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Error "cannot be solved" with Reduce[ ]?

Hessam mohammadi
Posted 3 years ago

I have a few expressions and need to get a final simplified (reduced,solved) results. I am new and tried/searched different ways for my problem but couldn't find the reason. These are my expressions:

R0 = FractionalPart[Log[5, x + (3/ 4)]] ==  FractionalPart[Log[5, xP + (3/ 4)]]
R1  = Log[5, xP + (3 / 4)] - Log[3, yP + ( 5 / 2)] == Log[5, x + (3 / 4)] - Log[3, y + ( 5 / 2)]
F = ( (x == xP && y == yP) || ( Exists[{xPP, yPP},  xPP < yPP && xP == 5 xPP + 3 && yP == 3 yPP + 5]))
R = R0 && R1 && F
Reduce[R && xP >= yP , {x, y, xP, yP}, Reals]

when I run this it takes a long time and seems infinite loop or something and never finishes. I also tried this 

Reduce[R , Element[{x, y, xP, yP}, Reals]]

and it returns the error: Reduce : This system cannot be solved with the methods available to Reduce We expect two functions for xP and yP based on x and y, for each. Please help me through this.

2- How can I make sure that each single expression is error free? Thanks a lot
POSTED BY: Hessam mohammadi
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Bill Nelson
Bill Nelson
Posted 3 years ago

If I change your approach to the problem

R0=(FractionalPart[Log[5,x+(3/4)]]-FractionalPart[Log[5,xP+(3/4)]])^2;
R1=((Log[5,xP+(3/4)]-Log[3,yP+(5/2)])-(Log[5,x+(3/4)]-Log[3,y+(5/2)]))^2;
F=(x-xP)^2+(y-yP)^2;
R=R0+R1+F;
NMinimize[{R,xP>=yP},{x,y,xP,yP}]

then it almost instantly returns

{1.05676*^-21,{x->0.492936,y->-0.138782,xP->0.492936,yP->-0.138782}}

and that is close enough to zero that this might be an acceptable solution for you.

You must check all this very carefully to try to make certain that I haven't made any mistakes.

If I start from a different position and see from the statement of your problem that x==xP and y==yP then it seems that R0, R1 and F are always true for any values of x and y. Thus the only condition seems to be that x>y. Is that correct?
POSTED BY: Bill Nelson

I changed it a bit and now got this error, I need two formula as a function for xp and yp, reaching to two final answer is not enough. thanks a lot

EDITED : this new gives me an error, "This system cannot be solved with the methods available to Solve" does someone know why ?

Solve[FractionalPart (Log[5, x + (3/4)]) == FractionalPart (Log[5, X + (3/4)]) && 
Log[5, x + (3/4)] - Log[3, y + (5/2)] == Log[5, X + (3/4)] - Log[3, Y + (5/2)] &&
    ((x == X && y == Y)||(x < y && (X - 3)/5 < (Y - 5)/3)) && X >= Y,{X,Y}, Integers]
POSTED BY: Hessam mohammadi

This subsystem gives a solution quickly:

Solve[FractionalPart[Log[5, x + (3/4)]] == 
   FractionalPart[Log[5, X + (3/4)]] && 
  Log[5, x + (3/4)] - Log[3, y + (5/2)] == 
   Log[5, X + (3/4)] - Log[3, Y + (5/2)] &&
  ((x == X && y == Y)) && 
  X >= Y,
 {X, Y}, Integers]

Here is one solution of the other subsystem:

In[24]:= FindInstance[
 FractionalPart[Log[5, x + (3/4)]] == 
   FractionalPart[Log[5, X + (3/4)]] && 
  Log[5, x + (3/4)] - Log[3, y + (5/2)] == 
   Log[5, X + (3/4)] - 
    Log[3, Y + (5/2)] && (x < y && (X - 3)/5 < (Y - 5)/3) && X >= Y &&
   Element[X | Y, Integers], {X, Y, x, y}]

Out[24]= {{X -> 13, Y -> 12, x -> 2, y -> 7/3}}
POSTED BY: Gianluca Gorni

Thanks, but we cannot change the specification.
POSTED BY: Hessam mohammadi
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