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Re: Re: Self-teaching snag
*To*: mathgroup at smc.vnet.net
*Subject*: [mg74611] Re: [mg74583] Re: [mg74556] Self-teaching snag
*From*: Bob Hanlon <hanlonr at cox.net>
*Date*: Wed, 28 Mar 2007 01:42:55 -0500 (EST)
*Reply-to*: hanlonr at cox.net
The difference is use of
ch[1]==1
rather than ch[0]==1
Bob Hanlon
---- Chris Chiasson <chris at chiasson.name> wrote:
> 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/
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