$t=exp(x-u) quad u=exp(x-t)$ for some $x =0..1$ Can $u ne t$?












1












$begingroup$


From deeper discussion of this question I have the pair of equations
$$ t = exp(x - u) \
u = exp(x - t) $$
Rearranging
$$ t/u = exp(t)/exp(u) qquad text{ or }\
u/exp(u) = t/exp(t) $$



I'm looking at $0 le x le 1$ if this is simpler. My hypothese is that
$t=u$ is required.

Trying with arithmetic or geometric mean of $u$ and $t$ didn't lead easily to a solution, also the use of the taylor-series for $exp()$ didn't give me the idea.



I think however it must be somehow easy to be proved, but after a time fiddling with it the penny still didn't drop...




Q: does the hypothese hold that for all $x$ in the interval $t=u$ is required?







algebra-precalculus elementary-number-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    @coffeemath, you have the right idea but, unless I misread the question, you should be using the reciprocal of the function you specify.
    $endgroup$
    – John Omielan
    Dec 24 '18 at 0:16










  • $begingroup$
    @JohnOmielan But $f$ is 1-1 iff $1/f$ is.
    $endgroup$
    – coffeemath
    Dec 24 '18 at 0:17






  • 1




    $begingroup$
    @coffeemath Thanks for explaining. You are right. I did not think this is what you were doing.
    $endgroup$
    – John Omielan
    Dec 24 '18 at 0:19
















1












$begingroup$


From deeper discussion of this question I have the pair of equations
$$ t = exp(x - u) \
u = exp(x - t) $$
Rearranging
$$ t/u = exp(t)/exp(u) qquad text{ or }\
u/exp(u) = t/exp(t) $$



I'm looking at $0 le x le 1$ if this is simpler. My hypothese is that
$t=u$ is required.

Trying with arithmetic or geometric mean of $u$ and $t$ didn't lead easily to a solution, also the use of the taylor-series for $exp()$ didn't give me the idea.



I think however it must be somehow easy to be proved, but after a time fiddling with it the penny still didn't drop...




Q: does the hypothese hold that for all $x$ in the interval $t=u$ is required?







algebra-precalculus elementary-number-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    @coffeemath, you have the right idea but, unless I misread the question, you should be using the reciprocal of the function you specify.
    $endgroup$
    – John Omielan
    Dec 24 '18 at 0:16










  • $begingroup$
    @JohnOmielan But $f$ is 1-1 iff $1/f$ is.
    $endgroup$
    – coffeemath
    Dec 24 '18 at 0:17






  • 1




    $begingroup$
    @coffeemath Thanks for explaining. You are right. I did not think this is what you were doing.
    $endgroup$
    – John Omielan
    Dec 24 '18 at 0:19














1












1








1


0



$begingroup$


From deeper discussion of this question I have the pair of equations
$$ t = exp(x - u) \
u = exp(x - t) $$
Rearranging
$$ t/u = exp(t)/exp(u) qquad text{ or }\
u/exp(u) = t/exp(t) $$



I'm looking at $0 le x le 1$ if this is simpler. My hypothese is that
$t=u$ is required.

Trying with arithmetic or geometric mean of $u$ and $t$ didn't lead easily to a solution, also the use of the taylor-series for $exp()$ didn't give me the idea.



I think however it must be somehow easy to be proved, but after a time fiddling with it the penny still didn't drop...




Q: does the hypothese hold that for all $x$ in the interval $t=u$ is required?







algebra-precalculus elementary-number-theory







share|cite|improve this question









$endgroup$




From deeper discussion of this question I have the pair of equations
$$ t = exp(x - u) \
u = exp(x - t) $$
Rearranging
$$ t/u = exp(t)/exp(u) qquad text{ or }\
u/exp(u) = t/exp(t) $$



I'm looking at $0 le x le 1$ if this is simpler. My hypothese is that
$t=u$ is required.

Trying with arithmetic or geometric mean of $u$ and $t$ didn't lead easily to a solution, also the use of the taylor-series for $exp()$ didn't give me the idea.



I think however it must be somehow easy to be proved, but after a time fiddling with it the penny still didn't drop...




Q: does the hypothese hold that for all $x$ in the interval $t=u$ is required?







algebra-precalculus elementary-number-theory




algebra-precalculus elementary-number-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 24 '18 at 0:03









Gottfried HelmsGottfried Helms

23.7k245101




23.7k245101












  • $begingroup$
    @coffeemath, you have the right idea but, unless I misread the question, you should be using the reciprocal of the function you specify.
    $endgroup$
    – John Omielan
    Dec 24 '18 at 0:16










  • $begingroup$
    @JohnOmielan But $f$ is 1-1 iff $1/f$ is.
    $endgroup$
    – coffeemath
    Dec 24 '18 at 0:17






  • 1




    $begingroup$
    @coffeemath Thanks for explaining. You are right. I did not think this is what you were doing.
    $endgroup$
    – John Omielan
    Dec 24 '18 at 0:19


















  • $begingroup$
    @coffeemath, you have the right idea but, unless I misread the question, you should be using the reciprocal of the function you specify.
    $endgroup$
    – John Omielan
    Dec 24 '18 at 0:16










  • $begingroup$
    @JohnOmielan But $f$ is 1-1 iff $1/f$ is.
    $endgroup$
    – coffeemath
    Dec 24 '18 at 0:17






  • 1




    $begingroup$
    @coffeemath Thanks for explaining. You are right. I did not think this is what you were doing.
    $endgroup$
    – John Omielan
    Dec 24 '18 at 0:19
















$begingroup$
@coffeemath, you have the right idea but, unless I misread the question, you should be using the reciprocal of the function you specify.
$endgroup$
– John Omielan
Dec 24 '18 at 0:16




$begingroup$
@coffeemath, you have the right idea but, unless I misread the question, you should be using the reciprocal of the function you specify.
$endgroup$
– John Omielan
Dec 24 '18 at 0:16












$begingroup$
@JohnOmielan But $f$ is 1-1 iff $1/f$ is.
$endgroup$
– coffeemath
Dec 24 '18 at 0:17




$begingroup$
@JohnOmielan But $f$ is 1-1 iff $1/f$ is.
$endgroup$
– coffeemath
Dec 24 '18 at 0:17




1




1




$begingroup$
@coffeemath Thanks for explaining. You are right. I did not think this is what you were doing.
$endgroup$
– John Omielan
Dec 24 '18 at 0:19




$begingroup$
@coffeemath Thanks for explaining. You are right. I did not think this is what you were doing.
$endgroup$
– John Omielan
Dec 24 '18 at 0:19










2 Answers
2






active

oldest

votes


















2












$begingroup$

Rearranging, you get:
$$ tag{1} te^u = e^x = ue^t $$
and from there
$$ tag{2} frac{t}{e^t} = frac{u}{e^u} $$
The function $tmapsto t/e^t$ increases from $t=0$ to $t=1$, but after that it starts falling again -- so you can find $tne u$ that satisfies $(2)$ quite fine, and from there use $(1)$ to compute $x=log(t)+u$.



The only question is whether $x$ will end up in your range from $0$ to $1$.
Unfortunately it never does. I don't have a slick theoretical proof of this, but by numerical experiment, as $t$ increases from $1$ to $e$ (after which point there is clearly no hope because the $log t$ term alone will be too large), the required $x$ seems to increase monotonically from $1$ to about $1.22$.



(For $tge 1$ solving for $u$ gives $u=-W(-t/e^t)$, so we can get Wolfram Alpha to plot $x$ as a function of $t$).



So there is no solution with $tne u$ given your constraints on $x$ -- but for every $x>1$ there is one.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hmm, this seems to be rather complete now. I'll try to apply this problem of when $x gt 1$ to the basic question which I've linked to.
    $endgroup$
    – Gottfried Helms
    Dec 24 '18 at 1:37










  • $begingroup$
    Next idea: I get that $u=W(exp(x+a))$ and $t=W(exp(x-a))$ . It has to be shown that $a=0$ is required when $x in 0..1$ . Perhaps this can make it smoother, but yet I do not see how....
    $endgroup$
    – Gottfried Helms
    Dec 24 '18 at 8:28



















0












$begingroup$

I've got possibly an ansatz, but get stuck at the end.



$$ begin{array} {rl|rl|l}
t &= exp(x-u) & u &= exp(x-t) & implies \
t exp(u) &= exp(x) & u exp(t) &= exp(x) & \
end{array}
$$ so we have
$$ t exp(u) = u exp(t) qquad gt 0 $$
See a contourplot for $log(t)+u = log(u)+t$ drawn by W/A:
image



Assume now, that $u=t+d$ then we get
$$ t exp(t) exp(d) = (t+d)exp(t) \
t exp(d) = t+d \
t (1+d+d^2/2+ cdots) = t+d qquad text{ we know $t gt 0$ so we can cancel}\
(1+d+d^2/2+ cdots) = 1+d/t \
d+d^2/2+ cdots = d/t \
$$
Hmm. If $t ge 1$ we have our contradiction and $d=0$ is required. But if $t lt d$ I'm again stuck....






share|cite|improve this answer











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    2 Answers
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    2 Answers
    2






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    active

    oldest

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    2












    $begingroup$

    Rearranging, you get:
    $$ tag{1} te^u = e^x = ue^t $$
    and from there
    $$ tag{2} frac{t}{e^t} = frac{u}{e^u} $$
    The function $tmapsto t/e^t$ increases from $t=0$ to $t=1$, but after that it starts falling again -- so you can find $tne u$ that satisfies $(2)$ quite fine, and from there use $(1)$ to compute $x=log(t)+u$.



    The only question is whether $x$ will end up in your range from $0$ to $1$.
    Unfortunately it never does. I don't have a slick theoretical proof of this, but by numerical experiment, as $t$ increases from $1$ to $e$ (after which point there is clearly no hope because the $log t$ term alone will be too large), the required $x$ seems to increase monotonically from $1$ to about $1.22$.



    (For $tge 1$ solving for $u$ gives $u=-W(-t/e^t)$, so we can get Wolfram Alpha to plot $x$ as a function of $t$).



    So there is no solution with $tne u$ given your constraints on $x$ -- but for every $x>1$ there is one.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Hmm, this seems to be rather complete now. I'll try to apply this problem of when $x gt 1$ to the basic question which I've linked to.
      $endgroup$
      – Gottfried Helms
      Dec 24 '18 at 1:37










    • $begingroup$
      Next idea: I get that $u=W(exp(x+a))$ and $t=W(exp(x-a))$ . It has to be shown that $a=0$ is required when $x in 0..1$ . Perhaps this can make it smoother, but yet I do not see how....
      $endgroup$
      – Gottfried Helms
      Dec 24 '18 at 8:28
















    2












    $begingroup$

    Rearranging, you get:
    $$ tag{1} te^u = e^x = ue^t $$
    and from there
    $$ tag{2} frac{t}{e^t} = frac{u}{e^u} $$
    The function $tmapsto t/e^t$ increases from $t=0$ to $t=1$, but after that it starts falling again -- so you can find $tne u$ that satisfies $(2)$ quite fine, and from there use $(1)$ to compute $x=log(t)+u$.



    The only question is whether $x$ will end up in your range from $0$ to $1$.
    Unfortunately it never does. I don't have a slick theoretical proof of this, but by numerical experiment, as $t$ increases from $1$ to $e$ (after which point there is clearly no hope because the $log t$ term alone will be too large), the required $x$ seems to increase monotonically from $1$ to about $1.22$.



    (For $tge 1$ solving for $u$ gives $u=-W(-t/e^t)$, so we can get Wolfram Alpha to plot $x$ as a function of $t$).



    So there is no solution with $tne u$ given your constraints on $x$ -- but for every $x>1$ there is one.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Hmm, this seems to be rather complete now. I'll try to apply this problem of when $x gt 1$ to the basic question which I've linked to.
      $endgroup$
      – Gottfried Helms
      Dec 24 '18 at 1:37










    • $begingroup$
      Next idea: I get that $u=W(exp(x+a))$ and $t=W(exp(x-a))$ . It has to be shown that $a=0$ is required when $x in 0..1$ . Perhaps this can make it smoother, but yet I do not see how....
      $endgroup$
      – Gottfried Helms
      Dec 24 '18 at 8:28














    2












    2








    2





    $begingroup$

    Rearranging, you get:
    $$ tag{1} te^u = e^x = ue^t $$
    and from there
    $$ tag{2} frac{t}{e^t} = frac{u}{e^u} $$
    The function $tmapsto t/e^t$ increases from $t=0$ to $t=1$, but after that it starts falling again -- so you can find $tne u$ that satisfies $(2)$ quite fine, and from there use $(1)$ to compute $x=log(t)+u$.



    The only question is whether $x$ will end up in your range from $0$ to $1$.
    Unfortunately it never does. I don't have a slick theoretical proof of this, but by numerical experiment, as $t$ increases from $1$ to $e$ (after which point there is clearly no hope because the $log t$ term alone will be too large), the required $x$ seems to increase monotonically from $1$ to about $1.22$.



    (For $tge 1$ solving for $u$ gives $u=-W(-t/e^t)$, so we can get Wolfram Alpha to plot $x$ as a function of $t$).



    So there is no solution with $tne u$ given your constraints on $x$ -- but for every $x>1$ there is one.






    share|cite|improve this answer











    $endgroup$



    Rearranging, you get:
    $$ tag{1} te^u = e^x = ue^t $$
    and from there
    $$ tag{2} frac{t}{e^t} = frac{u}{e^u} $$
    The function $tmapsto t/e^t$ increases from $t=0$ to $t=1$, but after that it starts falling again -- so you can find $tne u$ that satisfies $(2)$ quite fine, and from there use $(1)$ to compute $x=log(t)+u$.



    The only question is whether $x$ will end up in your range from $0$ to $1$.
    Unfortunately it never does. I don't have a slick theoretical proof of this, but by numerical experiment, as $t$ increases from $1$ to $e$ (after which point there is clearly no hope because the $log t$ term alone will be too large), the required $x$ seems to increase monotonically from $1$ to about $1.22$.



    (For $tge 1$ solving for $u$ gives $u=-W(-t/e^t)$, so we can get Wolfram Alpha to plot $x$ as a function of $t$).



    So there is no solution with $tne u$ given your constraints on $x$ -- but for every $x>1$ there is one.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 24 '18 at 1:29

























    answered Dec 24 '18 at 0:59









    Henning MakholmHenning Makholm

    243k17308553




    243k17308553












    • $begingroup$
      Hmm, this seems to be rather complete now. I'll try to apply this problem of when $x gt 1$ to the basic question which I've linked to.
      $endgroup$
      – Gottfried Helms
      Dec 24 '18 at 1:37










    • $begingroup$
      Next idea: I get that $u=W(exp(x+a))$ and $t=W(exp(x-a))$ . It has to be shown that $a=0$ is required when $x in 0..1$ . Perhaps this can make it smoother, but yet I do not see how....
      $endgroup$
      – Gottfried Helms
      Dec 24 '18 at 8:28


















    • $begingroup$
      Hmm, this seems to be rather complete now. I'll try to apply this problem of when $x gt 1$ to the basic question which I've linked to.
      $endgroup$
      – Gottfried Helms
      Dec 24 '18 at 1:37










    • $begingroup$
      Next idea: I get that $u=W(exp(x+a))$ and $t=W(exp(x-a))$ . It has to be shown that $a=0$ is required when $x in 0..1$ . Perhaps this can make it smoother, but yet I do not see how....
      $endgroup$
      – Gottfried Helms
      Dec 24 '18 at 8:28
















    $begingroup$
    Hmm, this seems to be rather complete now. I'll try to apply this problem of when $x gt 1$ to the basic question which I've linked to.
    $endgroup$
    – Gottfried Helms
    Dec 24 '18 at 1:37




    $begingroup$
    Hmm, this seems to be rather complete now. I'll try to apply this problem of when $x gt 1$ to the basic question which I've linked to.
    $endgroup$
    – Gottfried Helms
    Dec 24 '18 at 1:37












    $begingroup$
    Next idea: I get that $u=W(exp(x+a))$ and $t=W(exp(x-a))$ . It has to be shown that $a=0$ is required when $x in 0..1$ . Perhaps this can make it smoother, but yet I do not see how....
    $endgroup$
    – Gottfried Helms
    Dec 24 '18 at 8:28




    $begingroup$
    Next idea: I get that $u=W(exp(x+a))$ and $t=W(exp(x-a))$ . It has to be shown that $a=0$ is required when $x in 0..1$ . Perhaps this can make it smoother, but yet I do not see how....
    $endgroup$
    – Gottfried Helms
    Dec 24 '18 at 8:28











    0












    $begingroup$

    I've got possibly an ansatz, but get stuck at the end.



    $$ begin{array} {rl|rl|l}
    t &= exp(x-u) & u &= exp(x-t) & implies \
    t exp(u) &= exp(x) & u exp(t) &= exp(x) & \
    end{array}
    $$     so we have
    $$ t exp(u) = u exp(t) qquad gt 0 $$
    See a contourplot for $log(t)+u = log(u)+t$ drawn by W/A:
    image



    Assume now, that $u=t+d$ then we get
    $$ t exp(t) exp(d) = (t+d)exp(t) \
    t exp(d) = t+d \
    t (1+d+d^2/2+ cdots) = t+d qquad text{ we know $t gt 0$ so we can cancel}\
    (1+d+d^2/2+ cdots) = 1+d/t \
    d+d^2/2+ cdots = d/t \
    $$
    Hmm. If $t ge 1$ we have our contradiction and $d=0$ is required. But if $t lt d$ I'm again stuck....






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      I've got possibly an ansatz, but get stuck at the end.



      $$ begin{array} {rl|rl|l}
      t &= exp(x-u) & u &= exp(x-t) & implies \
      t exp(u) &= exp(x) & u exp(t) &= exp(x) & \
      end{array}
      $$       so we have
      $$ t exp(u) = u exp(t) qquad gt 0 $$
      See a contourplot for $log(t)+u = log(u)+t$ drawn by W/A:
      image



      Assume now, that $u=t+d$ then we get
      $$ t exp(t) exp(d) = (t+d)exp(t) \
      t exp(d) = t+d \
      t (1+d+d^2/2+ cdots) = t+d qquad text{ we know $t gt 0$ so we can cancel}\
      (1+d+d^2/2+ cdots) = 1+d/t \
      d+d^2/2+ cdots = d/t \
      $$
      Hmm. If $t ge 1$ we have our contradiction and $d=0$ is required. But if $t lt d$ I'm again stuck....






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        I've got possibly an ansatz, but get stuck at the end.



        $$ begin{array} {rl|rl|l}
        t &= exp(x-u) & u &= exp(x-t) & implies \
        t exp(u) &= exp(x) & u exp(t) &= exp(x) & \
        end{array}
        $$         so we have
        $$ t exp(u) = u exp(t) qquad gt 0 $$
        See a contourplot for $log(t)+u = log(u)+t$ drawn by W/A:
        image



        Assume now, that $u=t+d$ then we get
        $$ t exp(t) exp(d) = (t+d)exp(t) \
        t exp(d) = t+d \
        t (1+d+d^2/2+ cdots) = t+d qquad text{ we know $t gt 0$ so we can cancel}\
        (1+d+d^2/2+ cdots) = 1+d/t \
        d+d^2/2+ cdots = d/t \
        $$
        Hmm. If $t ge 1$ we have our contradiction and $d=0$ is required. But if $t lt d$ I'm again stuck....






        share|cite|improve this answer











        $endgroup$



        I've got possibly an ansatz, but get stuck at the end.



        $$ begin{array} {rl|rl|l}
        t &= exp(x-u) & u &= exp(x-t) & implies \
        t exp(u) &= exp(x) & u exp(t) &= exp(x) & \
        end{array}
        $$         so we have
        $$ t exp(u) = u exp(t) qquad gt 0 $$
        See a contourplot for $log(t)+u = log(u)+t$ drawn by W/A:
        image



        Assume now, that $u=t+d$ then we get
        $$ t exp(t) exp(d) = (t+d)exp(t) \
        t exp(d) = t+d \
        t (1+d+d^2/2+ cdots) = t+d qquad text{ we know $t gt 0$ so we can cancel}\
        (1+d+d^2/2+ cdots) = 1+d/t \
        d+d^2/2+ cdots = d/t \
        $$
        Hmm. If $t ge 1$ we have our contradiction and $d=0$ is required. But if $t lt d$ I'm again stuck....







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 24 '18 at 2:33

























        answered Dec 24 '18 at 1:05









        Gottfried HelmsGottfried Helms

        23.7k245101




        23.7k245101






























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