course, when you complete the number system, you get something different from the real numbers - in fact its topology looks rather like a Cantor set. But this object, the set of 2-adic numbers, turns out to be a very useful object in number theory - for example, "closeness" means congruence modulo a high power of two.
Posted by: Anne | March 3, 2009 10:35 AM
Neat post. I always enjoy the Sunday function. This graph is hard to read for us color blind folks, though. Those 2 colors are indistinguishable. Different weights, or dashed and solid lines would be much easier to read.
Posted by: Essex | March 6, 2009 1:23 PM
Why no proof that the series converges to 1/(1-x) when |x| = 1?
There is one more (mildly) interesting thing about the function.
y = 1 + x + x^2, is a parabola which intersects the point (0, 1).
y = 1 + x + x^2 + x^3, is a curve with a point of inflection at (0, 1).
If you continue to add the next term and then plot, each curve ending with an even power of x resembles the first curve as a distorted parabola intersecting (0, 1) and y has no negative values. Each curve ending with an odd power of x resembles the second curve and has the same point of inflection and y has both positive and negative values.
Posted by: Joe Doakes | March 7, 2009 3:46 PM