Re: Replacements and NIntegrate
- To: mathgroup at smc.vnet.net
- Subject: [mg118588] Re: Replacements and NIntegrate
- From: Andrew Moylan <amoylan at wolfram.com>
- Date: Tue, 3 May 2011 08:22:36 -0400 (EDT)
There are different ways to do it. As an example: In[1]:= h[z_] := a + b + z In[2]:= vals = {a -> 1, b -> 2, L -> 3}; The easiest way is to use Block instead of replacement rules: In[3]:= Block[{a = 1, b = 2, L = 3}, NIntegrate[h[z] z/Sqrt[L^2 + z^2], {z, -L, L}]] Out[3]= 4.79556 A good way is to make h depend on the parameters a and b: In[4]:= h[{a_, b_}, z_] := a + b + z Then you want NIntegrate to not be evaluated until after the replacement rules are applied by ReplaceAll. Thus: In[5]:= Unevaluated[ NIntegrate[h[{a, b}, z] z/Sqrt[L^2 + z^2], {z, -L, L}]] /. vals Out[5]= 4.79556 If you don't want to make either of these changes, you can construct the Block from the replacement rules: In[6]:= Block @@ Append[Apply[Set, Hold[Evaluate[vals]], {2}], Unevaluated[NIntegrate[h[z] z/Sqrt[L^2 + z^2], {z, -L, L}]]] Out[6]= 4.79556 Take it apart piece by piece to see how it works: In[7]:= Hold[Evaluate[vals]] Out[7]= Hold[{a -> 1, b -> 2, L -> 3}] In[8]:= Apply[Set, Hold[Evaluate[vals]], {2}] Out[8]= Hold[{a = 1, b = 2, L = 3}] In[9]:= Append[Apply[Set, Hold[Evaluate[vals]], {2}], Unevaluated[NIntegrate[h[z] z/Sqrt[L^2 + z^2], {z, -L, L}]]] Out[9]= Hold[{a = 1, b = 2, L = 3}, NIntegrate[(h[z] z)/Sqrt[L^2 + z^2], {z, -L, L}]] You can easily make a function to do this: In[10]:= SetAttributes[BlockRules, HoldRest]; In[11]:= BlockRules[rules_, expr_] := Block @@ Append[Apply[Set, Hold@rules, {2}], Unevaluated[expr]] Use it like this: In[12]:= BlockRules[vals, NIntegrate[h[z] z/Sqrt[L^2 + z^2], {z, -L, L}]] Out[12]= 4.79556 You can also make another useful variant, WithRules, which works the same way but with With instead of Block. Hope this helps. Andrew Moylan Wolfram Research ----- Original Message ----- > From: "Giacomo Ciani" <jackspam79 at gmail.com> > To: mathgroup at smc.vnet.net > Sent: Tuesday, May 3, 2011 7:44:43 PM > Subject: [mg118561] Replacements and NIntegrate > > Hi all, > > I've been reading quite a bit in the Mathematica docs and in this > newsgroup, but didn't find (or didn't recognize...) an answer to my > problem. > > I want to evaluate the following expression: > > NIntegrate[h[z] z/Sqrt[L^2 + z^2], {z, -L, L}] > > where h[z] has a delayed value set previously in the notebook. Also, > I > have previously defined a set of replacement rules in the form: > > vals = {a->1, b->2, ec....} > > to be used to specify the numerical values of the various parameters > (including those present in the delayed value of h[z]). > > As for now, the only (brute force) way I found to have my expression > correctly evaluated is to apply replacement rules separately to each > argument of NIntegrate (including integration limits): > > NIntegrate[ h[z] z / Sqrt[L^2 + z^2] /.vals, {z, -L /.vals, > L/.vals}] > > I think you agree with me that this does not look very elegant. > Instead, I would like to be able to write something like this: > > NIntegrate[h[z] z/Sqrt[L^2 + z^2], {z, -L, L}]/.vals > > I know this can't work, as Mathematica tries to evaluate NIntegrate > and then apply the replacement rules... but how can I ask Mathematica > to apply all the replacement rules and delayed values to an > expression > without (or before) actually trying to evaluate it? > > I found a lot of commands to hold the function from evaluating the > arguments, while I need pretty much the opposite... > > Maybe there is something very simple I am overlooking... > > Thanks > > Giacomo > >