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4

1

$begingroup$

In an optimiation problem I came across the following daunting equation:
$$
log left(frac1-t_21-yright) log left(frac(1-x) (t_1-t_2)(1-t_1) (x-y)right) log left(fract_2 xt_1 yright)\=log left(fract_2yright) log left(fracx (t_1-t_2)t_1 (x-y)right) log left(frac(1-t_2) (1-x)(1-t_1) (1-y)right)
$$
in which $x,yin[0,1]$ are given constants and $t_1,t_2in[0,1]$ are the variables.

I ended up plotting it numerically and noting that at the solutions (over $t_1,t_2$) the logarithmic factors would be pairwise equal, which in particular implied the hyperbola $$t_1 (1-x) y=t_1 t_2 (y-x)+t_2 x (1-y)$$ along with a few trivial solutions ($t_2=y$, $t_1=x$, $x t_2 = t_1 y$ and $t_2=t_1$).
Checking that these are indeed solutions is easy, but coming up with them was not.

I wonder if there was a more structural approach I could have taken to come up with this result?

In fact I have some even more formidable equations which I now hope have simple polynomial solutions, but those I haven't been lucky enough to guess.

real-analysis calculus polynomials logarithms alternative-proof

share|cite|improve this question

edited 17 hours ago

Thomas Ahle

asked Mar 11 at 10:34

Thomas AhleThomas Ahle

1,4541321

$endgroup$

This question has an open bounty worth +50
reputation from Thomas Ahle ending ending at 2019-03-20 10:44:37Z">in 5 days.

The current answers do not contain enough detail.

I would like an intuition for why the solution came out the way it did. Preferably one that I can apply to similar problems to find the answer more easily.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Saad Thank you, I have tried to clarify now. By pairwise equality, I mean that each factor on the left-hand side equals a factor on the right-hand side. It seems odd to me why this is the case.
  $endgroup$
  – Thomas Ahle
  17 hours ago
  

  
  
  
  

add a comment | 

4

1

$begingroup$

In an optimiation problem I came across the following daunting equation:
$$
log left(frac1-t_21-yright) log left(frac(1-x) (t_1-t_2)(1-t_1) (x-y)right) log left(fract_2 xt_1 yright)\=log left(fract_2yright) log left(fracx (t_1-t_2)t_1 (x-y)right) log left(frac(1-t_2) (1-x)(1-t_1) (1-y)right)
$$
in which $x,yin[0,1]$ are given constants and $t_1,t_2in[0,1]$ are the variables.

I ended up plotting it numerically and noting that at the solutions (over $t_1,t_2$) the logarithmic factors would be pairwise equal, which in particular implied the hyperbola $$t_1 (1-x) y=t_1 t_2 (y-x)+t_2 x (1-y)$$ along with a few trivial solutions ($t_2=y$, $t_1=x$, $x t_2 = t_1 y$ and $t_2=t_1$).
Checking that these are indeed solutions is easy, but coming up with them was not.

I wonder if there was a more structural approach I could have taken to come up with this result?

In fact I have some even more formidable equations which I now hope have simple polynomial solutions, but those I haven't been lucky enough to guess.

real-analysis calculus polynomials logarithms alternative-proof

share|cite|improve this question

edited 17 hours ago

Thomas Ahle

asked Mar 11 at 10:34

Thomas AhleThomas Ahle

1,4541321

$endgroup$

This question has an open bounty worth +50
reputation from Thomas Ahle ending ending at 2019-03-20 10:44:37Z">in 5 days.

The current answers do not contain enough detail.

I would like an intuition for why the solution came out the way it did. Preferably one that I can apply to similar problems to find the answer more easily.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Saad Thank you, I have tried to clarify now. By pairwise equality, I mean that each factor on the left-hand side equals a factor on the right-hand side. It seems odd to me why this is the case.
  $endgroup$
  – Thomas Ahle
  17 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

In an optimiation problem I came across the following daunting equation:
$$
log left(frac1-t_21-yright) log left(frac(1-x) (t_1-t_2)(1-t_1) (x-y)right) log left(fract_2 xt_1 yright)\=log left(fract_2yright) log left(fracx (t_1-t_2)t_1 (x-y)right) log left(frac(1-t_2) (1-x)(1-t_1) (1-y)right)
$$
in which $x,yin[0,1]$ are given constants and $t_1,t_2in[0,1]$ are the variables.

I ended up plotting it numerically and noting that at the solutions (over $t_1,t_2$) the logarithmic factors would be pairwise equal, which in particular implied the hyperbola $$t_1 (1-x) y=t_1 t_2 (y-x)+t_2 x (1-y)$$ along with a few trivial solutions ($t_2=y$, $t_1=x$, $x t_2 = t_1 y$ and $t_2=t_1$).
Checking that these are indeed solutions is easy, but coming up with them was not.

I wonder if there was a more structural approach I could have taken to come up with this result?

In fact I have some even more formidable equations which I now hope have simple polynomial solutions, but those I haven't been lucky enough to guess.

real-analysis calculus polynomials logarithms alternative-proof

share|cite|improve this question

edited 17 hours ago

Thomas Ahle

asked Mar 11 at 10:34

Thomas AhleThomas Ahle

1,4541321

$endgroup$

share|cite|improve this question

edited 17 hours ago

Thomas Ahle

asked Mar 11 at 10:34

Thomas AhleThomas Ahle

1,4541321

share|cite|improve this question

edited 17 hours ago

Thomas Ahle

asked Mar 11 at 10:34

Thomas AhleThomas Ahle

1,4541321

asked Mar 11 at 10:34

Thomas AhleThomas Ahle

1,4541321

asked Mar 11 at 10:34

Thomas AhleThomas Ahle

1,4541321