Find a number having minimum sum of distances between a set of numbers












5












$begingroup$


Lets say we have a set of numbers ${ 5, 7, 1, 2, 5, 100 }$. I want to find a number $x$ such that the sum of distances of every number from the set to $x$ is minimal.



My first thought was that $x$ is the average of all elements of the set: $frac{5+7+1+2+5+100}{6}$, but it is not true, it fails the above example.



Any help or hint will be appriciated, thanks.






optimization average median







share|cite|improve this question











$endgroup$












  • $begingroup$
    The answer is wrong. For example, $100$ is closer to each element than $104$ is.
    $endgroup$
    – Harnak
    Jan 29 at 11:01






  • 2




    $begingroup$
    math.stackexchange.com/questions/113270/…
    $endgroup$
    – Matti P.
    Jan 29 at 11:04










  • $begingroup$
    @Harnak Sorry but I don't quite understand your statement. $104$ is the total distance $sum_{x in S} |x - n|$ where $n$ is the desired number and $S = {1, 2, 5, 5, 7, 100}$. How do you achieve $100$? See my proof.
    $endgroup$
    – L. F.
    Jan 29 at 11:12










  • $begingroup$
    @L.F., I was just answering to the OP before editing. He stated that the solution was 104, but I thought he referred to the minimizer and not to the distance. So, that's why I provided an example of why 104 couldn't be a minimizer. I think I just misinterpreted what he meant.
    $endgroup$
    – Harnak
    Jan 29 at 11:16










  • $begingroup$
    @Harnak Oh, never mind. Everybody makes mistakes :)
    $endgroup$
    – L. F.
    Jan 29 at 11:17
















5












$begingroup$


Lets say we have a set of numbers ${ 5, 7, 1, 2, 5, 100 }$. I want to find a number $x$ such that the sum of distances of every number from the set to $x$ is minimal.



My first thought was that $x$ is the average of all elements of the set: $frac{5+7+1+2+5+100}{6}$, but it is not true, it fails the above example.



Any help or hint will be appriciated, thanks.






optimization average median







share|cite|improve this question











$endgroup$












  • $begingroup$
    The answer is wrong. For example, $100$ is closer to each element than $104$ is.
    $endgroup$
    – Harnak
    Jan 29 at 11:01






  • 2




    $begingroup$
    math.stackexchange.com/questions/113270/…
    $endgroup$
    – Matti P.
    Jan 29 at 11:04










  • $begingroup$
    @Harnak Sorry but I don't quite understand your statement. $104$ is the total distance $sum_{x in S} |x - n|$ where $n$ is the desired number and $S = {1, 2, 5, 5, 7, 100}$. How do you achieve $100$? See my proof.
    $endgroup$
    – L. F.
    Jan 29 at 11:12










  • $begingroup$
    @L.F., I was just answering to the OP before editing. He stated that the solution was 104, but I thought he referred to the minimizer and not to the distance. So, that's why I provided an example of why 104 couldn't be a minimizer. I think I just misinterpreted what he meant.
    $endgroup$
    – Harnak
    Jan 29 at 11:16










  • $begingroup$
    @Harnak Oh, never mind. Everybody makes mistakes :)
    $endgroup$
    – L. F.
    Jan 29 at 11:17














5












5








5





$begingroup$


Lets say we have a set of numbers ${ 5, 7, 1, 2, 5, 100 }$. I want to find a number $x$ such that the sum of distances of every number from the set to $x$ is minimal.



My first thought was that $x$ is the average of all elements of the set: $frac{5+7+1+2+5+100}{6}$, but it is not true, it fails the above example.



Any help or hint will be appriciated, thanks.






optimization average median







share|cite|improve this question











$endgroup$




Lets say we have a set of numbers ${ 5, 7, 1, 2, 5, 100 }$. I want to find a number $x$ such that the sum of distances of every number from the set to $x$ is minimal.



My first thought was that $x$ is the average of all elements of the set: $frac{5+7+1+2+5+100}{6}$, but it is not true, it fails the above example.



Any help or hint will be appriciated, thanks.






optimization average median




optimization average median






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 30 at 10:40







Abdenaceur Lichiheb

















asked Jan 29 at 10:58









Abdenaceur LichihebAbdenaceur Lichiheb

1436




1436












  • $begingroup$
    The answer is wrong. For example, $100$ is closer to each element than $104$ is.
    $endgroup$
    – Harnak
    Jan 29 at 11:01






  • 2




    $begingroup$
    math.stackexchange.com/questions/113270/…
    $endgroup$
    – Matti P.
    Jan 29 at 11:04










  • $begingroup$
    @Harnak Sorry but I don't quite understand your statement. $104$ is the total distance $sum_{x in S} |x - n|$ where $n$ is the desired number and $S = {1, 2, 5, 5, 7, 100}$. How do you achieve $100$? See my proof.
    $endgroup$
    – L. F.
    Jan 29 at 11:12










  • $begingroup$
    @L.F., I was just answering to the OP before editing. He stated that the solution was 104, but I thought he referred to the minimizer and not to the distance. So, that's why I provided an example of why 104 couldn't be a minimizer. I think I just misinterpreted what he meant.
    $endgroup$
    – Harnak
    Jan 29 at 11:16










  • $begingroup$
    @Harnak Oh, never mind. Everybody makes mistakes :)
    $endgroup$
    – L. F.
    Jan 29 at 11:17


















  • $begingroup$
    The answer is wrong. For example, $100$ is closer to each element than $104$ is.
    $endgroup$
    – Harnak
    Jan 29 at 11:01






  • 2




    $begingroup$
    math.stackexchange.com/questions/113270/…
    $endgroup$
    – Matti P.
    Jan 29 at 11:04










  • $begingroup$
    @Harnak Sorry but I don't quite understand your statement. $104$ is the total distance $sum_{x in S} |x - n|$ where $n$ is the desired number and $S = {1, 2, 5, 5, 7, 100}$. How do you achieve $100$? See my proof.
    $endgroup$
    – L. F.
    Jan 29 at 11:12










  • $begingroup$
    @L.F., I was just answering to the OP before editing. He stated that the solution was 104, but I thought he referred to the minimizer and not to the distance. So, that's why I provided an example of why 104 couldn't be a minimizer. I think I just misinterpreted what he meant.
    $endgroup$
    – Harnak
    Jan 29 at 11:16










  • $begingroup$
    @Harnak Oh, never mind. Everybody makes mistakes :)
    $endgroup$
    – L. F.
    Jan 29 at 11:17
















$begingroup$
The answer is wrong. For example, $100$ is closer to each element than $104$ is.
$endgroup$
– Harnak
Jan 29 at 11:01




$begingroup$
The answer is wrong. For example, $100$ is closer to each element than $104$ is.
$endgroup$
– Harnak
Jan 29 at 11:01




2




2




$begingroup$
math.stackexchange.com/questions/113270/…
$endgroup$
– Matti P.
Jan 29 at 11:04




$begingroup$
math.stackexchange.com/questions/113270/…
$endgroup$
– Matti P.
Jan 29 at 11:04












$begingroup$
@Harnak Sorry but I don't quite understand your statement. $104$ is the total distance $sum_{x in S} |x - n|$ where $n$ is the desired number and $S = {1, 2, 5, 5, 7, 100}$. How do you achieve $100$? See my proof.
$endgroup$
– L. F.
Jan 29 at 11:12




$begingroup$
@Harnak Sorry but I don't quite understand your statement. $104$ is the total distance $sum_{x in S} |x - n|$ where $n$ is the desired number and $S = {1, 2, 5, 5, 7, 100}$. How do you achieve $100$? See my proof.
$endgroup$
– L. F.
Jan 29 at 11:12












$begingroup$
@L.F., I was just answering to the OP before editing. He stated that the solution was 104, but I thought he referred to the minimizer and not to the distance. So, that's why I provided an example of why 104 couldn't be a minimizer. I think I just misinterpreted what he meant.
$endgroup$
– Harnak
Jan 29 at 11:16




$begingroup$
@L.F., I was just answering to the OP before editing. He stated that the solution was 104, but I thought he referred to the minimizer and not to the distance. So, that's why I provided an example of why 104 couldn't be a minimizer. I think I just misinterpreted what he meant.
$endgroup$
– Harnak
Jan 29 at 11:16












$begingroup$
@Harnak Oh, never mind. Everybody makes mistakes :)
$endgroup$
– L. F.
Jan 29 at 11:17




$begingroup$
@Harnak Oh, never mind. Everybody makes mistakes :)
$endgroup$
– L. F.
Jan 29 at 11:17










2 Answers
2






active

oldest

votes


















9












$begingroup$

You are looking to minimize $$sum_{y in A} |y - x|$$
with respect to $x$ where $A$ is your set.



It can be proved that any median minimizes this problem. In your case, the only median is $5$, so that's the result.






share|cite|improve this answer











$endgroup$









  • 9




    $begingroup$
    I upvoted your answer because it's correct, but I'd like to make a suggestion about the phrase "this is a well known result": it appears to serve the purpose of making you look well-read, and the OP look ignorant, which might both be true, but does saying it help? Probably not. If you'd included a reference, that'd be great -- OP could learn something. Most likely, this exercise is leading up to the very result you cited, and explaining the computation in this case would be more educational than citing the very thing that can be proved in general from similar computations.
    $endgroup$
    – John Hughes
    Jan 29 at 13:26










  • $begingroup$
    You're right. However, that was not my intent. I'll edit this. Thanks for the suggestion :)
    $endgroup$
    – Harnak
    Jan 29 at 23:15



















4












$begingroup$

First sort your [multi]set: ${1, 2, 5, 5, 7, 100}$. The number you want is $5$. The sum is $4 + 3 + 0 + 0 + 2 + 95 = 104.$



Proof: suppose you have another number $n neq 5$.



Note that for any number $x$, $|x - a| + |x - b| le |a - b|$.
Hence, it must hold that the sum of its distances to the two $5$s, i.e. $$|n - 5| + |n - 5| ge |5 - 5| + |5 - 5| = 0.$$ Similarly, $$|n - 2| + |n - 7| ge |5 - 2| + |5 - 7| = 5,$$
$$|n - 1| + |n - 100| ge |5 - 1| + |5 - 100| = 99.$$



You can't have the total distance any lower.
Q.E.D.



In general, first sort your set, then any number between (including) the middle two numbers will do. For example, for set ${1, 2, 3, 4, 5, 6}$, any $x$ such that $3 le x le 4$ does.






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%2f3092033%2ffind-a-number-having-minimum-sum-of-distances-between-a-set-of-numbers%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









    9












    $begingroup$

    You are looking to minimize $$sum_{y in A} |y - x|$$
    with respect to $x$ where $A$ is your set.



    It can be proved that any median minimizes this problem. In your case, the only median is $5$, so that's the result.






    share|cite|improve this answer











    $endgroup$









    • 9




      $begingroup$
      I upvoted your answer because it's correct, but I'd like to make a suggestion about the phrase "this is a well known result": it appears to serve the purpose of making you look well-read, and the OP look ignorant, which might both be true, but does saying it help? Probably not. If you'd included a reference, that'd be great -- OP could learn something. Most likely, this exercise is leading up to the very result you cited, and explaining the computation in this case would be more educational than citing the very thing that can be proved in general from similar computations.
      $endgroup$
      – John Hughes
      Jan 29 at 13:26










    • $begingroup$
      You're right. However, that was not my intent. I'll edit this. Thanks for the suggestion :)
      $endgroup$
      – Harnak
      Jan 29 at 23:15
















    9












    $begingroup$

    You are looking to minimize $$sum_{y in A} |y - x|$$
    with respect to $x$ where $A$ is your set.



    It can be proved that any median minimizes this problem. In your case, the only median is $5$, so that's the result.






    share|cite|improve this answer











    $endgroup$









    • 9




      $begingroup$
      I upvoted your answer because it's correct, but I'd like to make a suggestion about the phrase "this is a well known result": it appears to serve the purpose of making you look well-read, and the OP look ignorant, which might both be true, but does saying it help? Probably not. If you'd included a reference, that'd be great -- OP could learn something. Most likely, this exercise is leading up to the very result you cited, and explaining the computation in this case would be more educational than citing the very thing that can be proved in general from similar computations.
      $endgroup$
      – John Hughes
      Jan 29 at 13:26










    • $begingroup$
      You're right. However, that was not my intent. I'll edit this. Thanks for the suggestion :)
      $endgroup$
      – Harnak
      Jan 29 at 23:15














    9












    9








    9





    $begingroup$

    You are looking to minimize $$sum_{y in A} |y - x|$$
    with respect to $x$ where $A$ is your set.



    It can be proved that any median minimizes this problem. In your case, the only median is $5$, so that's the result.






    share|cite|improve this answer











    $endgroup$



    You are looking to minimize $$sum_{y in A} |y - x|$$
    with respect to $x$ where $A$ is your set.



    It can be proved that any median minimizes this problem. In your case, the only median is $5$, so that's the result.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 29 at 23:50

























    answered Jan 29 at 11:06









    HarnakHarnak

    1,309512




    1,309512








    • 9




      $begingroup$
      I upvoted your answer because it's correct, but I'd like to make a suggestion about the phrase "this is a well known result": it appears to serve the purpose of making you look well-read, and the OP look ignorant, which might both be true, but does saying it help? Probably not. If you'd included a reference, that'd be great -- OP could learn something. Most likely, this exercise is leading up to the very result you cited, and explaining the computation in this case would be more educational than citing the very thing that can be proved in general from similar computations.
      $endgroup$
      – John Hughes
      Jan 29 at 13:26










    • $begingroup$
      You're right. However, that was not my intent. I'll edit this. Thanks for the suggestion :)
      $endgroup$
      – Harnak
      Jan 29 at 23:15














    • 9




      $begingroup$
      I upvoted your answer because it's correct, but I'd like to make a suggestion about the phrase "this is a well known result": it appears to serve the purpose of making you look well-read, and the OP look ignorant, which might both be true, but does saying it help? Probably not. If you'd included a reference, that'd be great -- OP could learn something. Most likely, this exercise is leading up to the very result you cited, and explaining the computation in this case would be more educational than citing the very thing that can be proved in general from similar computations.
      $endgroup$
      – John Hughes
      Jan 29 at 13:26










    • $begingroup$
      You're right. However, that was not my intent. I'll edit this. Thanks for the suggestion :)
      $endgroup$
      – Harnak
      Jan 29 at 23:15








    9




    9




    $begingroup$
    I upvoted your answer because it's correct, but I'd like to make a suggestion about the phrase "this is a well known result": it appears to serve the purpose of making you look well-read, and the OP look ignorant, which might both be true, but does saying it help? Probably not. If you'd included a reference, that'd be great -- OP could learn something. Most likely, this exercise is leading up to the very result you cited, and explaining the computation in this case would be more educational than citing the very thing that can be proved in general from similar computations.
    $endgroup$
    – John Hughes
    Jan 29 at 13:26




    $begingroup$
    I upvoted your answer because it's correct, but I'd like to make a suggestion about the phrase "this is a well known result": it appears to serve the purpose of making you look well-read, and the OP look ignorant, which might both be true, but does saying it help? Probably not. If you'd included a reference, that'd be great -- OP could learn something. Most likely, this exercise is leading up to the very result you cited, and explaining the computation in this case would be more educational than citing the very thing that can be proved in general from similar computations.
    $endgroup$
    – John Hughes
    Jan 29 at 13:26












    $begingroup$
    You're right. However, that was not my intent. I'll edit this. Thanks for the suggestion :)
    $endgroup$
    – Harnak
    Jan 29 at 23:15




    $begingroup$
    You're right. However, that was not my intent. I'll edit this. Thanks for the suggestion :)
    $endgroup$
    – Harnak
    Jan 29 at 23:15











    4












    $begingroup$

    First sort your [multi]set: ${1, 2, 5, 5, 7, 100}$. The number you want is $5$. The sum is $4 + 3 + 0 + 0 + 2 + 95 = 104.$



    Proof: suppose you have another number $n neq 5$.



    Note that for any number $x$, $|x - a| + |x - b| le |a - b|$.
    Hence, it must hold that the sum of its distances to the two $5$s, i.e. $$|n - 5| + |n - 5| ge |5 - 5| + |5 - 5| = 0.$$ Similarly, $$|n - 2| + |n - 7| ge |5 - 2| + |5 - 7| = 5,$$
    $$|n - 1| + |n - 100| ge |5 - 1| + |5 - 100| = 99.$$



    You can't have the total distance any lower.
    Q.E.D.



    In general, first sort your set, then any number between (including) the middle two numbers will do. For example, for set ${1, 2, 3, 4, 5, 6}$, any $x$ such that $3 le x le 4$ does.






    share|cite|improve this answer











    $endgroup$


















      4












      $begingroup$

      First sort your [multi]set: ${1, 2, 5, 5, 7, 100}$. The number you want is $5$. The sum is $4 + 3 + 0 + 0 + 2 + 95 = 104.$



      Proof: suppose you have another number $n neq 5$.



      Note that for any number $x$, $|x - a| + |x - b| le |a - b|$.
      Hence, it must hold that the sum of its distances to the two $5$s, i.e. $$|n - 5| + |n - 5| ge |5 - 5| + |5 - 5| = 0.$$ Similarly, $$|n - 2| + |n - 7| ge |5 - 2| + |5 - 7| = 5,$$
      $$|n - 1| + |n - 100| ge |5 - 1| + |5 - 100| = 99.$$



      You can't have the total distance any lower.
      Q.E.D.



      In general, first sort your set, then any number between (including) the middle two numbers will do. For example, for set ${1, 2, 3, 4, 5, 6}$, any $x$ such that $3 le x le 4$ does.






      share|cite|improve this answer











      $endgroup$
















        4












        4








        4





        $begingroup$

        First sort your [multi]set: ${1, 2, 5, 5, 7, 100}$. The number you want is $5$. The sum is $4 + 3 + 0 + 0 + 2 + 95 = 104.$



        Proof: suppose you have another number $n neq 5$.



        Note that for any number $x$, $|x - a| + |x - b| le |a - b|$.
        Hence, it must hold that the sum of its distances to the two $5$s, i.e. $$|n - 5| + |n - 5| ge |5 - 5| + |5 - 5| = 0.$$ Similarly, $$|n - 2| + |n - 7| ge |5 - 2| + |5 - 7| = 5,$$
        $$|n - 1| + |n - 100| ge |5 - 1| + |5 - 100| = 99.$$



        You can't have the total distance any lower.
        Q.E.D.



        In general, first sort your set, then any number between (including) the middle two numbers will do. For example, for set ${1, 2, 3, 4, 5, 6}$, any $x$ such that $3 le x le 4$ does.






        share|cite|improve this answer











        $endgroup$



        First sort your [multi]set: ${1, 2, 5, 5, 7, 100}$. The number you want is $5$. The sum is $4 + 3 + 0 + 0 + 2 + 95 = 104.$



        Proof: suppose you have another number $n neq 5$.



        Note that for any number $x$, $|x - a| + |x - b| le |a - b|$.
        Hence, it must hold that the sum of its distances to the two $5$s, i.e. $$|n - 5| + |n - 5| ge |5 - 5| + |5 - 5| = 0.$$ Similarly, $$|n - 2| + |n - 7| ge |5 - 2| + |5 - 7| = 5,$$
        $$|n - 1| + |n - 100| ge |5 - 1| + |5 - 100| = 99.$$



        You can't have the total distance any lower.
        Q.E.D.



        In general, first sort your set, then any number between (including) the middle two numbers will do. For example, for set ${1, 2, 3, 4, 5, 6}$, any $x$ such that $3 le x le 4$ does.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 29 at 11:13

























        answered Jan 29 at 11:06









        L. F.L. F.

        18511




        18511






























            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%2f3092033%2ffind-a-number-having-minimum-sum-of-distances-between-a-set-of-numbers%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

            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