Re: Self-teaching snag

*To*: mathgroup at smc.vnet.net*Subject*: [mg74583] Re: [mg74556] Self-teaching snag*From*: Daniel Lichtblau <danl at wolfram.com>*Date*: Tue, 27 Mar 2007 04:00:53 -0500 (EST)*References*: <200703260704.CAA11373@smc.vnet.net>

Todd Allen wrote: > Hi All, > > I am trying to refresh my skills in basic problem > solving using Mathematica, but am running into some > difficulties which are beginning to make me suspicious > of Mathematica itself. (I probably should be > suspicious of my own brain...but you know how that is > :-) > > Here is the scenario: I have written a basic function > to tell me what percentage of battery power will > remain in a battery after x number of days, provided > that we start with a full charge and lose 5% of that > charge per day. > > If you execute the following code in Mathematica > (V5.1): > > charge[0]=1.0 (* 100% *); > charge[day_]:=(charge[day-1]-(0.05*charge[day-1])); > charge[20] > > I receive an output of 0.358486 for my query at the 20 > day mark.....so, no problem so far. > > However, when I try to ask for the output at > charge[35], mathematica seems to enter an endless > calculation. I've let the computer run for as long as > 5 minutes without getting an answer. Is there > something wrong with my function, my version of > Mathematica or something else I haven't considered? > > > Additionally, > > When I try the following: > > In[145]:= > Solve[charge[day]==0.15,day]; > > Mathematica gives me the error: > "$RecursionLimit::reclim: Recursion depth of 256 > exceeded." > > I am trying to ask Mathematica to tell my how many > days it takes to reduce the battery power to 15 > percent, but I must be messing something up?? > > If anyone has any pointers, I'd certainly appreciate > it, because I am a little stuck right now. > > Best regards, > Todd Allen I expect you will receive several reasonable responses to this. I wanted to address what I think is a subtlety. I apologize in advance if this drawn out explanation bores anyone beyond the usual level of tears. First let me cover some of the basics. I will use exact arithmetic because in some steps it may make sense to avoid approximations. I define the recursive computation as below. I use a common caching technique many others will likely also mention, memoization (also called "memorization"), so as to avoid recomputations. In[54]:= charge0[0] = 1; In[55]:= charge0[day_] := charge0[day] = 19/20*charge0[day-1] Note the computation is virtually instantaneous. In[56]:= InputForm[Timing[charge0[35]]] Out[56]//InputForm= {0., 570658162108627174778971075491512021856922699/ 3435973836800000000000000000000000000000000000} I'll redo this but without memoization. In[58]:= charge1[0] = 1; In[59]:= charge1[day_] := 19/20*charge1[day-1] In[60]:= InputForm[Timing[charge1[35]]] Out[60]//InputForm= {0., 570658162108627174778971075491512021856922699/ 3435973836800000000000000000000000000000000000} Well, that too was quite fast. Now let's look at your definition (changed to use exact arithmetic, but not altered in any way that makes an essential difference). First I show a red herring. In[69]:= InputForm[charge2[day-1] - 1/20*charge2[day-1]] Out[69]//InputForm= (19*charge2[-1 + day])/20 Now let's actually define the function. In[70]:= charge2[0] = 1; In[71]:= charge2[day_] := charge2[day-1] - 1/20*charge2[day-1] Is this different from charge1 above? Yes, considerably. First check the speed of evaluating charge2[15]. In[72]:= InputForm[Timing[charge2[15]]] Out[72]//InputForm= {0.21201399999999962, 15181127029874798299/32768000000000000000} So how is this different from charge1? And why was charge1 fast to compute, just like charge0? We look at how charge2 is actually stored. Since we use SetDelayed (the "colon-equal" assignment), it is not actually evalauted as in "red herring" above. In[73]:= ??charge2 Global`charge2 charge2[0] = 1 charge2[day_] := charge2[day - 1] - (1*charge2[day - 1])/20 We see the terms are not combined. So a tremendous amount of reevaluation will take place. How to cure this? Again, we can use memoization. In[74]:= charge3[0] = 1; In[75]:= charge3[day_] := charge3[day] = charge3[day-1] - 1/20*charge3[day-1] In[76]:= InputForm[Timing[charge3[35]]] Out[76]//InputForm= {3.2578106878844437*^-15, 570658162108627174778971075491512021856922699/ 3435973836800000000000000000000000000000000000} Why was charge1, which did not use memoization, so fast? Because it was written in a way that is tail recursive, hence avoids massive recomputation. Since it is not always easy to write arbitrary recursions as tail recursive, I'd recommend the memoization approach as a general tactic (but then be aware of things like $RecursionLimit). To address your second question one can use RSolve to get a closed form of the recurrence solution, then use Solve to find where we hit the 15% mark. In[77]:= InputForm[chargefunc = ch[d] /. RSolve[{ch[d]==19/20*ch[d-1],ch[1]==1}, ch[d], d]] Out[77]//InputForm= {(20/19)^(1 - d)} In[79]:= InputForm[soln = d /. Solve[chargefunc[[1]]==3/20, d]] Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. Out[79]//InputForm= {(Log[20/19] + Log[20/3])/Log[20/19]} Turns out to be around 38 days. In[81]:= N[soln[[1]]] Out[81]= 37.9857 Adequate for most uses, I should think, short of say FOBP ("Floods of Biblical Proportions"). Daniel Lichtblau Wolfram Research

**References**:**Self-teaching snag***From:*Todd Allen <genesplicer28@yahoo.com>

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