Infinite sum of Gaussian functions convergence to a constant











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I have an application where I get following function as a result:
$$f(z;sigma) = sum_{k in mathbb{Z}} frac{1}{sqrt{2 pi} , sigma} textrm{e}^{-frac{(z - k)^2}{2 {sigma}^{2}}}$$



It appears that
$$lim_{sigma rightarrow infty} f(z;sigma) = 1$$
but I can't currently find a way to prove this.



Is this property of the sum true, and if it is, why? Any references would be greatly appreciated.



Sum of Gaussian functions










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  • I think this is an (almost) immediate consequence of Poisson Summation Formula. en.wikipedia.org/wiki/Poisson_summation_formula
    – Kavi Rama Murthy
    52 mins ago

















up vote
6
down vote

favorite












I have an application where I get following function as a result:
$$f(z;sigma) = sum_{k in mathbb{Z}} frac{1}{sqrt{2 pi} , sigma} textrm{e}^{-frac{(z - k)^2}{2 {sigma}^{2}}}$$



It appears that
$$lim_{sigma rightarrow infty} f(z;sigma) = 1$$
but I can't currently find a way to prove this.



Is this property of the sum true, and if it is, why? Any references would be greatly appreciated.



Sum of Gaussian functions










share|cite|improve this question







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Arnfinn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • I think this is an (almost) immediate consequence of Poisson Summation Formula. en.wikipedia.org/wiki/Poisson_summation_formula
    – Kavi Rama Murthy
    52 mins ago















up vote
6
down vote

favorite









up vote
6
down vote

favorite











I have an application where I get following function as a result:
$$f(z;sigma) = sum_{k in mathbb{Z}} frac{1}{sqrt{2 pi} , sigma} textrm{e}^{-frac{(z - k)^2}{2 {sigma}^{2}}}$$



It appears that
$$lim_{sigma rightarrow infty} f(z;sigma) = 1$$
but I can't currently find a way to prove this.



Is this property of the sum true, and if it is, why? Any references would be greatly appreciated.



Sum of Gaussian functions










share|cite|improve this question







New contributor




Arnfinn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I have an application where I get following function as a result:
$$f(z;sigma) = sum_{k in mathbb{Z}} frac{1}{sqrt{2 pi} , sigma} textrm{e}^{-frac{(z - k)^2}{2 {sigma}^{2}}}$$



It appears that
$$lim_{sigma rightarrow infty} f(z;sigma) = 1$$
but I can't currently find a way to prove this.



Is this property of the sum true, and if it is, why? Any references would be greatly appreciated.



Sum of Gaussian functions







sequences-and-series exponential-function






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  • I think this is an (almost) immediate consequence of Poisson Summation Formula. en.wikipedia.org/wiki/Poisson_summation_formula
    – Kavi Rama Murthy
    52 mins ago




















  • I think this is an (almost) immediate consequence of Poisson Summation Formula. en.wikipedia.org/wiki/Poisson_summation_formula
    – Kavi Rama Murthy
    52 mins ago


















I think this is an (almost) immediate consequence of Poisson Summation Formula. en.wikipedia.org/wiki/Poisson_summation_formula
– Kavi Rama Murthy
52 mins ago






I think this is an (almost) immediate consequence of Poisson Summation Formula. en.wikipedia.org/wiki/Poisson_summation_formula
– Kavi Rama Murthy
52 mins ago












3 Answers
3






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1
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As @Kavi Rama Murthy commented, this is an immediate consequence of the Poisson summation formula which is saying that
$$
sum_{kin mathbb{Z}} f(x+k) = sum_{jinmathbb{Z}} hat{f}(j)e^{2pi ijx},quad forall xin mathbb{R}
$$
for all Schwartz function $f$. Here, $hat{f}$ is the Fourier transform of $f$ on $mathbb{R}$. In this case, let $$f_sigma(x) = frac{1}{sqrt{2pi}sigma}e^{-frac{x^2}{2sigma^2}}= D^1_sigma f_1(x)$$ where $D^s_alpha g(x) = frac{1}{alpha^{frac{1}{s}}}g(frac{x}{alpha})$ is a dilation operator. Then, it holds that
$$
widehat{f_sigma}(xi) =widehat{D^1_sigma f_1}(xi) = D^infty_{1/sigma}widehat{f_1}(xi)=e^{-2pi^2sigma^2xi^2},quadforall xiinmathbb{R}.
$$
Hence the given sum is
$$
sum_{kin mathbb{Z}} f_sigma(x+k) = sum_{jinmathbb{Z}} widehat{f_sigma}(j)e^{2pi ijx}=sum_{jinmathbb{Z}} e^{-2pi^2sigma^2j^2}e^{2pi ijx} = 1+sum_{jneq 0} e^{-2pi^2sigma^2j^2}e^{2pi ijx}.
$$
For $sigma>1$, we have
$$
|e^{-2pi^2sigma^2j^2}e^{2pi ijx}|leq e^{-2pi^2j^2} in l^1(mathbb{Z}).
$$
Thus, by Lebesgue's dominated convergence theorem, as $sigma toinfty$, we get
$$
sum_{jneq 0} e^{-2pi^2sigma^2j^2}e^{2pi ijx} to 0,
$$
and as a result
$$
lim_{sigmatoinfty}sum_{kin mathbb{Z}} f_sigma(x+k) = 1,quad forall xin mathbb{R}.
$$






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    Hint. Consider $h(w)=frac{1}{sqrt{2 pi} , sigma} e^{-frac{w^2}{2 {sigma}^{2}}}$
    Then
    $$sum_{k in mathbb{Z}} frac{1}{sqrt{2 pi} , sigma} e^{-frac{(z - k)^2}{2 {sigma}^{2}}}-frac{1}{sqrt{2 pi} , sigma} e^{-frac{(z)^2}{2 {sigma}^{2}}}=sum_{k in mathbb{Z}setminus {0}} h(z-k)\leq int_{-infty}^{infty}h(w),dw=1$$
    where the sum on the left is the sum of the areas of rectangles with bases $[z-k-1,z-k]$ and $k in mathbb{Z}$ under the graph of the function $h$.
    In a similar way we have that
    $$sum_{k in mathbb{Z}} frac{1}{sqrt{2 pi} , sigma} e^{-frac{(z - k)^2}{2 {sigma}^{2}}}+frac{1}{sqrt{2 pi} , sigma} e^{-frac{(z)^2}{2 {sigma}^{2}}}=2h(0)+sum_{k in mathbb{Z}setminus {0}} h(z-k)\geq int_{-infty}^{infty}h(w),dw=1$$
    where this time the union of the rectangles contains the area under the graph of $h$.






    share|cite|improve this answer






























      up vote
      0
      down vote













      While other people gave you mathematically rigorous solution, here is a more intuitive one:



      Let's go in the other limit, $sigmato 0$. Then what you have is a sum of delta functions at each integer. Let's calculate the area between half integers: $$int_{-0.5}^{0.5}f(z,0)dz=1$$
      When you increase $sigma$, similar to melting peaks, some of the area will "flow out" from the central delta function into the adjacent intervals. But an equal area will "flow in". The area in each interval is conserved. So in the limit $sigmatoinfty$ each Gaussian becomes flat, so the sum is a constant. But in each interval $$int_{-0.5}^{0.5}f(z,infty)dz=int_{-0.5}^{0.5}Cdz=C=1$$






      share|cite|improve this answer





















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        3 Answers
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        3 Answers
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        As @Kavi Rama Murthy commented, this is an immediate consequence of the Poisson summation formula which is saying that
        $$
        sum_{kin mathbb{Z}} f(x+k) = sum_{jinmathbb{Z}} hat{f}(j)e^{2pi ijx},quad forall xin mathbb{R}
        $$
        for all Schwartz function $f$. Here, $hat{f}$ is the Fourier transform of $f$ on $mathbb{R}$. In this case, let $$f_sigma(x) = frac{1}{sqrt{2pi}sigma}e^{-frac{x^2}{2sigma^2}}= D^1_sigma f_1(x)$$ where $D^s_alpha g(x) = frac{1}{alpha^{frac{1}{s}}}g(frac{x}{alpha})$ is a dilation operator. Then, it holds that
        $$
        widehat{f_sigma}(xi) =widehat{D^1_sigma f_1}(xi) = D^infty_{1/sigma}widehat{f_1}(xi)=e^{-2pi^2sigma^2xi^2},quadforall xiinmathbb{R}.
        $$
        Hence the given sum is
        $$
        sum_{kin mathbb{Z}} f_sigma(x+k) = sum_{jinmathbb{Z}} widehat{f_sigma}(j)e^{2pi ijx}=sum_{jinmathbb{Z}} e^{-2pi^2sigma^2j^2}e^{2pi ijx} = 1+sum_{jneq 0} e^{-2pi^2sigma^2j^2}e^{2pi ijx}.
        $$
        For $sigma>1$, we have
        $$
        |e^{-2pi^2sigma^2j^2}e^{2pi ijx}|leq e^{-2pi^2j^2} in l^1(mathbb{Z}).
        $$
        Thus, by Lebesgue's dominated convergence theorem, as $sigma toinfty$, we get
        $$
        sum_{jneq 0} e^{-2pi^2sigma^2j^2}e^{2pi ijx} to 0,
        $$
        and as a result
        $$
        lim_{sigmatoinfty}sum_{kin mathbb{Z}} f_sigma(x+k) = 1,quad forall xin mathbb{R}.
        $$






        share|cite|improve this answer

























          up vote
          1
          down vote













          As @Kavi Rama Murthy commented, this is an immediate consequence of the Poisson summation formula which is saying that
          $$
          sum_{kin mathbb{Z}} f(x+k) = sum_{jinmathbb{Z}} hat{f}(j)e^{2pi ijx},quad forall xin mathbb{R}
          $$
          for all Schwartz function $f$. Here, $hat{f}$ is the Fourier transform of $f$ on $mathbb{R}$. In this case, let $$f_sigma(x) = frac{1}{sqrt{2pi}sigma}e^{-frac{x^2}{2sigma^2}}= D^1_sigma f_1(x)$$ where $D^s_alpha g(x) = frac{1}{alpha^{frac{1}{s}}}g(frac{x}{alpha})$ is a dilation operator. Then, it holds that
          $$
          widehat{f_sigma}(xi) =widehat{D^1_sigma f_1}(xi) = D^infty_{1/sigma}widehat{f_1}(xi)=e^{-2pi^2sigma^2xi^2},quadforall xiinmathbb{R}.
          $$
          Hence the given sum is
          $$
          sum_{kin mathbb{Z}} f_sigma(x+k) = sum_{jinmathbb{Z}} widehat{f_sigma}(j)e^{2pi ijx}=sum_{jinmathbb{Z}} e^{-2pi^2sigma^2j^2}e^{2pi ijx} = 1+sum_{jneq 0} e^{-2pi^2sigma^2j^2}e^{2pi ijx}.
          $$
          For $sigma>1$, we have
          $$
          |e^{-2pi^2sigma^2j^2}e^{2pi ijx}|leq e^{-2pi^2j^2} in l^1(mathbb{Z}).
          $$
          Thus, by Lebesgue's dominated convergence theorem, as $sigma toinfty$, we get
          $$
          sum_{jneq 0} e^{-2pi^2sigma^2j^2}e^{2pi ijx} to 0,
          $$
          and as a result
          $$
          lim_{sigmatoinfty}sum_{kin mathbb{Z}} f_sigma(x+k) = 1,quad forall xin mathbb{R}.
          $$






          share|cite|improve this answer























            up vote
            1
            down vote










            up vote
            1
            down vote









            As @Kavi Rama Murthy commented, this is an immediate consequence of the Poisson summation formula which is saying that
            $$
            sum_{kin mathbb{Z}} f(x+k) = sum_{jinmathbb{Z}} hat{f}(j)e^{2pi ijx},quad forall xin mathbb{R}
            $$
            for all Schwartz function $f$. Here, $hat{f}$ is the Fourier transform of $f$ on $mathbb{R}$. In this case, let $$f_sigma(x) = frac{1}{sqrt{2pi}sigma}e^{-frac{x^2}{2sigma^2}}= D^1_sigma f_1(x)$$ where $D^s_alpha g(x) = frac{1}{alpha^{frac{1}{s}}}g(frac{x}{alpha})$ is a dilation operator. Then, it holds that
            $$
            widehat{f_sigma}(xi) =widehat{D^1_sigma f_1}(xi) = D^infty_{1/sigma}widehat{f_1}(xi)=e^{-2pi^2sigma^2xi^2},quadforall xiinmathbb{R}.
            $$
            Hence the given sum is
            $$
            sum_{kin mathbb{Z}} f_sigma(x+k) = sum_{jinmathbb{Z}} widehat{f_sigma}(j)e^{2pi ijx}=sum_{jinmathbb{Z}} e^{-2pi^2sigma^2j^2}e^{2pi ijx} = 1+sum_{jneq 0} e^{-2pi^2sigma^2j^2}e^{2pi ijx}.
            $$
            For $sigma>1$, we have
            $$
            |e^{-2pi^2sigma^2j^2}e^{2pi ijx}|leq e^{-2pi^2j^2} in l^1(mathbb{Z}).
            $$
            Thus, by Lebesgue's dominated convergence theorem, as $sigma toinfty$, we get
            $$
            sum_{jneq 0} e^{-2pi^2sigma^2j^2}e^{2pi ijx} to 0,
            $$
            and as a result
            $$
            lim_{sigmatoinfty}sum_{kin mathbb{Z}} f_sigma(x+k) = 1,quad forall xin mathbb{R}.
            $$






            share|cite|improve this answer












            As @Kavi Rama Murthy commented, this is an immediate consequence of the Poisson summation formula which is saying that
            $$
            sum_{kin mathbb{Z}} f(x+k) = sum_{jinmathbb{Z}} hat{f}(j)e^{2pi ijx},quad forall xin mathbb{R}
            $$
            for all Schwartz function $f$. Here, $hat{f}$ is the Fourier transform of $f$ on $mathbb{R}$. In this case, let $$f_sigma(x) = frac{1}{sqrt{2pi}sigma}e^{-frac{x^2}{2sigma^2}}= D^1_sigma f_1(x)$$ where $D^s_alpha g(x) = frac{1}{alpha^{frac{1}{s}}}g(frac{x}{alpha})$ is a dilation operator. Then, it holds that
            $$
            widehat{f_sigma}(xi) =widehat{D^1_sigma f_1}(xi) = D^infty_{1/sigma}widehat{f_1}(xi)=e^{-2pi^2sigma^2xi^2},quadforall xiinmathbb{R}.
            $$
            Hence the given sum is
            $$
            sum_{kin mathbb{Z}} f_sigma(x+k) = sum_{jinmathbb{Z}} widehat{f_sigma}(j)e^{2pi ijx}=sum_{jinmathbb{Z}} e^{-2pi^2sigma^2j^2}e^{2pi ijx} = 1+sum_{jneq 0} e^{-2pi^2sigma^2j^2}e^{2pi ijx}.
            $$
            For $sigma>1$, we have
            $$
            |e^{-2pi^2sigma^2j^2}e^{2pi ijx}|leq e^{-2pi^2j^2} in l^1(mathbb{Z}).
            $$
            Thus, by Lebesgue's dominated convergence theorem, as $sigma toinfty$, we get
            $$
            sum_{jneq 0} e^{-2pi^2sigma^2j^2}e^{2pi ijx} to 0,
            $$
            and as a result
            $$
            lim_{sigmatoinfty}sum_{kin mathbb{Z}} f_sigma(x+k) = 1,quad forall xin mathbb{R}.
            $$







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            answered 31 mins ago









            Song

            2,215112




            2,215112






















                up vote
                1
                down vote













                Hint. Consider $h(w)=frac{1}{sqrt{2 pi} , sigma} e^{-frac{w^2}{2 {sigma}^{2}}}$
                Then
                $$sum_{k in mathbb{Z}} frac{1}{sqrt{2 pi} , sigma} e^{-frac{(z - k)^2}{2 {sigma}^{2}}}-frac{1}{sqrt{2 pi} , sigma} e^{-frac{(z)^2}{2 {sigma}^{2}}}=sum_{k in mathbb{Z}setminus {0}} h(z-k)\leq int_{-infty}^{infty}h(w),dw=1$$
                where the sum on the left is the sum of the areas of rectangles with bases $[z-k-1,z-k]$ and $k in mathbb{Z}$ under the graph of the function $h$.
                In a similar way we have that
                $$sum_{k in mathbb{Z}} frac{1}{sqrt{2 pi} , sigma} e^{-frac{(z - k)^2}{2 {sigma}^{2}}}+frac{1}{sqrt{2 pi} , sigma} e^{-frac{(z)^2}{2 {sigma}^{2}}}=2h(0)+sum_{k in mathbb{Z}setminus {0}} h(z-k)\geq int_{-infty}^{infty}h(w),dw=1$$
                where this time the union of the rectangles contains the area under the graph of $h$.






                share|cite|improve this answer



























                  up vote
                  1
                  down vote













                  Hint. Consider $h(w)=frac{1}{sqrt{2 pi} , sigma} e^{-frac{w^2}{2 {sigma}^{2}}}$
                  Then
                  $$sum_{k in mathbb{Z}} frac{1}{sqrt{2 pi} , sigma} e^{-frac{(z - k)^2}{2 {sigma}^{2}}}-frac{1}{sqrt{2 pi} , sigma} e^{-frac{(z)^2}{2 {sigma}^{2}}}=sum_{k in mathbb{Z}setminus {0}} h(z-k)\leq int_{-infty}^{infty}h(w),dw=1$$
                  where the sum on the left is the sum of the areas of rectangles with bases $[z-k-1,z-k]$ and $k in mathbb{Z}$ under the graph of the function $h$.
                  In a similar way we have that
                  $$sum_{k in mathbb{Z}} frac{1}{sqrt{2 pi} , sigma} e^{-frac{(z - k)^2}{2 {sigma}^{2}}}+frac{1}{sqrt{2 pi} , sigma} e^{-frac{(z)^2}{2 {sigma}^{2}}}=2h(0)+sum_{k in mathbb{Z}setminus {0}} h(z-k)\geq int_{-infty}^{infty}h(w),dw=1$$
                  where this time the union of the rectangles contains the area under the graph of $h$.






                  share|cite|improve this answer

























                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    Hint. Consider $h(w)=frac{1}{sqrt{2 pi} , sigma} e^{-frac{w^2}{2 {sigma}^{2}}}$
                    Then
                    $$sum_{k in mathbb{Z}} frac{1}{sqrt{2 pi} , sigma} e^{-frac{(z - k)^2}{2 {sigma}^{2}}}-frac{1}{sqrt{2 pi} , sigma} e^{-frac{(z)^2}{2 {sigma}^{2}}}=sum_{k in mathbb{Z}setminus {0}} h(z-k)\leq int_{-infty}^{infty}h(w),dw=1$$
                    where the sum on the left is the sum of the areas of rectangles with bases $[z-k-1,z-k]$ and $k in mathbb{Z}$ under the graph of the function $h$.
                    In a similar way we have that
                    $$sum_{k in mathbb{Z}} frac{1}{sqrt{2 pi} , sigma} e^{-frac{(z - k)^2}{2 {sigma}^{2}}}+frac{1}{sqrt{2 pi} , sigma} e^{-frac{(z)^2}{2 {sigma}^{2}}}=2h(0)+sum_{k in mathbb{Z}setminus {0}} h(z-k)\geq int_{-infty}^{infty}h(w),dw=1$$
                    where this time the union of the rectangles contains the area under the graph of $h$.






                    share|cite|improve this answer














                    Hint. Consider $h(w)=frac{1}{sqrt{2 pi} , sigma} e^{-frac{w^2}{2 {sigma}^{2}}}$
                    Then
                    $$sum_{k in mathbb{Z}} frac{1}{sqrt{2 pi} , sigma} e^{-frac{(z - k)^2}{2 {sigma}^{2}}}-frac{1}{sqrt{2 pi} , sigma} e^{-frac{(z)^2}{2 {sigma}^{2}}}=sum_{k in mathbb{Z}setminus {0}} h(z-k)\leq int_{-infty}^{infty}h(w),dw=1$$
                    where the sum on the left is the sum of the areas of rectangles with bases $[z-k-1,z-k]$ and $k in mathbb{Z}$ under the graph of the function $h$.
                    In a similar way we have that
                    $$sum_{k in mathbb{Z}} frac{1}{sqrt{2 pi} , sigma} e^{-frac{(z - k)^2}{2 {sigma}^{2}}}+frac{1}{sqrt{2 pi} , sigma} e^{-frac{(z)^2}{2 {sigma}^{2}}}=2h(0)+sum_{k in mathbb{Z}setminus {0}} h(z-k)\geq int_{-infty}^{infty}h(w),dw=1$$
                    where this time the union of the rectangles contains the area under the graph of $h$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 10 mins ago

























                    answered 30 mins ago









                    Robert Z

                    92k1058129




                    92k1058129






















                        up vote
                        0
                        down vote













                        While other people gave you mathematically rigorous solution, here is a more intuitive one:



                        Let's go in the other limit, $sigmato 0$. Then what you have is a sum of delta functions at each integer. Let's calculate the area between half integers: $$int_{-0.5}^{0.5}f(z,0)dz=1$$
                        When you increase $sigma$, similar to melting peaks, some of the area will "flow out" from the central delta function into the adjacent intervals. But an equal area will "flow in". The area in each interval is conserved. So in the limit $sigmatoinfty$ each Gaussian becomes flat, so the sum is a constant. But in each interval $$int_{-0.5}^{0.5}f(z,infty)dz=int_{-0.5}^{0.5}Cdz=C=1$$






                        share|cite|improve this answer

























                          up vote
                          0
                          down vote













                          While other people gave you mathematically rigorous solution, here is a more intuitive one:



                          Let's go in the other limit, $sigmato 0$. Then what you have is a sum of delta functions at each integer. Let's calculate the area between half integers: $$int_{-0.5}^{0.5}f(z,0)dz=1$$
                          When you increase $sigma$, similar to melting peaks, some of the area will "flow out" from the central delta function into the adjacent intervals. But an equal area will "flow in". The area in each interval is conserved. So in the limit $sigmatoinfty$ each Gaussian becomes flat, so the sum is a constant. But in each interval $$int_{-0.5}^{0.5}f(z,infty)dz=int_{-0.5}^{0.5}Cdz=C=1$$






                          share|cite|improve this answer























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            While other people gave you mathematically rigorous solution, here is a more intuitive one:



                            Let's go in the other limit, $sigmato 0$. Then what you have is a sum of delta functions at each integer. Let's calculate the area between half integers: $$int_{-0.5}^{0.5}f(z,0)dz=1$$
                            When you increase $sigma$, similar to melting peaks, some of the area will "flow out" from the central delta function into the adjacent intervals. But an equal area will "flow in". The area in each interval is conserved. So in the limit $sigmatoinfty$ each Gaussian becomes flat, so the sum is a constant. But in each interval $$int_{-0.5}^{0.5}f(z,infty)dz=int_{-0.5}^{0.5}Cdz=C=1$$






                            share|cite|improve this answer












                            While other people gave you mathematically rigorous solution, here is a more intuitive one:



                            Let's go in the other limit, $sigmato 0$. Then what you have is a sum of delta functions at each integer. Let's calculate the area between half integers: $$int_{-0.5}^{0.5}f(z,0)dz=1$$
                            When you increase $sigma$, similar to melting peaks, some of the area will "flow out" from the central delta function into the adjacent intervals. But an equal area will "flow in". The area in each interval is conserved. So in the limit $sigmatoinfty$ each Gaussian becomes flat, so the sum is a constant. But in each interval $$int_{-0.5}^{0.5}f(z,infty)dz=int_{-0.5}^{0.5}Cdz=C=1$$







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                            answered 14 mins ago









                            Andrei

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