[Date Index]
[Thread Index]
[Author Index]
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
Prev by Date:
**Re: Self-teaching snag**
Next by Date:
**Re: Definite Integration in Mathematica**
Previous by thread:
**Re: Re: Self-teaching snag**
Next by thread:
**Re: Re: Self-teaching snag**
| |