Function such that $f(frac{1}{n}) = n/(n + 1)$











up vote
0
down vote

favorite












I want to prove something about functions that satisfy



$$fleft(frac{1}{n}right) = frac{n}{n + 1}.$$



Obviously, one thing that we can do is



$$f(x) = frac{1}{1 + x} $$



because we can write $frac{n}{n + 1} = frac{1}{1 + frac{1}{n}}$.



But is there any other function that can satisfy this property? I think the answer is no, but how can I show it? Thanks






algebra-precalculus







share|cite|improve this question


















  • 2




    If you are not asking for any special property of $f$ (continuity etc) then you can define $f$ quite arbitrarily when $x$ is not of the type $frac 1 n$.
    – Kavi Rama Murthy
    2 hours ago










  • Domain of function?
    – edm
    2 hours ago










  • Domain $mathbb{R}$
    – joseph
    2 hours ago










  • I also need $f$ analytic if it matters
    – joseph
    2 hours ago










  • Ok if $f$ is analytic, how can I show it is unique ?
    – joseph
    2 hours ago















up vote
0
down vote

favorite












I want to prove something about functions that satisfy



$$fleft(frac{1}{n}right) = frac{n}{n + 1}.$$



Obviously, one thing that we can do is



$$f(x) = frac{1}{1 + x} $$



because we can write $frac{n}{n + 1} = frac{1}{1 + frac{1}{n}}$.



But is there any other function that can satisfy this property? I think the answer is no, but how can I show it? Thanks






algebra-precalculus







share|cite|improve this question


















  • 2




    If you are not asking for any special property of $f$ (continuity etc) then you can define $f$ quite arbitrarily when $x$ is not of the type $frac 1 n$.
    – Kavi Rama Murthy
    2 hours ago










  • Domain of function?
    – edm
    2 hours ago










  • Domain $mathbb{R}$
    – joseph
    2 hours ago










  • I also need $f$ analytic if it matters
    – joseph
    2 hours ago










  • Ok if $f$ is analytic, how can I show it is unique ?
    – joseph
    2 hours ago













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I want to prove something about functions that satisfy



$$fleft(frac{1}{n}right) = frac{n}{n + 1}.$$



Obviously, one thing that we can do is



$$f(x) = frac{1}{1 + x} $$



because we can write $frac{n}{n + 1} = frac{1}{1 + frac{1}{n}}$.



But is there any other function that can satisfy this property? I think the answer is no, but how can I show it? Thanks






algebra-precalculus







share|cite|improve this question













I want to prove something about functions that satisfy



$$fleft(frac{1}{n}right) = frac{n}{n + 1}.$$



Obviously, one thing that we can do is



$$f(x) = frac{1}{1 + x} $$



because we can write $frac{n}{n + 1} = frac{1}{1 + frac{1}{n}}$.



But is there any other function that can satisfy this property? I think the answer is no, but how can I show it? Thanks






algebra-precalculus




algebra-precalculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 2 hours ago









joseph

335




335








  • 2




    If you are not asking for any special property of $f$ (continuity etc) then you can define $f$ quite arbitrarily when $x$ is not of the type $frac 1 n$.
    – Kavi Rama Murthy
    2 hours ago










  • Domain of function?
    – edm
    2 hours ago










  • Domain $mathbb{R}$
    – joseph
    2 hours ago










  • I also need $f$ analytic if it matters
    – joseph
    2 hours ago










  • Ok if $f$ is analytic, how can I show it is unique ?
    – joseph
    2 hours ago














  • 2




    If you are not asking for any special property of $f$ (continuity etc) then you can define $f$ quite arbitrarily when $x$ is not of the type $frac 1 n$.
    – Kavi Rama Murthy
    2 hours ago










  • Domain of function?
    – edm
    2 hours ago










  • Domain $mathbb{R}$
    – joseph
    2 hours ago










  • I also need $f$ analytic if it matters
    – joseph
    2 hours ago










  • Ok if $f$ is analytic, how can I show it is unique ?
    – joseph
    2 hours ago








2




2




If you are not asking for any special property of $f$ (continuity etc) then you can define $f$ quite arbitrarily when $x$ is not of the type $frac 1 n$.
– Kavi Rama Murthy
2 hours ago




If you are not asking for any special property of $f$ (continuity etc) then you can define $f$ quite arbitrarily when $x$ is not of the type $frac 1 n$.
– Kavi Rama Murthy
2 hours ago












Domain of function?
– edm
2 hours ago




Domain of function?
– edm
2 hours ago












Domain $mathbb{R}$
– joseph
2 hours ago




Domain $mathbb{R}$
– joseph
2 hours ago












I also need $f$ analytic if it matters
– joseph
2 hours ago




I also need $f$ analytic if it matters
– joseph
2 hours ago












Ok if $f$ is analytic, how can I show it is unique ?
– joseph
2 hours ago




Ok if $f$ is analytic, how can I show it is unique ?
– joseph
2 hours ago










1 Answer
1






active

oldest

votes

















up vote
2
down vote













If $f$ is analytic in ${|x| <1}$ and $f(frac 1 n)=frac n {n+1}$ for all $n in mathbb N$ then $f(x)-frac 1 {1+x}$ is also analytic in the same disc and it vanishes at the points $frac 1 n$. Hence it must be identically $0$ so $f(x)=frac 1 {1+x}$ for $|x|<1$.






share|cite|improve this answer





















    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',
    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%2f2997925%2ffunction-such-that-f-frac1n-n-n-1%23new-answer', 'question_page');
    }
    );

    Post as a guest


















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote













    If $f$ is analytic in ${|x| <1}$ and $f(frac 1 n)=frac n {n+1}$ for all $n in mathbb N$ then $f(x)-frac 1 {1+x}$ is also analytic in the same disc and it vanishes at the points $frac 1 n$. Hence it must be identically $0$ so $f(x)=frac 1 {1+x}$ for $|x|<1$.






    share|cite|improve this answer

























      up vote
      2
      down vote













      If $f$ is analytic in ${|x| <1}$ and $f(frac 1 n)=frac n {n+1}$ for all $n in mathbb N$ then $f(x)-frac 1 {1+x}$ is also analytic in the same disc and it vanishes at the points $frac 1 n$. Hence it must be identically $0$ so $f(x)=frac 1 {1+x}$ for $|x|<1$.






      share|cite|improve this answer























        up vote
        2
        down vote










        up vote
        2
        down vote









        If $f$ is analytic in ${|x| <1}$ and $f(frac 1 n)=frac n {n+1}$ for all $n in mathbb N$ then $f(x)-frac 1 {1+x}$ is also analytic in the same disc and it vanishes at the points $frac 1 n$. Hence it must be identically $0$ so $f(x)=frac 1 {1+x}$ for $|x|<1$.






        share|cite|improve this answer












        If $f$ is analytic in ${|x| <1}$ and $f(frac 1 n)=frac n {n+1}$ for all $n in mathbb N$ then $f(x)-frac 1 {1+x}$ is also analytic in the same disc and it vanishes at the points $frac 1 n$. Hence it must be identically $0$ so $f(x)=frac 1 {1+x}$ for $|x|<1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 hours ago









        Kavi Rama Murthy

        38.5k31747




        38.5k31747






























             

            draft saved


            draft discarded



















































             


            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2997925%2ffunction-such-that-f-frac1n-n-n-1%23new-answer', 'question_page');
            }
            );

            Post as a guest






































































            -

            Popular posts from this blog

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
            Read more

            Puebla de Zaragoza

            Image
            Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
            Read more

            Musa

            Image
            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
            Read more