Re: Re: Self-teaching snag

*To*: mathgroup at smc.vnet.net*Subject*: [mg74618] Re: [mg74583] Re: [mg74556] Self-teaching snag*From*: "Chris Chiasson" <chris at chiasson.name>*Date*: Wed, 28 Mar 2007 01:46:30 -0500 (EST)*References*: <200703260704.CAA11373@smc.vnet.net>

Some of us ended up with 37 days and some with 38 days. This is bad. On 3/27/07, Daniel Lichtblau <danl at wolfram.com> wrote: > 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 > > -- http://chris.chiasson.name/

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

**Re: Re: Self-teaching snag**

**Re: Re: Self-teaching snag**

**Re: Self-teaching snag**

**Re: Self-teaching snag**