Show that two binary expansion are equal [closed]

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I am new to analysis and dyadic expansion. One of the problem I came across is



If $x_n$, $y_n$ are two sequences of zeros and ones, show that



$sum_n=1^inftyx_n/2^n = sum_n=1^inftyy_n/2^n$



if and only if there is an integer $n$ such that $x_k=0$ and $y_k=1$ for all $kgeq n$.



I understand that every number has a unique binary expansion, but this problem just not intuitively make sense to me.






real-analysis binary







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closed as off-topic by Did, José Carlos Santos, Theoretical Economist, Adrian Keister, amWhy Sep 11 at 0:22


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, José Carlos Santos, Theoretical Economist, Adrian Keister, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
















    up vote
    -1
    down vote

    favorite












    I am new to analysis and dyadic expansion. One of the problem I came across is



    If $x_n$, $y_n$ are two sequences of zeros and ones, show that



    $sum_n=1^inftyx_n/2^n = sum_n=1^inftyy_n/2^n$



    if and only if there is an integer $n$ such that $x_k=0$ and $y_k=1$ for all $kgeq n$.



    I understand that every number has a unique binary expansion, but this problem just not intuitively make sense to me.






    real-analysis binary







    share|cite|improve this question













    closed as off-topic by Did, José Carlos Santos, Theoretical Economist, Adrian Keister, amWhy Sep 11 at 0:22


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, José Carlos Santos, Theoretical Economist, Adrian Keister, amWhy
    If this question can be reworded to fit the rules in the help center, please edit the question.














      up vote
      -1
      down vote

      favorite









      up vote
      -1
      down vote

      favorite











      I am new to analysis and dyadic expansion. One of the problem I came across is



      If $x_n$, $y_n$ are two sequences of zeros and ones, show that



      $sum_n=1^inftyx_n/2^n = sum_n=1^inftyy_n/2^n$



      if and only if there is an integer $n$ such that $x_k=0$ and $y_k=1$ for all $kgeq n$.



      I understand that every number has a unique binary expansion, but this problem just not intuitively make sense to me.






      real-analysis binary







      share|cite|improve this question













      I am new to analysis and dyadic expansion. One of the problem I came across is



      If $x_n$, $y_n$ are two sequences of zeros and ones, show that



      $sum_n=1^inftyx_n/2^n = sum_n=1^inftyy_n/2^n$



      if and only if there is an integer $n$ such that $x_k=0$ and $y_k=1$ for all $kgeq n$.



      I understand that every number has a unique binary expansion, but this problem just not intuitively make sense to me.






      real-analysis binary




      real-analysis binary






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      asked Sep 10 at 18:23









      Neyo Yang

      13




      13




      closed as off-topic by Did, José Carlos Santos, Theoretical Economist, Adrian Keister, amWhy Sep 11 at 0:22


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, José Carlos Santos, Theoretical Economist, Adrian Keister, amWhy
      If this question can be reworded to fit the rules in the help center, please edit the question.




      closed as off-topic by Did, José Carlos Santos, Theoretical Economist, Adrian Keister, amWhy Sep 11 at 0:22


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, José Carlos Santos, Theoretical Economist, Adrian Keister, amWhy
      If this question can be reworded to fit the rules in the help center, please edit the question.




















          1 Answer
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          $$sum_n=1^inftyx_n/2^n = sum_n=1^inftyy_n/2^n$$ will not hold if you have all your $x_n =1$ and all you $y_n=0$ for $nge 1$



          On the other hand if you have $x_1=0, x_n=1$ for $nge 2$ we will get



          $$sum_n=1^inftyx_n/2^n=1/2$$



          And if we let $y_1=0$ and $y_n=1$ for all $nge 2$ we get $$ sum_n=1^inftyy_n/2^n=1/2$$



          Thus we have to modify the problem so it makes sense.






          share|cite|improve this answer



























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

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            active

            oldest

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            up vote
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            down vote













            $$sum_n=1^inftyx_n/2^n = sum_n=1^inftyy_n/2^n$$ will not hold if you have all your $x_n =1$ and all you $y_n=0$ for $nge 1$



            On the other hand if you have $x_1=0, x_n=1$ for $nge 2$ we will get



            $$sum_n=1^inftyx_n/2^n=1/2$$



            And if we let $y_1=0$ and $y_n=1$ for all $nge 2$ we get $$ sum_n=1^inftyy_n/2^n=1/2$$



            Thus we have to modify the problem so it makes sense.






            share|cite|improve this answer
























              up vote
              1
              down vote













              $$sum_n=1^inftyx_n/2^n = sum_n=1^inftyy_n/2^n$$ will not hold if you have all your $x_n =1$ and all you $y_n=0$ for $nge 1$



              On the other hand if you have $x_1=0, x_n=1$ for $nge 2$ we will get



              $$sum_n=1^inftyx_n/2^n=1/2$$



              And if we let $y_1=0$ and $y_n=1$ for all $nge 2$ we get $$ sum_n=1^inftyy_n/2^n=1/2$$



              Thus we have to modify the problem so it makes sense.






              share|cite|improve this answer






















                up vote
                1
                down vote










                up vote
                1
                down vote









                $$sum_n=1^inftyx_n/2^n = sum_n=1^inftyy_n/2^n$$ will not hold if you have all your $x_n =1$ and all you $y_n=0$ for $nge 1$



                On the other hand if you have $x_1=0, x_n=1$ for $nge 2$ we will get



                $$sum_n=1^inftyx_n/2^n=1/2$$



                And if we let $y_1=0$ and $y_n=1$ for all $nge 2$ we get $$ sum_n=1^inftyy_n/2^n=1/2$$



                Thus we have to modify the problem so it makes sense.






                share|cite|improve this answer












                $$sum_n=1^inftyx_n/2^n = sum_n=1^inftyy_n/2^n$$ will not hold if you have all your $x_n =1$ and all you $y_n=0$ for $nge 1$



                On the other hand if you have $x_1=0, x_n=1$ for $nge 2$ we will get



                $$sum_n=1^inftyx_n/2^n=1/2$$



                And if we let $y_1=0$ and $y_n=1$ for all $nge 2$ we get $$ sum_n=1^inftyy_n/2^n=1/2$$



                Thus we have to modify the problem so it makes sense.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Sep 10 at 18:43









                Mohammad Riazi-Kermani

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                32.2k41853












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