{"id":9506,"date":"2017-06-30T11:12:27","date_gmt":"2017-06-30T09:12:27","guid":{"rendered":"http:\/\/djalil.chafai.net\/blog\/?p=9506"},"modified":"2017-07-20T17:55:47","modified_gmt":"2017-07-20T15:55:47","slug":"about-the-exponential-series","status":"publish","type":"post","link":"https:\/\/djalil.chafai.net\/blog\/2017\/06\/30\/about-the-exponential-series\/","title":{"rendered":"About the exponential series"},"content":{"rendered":"
\"Leonhard<\/a>
Leonhard Euler, the great master<\/figcaption><\/figure>\n

This post is about some aspects of the exponential series<\/p>\n

\\[ \\mathrm{e}_N(z):=\\sum_{\\ell=0}^{N-1}\\frac{z^\\ell}{\\ell!}. \\]<\/p>\n

A bit of probabilistic intuition suggests that when \\( {z=r>0} \\), the behavior of \\( {\\mathrm{e}^{-rN}\\mathrm{e}_N(Nr)} \\) as \\( {N\\rightarrow\\infty} \\) has a critical value \\( {r=1} \\). Namely, if \\( {X_1,\\ldots,X_N} \\) are iid random variables following the Poisson distribution of mean \\( {r} \\), then<\/p>\n

\\[ \\mathrm{e}^{-rN}\\mathrm{e}_N(Nr) =\\mathbb{P}(X_1+\\cdots+X_N<N) =\\mathbb{P}\\left(\\frac{X_1+\\cdots+X_N}{N}<1\\right). \\]<\/p>\n

Now as \\( {N\\rightarrow\\infty} \\), the right hand side tends to \\( {0} \\) if \\( {r>1} \\) and to \\( {1} \\) if \\( {r<1} \\), since \\( {\\lim_{N\\rightarrow\\infty}\\frac{X_1+\\cdots+X_N}{N}=r} \\) almost surely by the law of large numbers. We have shown that<\/p>\n

\\[ \\lim_{N\\rightarrow\\infty}\\mathrm{e}^{-rN}\\mathrm{e}_N(Nr)=\\mathbf{1}_{r\\leq 1} \\]<\/p>\n

for any \\( {r\\neq 1} \\), but the convergence to this indicator does not hold for \\( {r=1} \\) since by the central limit theorem, in this case,<\/p>\n

\\[ \\mathbb{P}(X_1+\\cdots+X_N<N) =\\mathbb{P}\\left(\\frac{X_1+\\cdots+X_N-N}{\\sqrt{N}}<0\\right) \\rightarrow\\frac{1}{2}. \\]<\/p>\n

The same argument works with the Gamma distribution: namely, for any \\( {r>0} \\),<\/p>\n

\\[ \\frac{1}{(N-1)!}\\int_r^\\infty t^{N-1}\\mathrm{e}^{-t}\\,\\mathrm{d}t =\\mathrm{e}^{-r}\\sum_{\\ell=0}^{N-1}\\frac{r^\\ell}{\\ell!}, \\]<\/p>\n

in other words, if \\( {Y_1,\\ldots,Y_N} \\) are iid random variables with law \\( {\\mathrm{Gamma}(1,1)} \\), then<\/p>\n

\\[ \\mathbb{P}(Y_1+\\cdots+Y_N>r)=\\mathrm{e}^{-r}\\mathrm{e}_N(r). \\]<\/p>\n

Bayesian statisticians are quite familiar with these Gamma-Poisson games.<\/p>\n

Error.<\/b> For every \\( {N\\geq1} \\) and \\( {z\\in\\mathbb{C}} \\),<\/p>\n

\\[ |\\mathrm{e}_N(Nz)-\\mathrm{e}^{Nz}\\mathbf{1}_{|z|\\leq1}|\\leq r_N(z) \\]<\/p>\n

where<\/p>\n

\\[ r_N(z):= \\frac{\\mathrm{e}^N}{\\sqrt{2\\pi N}}|z|^N\\left(\\frac{N+1}{N(1-|z|)+1} \\mathbf{1}_{|z|\\leq1} +\\frac{N}{N(|z|-1)+1}\\mathbf{1}_{|z|>1}\\right). \\]<\/p>\n

In particular, if we define<\/p>\n

\\[ \\varphi^{1,N}(z):=\\frac{\\mathrm{e}^{-N|z|^2}}{\\pi}\\mathrm{e}_N(N|z|^2) \\]<\/p>\n

then, for any compact subset \\( {K\\subset\\mathbb{C}\\setminus\\{z\\in\\mathbb{C}:|z|=1\\}} \\),<\/p>\n

\\[ \\lim_{N\\rightarrow\\infty}\\sup_{z\\in K} \\left|\\varphi^{1,N}(z)-\\frac{\\mathbf{1}_{|z|\\leq1}}{\\pi}\\right| = \\pi^{-1}\\lim_{N\\rightarrow\\infty}\\sup_{z\\in K} \\left|\\mathrm{e}^{-N|z|^2}\\mathrm{e}_N(N|z|^2)-\\mathbf{1}_{|z|\\leq1}\\right| =0. \\]<\/p>\n

Proof.<\/b> Following Mehta, for every \\( {N\\geq1} \\), \\( {z\\in\\mathbb{C}} \\), if \\( {|z|\\leq N} \\) then<\/p>\n

\\[ \\left|\\mathrm{e}^{z}-\\mathrm{e}_N(z)\\right| =\\left|\\sum_{\\ell=N}^\\infty\\frac{z^\\ell}{\\ell!}\\right| \\leq\\frac{|z|^N}{N!}\\sum_{\\ell=0}^\\infty\\frac{|z|^\\ell}{(N+1)^\\ell} =\\frac{|z|^N}{N!}\\frac{N+1}{N+1-|z|}, \\]<\/p>\n

while if \\( {|z|>N} \\) then<\/p>\n

\\[ |\\mathrm{e}_N(z)| \\leq \\sum_{\\ell=0}^{N-1}\\frac{|z|^\\ell}{\\ell!} \\leq \\frac{|z|^{N-1}}{(N-1)!}\\sum_{\\ell=0}^{N-1}\\frac{(N-1)^{\\ell}}{|z|^\\ell} \\leq\\frac{|z|^{N-1}}{(N-1)!}\\frac{|z|}{|z|-N+1}. \\]<\/p>\n

Therefore, for every \\( {N\\geq1} \\) and \\( {z\\in\\mathbb{C}} \\),<\/p>\n

\\[ |\\mathrm{e}_N(Nz)-\\mathrm{e}^{Nz}\\mathbf{1}_{|z|\\leq1}| \\leq \\frac{N^N}{N!}\\left(|z|^N\\frac{N+1}{N+1-|Nz|}\\mathbf{1}_{|z|\\leq1} +|z|^{N-1}\\frac{|Nz|}{|Nz|-N+1}\\mathbf{1}_{|z|>1}\\right). \\]<\/p>\n

It remains to use the Stirling bound<\/p>\n

\\[ \\sqrt{2\\pi N}N^N\\leq N!\\mathrm{e}^N. \\]<\/p>\n

Ginibre.<\/b> Let us give an application to random matrices. Let \\( {M} \\) be a random \\( {N\\times N} \\) complex matrix with independent and identically distributed Gaussian entries on \\( {\\mathbb{C}} \\) of mean \\( {0} \\) and variance \\( {1\/N} \\) with density \\( {z\\in\\mathbb{C}\\mapsto\\pi^{-1}N\\exp(-N|z|^2)} \\). The variance scaling is chosen so that by the law of large numbers, asymptotically as \\( {N\\rightarrow\\infty} \\), the rows and the columns of \\( {M} \\) are stabilized: they have unit norm and are orthogonal in \\( {\\mathbb{C}^N} \\). The density of the random matrix \\( {M} \\) is proportional to<\/p>\n

\\[ M\\mapsto \\prod_{1\\leq j,k\\leq N}\\exp\\Bigr(-N|M_{jk}|^2\\Bigr) =\\exp\\left(-N\\mathrm{Tr}(MM^*)\\right). \\]<\/p>\n

The spectral change of variables \\( {M=U(D+N)U^*} \\), which is the Schur unitary decomposition, gives that the joint law of the eigenvalues of \\( {M} \\) has density<\/p>\n

\\[ \\varphi^{N,N}(z_1,\\ldots,z_n) :=\\frac{N^{\\frac{N(N+1)}{2}}}{1!2!\\cdots N!} \\frac{\\mathrm{e}^{-\\sum_{i=1}^N N|z_i|^2}}{\\pi^N} \\prod_{i<j}|z_i-z_j|^2 \\]<\/p>\n

with respect to the Lebesgue measure on \\( {\\mathbb{C}^N} \\). This law is usually referred to as the ``complex Ginibre Ensemble''. It is a well known fact that for every \\( {1\\leq k\\leq N} \\), the \\( {k} \\)-th dimensional marginal distribution has density <\/a><\/p>\n

\\[ \\begin{array}{rcl} \\varphi^{k,N}(z_1,\\ldots,z_k) &=&\\int_{\\mathbb{C}^{N-k}}\\!\\varphi^{N,N}(z_1,\\ldots,z_N)\\,\\mathrm{d} z_{k+1}\\cdots \\mathrm{d} z_N \\nonumber\\\\ &=&\\frac{(N-k)!}{N!}\\frac{\\mathrm{e}^{-N(|z_1|^2+\\cdots+|z_k|^2)}}{\\pi^kN^{-k}} \\det\\left[(\\mathrm{e}_N(Nz_i\\overline{z}_j))_{1\\leq i,j\\leq k}\\right], \\end{array} \\]<\/p>\n

where \\( {\\mathrm{e}_N(w):=\\sum_{\\ell=0}^{N-1}w^\\ell\/\\ell!} \\) is the truncated exponential series.<\/p>\n

We have seen above that uniformly on compact subsets of \\( {\\{z\\in\\mathbb{C}:|z|\\neq1\\}} \\),<\/p>\n

\\[ \\lim_{N\\rightarrow\\infty}\\varphi^{1,N}(z)=\\frac{\\mathbf{1}_{|z|\\leq1}}{\\pi}. \\]<\/p>\n

Let us show now that<\/p>\n

\\[ \\lim_{N\\rightarrow\\infty}(\\varphi^{2,N}-(\\varphi^{1,N})^{\\otimes 2})=0 \\]<\/p>\n

uniformly on compact subsets of \\( {\\{(z_1,z_2)\\in\\mathbb{C}^2:|z_1|\\neq1,|z_2|\\neq1,z_1\\neq z_2\\}} \\).<\/p>\n

Note that this cannot hold on \\( {\\{(z,z):z\\in\\mathbb{C}, |z|\\leq1\\}} \\) since \\( {\\varphi^{2,N}(z,z)=0} \\) for any \\( {N\\geq2} \\) and \\( {z\\in\\mathbb{C}} \\) while \\( {\\lim_{N\\rightarrow\\infty}\\varphi^{1,N}(z)\\varphi^{1,N}(z)=1\/\\pi^2\\neq0} \\) when \\( {|z|\\leq1} \\), and this phenomenon is due to the singularity of the interaction.<\/p>\n

From the determinantal formula above \\( {\\varphi^{1,N}} \\) and \\( {\\varphi^{2,N}} \\) are given by<\/p>\n

\\[ \\varphi^{1,N}(z)= \\frac{\\mathrm{e}^{-N|z|^2}}{\\pi}\\mathrm{e}_N(N|z|^2),\\quad z\\in\\mathbb{C}, \\]<\/p>\n

where \\( {\\mathrm{e}_N(w):=\\sum_{\\ell=0}^{N-1}w^\\ell\/\\ell!} \\) as usual, and, for every \\( {z_1,z_2\\in\\mathbb{C}} \\),<\/p>\n

\\[ \\begin{array}{rcl} \\varphi^{2,N}(z_1,z_2) &=& \\frac{N}{N-1} \\frac{\\mathrm{e}^{-N(|z_1|^2+|z_2|^2)}}{\\pi^2} \\big( \\mathrm{e}_N(N|z_1|^2)\\mathrm{e}_N(N|z_2|^2)-|\\mathrm{e}_N(Nz_1\\overline{z}_2)|^2 \\big)\\\\ &=&\\frac{N}{N-1}\\varphi^{1,N}(z_1)\\varphi^{1,N}(z_2) - \\frac{N}{N-1} \\frac{\\mathrm{e}^{-N(|z_1|^2+|z_2|^2)}}{\\pi^2} |\\mathrm{e}_N(Nz_1\\overline{z}_2)|^2. \\end{array} \\]<\/p>\n

It follows that for any \\( {N\\geq2} \\) and \\( {z_1,z_2\\in\\mathbb{C}} \\),<\/p>\n

\\[ \\begin{array}{rcl} \\Delta_N(z_1,z_2) &=& \\varphi^{2,N}(z_1,z_2)-\\varphi^{1,N}(z_1)\\varphi^{1,N}(z_2)\\\\ &=& \\frac{1}{N-1}\\varphi^{1,N}(z_1)\\varphi^{1,N}(z_2) -\\frac{N}{N-1}\\frac{\\mathrm{e}^{-N(|z_1|^2+|z_2|^2)}}{\\pi^2}|\\mathrm{e}_N(Nz_1\\overline{z}_2)|^2. \\end{array} \\]<\/p>\n

In particular, using \\( {\\varphi^{2,N}\\geq0} \\) for the lower bound,<\/p>\n

\\[ -\\varphi^{1,N}(z_1)\\varphi^{1,N}(z_2) \\leq \\Delta_N(z_1,z_2) \\leq \\frac{1}{N-1}\\varphi^{1,N}(z_1)\\varphi^{1,N}(z_2). \\]<\/p>\n

From this and the error control for the exponential series above, we first deduce that for any compact subset \\( {K} \\) of \\( {\\{z\\in\\mathbb{C}:|z|>1\\}} \\)<\/p>\n

\\[ \\lim_{N\\rightarrow\\infty} \\sup_{\\substack{z_1\\in\\mathbb{C}\\\\z_2\\in K}}|\\Delta_N(z_1,z_2)| = \\lim_{N\\rightarrow\\infty} \\sup_{\\substack{z_1\\in K\\\\z_2\\in\\mathbb{C}}}|\\Delta_N(z_1,z_2)| =0. \\]<\/p>\n

It would remain to show that \\( {\\lim_{N\\rightarrow\\infty}\\Delta_N(z_1,z_2)=0} \\) when in the same time \\( {|z_1|\\leq1} \\) and \\( {|z_2|\\leq1} \\). In this case \\( {|z_1\\overline{z}_2|\\leq1} \\), and<\/p>\n

\\[ |\\mathrm{e}_N(Nz_1\\overline{z}_2)|^2 \\leq 2\\mathrm{e}^{2N\\Re(z_1\\overline{z}_2)} + 2r_N^2(z_1\\overline{z}_2). \\]<\/p>\n

Next, using the elementary identity<\/p>\n

\\[ 2\\Re(z_1\\overline{z}_2)=|z_1|^2+|z_2|^2-|z_1-z_2|^2, \\]<\/p>\n

we get<\/p>\n

\\[ \\mathrm{e}^{-N(|z_1|^2+|z_2|^2)}| \\mathrm{e}_N(Nz_1\\overline{z}_2)|^2 \\leq 2\\mathrm{e}^{-N|z_1-z_2|^2} +2\\mathrm{e}^{-N(|z_1|^2+|z_2|^2)}r_N^2(z_1\\overline{z}_2). \\]<\/p>\n

Since \\( {|z_1\\overline{z}_2|\\leq1} \\), the formula for \\( {r_N} \\) gives<\/p>\n

\\[ \\mathrm{e}^{-N(|z_1|^2+|z_2|^2)}r^2_N(z_1\\overline{z}_2) \\leq \\mathrm{e}^{-N(|z_1|^2+|z_2|^2-2-\\log|z_1|^2-\\log|z_2|^2)} \\frac{(N+1)^2}{2\\pi N}. \\]<\/p>\n

Therefore, using the bounds \\( {\\varphi^{1,N} \\leq 1\/\\pi} \\) and \\( {u-1 - \\log u >0} \\) for \\( {0<u<1} \\), it follows that \\( {\\lim_{N\\rightarrow\\infty}\\Delta_N(z_1, z_2)=0} \\) uniformly in \\( {z_1,z_2} \\) on compact subsets of<\/p>\n

\\[ \\{(z_1,z_2)\\in\\mathbb{C}^2:|z_1|<1,|z_2|<1,z_1\\neq z_2\\}. \\]<\/p>\n

Further reading.<\/b><\/p>\n