[math-fun] oscillating series
Define f(x) = x-x^2+x^4-x^8+... As x->1, f(x) becomes Grandi's series 1-1+1-1+..., and it oscillates more and more rapidly around 1/2. These oscillations are periodic in log(1-x): one way to see that is to notice that f(x) = x-f(x^2) or roughly f(x) = 1-f(x^2). So squaring x, which (when x is close to 1) increments log(1-x) by log 2, flips f(x) around 1/2. Hardy knew about these oscillations (see http://en.wikipedia.org/wiki/Summation_of_Grandi%27s_series). But my question is: how do we calculate their amplitude? Numerically, f(x) ranges from 1/2+delta to 1/2-delta, where delta = 0.00274 or so. Does anyone know how to obtain this number? - Cris
Noam Elkies doesn't answer Cris's question, but has a nice little writeup about this series in the solution to question #8, here. (Or paste http://www.math.harvard.edu/~elkies/Misc/index.html into browser.) --Dan On 2014-02-01, at 3:02 PM, Cris Moore wrote:
Define f(x) = x-x^2+x^4-x^8+...
As x->1, f(x) becomes Grandi's series 1-1+1-1+..., and it oscillates more and more rapidly around 1/2. These oscillations are periodic in log(1-x): one way to see that is to notice that
f(x) = x-f(x^2)
or roughly
f(x) = 1-f(x^2).
So squaring x, which (when x is close to 1) increments log(1-x) by log 2, flips f(x) around 1/2.
Hardy knew about these oscillations (see http://en.wikipedia.org/wiki/Summation_of_Grandi%27s_series). But my question is: how do we calculate their amplitude?
Numerically, f(x) ranges from 1/2+delta to 1/2-delta, where delta = 0.00274 or so. Does anyone know how to obtain this number?
- Cris
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participants (2)
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Cris Moore -
Dan Asimov