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http://www.mapleprimes.com/tags/asymptotic
Hankel function, series, error bound
http://www.mapleprimes.com/questions/209705-Hankel-Function-Series-Error-Bound?ref=Feed:MaplePrimes:Tagged With asymptotic
<p>Dear all;</p>
<p>I open this good discussion, and hope can get a nice and strong idea in this domain of approximation of Hankel funciton and order truncation of infinite series. Thanks for all idea, can improve the discussion. </p>
<p>Using the asymptotics of Hankel function for large argument and large orders ( both together) and <br>find an order of truncation N of the obove series so that we can ensure an error bound of epsilon( epsilon very small given). abs(sum ( c[m]*HankelH1(m,x)*exp(I*m*theta), m=-infinity..infinity)-sum ( c[m]*HankelH1(m,x)*exp(I*m*theta), m=-N..N))<epsilon.</p>
<p>A first idea come in mind: the series converge, so that the general terms of this series converge to zero, and in particularity, abs(c[m]* HankelH1(m,x))<1: <br>then abs(c[m])<1/ abs(HankelH1(m,x)). <br>So we can ensure an error bound of epsilon on the coefficient c[m] by imposing (HankelH1(m,x))<epsilon this lead to abs(c[m])<epsilon. <br>I consider the case where m and x are very large, I can suppose for example m=x*(1+zeta), with 0<zeta<1. zeta parameter. So that our truncation N depend on zeta. <br>And then how can I find and approximation of the inverse of Hankel function for large argument and large order. using m=x*(1+zeta). I think this give us N the truncation order. <br>I hope get a good discussion in this subjet. <br>Of course maybe there are other strong idea to compute the truncation series. <br>I get the following error in the code:</p>
<p>Error, (in MultiSeries:-multiseries) unable to expand with respect to parameter</p>
<p> </p>
<p> </p>
<p> </p>
<p>###### Code### and error <br>restart:</p>
<p>with(MultiSeries):<br>assume(0 <= x);</p>
<p>assume(0 <= zeta<1);</p>
<p>HankelH1(v,x):</p>
<p>sum(c[m]*HankelH1(m, x)*exp(I*m*theta),m=-infinity..infinity);</p>
<p>1/HankelH1(x*(1+zeta),x);</p>
<p>MultiSeries:-asympt(%,x, 4);</p>
<p>eval(%, O=0);</p>
<p>convert(%,exp);</p>
<p>simplify(%);</p>
<p><a href="/view.aspx?sf=209705_question/truncated_series.mw">truncated_series.mw</a></p><p>Dear all;</p>
<p>I open this good discussion, and hope can get a nice and strong idea in this domain of approximation of Hankel funciton and order truncation of infinite series. Thanks for all idea, can improve the discussion. </p>
<p>Using the asymptotics of Hankel function for large argument and large orders ( both together) and <br>find an order of truncation N of the obove series so that we can ensure an error bound of epsilon( epsilon very small given). abs(sum ( c[m]*HankelH1(m,x)*exp(I*m*theta), m=-infinity..infinity)-sum ( c[m]*HankelH1(m,x)*exp(I*m*theta), m=-N..N))<epsilon.</p>
<p>A first idea come in mind: the series converge, so that the general terms of this series converge to zero, and in particularity, abs(c[m]* HankelH1(m,x))<1: <br>then abs(c[m])<1/ abs(HankelH1(m,x)). <br>So we can ensure an error bound of epsilon on the coefficient c[m] by imposing (HankelH1(m,x))<epsilon this lead to abs(c[m])<epsilon. <br>I consider the case where m and x are very large, I can suppose for example m=x*(1+zeta), with 0<zeta<1. zeta parameter. So that our truncation N depend on zeta. <br>And then how can I find and approximation of the inverse of Hankel function for large argument and large order. using m=x*(1+zeta). I think this give us N the truncation order. <br>I hope get a good discussion in this subjet. <br>Of course maybe there are other strong idea to compute the truncation series. <br>I get the following error in the code:</p>
<p>Error, (in MultiSeries:-multiseries) unable to expand with respect to parameter</p>
<p> </p>
<p> </p>
<p> </p>
<p>###### Code### and error <br>restart:</p>
<p>with(MultiSeries):<br>assume(0 <= x);</p>
<p>assume(0 <= zeta<1);</p>
<p>HankelH1(v,x):</p>
<p>sum(c[m]*HankelH1(m, x)*exp(I*m*theta),m=-infinity..infinity);</p>
<p>1/HankelH1(x*(1+zeta),x);</p>
<p>MultiSeries:-asympt(%,x, 4);</p>
<p>eval(%, O=0);</p>
<p>convert(%,exp);</p>
<p>simplify(%);</p>
<p><a href="/view.aspx?sf=209705_question/truncated_series.mw">truncated_series.mw</a></p>209705Tue, 23 Feb 2016 22:29:08 ZsarrasarraAsymptotic behaviour
http://www.mapleprimes.com/questions/204273-Asymptotic-Behaviour?ref=Feed:MaplePrimes:Tagged With asymptotic
<p>How to find asymptotic behaviour of a function.</p>
<p>For example at infinity</p>
<p>sinh(x) behaves as 1/2*exp(x)</p>
<p>1/sinh(x) behaves as 2*exp(-x)</p>
<p>exp(-x)*(exp(-x)+1) behaves as exp(-x)</p>
<p>so that it works with a more complex expression.</p><p>How to find asymptotic behaviour of a function.</p>
<p>For example at infinity</p>
<p>sinh(x) behaves as 1/2*exp(x)</p>
<p>1/sinh(x) behaves as 2*exp(-x)</p>
<p>exp(-x)*(exp(-x)+1) behaves as exp(-x)</p>
<p>so that it works with a more complex expression.</p>204273Thu, 21 May 2015 09:27:47 ZmprempreEuler-Maclaurin Summation
http://www.mapleprimes.com/posts/200229-EulerMaclaurin-Summation?ref=Feed:MaplePrimes:Tagged With asymptotic
<p>Greetings to all.</p>
<p>I would like to share a brief observation concerning my experiences with the Euler-Maclaurin summation routine in <strong>Maple 17 (X86 64 LINUX)</strong>. The following <a href="http://math.stackexchange.com/questions/740088/use-the-euler-maclaurin-summation-formula-to-estimate-a-summation">Math StackExchange Link</a> shows how to compute a certain Euler-MacLaurin type asymptotic expansion using highly unorthodox divergent series summation techniques. The result that was obtained matches the output from <strong>eulermac</strong> which is definitely good to know. What follows is the output from said routine.</p>
<pre>> eulermac(1/(1+k/n),k=0..n,18);
1 929569 3202291 691 O(1)
O(- ---) - ----------- + ----------- - --------- + 1/1048576 ----
19 15 17 11 19
n 2097152 n 1048576 n 32768 n n
n
/
174611 5461 31 | 1 17 1
- -------- + --------- + ------- + | ------- dk - ------- + ------
19 13 9 | 1 + k/n 7 5
6600 n 65536 n 4096 n / 4096 n 256 n
0
1 1
- ------ + ---- + 3/4
3 16 n
128 n
</pre>
<p>While I realize that this is good enough for most purposes I have two minor issues.</p>
<ul>
<li>One could certainly evaluate the integral without leaving it to the user to force evaluation with the <strong>AllSolutions</strong> option. One can and should make use of what is known about n and k. In particular one can check whether there are singularities on the integration path because we know the range of <em>k/n</em>.</li>
<li>Why are there two order terms for the order of the remainder term? There should be at most one and a coefficient times an <em>O(1)</em> term makes little sense as the coefficient would be absorbed.</li>
</ul>
<p>You might want to fix these so that the output looks a bit more professional which does enter into play when potential future users decide on what CAS to commit to. Other than that it is a very useful routine even for certain harmonic sum computations where one can use Euler-Maclaurin to verify results.</p>
<p>Best regards,</p>
<p>Marko Riedel</p><p>Greetings to all.</p>
<p>I would like to share a brief observation concerning my experiences with the Euler-Maclaurin summation routine in <strong>Maple 17 (X86 64 LINUX)</strong>. The following <a href="http://math.stackexchange.com/questions/740088/use-the-euler-maclaurin-summation-formula-to-estimate-a-summation">Math StackExchange Link</a> shows how to compute a certain Euler-MacLaurin type asymptotic expansion using highly unorthodox divergent series summation techniques. The result that was obtained matches the output from <strong>eulermac</strong> which is definitely good to know. What follows is the output from said routine.</p>
<pre>> eulermac(1/(1+k/n),k=0..n,18);
1 929569 3202291 691 O(1)
O(- ---) - ----------- + ----------- - --------- + 1/1048576 ----
19 15 17 11 19
n 2097152 n 1048576 n 32768 n n
n
/
174611 5461 31 | 1 17 1
- -------- + --------- + ------- + | ------- dk - ------- + ------
19 13 9 | 1 + k/n 7 5
6600 n 65536 n 4096 n / 4096 n 256 n
0
1 1
- ------ + ---- + 3/4
3 16 n
128 n
</pre>
<p>While I realize that this is good enough for most purposes I have two minor issues.</p>
<ul>
<li>One could certainly evaluate the integral without leaving it to the user to force evaluation with the <strong>AllSolutions</strong> option. One can and should make use of what is known about n and k. In particular one can check whether there are singularities on the integration path because we know the range of <em>k/n</em>.</li>
<li>Why are there two order terms for the order of the remainder term? There should be at most one and a coefficient times an <em>O(1)</em> term makes little sense as the coefficient would be absorbed.</li>
</ul>
<p>You might want to fix these so that the output looks a bit more professional which does enter into play when potential future users decide on what CAS to commit to. Other than that it is a very useful routine even for certain harmonic sum computations where one can use Euler-Maclaurin to verify results.</p>
<p>Best regards,</p>
<p>Marko Riedel</p>200229Sun, 06 Apr 2014 21:18:57 Zmriedelmriedelsimple asymptotic expansion
http://www.mapleprimes.com/questions/144127-Simple-Asymptotic-Expansion?ref=Feed:MaplePrimes:Tagged With asymptotic
<p>Dear friends,</p>
<p>I wonder how I would go about calculating the asymptotic expansion of</p>
<pre>sum(5^j/j, j=1..m+1)?</pre>
<p>The motivation for this calculation can be found <a href="http://math.stackexchange.com/questions/318247/tn-t-fracn5-frac-n-log-n-solving">here</a>. The correct answer is</p>
<pre>5/4 5^(m+1)/(m+1).</pre>
<p>The classic asympt and the one from multiseries both fail on this one.</p>
<p>Thanks,</p>
<p>Marko Riedel</p><p>Dear friends,</p>
<p>I wonder how I would go about calculating the asymptotic expansion of</p>
<pre>sum(5^j/j, j=1..m+1)?</pre>
<p>The motivation for this calculation can be found <a href="http://math.stackexchange.com/questions/318247/tn-t-fracn5-frac-n-log-n-solving">here</a>. The correct answer is</p>
<pre>5/4 5^(m+1)/(m+1).</pre>
<p>The classic asympt and the one from multiseries both fail on this one.</p>
<p>Thanks,</p>
<p>Marko Riedel</p>144127Sun, 03 Mar 2013 02:42:50 Zmriedelmriedelhow to plot Sigma(R30)?
http://www.mapleprimes.com/questions/124946-How-To-Plot-SigmaR30?ref=Feed:MaplePrimes:Tagged With asymptotic
<p>Hi,</p>
<p>I am trying to plot this function Sigma(R30) but I get failed to do so. Any one would like to try to help me out?</p>
<p>The attached maple sheet contains the asymptotic solution of the huge equation in .txt file. </p>
<p>thx.</p>
<p> <a href="/ViewTemp.ashx?f=68690_1313884198/20110819_doodles2.mws">20110819_doodles2.mws</a></p>
<p><a href="/ViewTemp.ashx?f=68690_1313884198/20110818_section4-5a.txt">20110818_section4-5a.txt</a></p><p>Hi,</p>
<p>I am trying to plot this function Sigma(R30) but I get failed to do so. Any one would like to try to help me out?</p>
<p>The attached maple sheet contains the asymptotic solution of the huge equation in .txt file. </p>
<p>thx.</p>
<p> <a href="/view.aspx?sf=124946/419514/20110819_doodles2.mws">20110819_doodles2.mws</a></p>
<p><a href="/view.aspx?sf=124946/419514/20110818_section4-5a.txt">20110818_section4-5a.txt</a></p>124946Sun, 21 Aug 2011 03:57:05 Zkh2nkh2nOnce again an Asymptotic series problem
http://www.mapleprimes.com/questions/123502-Once-Again-An-Asymptotic-Series-Problem?ref=Feed:MaplePrimes:Tagged With asymptotic
<p>I am sorry for bothering you all with the asymptotic again and again. Actually I am unable to find a magic way to evaluate an asymptotic expension. </p>
<p><a href="/ViewTemp.ashx?f=68690_1309598042/R3Infintyplots_S0=.mws">R3Infintyplots_S0=.mws</a></p><p>I am sorry for bothering you all with the asymptotic again and again. Actually I am unable to find a magic way to evaluate an asymptotic expension. </p>
<p><a href="/view.aspx?sf=123502/416499/R3Infintyplots_S0=.mws">R3Infintyplots_S0=.mws</a></p>123502Sat, 02 Jul 2011 13:16:56 Zkh2nkh2n