Tetration of non-integers: is there something wrong with this approach?












3












$begingroup$


I'm trying to figure out a formula for tetration that will work for
non-integer heights.



I know the usual recurrence relation for tetration
($x in mathbb{R}, text{ }n in mathbb{N})$:



$${^{n}x} = begin{cases} 1 &text{if }n=0 \ \ x^{left(^{(n-1)}xright)} &text{if }n>0 end{cases}$$



I also know that $x^y=e^{y ln x}$ for positive $x$.



I combined these two and formed this recurrence:
$$
{^y}x =
f(x,y) =
begin{cases}
e^{y ln x} & text{if }0 lt y le 1 \
\
e^{f(x,text{ }y-1) ln x} & text{if }1 lt y
end{cases}
$$



Playing around with this in Maxima, I got correct answers for integer $y$, and reasonable-looking answers for non-integers. Yet I have read
numerous sources stating that a general formula for tetration is very difficult.



So, my question: have I a correct solution for a limited domain, or am I
off in the weeds and it just happens to work for integers?



Thank you.






functions exponentiation tetration







share|cite|improve this question









$endgroup$












  • $begingroup$
    it seems fine to me. I guess that the sources that you read state that it is difficult to find an explicit formula for tetration, that is, a formula for a direct computation, without the need of a recurrence
    $endgroup$
    – Masacroso
    Dec 5 '18 at 2:51
















3












$begingroup$


I'm trying to figure out a formula for tetration that will work for
non-integer heights.



I know the usual recurrence relation for tetration
($x in mathbb{R}, text{ }n in mathbb{N})$:



$${^{n}x} = begin{cases} 1 &text{if }n=0 \ \ x^{left(^{(n-1)}xright)} &text{if }n>0 end{cases}$$



I also know that $x^y=e^{y ln x}$ for positive $x$.



I combined these two and formed this recurrence:
$$
{^y}x =
f(x,y) =
begin{cases}
e^{y ln x} & text{if }0 lt y le 1 \
\
e^{f(x,text{ }y-1) ln x} & text{if }1 lt y
end{cases}
$$



Playing around with this in Maxima, I got correct answers for integer $y$, and reasonable-looking answers for non-integers. Yet I have read
numerous sources stating that a general formula for tetration is very difficult.



So, my question: have I a correct solution for a limited domain, or am I
off in the weeds and it just happens to work for integers?



Thank you.






functions exponentiation tetration







share|cite|improve this question









$endgroup$












  • $begingroup$
    it seems fine to me. I guess that the sources that you read state that it is difficult to find an explicit formula for tetration, that is, a formula for a direct computation, without the need of a recurrence
    $endgroup$
    – Masacroso
    Dec 5 '18 at 2:51














3












3








3


1



$begingroup$


I'm trying to figure out a formula for tetration that will work for
non-integer heights.



I know the usual recurrence relation for tetration
($x in mathbb{R}, text{ }n in mathbb{N})$:



$${^{n}x} = begin{cases} 1 &text{if }n=0 \ \ x^{left(^{(n-1)}xright)} &text{if }n>0 end{cases}$$



I also know that $x^y=e^{y ln x}$ for positive $x$.



I combined these two and formed this recurrence:
$$
{^y}x =
f(x,y) =
begin{cases}
e^{y ln x} & text{if }0 lt y le 1 \
\
e^{f(x,text{ }y-1) ln x} & text{if }1 lt y
end{cases}
$$



Playing around with this in Maxima, I got correct answers for integer $y$, and reasonable-looking answers for non-integers. Yet I have read
numerous sources stating that a general formula for tetration is very difficult.



So, my question: have I a correct solution for a limited domain, or am I
off in the weeds and it just happens to work for integers?



Thank you.






functions exponentiation tetration







share|cite|improve this question









$endgroup$




I'm trying to figure out a formula for tetration that will work for
non-integer heights.



I know the usual recurrence relation for tetration
($x in mathbb{R}, text{ }n in mathbb{N})$:



$${^{n}x} = begin{cases} 1 &text{if }n=0 \ \ x^{left(^{(n-1)}xright)} &text{if }n>0 end{cases}$$



I also know that $x^y=e^{y ln x}$ for positive $x$.



I combined these two and formed this recurrence:
$$
{^y}x =
f(x,y) =
begin{cases}
e^{y ln x} & text{if }0 lt y le 1 \
\
e^{f(x,text{ }y-1) ln x} & text{if }1 lt y
end{cases}
$$



Playing around with this in Maxima, I got correct answers for integer $y$, and reasonable-looking answers for non-integers. Yet I have read
numerous sources stating that a general formula for tetration is very difficult.



So, my question: have I a correct solution for a limited domain, or am I
off in the weeds and it just happens to work for integers?



Thank you.






functions exponentiation tetration




functions exponentiation tetration






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 5 '18 at 2:34









user3412516user3412516

183




183












  • $begingroup$
    it seems fine to me. I guess that the sources that you read state that it is difficult to find an explicit formula for tetration, that is, a formula for a direct computation, without the need of a recurrence
    $endgroup$
    – Masacroso
    Dec 5 '18 at 2:51


















  • $begingroup$
    it seems fine to me. I guess that the sources that you read state that it is difficult to find an explicit formula for tetration, that is, a formula for a direct computation, without the need of a recurrence
    $endgroup$
    – Masacroso
    Dec 5 '18 at 2:51
















$begingroup$
it seems fine to me. I guess that the sources that you read state that it is difficult to find an explicit formula for tetration, that is, a formula for a direct computation, without the need of a recurrence
$endgroup$
– Masacroso
Dec 5 '18 at 2:51




$begingroup$
it seems fine to me. I guess that the sources that you read state that it is difficult to find an explicit formula for tetration, that is, a formula for a direct computation, without the need of a recurrence
$endgroup$
– Masacroso
Dec 5 '18 at 2:51










2 Answers
2






active

oldest

votes


















2












$begingroup$

The real question here is what various people consider to be necessary for a formula "that will work." The first requirement you give is very unobjectionable - there is some recurrence relation that tetration should satisfy. This recurrence can be used to take a definition of $^yx$ for $yin [0,1)$ and extend it to work for all positive $y$. However, the issue is not that it is hard to find a function satisfying the laid out condition: The issue is that there are a lot of functions that work - I could define $^yx$ to be anything I want in that interval and there's no clear reason to take $^yx=x^y$ for $0<yleq 1$ as you do - it makes the function continuous, but I could just as easily take $^yx=y(x-1)+1$ to get a continuous answer that will make results that look nice for non-integers.



Usually, what makes things hard, is that people want conditions like differentiability or convexity in their definition of tetration - this greatly restricts your possibilities. Unfortunately, there's not real consensus on what properties one would like - so there are a number of different functions that might claim to extend tetration.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    ...Mathematica is being annoying right now, but I'll plot the function once it behaves; the function you behave has some noticeable corners in it - it's not differentiable.
    $endgroup$
    – Milo Brandt
    Dec 5 '18 at 3:02










  • $begingroup$
    I figured out how to get Maxima on Android to plot the function. It DOES have noticeable "kinks" as the curve crosses each integer. Visually, it appears like the cables of a suspension bridge with each pillar higher than the last. I would have expected tetration to smoothly cross the integers. I think my function is wrong: it yields the correct answer at the integers, but it is slightly "low" in between the integers.
    $endgroup$
    – user3412516
    Dec 10 '18 at 19:32



















1












$begingroup$

Just to add more visual explanation to the answer of @MiloBrandt you might look at an older casual essay of mine. There I show the effect of setting an individual value is initially interpolated and then exponentiated a small number of $n$. With any selection of the initial the resulting curve is edgy except of one - and not only you need to find this but also some formula by which it depends on $x$. (The pages are made by Excel and are thus imperfect - if you want to play with this you can mail me for the file (Excel 2000 with modules)) The method I used here was already known to and described by G.H. Hardy in a discussion of a function with an interpolated growth-rate, but I don't have the reference at hand.






share|cite|improve this answer









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026519%2ftetration-of-non-integers-is-there-something-wrong-with-this-approach%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    The real question here is what various people consider to be necessary for a formula "that will work." The first requirement you give is very unobjectionable - there is some recurrence relation that tetration should satisfy. This recurrence can be used to take a definition of $^yx$ for $yin [0,1)$ and extend it to work for all positive $y$. However, the issue is not that it is hard to find a function satisfying the laid out condition: The issue is that there are a lot of functions that work - I could define $^yx$ to be anything I want in that interval and there's no clear reason to take $^yx=x^y$ for $0<yleq 1$ as you do - it makes the function continuous, but I could just as easily take $^yx=y(x-1)+1$ to get a continuous answer that will make results that look nice for non-integers.



    Usually, what makes things hard, is that people want conditions like differentiability or convexity in their definition of tetration - this greatly restricts your possibilities. Unfortunately, there's not real consensus on what properties one would like - so there are a number of different functions that might claim to extend tetration.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      ...Mathematica is being annoying right now, but I'll plot the function once it behaves; the function you behave has some noticeable corners in it - it's not differentiable.
      $endgroup$
      – Milo Brandt
      Dec 5 '18 at 3:02










    • $begingroup$
      I figured out how to get Maxima on Android to plot the function. It DOES have noticeable "kinks" as the curve crosses each integer. Visually, it appears like the cables of a suspension bridge with each pillar higher than the last. I would have expected tetration to smoothly cross the integers. I think my function is wrong: it yields the correct answer at the integers, but it is slightly "low" in between the integers.
      $endgroup$
      – user3412516
      Dec 10 '18 at 19:32
















    2












    $begingroup$

    The real question here is what various people consider to be necessary for a formula "that will work." The first requirement you give is very unobjectionable - there is some recurrence relation that tetration should satisfy. This recurrence can be used to take a definition of $^yx$ for $yin [0,1)$ and extend it to work for all positive $y$. However, the issue is not that it is hard to find a function satisfying the laid out condition: The issue is that there are a lot of functions that work - I could define $^yx$ to be anything I want in that interval and there's no clear reason to take $^yx=x^y$ for $0<yleq 1$ as you do - it makes the function continuous, but I could just as easily take $^yx=y(x-1)+1$ to get a continuous answer that will make results that look nice for non-integers.



    Usually, what makes things hard, is that people want conditions like differentiability or convexity in their definition of tetration - this greatly restricts your possibilities. Unfortunately, there's not real consensus on what properties one would like - so there are a number of different functions that might claim to extend tetration.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      ...Mathematica is being annoying right now, but I'll plot the function once it behaves; the function you behave has some noticeable corners in it - it's not differentiable.
      $endgroup$
      – Milo Brandt
      Dec 5 '18 at 3:02










    • $begingroup$
      I figured out how to get Maxima on Android to plot the function. It DOES have noticeable "kinks" as the curve crosses each integer. Visually, it appears like the cables of a suspension bridge with each pillar higher than the last. I would have expected tetration to smoothly cross the integers. I think my function is wrong: it yields the correct answer at the integers, but it is slightly "low" in between the integers.
      $endgroup$
      – user3412516
      Dec 10 '18 at 19:32














    2












    2








    2





    $begingroup$

    The real question here is what various people consider to be necessary for a formula "that will work." The first requirement you give is very unobjectionable - there is some recurrence relation that tetration should satisfy. This recurrence can be used to take a definition of $^yx$ for $yin [0,1)$ and extend it to work for all positive $y$. However, the issue is not that it is hard to find a function satisfying the laid out condition: The issue is that there are a lot of functions that work - I could define $^yx$ to be anything I want in that interval and there's no clear reason to take $^yx=x^y$ for $0<yleq 1$ as you do - it makes the function continuous, but I could just as easily take $^yx=y(x-1)+1$ to get a continuous answer that will make results that look nice for non-integers.



    Usually, what makes things hard, is that people want conditions like differentiability or convexity in their definition of tetration - this greatly restricts your possibilities. Unfortunately, there's not real consensus on what properties one would like - so there are a number of different functions that might claim to extend tetration.






    share|cite|improve this answer









    $endgroup$



    The real question here is what various people consider to be necessary for a formula "that will work." The first requirement you give is very unobjectionable - there is some recurrence relation that tetration should satisfy. This recurrence can be used to take a definition of $^yx$ for $yin [0,1)$ and extend it to work for all positive $y$. However, the issue is not that it is hard to find a function satisfying the laid out condition: The issue is that there are a lot of functions that work - I could define $^yx$ to be anything I want in that interval and there's no clear reason to take $^yx=x^y$ for $0<yleq 1$ as you do - it makes the function continuous, but I could just as easily take $^yx=y(x-1)+1$ to get a continuous answer that will make results that look nice for non-integers.



    Usually, what makes things hard, is that people want conditions like differentiability or convexity in their definition of tetration - this greatly restricts your possibilities. Unfortunately, there's not real consensus on what properties one would like - so there are a number of different functions that might claim to extend tetration.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 5 '18 at 2:58









    Milo BrandtMilo Brandt

    39.5k475139




    39.5k475139












    • $begingroup$
      ...Mathematica is being annoying right now, but I'll plot the function once it behaves; the function you behave has some noticeable corners in it - it's not differentiable.
      $endgroup$
      – Milo Brandt
      Dec 5 '18 at 3:02










    • $begingroup$
      I figured out how to get Maxima on Android to plot the function. It DOES have noticeable "kinks" as the curve crosses each integer. Visually, it appears like the cables of a suspension bridge with each pillar higher than the last. I would have expected tetration to smoothly cross the integers. I think my function is wrong: it yields the correct answer at the integers, but it is slightly "low" in between the integers.
      $endgroup$
      – user3412516
      Dec 10 '18 at 19:32


















    • $begingroup$
      ...Mathematica is being annoying right now, but I'll plot the function once it behaves; the function you behave has some noticeable corners in it - it's not differentiable.
      $endgroup$
      – Milo Brandt
      Dec 5 '18 at 3:02










    • $begingroup$
      I figured out how to get Maxima on Android to plot the function. It DOES have noticeable "kinks" as the curve crosses each integer. Visually, it appears like the cables of a suspension bridge with each pillar higher than the last. I would have expected tetration to smoothly cross the integers. I think my function is wrong: it yields the correct answer at the integers, but it is slightly "low" in between the integers.
      $endgroup$
      – user3412516
      Dec 10 '18 at 19:32
















    $begingroup$
    ...Mathematica is being annoying right now, but I'll plot the function once it behaves; the function you behave has some noticeable corners in it - it's not differentiable.
    $endgroup$
    – Milo Brandt
    Dec 5 '18 at 3:02




    $begingroup$
    ...Mathematica is being annoying right now, but I'll plot the function once it behaves; the function you behave has some noticeable corners in it - it's not differentiable.
    $endgroup$
    – Milo Brandt
    Dec 5 '18 at 3:02












    $begingroup$
    I figured out how to get Maxima on Android to plot the function. It DOES have noticeable "kinks" as the curve crosses each integer. Visually, it appears like the cables of a suspension bridge with each pillar higher than the last. I would have expected tetration to smoothly cross the integers. I think my function is wrong: it yields the correct answer at the integers, but it is slightly "low" in between the integers.
    $endgroup$
    – user3412516
    Dec 10 '18 at 19:32




    $begingroup$
    I figured out how to get Maxima on Android to plot the function. It DOES have noticeable "kinks" as the curve crosses each integer. Visually, it appears like the cables of a suspension bridge with each pillar higher than the last. I would have expected tetration to smoothly cross the integers. I think my function is wrong: it yields the correct answer at the integers, but it is slightly "low" in between the integers.
    $endgroup$
    – user3412516
    Dec 10 '18 at 19:32











    1












    $begingroup$

    Just to add more visual explanation to the answer of @MiloBrandt you might look at an older casual essay of mine. There I show the effect of setting an individual value is initially interpolated and then exponentiated a small number of $n$. With any selection of the initial the resulting curve is edgy except of one - and not only you need to find this but also some formula by which it depends on $x$. (The pages are made by Excel and are thus imperfect - if you want to play with this you can mail me for the file (Excel 2000 with modules)) The method I used here was already known to and described by G.H. Hardy in a discussion of a function with an interpolated growth-rate, but I don't have the reference at hand.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Just to add more visual explanation to the answer of @MiloBrandt you might look at an older casual essay of mine. There I show the effect of setting an individual value is initially interpolated and then exponentiated a small number of $n$. With any selection of the initial the resulting curve is edgy except of one - and not only you need to find this but also some formula by which it depends on $x$. (The pages are made by Excel and are thus imperfect - if you want to play with this you can mail me for the file (Excel 2000 with modules)) The method I used here was already known to and described by G.H. Hardy in a discussion of a function with an interpolated growth-rate, but I don't have the reference at hand.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Just to add more visual explanation to the answer of @MiloBrandt you might look at an older casual essay of mine. There I show the effect of setting an individual value is initially interpolated and then exponentiated a small number of $n$. With any selection of the initial the resulting curve is edgy except of one - and not only you need to find this but also some formula by which it depends on $x$. (The pages are made by Excel and are thus imperfect - if you want to play with this you can mail me for the file (Excel 2000 with modules)) The method I used here was already known to and described by G.H. Hardy in a discussion of a function with an interpolated growth-rate, but I don't have the reference at hand.






        share|cite|improve this answer









        $endgroup$



        Just to add more visual explanation to the answer of @MiloBrandt you might look at an older casual essay of mine. There I show the effect of setting an individual value is initially interpolated and then exponentiated a small number of $n$. With any selection of the initial the resulting curve is edgy except of one - and not only you need to find this but also some formula by which it depends on $x$. (The pages are made by Excel and are thus imperfect - if you want to play with this you can mail me for the file (Excel 2000 with modules)) The method I used here was already known to and described by G.H. Hardy in a discussion of a function with an interpolated growth-rate, but I don't have the reference at hand.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 5 '18 at 7:22









        Gottfried HelmsGottfried Helms

        23.3k24498




        23.3k24498






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026519%2ftetration-of-non-integers-is-there-something-wrong-with-this-approach%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Probability when a professor distributes a quiz and homework assignment to a class of n students.

            Image
            0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
            Read more

            Aardman Animations

            Image
            Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
            Read more

            Are they similar matrix

            Image
            0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
            Read more