$sum exp(-x^2)$ vs $sum x^2 exp(-x^2)$












0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    These are related to the Jacobi theta function. I'd try Poisson summation.
    $endgroup$
    – Lord Shark the Unknown
    Nov 25 '18 at 7:12
















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    These are related to the Jacobi theta function. I'd try Poisson summation.
    $endgroup$
    – Lord Shark the Unknown
    Nov 25 '18 at 7:12














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 '18 at 8:25









Mason

1,9551530




1,9551530










asked Nov 25 '18 at 7:05









J RicheyJ Richey

351318




351318












  • $begingroup$
    These are related to the Jacobi theta function. I'd try Poisson summation.
    $endgroup$
    – Lord Shark the Unknown
    Nov 25 '18 at 7:12


















  • $begingroup$
    These are related to the Jacobi theta function. I'd try Poisson summation.
    $endgroup$
    – Lord Shark the Unknown
    Nov 25 '18 at 7:12
















$begingroup$
These are related to the Jacobi theta function. I'd try Poisson summation.
$endgroup$
– Lord Shark the Unknown
Nov 25 '18 at 7:12




$begingroup$
These are related to the Jacobi theta function. I'd try Poisson summation.
$endgroup$
– Lord Shark the Unknown
Nov 25 '18 at 7:12










1 Answer
1






active

oldest

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2












$begingroup$

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.






share|cite|improve this answer









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    1 Answer
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    1 Answer
    1






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    active

    oldest

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    2












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 25 '18 at 9:05









        Greg MartinGreg Martin

        34.9k23161




        34.9k23161






























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