Groups | Search | Server Info | Login | Register

Groups > comp.soft-sys.math.maple > #195

Re: Missing solution to (1-p)^n = 1-n*p

Show key headers only | View raw

Axel Vogt <&noreply@axelvogt.de> writes:

> On 23.06.2011 16:29, Ilmari Karonen wrote:
>> While I was playing around with probabilities, I noticed that I wasn't
>> able to get Maple 12 to figure out that
>>
>>    (1-p)^n = 1-n*p
>>
>> has a solution at n = 1.  Of course, it's trivial to verify by
>> substitution that this is indeed a solution for any p, but I wasn't
>> able to get solve() to cough it up:
>>
>>> assume(p>  0, p<  1, n>  0):
>>> interface(showassumed = 0):
>>> solve((1-p)^n = 1-n*p, n);
>>                                     0
>>
>> OK, n = 0 certainly is a solution, but I'd have liked the other one
>> too.  Let me see if solving for all variables would do any better...
>>
>>> solve((1-p)^n = 1-n*p);
>>      {n = n, p = -exp(RootOf(_Z n - ln(1 + n exp(_Z) - n))) + 1},
>>      {n = 1, p = 1}
>>
>> Well, that wasn't very useful.  I'm not sure offhand if that RootOf()
>> can actually take values other than 0, but either way I really wanted
>> a solution for n, not p.  Any why does Maple figure out the specific
>> solution n = 1, p = 1, but not the more general n = 1, p = p?
>>
>> Rewriting the left hand side as exp(n*log(1-p)) helps a little...
>>
>>> solve(exp(n*log(1-p)) = 1-n*p, n);
>> memory used=3.8MB, alloc=3.2MB, time=0.22
>> memory used=7.6MB, alloc=5.1MB, time=0.38
>>                                    ln(1 - p)
>>                      ln(1 - p) exp(---------)
>>                                        p
>>            -LambertW(------------------------) p + ln(1 - p)
>>                                 p
>>        exp(-------------------------------------------------) - 1
>>                                    p
>>      - ----------------------------------------------------------, 0
>>                                    p
>>
>> ...but unfortunately Maple does no better than me in simplifying that
>> awful nested expression, even though numerical testing does indeed
>> suggest that it is identically 1 at least for 0<  p<  1.
>>
>> Solving in the real domain does no better:
>>
>>> RealDomain[solve]((1-p)^n = 1-n*p, n);
>> memory used=11.4MB, alloc=5.9MB, time=0.60
>>                                             (1/p)
>>                            ln(1 - p) (1 - p)
>>            -LambertW(-_B3, ----------------------) p + ln(1 - p)
>>                                      p
>>        exp(-----------------------------------------------------) - 1
>>                                      p
>>      - --------------------------------------------------------------
>>                                      p
>>
>> So, now I'm curious.  Is this just a bug in Maple 12, or am I doing
>> something silly?  Or both?  And does it work any better in later
>> versions?
>>
>
> Generally if RootOf occurs, the 'allvalues' may be the way to go.
>
> With your assumptions and using Maple 15:
>
>   (1-p)^n = 1-n*p;
>   solve(%,n, allsolutions=true);
>   expand(%); simplify(%);
>   collect(%, LambertW);
>
>     -1/ln(1-p)*LambertW(_Z30,ln(1-p)/p*(1-p)^(1/p))+1/p
>
> Where a warning is given: "Warning, solve may be ignoring assumptions
> on the input variables."
>
> In older versions use RootOf(eq, n); allvalues(%);

The way to solve this is with solve/identity.

(**) eq := (1-p)^n = 1-n*p:
(**) solve(identity(eq,p),{n}); 
                               {n = 0}, {n = 1}


-- 
Joe Riel

Back to comp.soft-sys.math.maple | Previous | Next — Previous in thread | Next in thread | Find similar

Thread

Missing solution to (1-p)^n = 1-n*p Ilmari Karonen <usenet2@vyznev.invalid> - 2011-06-23 14:29 +0000
  Re: Missing solution to (1-p)^n = 1-n*p Axel Vogt <&noreply@axelvogt.de> - 2011-06-23 18:47 +0200
    Re: Missing solution to (1-p)^n = 1-n*p Joe Riel <joer@san.rr.com> - 2011-06-23 10:49 -0700
      Re: Missing solution to (1-p)^n = 1-n*p Ilmari Karonen <usenet2@vyznev.invalid> - 2011-06-23 20:12 +0000

csiph-web