Cris Moore [WDS editing a bit & changing to Elkies notation]: Define S(x) = x-x^2+x^4-x^8+... As x->1, S(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 S(x) = x-S(x^2) = x-x^2+S(x^4) = x-x^2+x^4-S(x^8) = etc 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? --That Elkies page DA pointed out http://www.math.harvard.edu/~elkies/Misc/sol8.html says the Fourier series is known in closed form, which should solve your problem and then some. Except he does not say what it is. S(x) is ultimately periodic as a function of y=log_4 ( log(1/x) ) with period=1 and amplitude (says Elkies) approx 0.00275. I'd be interested to know what this fourier series is & how to get it. Let T(y)=S(x) Then x=exp(-4^y) and T(y) = exp(-4^y) - T(y + 1/2) and we are interested in y-->+infinity and in what is the fourier series of T(y).
Here's something I sent directly to Cris: f = x-x^2+x^4-x^8+… let x=e^(-2^(-2y)) then f(y)=\sum_{n=0}^\infty (-1)^n g(n-2y) where g(z)=e^(-2^z) Now combine successive +/- terms: f(y)=\sum_{m=0}^\infty g(2m-2y)-g(2m+1-2y) =\sum_{m=0}^\infty h(m-y) where h(z)=e^(-2^(2z))-e^(-2^(2z+1)) Make a plot of h. Convince yourself it vanishes exponentially for z -> -\infty and super-exponentially for z -> +\infty. Now look back at the formula for f(y) as a semi-infinite sum of h(m-y) and convince yourself that f(y) approaches a periodic function of y for large y. In fact, the periodic limit is exactly the periodic function you get by extending the sum to be doubly-infinite: p(y)=\sum_{m=-\infty}^\infty h(m-y) This is where you use the Poisson summation formula, to express p(y) in terms of its Fourier amplitudes: p(y)=\sum_{k=-\infty}^\infty e^(i 2pi k y) a_k a_k=\int_{-\infty}^{+\infty} e^(i 2pi k y) h(y) So the Fourier amplitudes of the periodic limit (in the logarithmically stretched variable y) is just the Fourier transform of h(y) evaluated at integers. Apart from a_0=1/2, these turn out to be very small: a_1=0.0000789616 + 0.00137234 i -Veit On Feb 2, 2014, at 11:19 AM, Warren D Smith <warren.wds@gmail.com> wrote:
Cris Moore [WDS editing a bit & changing to Elkies notation]: Define S(x) = x-x^2+x^4-x^8+... As x->1, S(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 S(x) = x-S(x^2) = x-x^2+S(x^4) = x-x^2+x^4-S(x^8) = etc 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?
--That Elkies page DA pointed out http://www.math.harvard.edu/~elkies/Misc/sol8.html says the Fourier series is known in closed form, which should solve your problem and then some. Except he does not say what it is. S(x) is ultimately periodic as a function of y=log_4 ( log(1/x) ) with period=1 and amplitude (says Elkies) approx 0.00275.
I'd be interested to know what this fourier series is & how to get it. Let T(y)=S(x) Then x=exp(-4^y) and T(y) = exp(-4^y) - T(y + 1/2) and we are interested in y-->+infinity and in what is the fourier series of T(y).
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In a bit of a rush, but this looks very nice, Veit! I'm puzzled, though: How could a real function's oscillations have a complex amplitude in the limit? Oh -- because the imaginary parts of the a_n terms' limits and the a_(-n) terms' limits cancel each other? --Dan On 2014-02-02, at 12:35 PM, Veit Elser wrote:
Here's something I sent directly to Cris:
f = x-x^2+x^4-x^8+…
let
x=e^(-2^(-2y))
then
f(y)=\sum_{n=0}^\infty (-1)^n g(n-2y)
where
g(z)=e^(-2^z)
Now combine successive +/- terms:
f(y)=\sum_{m=0}^\infty g(2m-2y)-g(2m+1-2y) =\sum_{m=0}^\infty h(m-y)
where
h(z)=e^(-2^(2z))-e^(-2^(2z+1))
Make a plot of h. Convince yourself it vanishes exponentially for z -> -\infty and super-exponentially for z -> +\infty. Now look back at the formula for f(y) as a semi-infinite sum of h(m-y) and convince yourself that f(y) approaches a periodic function of y for large y. In fact, the periodic limit is exactly the periodic function you get by extending the sum to be doubly-infinite:
p(y)=\sum_{m=-\infty}^\infty h(m-y)
This is where you use the Poisson summation formula, to express p(y) in terms of its Fourier amplitudes:
p(y)=\sum_{k=-\infty}^\infty e^(i 2pi k y) a_k a_k=\int_{-\infty}^{+\infty} e^(i 2pi k y) h(y)
So the Fourier amplitudes of the periodic limit (in the logarithmically stretched variable y) is just the Fourier transform of h(y) evaluated at integers. Apart from a_0=1/2, these turn out to be very small:
a_1=0.0000789616 + 0.00137234 i
-Veit
On Feb 2, 2014, at 11:19 AM, Warren D Smith <warren.wds@gmail.com> wrote:
Cris Moore [WDS editing a bit & changing to Elkies notation]: Define S(x) = x-x^2+x^4-x^8+... As x->1, S(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 S(x) = x-S(x^2) = x-x^2+S(x^4) = x-x^2+x^4-S(x^8) = etc 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?
--That Elkies page DA pointed out http://www.math.harvard.edu/~elkies/Misc/sol8.html says the Fourier series is known in closed form, which should solve your problem and then some. Except he does not say what it is. S(x) is ultimately periodic as a function of y=log_4 ( log(1/x) ) with period=1 and amplitude (says Elkies) approx 0.00275.
I'd be interested to know what this fourier series is & how to get it. Let T(y)=S(x) Then x=exp(-4^y) and T(y) = exp(-4^y) - T(y + 1/2) and we are interested in y-->+infinity and in what is the fourier series of T(y).
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participants (3)
-
Dan Asimov -
Veit Elser -
Warren D Smith