Hi all, I noticed that my previous post can't be fully displayed, because the equations in it are too long. I uploaded the file onto my website: http://www.**--****.com/ ~ghyan/math-model.m . If anyone has time to take a look at it, I appreciate it a lot. many thanks in advance, -Ghyan Link to the forum page for this post: http://www.**--****.com/ :Forum_ViewTopic&pid=12071#p12071 Posted through http://www.**--****.com/ [[postId=12071]]
General--FindRoot::nlnum error (Updated)
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- 1. Symbolic Calculus: How to reduce a function to two functions
I use Collect[myfunction, ξ^(-_)] and mathematica gives me a nice solution but I can not have the output like A[l,n]/ξ + B[l,n]. Can you help me please? - 2. Repost --- Another limit problem
It seems the characters were fouled up (at least when I read my own posting), so here it is again, and hopefully the characters are being converted correctly this time. I am trying to evaluate the limit of the following expression as s goes to infinity, \!\(\((1 + \[Sum]\+\(k = 0\)\%\(s - 1\)\((\((s\ \ \[Rho])\)\^k\/\(k!\))\)/\((\((s\ \[Rho])\)\^s\/\(\(s!\) \((1 - \ \[Rho])\)\))\))\)\^\(-1\)\) where, \[Rho]\ (Real) < 1, s (Integer) > 0. I am quite sure that the limit is 0; but, I am unable to get this result using Mathematica 5.2. Any suggestions would be appreciated. --V. Stokes - 3. construct?
On 8 Jul 2006, at 09:57, gardyloo wrote: > > Hi, all, > > It's probably that I'm too tired, but I have a question about using > ReplacePart in an If[] construct. I've made up a (very) minimal > example: > > In[1]:= > testList = {{3, an, example, list}, > {4, another, example, list}} > > Out[1]= > {{3, an, example, list}, {4, another, example, list}} > > In[2]:= > (If[ (ListQ[#1] && First[#1] == 4), > ReplacePart[#1, replaced!, 3]; > ReplacePart[#1, replaced!, 2], > #1 (*otherwise*) > ] & ) /@ testList > > Out[2]= > {{3, an, example, list}, {4, replaced!, example, list}} > > > Can someone tell me why BOTH the second and third positions in > the second element in testList aren't turned to "replaced!" ? I have > plenty of other ways of doing this, but this seemed the most > straightforward, and I can't wrap my head around it, for some reason. They are indeed both "replaced", but only the second one is RETURNED. Note that ReplacePart does not change its first argument: a= {4, another, example, list}; ReplacePart[a,replaced!,3] {4,another,replaced!,list} a {4,another,example,list} Andrzej Kozlowski Oxford, UK - 4. ReplacePart in an Ifconstruct?
On 7/8/06 at 4:57 AM, XXXX@XXXXX.COM (gardyloo) wrote: >It's probably that I'm too tired, but I have a question about using >ReplacePart in an If[] construct. I've made up a (very) minimal >example: >In[1]:= testList = {{3, an, example, list}, >{4, another, example, list}} >In[2]:= >(If[ (ListQ[#1] && First[#1] == 4), > ReplacePart[#1, replaced!, 3]; > ReplacePart[#1, replaced!, 2], > #1 (*otherwise*) >] & ) /@ testList >Out[2]= {{3, an, example, list}, {4, replaced!, example, list}} >Can someone tell me why BOTH the second and third positions in >the second element in testList aren't turned to "replaced!" ? You have a block expression with two operations that execute with the first element of the list is 4. The first operation replaces the third element of the list with replaced!. But since this is in a block terminated with a semicolon, you will never see this result. You will see the results of the second operation which replaces the second element with replaced. For the test list given the conditional is exactly equivalent to doing In[8]:= ReplacePart[testList,replaced!,{2,3}]; ReplacePart[testList,replaced!,{2,2}] Out[9]= {{3, an, example, list}, {4, replaced!, example, list}} That is the first replacement occurs but simply isn't displayed or retained in any way.
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General--FindRoot::nlnum error (Updated)
by ghyan » Sun, 23 Jul 2006 07:01:06 GMT
Re: General--FindRoot::nlnum error (Updated)
by Valeri Astanoff » Fri, 28 Jul 2006 18:34:29 GMT
Hi Ghyan, Why not use Min and Log instead of min and log ? V.Astanoff
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