asymptotic behaviour of a sum - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T15:58:14Z http://mathoverflow.net/feeds/question/98668 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/98668/asymptotic-behaviour-of-a-sum asymptotic behaviour of a sum xn--qwertyuiop-86a 2012-06-02T17:08:46Z 2012-06-04T05:40:47Z <p>I'd like to know the asymptotic behaviour as $N\to\infty$ of the following sum</p> <p>$$Z_N(x) := 2^{-N/2} \sum_{k=0}^{N/2} \frac{N!}{k! (N-2k)!} (N-1)^{-k} (\sqrt{2} x)^{N-2k}$$</p> <p>in order to compute $p(x):=\lim_{N\to\infty} \frac{log(Z_N(x))}{N}$ with $x\geq0$ (I already know this limit exists) .</p> <p>I found the lower bound $p(x)\geq\log(x)$, that by graphical simulations seems to be very good when $x$ is big enough.</p> <p>Can you help me to compute $p(x)$?</p> http://mathoverflow.net/questions/98668/asymptotic-behaviour-of-a-sum/98676#98676 Answer by Alex Becker for asymptotic behaviour of a sum Alex Becker 2012-06-02T19:31:50Z 2012-06-04T05:40:47Z <p>Let $T_N(x)$ denote the largest term of the sum. Note that $$\frac{\log T_N(x)}{N}\leq \frac{\log (Z_N(x))}{N}\leq \frac{\log ((N/2)T_N(x))}{N}=\frac{\log T_N(x)}{N}+\frac{\log(N/2)}{N}$$ and that the limits of both sides are identical, so by Squeeze theorem $p(x)=\lim\limits_{N\to\infty} \frac{\log(T_N(x))}{N}$. To find the largest term, we4 want to minimize the expression $k!(N-2k)!(N-1)^k(\sqrt{2}x)^{2k-N}$. Note that $$\frac{(k+1)!(N-2k-2)!(N-1)^{k+1}(\sqrt{2}x)^{2k+2-N}}{k!(N-2k)!(N-1)^k(\sqrt{2}x)^{2k-N}}=\frac{2x^2(k+1)(N-1)}{(N-2k-1)(N-2k)}$$ and so we continue to make the denominator smaller by increasing $k$ so long as $2x^2(k+1)(N-1)&lt;(N-2k-1)(N-2k)$. Solving for $k$ (and making approximations valid for large $N$) gives \begin{align}k &amp;&lt; \frac{N x^2+2N-x^2-1-\sqrt{N^2x^4+4N^2x^2-2Nx^4+2Nx^2+x^4-6x^2+1}}{4}\\ &amp;&lt; \frac{N x^2+ 2N-\sqrt{x^4+4x^2}N}{4}=\frac{2+x^2-\sqrt{x^4+4x^2}}{4}N\\ \end{align} and so for large $N$ the largest term will be $k=f(x)N$, with $f(x)=\frac{2+x^2-\sqrt{x^4+4x^2}}{4}$. Note that we can assume equality (rather than accuracy to the nearest integer) without hurting our calculations since the limit exists and for any $\epsilon$ we have arbitrarily large $N$ such that this is within $\epsilon$ of an integer. Thus $$T_N(x) = 2^{-N/2}\frac{N!}{(f(x)N)!((1-2f(x))N!)}(N-1)^{-f(x)N}(\sqrt{2}x)^{(1-2f(x))N}$$ and so taking logarithms and applying Stirling's approximation ($\log(y!)\approx y\log(y)-y$) we get \begin{align}\log(T_N(x)) &amp;= -\frac{N}{2}\log(2)+\log(N!)-\log((f(x)N)!)-\log((1-2f(x))N!)\\ &amp; \;\;\;\;-f(x)N\log(N-1)+(1-2f(x))N\log(\sqrt{2}x)\\ &amp;\approx -\frac{N}{2}\log(2)+N\log(N)-N-f(x)N\log(f(x)N)+f(x)N\\ &amp; \;\;\;\;-(1-2f(x))N\log((1-2f(x))N)+(1-2f(x))N-f(x)N\log(N-1)\\ &amp; \;\;\;\;+(1-2f(x))N\log(\sqrt{2}x)\\ \end{align} and so (note we make the approximation $\log(N-1)\approx \log(N)$) \begin{align} \frac{\log(T_N(x))}{N}&amp;\approx -\frac{\log(2)}{2}-f(x)+\log(N)-f(x)\log(f(x)N) \\ &amp; \;\;\;\;-(1-2f(x))\log((1-2f(x))N)-f(x)\log(N-1)\\ &amp; \;\;\;\;+(1-2f(x))\log(\sqrt{2}x)\\ &amp;\approx -\frac{\log(2)}{2}-f(x)+\log(N)-f(x)\log(f(x))-f(x)\log(N) \\ &amp; \;\;\;\;-(1-2f(x))\log(1-2f(x))-(1-2f(x))\log(N)-f(x)\log(N-1)\\ &amp; \;\;\;\;+(1-2f(x))\log(\sqrt{2}x)\\ &amp;\approx -\frac{\log(2)}{2}-f(x)-f(x)\log(f(x))-(1-2f(x))\log(1-2f(x))\\ &amp; \;\;\;\;+(1-2f(x))\log(\sqrt{2}x)\\ \end{align} thus \begin{align} p(x) &amp;= \lim\limits_{N\to\infty} \frac{\log(T_N(x))}{N}\\ &amp;\approx -\frac{\log(2)}{2}-f(x)(1+\log(f(x)))+(1-2f(x))(\log(\sqrt{2}x)-\log(1-2f(x)))\\ \end{align} which should be quite close to the actual value due to the accuracy of Stirling's for large values.</p>