$sum exp(-x^2)$ vs $sum x^2 exp(-x^2)$
I am curious about the following sum, for $alpha in (0,1)$:
$$sum_{k = -infty}^{infty} (1-(2k - 1 + alpha)^2) exp(-frac{1}{2} (2k - 1 + alpha)^2)$$
I have reasons to believe sum should be zero when $alpha = 1/2$, but I don't know how to prove it. And according to Mathematica, the value at $alpha = 1/2$ is
$-2.6474039 times 10^{-7}$,
which is notably not zero (though I'm not convinced that this isn't a rounding error). So: prove me wrong! (Is it possible to show that the value of this sum is non-zero?)
For some 'intuition' about why the sum is so close to zero, note that
$$int_{-infty}^{infty} (1-(2x - alpha + 1)^2)exp(-frac{1}{2} (2x - alpha + 1)^2) , dx = 0$$
for all $alpha in mathbb{R}$. This is because the value of the integral doesn't depend on $alpha$, and computing the value at $alpha = 0$ is an easy exercise. The sum above is a Riemann sum for this integral. Of course, there is always error in changing from a Riemann sum to an integral -- I am just suspicious of how small the error is!
I would also like to know if there is a 'nicer' expression for the sum, or some relationship to well-known special functions. I suspect there are some clever Fourier-analysis type ideas to help evaluate this kind of sum.
sequences-and-series analysis riemann-sum
add a comment |
I am curious about the following sum, for $alpha in (0,1)$:
$$sum_{k = -infty}^{infty} (1-(2k - 1 + alpha)^2) exp(-frac{1}{2} (2k - 1 + alpha)^2)$$
I have reasons to believe sum should be zero when $alpha = 1/2$, but I don't know how to prove it. And according to Mathematica, the value at $alpha = 1/2$ is
$-2.6474039 times 10^{-7}$,
which is notably not zero (though I'm not convinced that this isn't a rounding error). So: prove me wrong! (Is it possible to show that the value of this sum is non-zero?)
For some 'intuition' about why the sum is so close to zero, note that
$$int_{-infty}^{infty} (1-(2x - alpha + 1)^2)exp(-frac{1}{2} (2x - alpha + 1)^2) , dx = 0$$
for all $alpha in mathbb{R}$. This is because the value of the integral doesn't depend on $alpha$, and computing the value at $alpha = 0$ is an easy exercise. The sum above is a Riemann sum for this integral. Of course, there is always error in changing from a Riemann sum to an integral -- I am just suspicious of how small the error is!
I would also like to know if there is a 'nicer' expression for the sum, or some relationship to well-known special functions. I suspect there are some clever Fourier-analysis type ideas to help evaluate this kind of sum.
sequences-and-series analysis riemann-sum
These are related to the Jacobi theta function. I'd try Poisson summation.
– Lord Shark the Unknown
Nov 25 at 7:12
add a comment |
I am curious about the following sum, for $alpha in (0,1)$:
$$sum_{k = -infty}^{infty} (1-(2k - 1 + alpha)^2) exp(-frac{1}{2} (2k - 1 + alpha)^2)$$
I have reasons to believe sum should be zero when $alpha = 1/2$, but I don't know how to prove it. And according to Mathematica, the value at $alpha = 1/2$ is
$-2.6474039 times 10^{-7}$,
which is notably not zero (though I'm not convinced that this isn't a rounding error). So: prove me wrong! (Is it possible to show that the value of this sum is non-zero?)
For some 'intuition' about why the sum is so close to zero, note that
$$int_{-infty}^{infty} (1-(2x - alpha + 1)^2)exp(-frac{1}{2} (2x - alpha + 1)^2) , dx = 0$$
for all $alpha in mathbb{R}$. This is because the value of the integral doesn't depend on $alpha$, and computing the value at $alpha = 0$ is an easy exercise. The sum above is a Riemann sum for this integral. Of course, there is always error in changing from a Riemann sum to an integral -- I am just suspicious of how small the error is!
I would also like to know if there is a 'nicer' expression for the sum, or some relationship to well-known special functions. I suspect there are some clever Fourier-analysis type ideas to help evaluate this kind of sum.
sequences-and-series analysis riemann-sum
I am curious about the following sum, for $alpha in (0,1)$:
$$sum_{k = -infty}^{infty} (1-(2k - 1 + alpha)^2) exp(-frac{1}{2} (2k - 1 + alpha)^2)$$
I have reasons to believe sum should be zero when $alpha = 1/2$, but I don't know how to prove it. And according to Mathematica, the value at $alpha = 1/2$ is
$-2.6474039 times 10^{-7}$,
which is notably not zero (though I'm not convinced that this isn't a rounding error). So: prove me wrong! (Is it possible to show that the value of this sum is non-zero?)
For some 'intuition' about why the sum is so close to zero, note that
$$int_{-infty}^{infty} (1-(2x - alpha + 1)^2)exp(-frac{1}{2} (2x - alpha + 1)^2) , dx = 0$$
for all $alpha in mathbb{R}$. This is because the value of the integral doesn't depend on $alpha$, and computing the value at $alpha = 0$ is an easy exercise. The sum above is a Riemann sum for this integral. Of course, there is always error in changing from a Riemann sum to an integral -- I am just suspicious of how small the error is!
I would also like to know if there is a 'nicer' expression for the sum, or some relationship to well-known special functions. I suspect there are some clever Fourier-analysis type ideas to help evaluate this kind of sum.
sequences-and-series analysis riemann-sum
sequences-and-series analysis riemann-sum
edited Nov 25 at 8:25
Mason
1,9591530
1,9591530
asked Nov 25 at 7:05
J Richey
351318
351318
These are related to the Jacobi theta function. I'd try Poisson summation.
– Lord Shark the Unknown
Nov 25 at 7:12
add a comment |
These are related to the Jacobi theta function. I'd try Poisson summation.
– Lord Shark the Unknown
Nov 25 at 7:12
These are related to the Jacobi theta function. I'd try Poisson summation.
– Lord Shark the Unknown
Nov 25 at 7:12
These are related to the Jacobi theta function. I'd try Poisson summation.
– Lord Shark the Unknown
Nov 25 at 7:12
add a comment |
1 Answer
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Let $f(k) = sum_{k = -infty}^{infty} (1-(2k - 1 + frac12)^2) exp(-frac{1}{2} (2k - 1 + frac12)^2)$, so that you are asking about the sum $S = sum_{k=-infty}^infty f(k)$.
We can calculate that
$$
sum_{k=-2}^3 f(k) approx -2.37104times10^{-7}
$$
(and presumably you are comfortable trusting this finite calculation). This proves that $S<0$, since the omitted summands $f(k)$ are negative for all $kle-3$ and all $kge4$.
Indeed, we can also calculate
$$
sum_{k=-4}^5 f(k) approx -2.647403940047578times10^{-7} tag{$*$}
$$
and, since $|f(t)|$ is increasing for $t<-2$ and decreasing for $t>2$,
$$
sum_{k=6}^infty |f(k)| < int_5^infty |f(t)|,dt = |F(5)| < 1.2times10^{-19}
$$
and
$$
sum_{k=-infty}^{-5} |f(k)| < int_{-infty}^{-4} |f(t)|,dt = |F(-4)| < 8.7times10^{-16},
$$
where
$$
F(t) = int f(t),dt = frac{1}{4} e^{-2t^2+t-frac{1}{8}} (4t-1);
$$
these calculations show that the right-hand side of $(*)$ is the correct value of the infinite sum $S$ to 7 or 8 significant digits.
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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votes
Let $f(k) = sum_{k = -infty}^{infty} (1-(2k - 1 + frac12)^2) exp(-frac{1}{2} (2k - 1 + frac12)^2)$, so that you are asking about the sum $S = sum_{k=-infty}^infty f(k)$.
We can calculate that
$$
sum_{k=-2}^3 f(k) approx -2.37104times10^{-7}
$$
(and presumably you are comfortable trusting this finite calculation). This proves that $S<0$, since the omitted summands $f(k)$ are negative for all $kle-3$ and all $kge4$.
Indeed, we can also calculate
$$
sum_{k=-4}^5 f(k) approx -2.647403940047578times10^{-7} tag{$*$}
$$
and, since $|f(t)|$ is increasing for $t<-2$ and decreasing for $t>2$,
$$
sum_{k=6}^infty |f(k)| < int_5^infty |f(t)|,dt = |F(5)| < 1.2times10^{-19}
$$
and
$$
sum_{k=-infty}^{-5} |f(k)| < int_{-infty}^{-4} |f(t)|,dt = |F(-4)| < 8.7times10^{-16},
$$
where
$$
F(t) = int f(t),dt = frac{1}{4} e^{-2t^2+t-frac{1}{8}} (4t-1);
$$
these calculations show that the right-hand side of $(*)$ is the correct value of the infinite sum $S$ to 7 or 8 significant digits.
add a comment |
Let $f(k) = sum_{k = -infty}^{infty} (1-(2k - 1 + frac12)^2) exp(-frac{1}{2} (2k - 1 + frac12)^2)$, so that you are asking about the sum $S = sum_{k=-infty}^infty f(k)$.
We can calculate that
$$
sum_{k=-2}^3 f(k) approx -2.37104times10^{-7}
$$
(and presumably you are comfortable trusting this finite calculation). This proves that $S<0$, since the omitted summands $f(k)$ are negative for all $kle-3$ and all $kge4$.
Indeed, we can also calculate
$$
sum_{k=-4}^5 f(k) approx -2.647403940047578times10^{-7} tag{$*$}
$$
and, since $|f(t)|$ is increasing for $t<-2$ and decreasing for $t>2$,
$$
sum_{k=6}^infty |f(k)| < int_5^infty |f(t)|,dt = |F(5)| < 1.2times10^{-19}
$$
and
$$
sum_{k=-infty}^{-5} |f(k)| < int_{-infty}^{-4} |f(t)|,dt = |F(-4)| < 8.7times10^{-16},
$$
where
$$
F(t) = int f(t),dt = frac{1}{4} e^{-2t^2+t-frac{1}{8}} (4t-1);
$$
these calculations show that the right-hand side of $(*)$ is the correct value of the infinite sum $S$ to 7 or 8 significant digits.
add a comment |
Let $f(k) = sum_{k = -infty}^{infty} (1-(2k - 1 + frac12)^2) exp(-frac{1}{2} (2k - 1 + frac12)^2)$, so that you are asking about the sum $S = sum_{k=-infty}^infty f(k)$.
We can calculate that
$$
sum_{k=-2}^3 f(k) approx -2.37104times10^{-7}
$$
(and presumably you are comfortable trusting this finite calculation). This proves that $S<0$, since the omitted summands $f(k)$ are negative for all $kle-3$ and all $kge4$.
Indeed, we can also calculate
$$
sum_{k=-4}^5 f(k) approx -2.647403940047578times10^{-7} tag{$*$}
$$
and, since $|f(t)|$ is increasing for $t<-2$ and decreasing for $t>2$,
$$
sum_{k=6}^infty |f(k)| < int_5^infty |f(t)|,dt = |F(5)| < 1.2times10^{-19}
$$
and
$$
sum_{k=-infty}^{-5} |f(k)| < int_{-infty}^{-4} |f(t)|,dt = |F(-4)| < 8.7times10^{-16},
$$
where
$$
F(t) = int f(t),dt = frac{1}{4} e^{-2t^2+t-frac{1}{8}} (4t-1);
$$
these calculations show that the right-hand side of $(*)$ is the correct value of the infinite sum $S$ to 7 or 8 significant digits.
Let $f(k) = sum_{k = -infty}^{infty} (1-(2k - 1 + frac12)^2) exp(-frac{1}{2} (2k - 1 + frac12)^2)$, so that you are asking about the sum $S = sum_{k=-infty}^infty f(k)$.
We can calculate that
$$
sum_{k=-2}^3 f(k) approx -2.37104times10^{-7}
$$
(and presumably you are comfortable trusting this finite calculation). This proves that $S<0$, since the omitted summands $f(k)$ are negative for all $kle-3$ and all $kge4$.
Indeed, we can also calculate
$$
sum_{k=-4}^5 f(k) approx -2.647403940047578times10^{-7} tag{$*$}
$$
and, since $|f(t)|$ is increasing for $t<-2$ and decreasing for $t>2$,
$$
sum_{k=6}^infty |f(k)| < int_5^infty |f(t)|,dt = |F(5)| < 1.2times10^{-19}
$$
and
$$
sum_{k=-infty}^{-5} |f(k)| < int_{-infty}^{-4} |f(t)|,dt = |F(-4)| < 8.7times10^{-16},
$$
where
$$
F(t) = int f(t),dt = frac{1}{4} e^{-2t^2+t-frac{1}{8}} (4t-1);
$$
these calculations show that the right-hand side of $(*)$ is the correct value of the infinite sum $S$ to 7 or 8 significant digits.
answered Nov 25 at 9:05
Greg Martin
34.8k23161
34.8k23161
add a comment |
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These are related to the Jacobi theta function. I'd try Poisson summation.
– Lord Shark the Unknown
Nov 25 at 7:12