If $|f(p+q)-f(q)|leq p/q$ for rational $p$ and $q$ (with $qneq 0$), then $sum_{i=0}^k |f(2^k) -f(2^i)| leq...












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If $left|f(p+q) - f(q)right|leq p/q$ for all rational $p$ and $q$, with $q neq 0$, then prove that
$$sum_{i=0}^k left|fleft(2^kright) -fleft(2^iright)right| leq frac12 k(k-1)$$




My try:



I consider the sum for $i=r $ which gives the inequality from given property of function $$ left |fleft(2^kright) -fleft(2^iright)right| leq 2^{k-i} - 1$$, and then summed it, but it doesn't work.






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If $left|f(p+q) - f(q)right|leq p/q$ for all rational $p$ and $q$, with $q neq 0$, then prove that
$$sum_{i=0}^k left|fleft(2^kright) -fleft(2^iright)right| leq frac12 k(k-1)$$




My try:



I consider the sum for $i=r $ which gives the inequality from given property of function $$ left |fleft(2^kright) -fleft(2^iright)right| leq 2^{k-i} - 1$$, and then summed it, but it doesn't work.






calculus functions summation







share|cite|improve this question
























  • $texttt{MathJax}$ Basic Tutorial and Quick Reference.
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  • Thanks for this
    – Keshav Sharma
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If $left|f(p+q) - f(q)right|leq p/q$ for all rational $p$ and $q$, with $q neq 0$, then prove that
$$sum_{i=0}^k left|fleft(2^kright) -fleft(2^iright)right| leq frac12 k(k-1)$$




My try:



I consider the sum for $i=r $ which gives the inequality from given property of function $$ left |fleft(2^kright) -fleft(2^iright)right| leq 2^{k-i} - 1$$, and then summed it, but it doesn't work.






calculus functions summation







share|cite|improve this question
















If $left|f(p+q) - f(q)right|leq p/q$ for all rational $p$ and $q$, with $q neq 0$, then prove that
$$sum_{i=0}^k left|fleft(2^kright) -fleft(2^iright)right| leq frac12 k(k-1)$$




My try:



I consider the sum for $i=r $ which gives the inequality from given property of function $$ left |fleft(2^kright) -fleft(2^iright)right| leq 2^{k-i} - 1$$, and then summed it, but it doesn't work.






calculus functions summation




calculus functions summation






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edited 20 hours ago

























asked yesterday









Keshav Sharma

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  • $texttt{MathJax}$ Basic Tutorial and Quick Reference.
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  • $texttt{MathJax}$ Basic Tutorial and Quick Reference.
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    yesterday












  • Thanks for this
    – Keshav Sharma
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$texttt{MathJax}$ Basic Tutorial and Quick Reference.
– Felix Marin
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$texttt{MathJax}$ Basic Tutorial and Quick Reference.
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Thanks for this
– Keshav Sharma
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Thanks for this
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By triangle inequality



$|f(2^k)-f(2^j)| leq |f(2^k)-f(2^{k-1})|+...|f(2^{j+1})-f(2^j)|leq k-j$



Hence $sum_{i=0}^{i=k}{|f(2^k)-f(2^i)|} leq sum_{i=0}^{i=k}{k-i} leq 0.5(k)(k+1)$



Still not as good as $0.5(k)(k-1)$






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    By triangle inequality



    $|f(2^k)-f(2^j)| leq |f(2^k)-f(2^{k-1})|+...|f(2^{j+1})-f(2^j)|leq k-j$



    Hence $sum_{i=0}^{i=k}{|f(2^k)-f(2^i)|} leq sum_{i=0}^{i=k}{k-i} leq 0.5(k)(k+1)$



    Still not as good as $0.5(k)(k-1)$






    share|cite|improve this answer


























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      By triangle inequality



      $|f(2^k)-f(2^j)| leq |f(2^k)-f(2^{k-1})|+...|f(2^{j+1})-f(2^j)|leq k-j$



      Hence $sum_{i=0}^{i=k}{|f(2^k)-f(2^i)|} leq sum_{i=0}^{i=k}{k-i} leq 0.5(k)(k+1)$



      Still not as good as $0.5(k)(k-1)$






      share|cite|improve this answer
























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        2








        2






        By triangle inequality



        $|f(2^k)-f(2^j)| leq |f(2^k)-f(2^{k-1})|+...|f(2^{j+1})-f(2^j)|leq k-j$



        Hence $sum_{i=0}^{i=k}{|f(2^k)-f(2^i)|} leq sum_{i=0}^{i=k}{k-i} leq 0.5(k)(k+1)$



        Still not as good as $0.5(k)(k-1)$






        share|cite|improve this answer












        By triangle inequality



        $|f(2^k)-f(2^j)| leq |f(2^k)-f(2^{k-1})|+...|f(2^{j+1})-f(2^j)|leq k-j$



        Hence $sum_{i=0}^{i=k}{|f(2^k)-f(2^i)|} leq sum_{i=0}^{i=k}{k-i} leq 0.5(k)(k+1)$



        Still not as good as $0.5(k)(k-1)$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 19 hours ago









        acreativename

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