Show that $|y-y_0|<text{min}(frac{|y_0|}{2},frac{epsilon|y_0|^2}{2}) implies leftvert...












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I need to show that if $|y-y_0|<text{min}(frac{|y_0|}{2},frac{epsilon|y_0|^2}{2})$ and $yneq 0$ and $y_0neq 0$ are true, then the following inequality is also true:
$$ leftvert frac{1}{y}-frac{1}{y_0} rightvert < epsilon.$$



I tried to check the information given by the inequalities $|y-y_0|<frac{|y_0|}{2}$ and $|y-y_0|<frac{epsilon|y_0|^2}{2}$. Then I arrived from those inequalities to these ones: $|frac{y}{y_0}-1|<frac{1}{2}$ and $|frac{1}{y} cdot frac{y^2}{y_0^2}-frac{1}{y_0}|<frac{epsilon}{2}$.



I multiplied the inequality $|y-y_0|<frac{epsilon|y_0|^2}{2}$ by $frac{1}{|y y_0|}$ , because $frac{1}{|yy_0|}>0$. It follows that $|frac{1}{y_0}-frac{1}{y}|<frac{epsilon|y_0|}{2|y|}$. Because $|frac{1}{y_0}-frac{1}{y}|=|frac{1}{y}-frac{1}{y_0}|$ is true for the absolute value function, we get $|frac{1}{y}-frac{1}{y_0}|<frac{epsilon|y_0|}{2|y|}$...



I am still trying to figure out, how I can prove $ leftvert frac{1}{y}-frac{1}{y_0} rightvert < epsilon.$



Any help is appreciated. Thanks in advance.



Regards,



Ahmed Hossam










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$endgroup$

















    0












    $begingroup$


    I need to show that if $|y-y_0|<text{min}(frac{|y_0|}{2},frac{epsilon|y_0|^2}{2})$ and $yneq 0$ and $y_0neq 0$ are true, then the following inequality is also true:
    $$ leftvert frac{1}{y}-frac{1}{y_0} rightvert < epsilon.$$



    I tried to check the information given by the inequalities $|y-y_0|<frac{|y_0|}{2}$ and $|y-y_0|<frac{epsilon|y_0|^2}{2}$. Then I arrived from those inequalities to these ones: $|frac{y}{y_0}-1|<frac{1}{2}$ and $|frac{1}{y} cdot frac{y^2}{y_0^2}-frac{1}{y_0}|<frac{epsilon}{2}$.



    I multiplied the inequality $|y-y_0|<frac{epsilon|y_0|^2}{2}$ by $frac{1}{|y y_0|}$ , because $frac{1}{|yy_0|}>0$. It follows that $|frac{1}{y_0}-frac{1}{y}|<frac{epsilon|y_0|}{2|y|}$. Because $|frac{1}{y_0}-frac{1}{y}|=|frac{1}{y}-frac{1}{y_0}|$ is true for the absolute value function, we get $|frac{1}{y}-frac{1}{y_0}|<frac{epsilon|y_0|}{2|y|}$...



    I am still trying to figure out, how I can prove $ leftvert frac{1}{y}-frac{1}{y_0} rightvert < epsilon.$



    Any help is appreciated. Thanks in advance.



    Regards,



    Ahmed Hossam










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I need to show that if $|y-y_0|<text{min}(frac{|y_0|}{2},frac{epsilon|y_0|^2}{2})$ and $yneq 0$ and $y_0neq 0$ are true, then the following inequality is also true:
      $$ leftvert frac{1}{y}-frac{1}{y_0} rightvert < epsilon.$$



      I tried to check the information given by the inequalities $|y-y_0|<frac{|y_0|}{2}$ and $|y-y_0|<frac{epsilon|y_0|^2}{2}$. Then I arrived from those inequalities to these ones: $|frac{y}{y_0}-1|<frac{1}{2}$ and $|frac{1}{y} cdot frac{y^2}{y_0^2}-frac{1}{y_0}|<frac{epsilon}{2}$.



      I multiplied the inequality $|y-y_0|<frac{epsilon|y_0|^2}{2}$ by $frac{1}{|y y_0|}$ , because $frac{1}{|yy_0|}>0$. It follows that $|frac{1}{y_0}-frac{1}{y}|<frac{epsilon|y_0|}{2|y|}$. Because $|frac{1}{y_0}-frac{1}{y}|=|frac{1}{y}-frac{1}{y_0}|$ is true for the absolute value function, we get $|frac{1}{y}-frac{1}{y_0}|<frac{epsilon|y_0|}{2|y|}$...



      I am still trying to figure out, how I can prove $ leftvert frac{1}{y}-frac{1}{y_0} rightvert < epsilon.$



      Any help is appreciated. Thanks in advance.



      Regards,



      Ahmed Hossam










      share|cite|improve this question











      $endgroup$




      I need to show that if $|y-y_0|<text{min}(frac{|y_0|}{2},frac{epsilon|y_0|^2}{2})$ and $yneq 0$ and $y_0neq 0$ are true, then the following inequality is also true:
      $$ leftvert frac{1}{y}-frac{1}{y_0} rightvert < epsilon.$$



      I tried to check the information given by the inequalities $|y-y_0|<frac{|y_0|}{2}$ and $|y-y_0|<frac{epsilon|y_0|^2}{2}$. Then I arrived from those inequalities to these ones: $|frac{y}{y_0}-1|<frac{1}{2}$ and $|frac{1}{y} cdot frac{y^2}{y_0^2}-frac{1}{y_0}|<frac{epsilon}{2}$.



      I multiplied the inequality $|y-y_0|<frac{epsilon|y_0|^2}{2}$ by $frac{1}{|y y_0|}$ , because $frac{1}{|yy_0|}>0$. It follows that $|frac{1}{y_0}-frac{1}{y}|<frac{epsilon|y_0|}{2|y|}$. Because $|frac{1}{y_0}-frac{1}{y}|=|frac{1}{y}-frac{1}{y_0}|$ is true for the absolute value function, we get $|frac{1}{y}-frac{1}{y_0}|<frac{epsilon|y_0|}{2|y|}$...



      I am still trying to figure out, how I can prove $ leftvert frac{1}{y}-frac{1}{y_0} rightvert < epsilon.$



      Any help is appreciated. Thanks in advance.



      Regards,



      Ahmed Hossam







      real-analysis calculus inequality






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      edited Dec 9 '18 at 20:14







      Ahmed Hossam

















      asked Dec 9 '18 at 19:48









      Ahmed HossamAhmed Hossam

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          $begingroup$

          Case 1:



          $$text{min}(frac{|y_0|}{2},frac{epsilon|y_0|^2}{2})=frac{|y_0|}{2}$$



          $$frac{|y_0|}{2}lefrac{epsilon|y_0|^2}{2}$$



          $$frac1{|y_0|}leepsilontag{1}$$



          In this particular case:



          $$|y-y_0|lt frac{|y_0|}{2}$$



          $$-frac{|y_0|}{2}<y-y_0lt frac{|y_0|}{2}$$



          $$y_0-frac{|y_0|}{2}<y< y_0+frac{|y_0|}{2}tag{2}$$



          If $y_0>0$, (2) becomes:



          $$frac{y_0}{2}<y< frac{3y_0}{2}$$



          $$frac{2}{y_0}>frac1y> frac{2}{3y_0}$$



          $$frac{2}{y_0}-frac{1}{y_0}>frac1y - frac{1}{y_0}> frac{2}{3y_0}-frac{1}{y_0}$$



          $$frac{1}{y_0}>frac1y - frac{1}{y_0}> -frac{1}{3y_0}$$



          $$frac{1}{y_0}>|frac1y - frac{1}{y_0}|tag{3}$$



          Because of (1) and (3) we have:



          $$epsilon>|frac1y - frac{1}{y_0}|$$



          If $y_0<0$, (2) becomes:



          $$y_0+frac{y_0}{2}<y< y_0-frac{y_0}{2}$$



          $$frac{3y_0}{2}<y< frac{y_0}{2}$$



          Notice that $y$ is squeezed between negative values so we can write:



          $$frac2{3y_0}>frac1y> frac2{y_0}$$



          $$frac2{3y_0}-frac{1}{y_0}>frac1y-frac1{y_0}> frac2{y_0}-frac1{y_0}$$



          $$-frac{1}{3y_0}>frac1y-frac1{y_0}> frac1{y_0}$$



          $$|frac1y-frac1{y_0}|< frac1{|y_0|}tag{4}$$



          Because of (1) and (4) we have:



          $$|frac1y-frac1{y_0}|< epsilon$$



          Case 2:



          $$text{min}(frac{|y_0|}{2},frac{epsilon|y_0|^2}{2})=frac{epsilon|y_0|^2}{2}$$



          $$frac{|y_0|}{2}gefrac{epsilon|y_0|^2}{2})$$



          $$1geepsilon|y_0|tag{5}$$



          In this particular case:



          $$|y-y_0|lt frac{epsilon|y_0|^2}{2}$$



          $$-frac{epsilon|y_0|^2}{2}lt y-y_0lt frac{epsilon|y_0|^2}{2}$$



          $$y_0-frac{epsilon|y_0|^2}{2}lt y lt y_0+frac{epsilon|y_0|^2}{2}tag{6}$$



          For $y_0>0$, (6) becomes:



          $$y_0-frac{epsilon y_0^2}{2}lt y lt y_0+frac{epsilon y_0^2}{2}$$



          Notice that because of (5) $y$ is squeezed between positive values so we can write:



          $$frac{1}{y_0-frac{epsilon y_0^2}{2}}gt frac1y gt frac1{y_0+frac{epsilon y_0^2}{2}}$$



          $$frac{1}{y_0-frac{epsilon y_0^2}{2}}-frac 1{y_0} gt frac1y-frac 1{y_0} gt frac1{y_0+frac{epsilon y_0^2}{2}}-frac 1{y_0}$$



          This evolves into:



          $$frac{epsilon}{2-epsilon y_0}>frac1y-frac 1{y_0}>-frac{epsilon}{2+epsilon y_0}tag{7}$$



          Notice that because of (5):



          $$2-epsilon y_0 ge 1 implies epsilon ge frac{epsilon}{2-epsilon y_0}$$



          $$2+epsilon y_0 ge 2 implies -frac{epsilon}{2+epsilon y_0}>-frac{epsilon}{2}$$



          With this in mind, (7) now becomes:



          $$epsilon >frac1y-frac 1{y_0}>-frac{epsilon}{2}$$



          $$epsilon >|frac1y-frac 1{y_0}|$$



          I'm leaving the last case ($y_0<0$) to you as an exercise.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for your answer. The equation $|y-y_0|<frac{epsilon|y_0|^2}{2}$ can be multiplied with $frac{1}{|y y_0|}$ , because $frac{1}{|yy_0|}>0$ . Thus $|frac{1}{y_0}-frac{1}{y}|<frac{epsilon|y_0|}{2|y|}$. It is $|frac{1}{y_0}-frac{1}{y}|=|frac{1}{y}-frac{1}{y_0}|$, That's why $|frac{1}{y}-frac{1}{y_0}|<frac{epsilon|y_0|}{2|y|}$ is true. $||y|-|y_0||leq |y-y_0|<frac{|y_0|}{2} Rightarrow |frac{|y|}{|y_0|}-1|<frac{1}{2} Rightarrow 2>frac{|y_0|}{|y|}>frac{2}{3}$. This means $|frac{1}{y}-frac{1}{y_0}|<frac{epsilon|y_0|}{2|y|}<epsilon$, because $2>frac{|y_0|}{|y|}$.
            $endgroup$
            – Ahmed Hossam
            Dec 10 '18 at 19:20













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          $begingroup$

          Case 1:



          $$text{min}(frac{|y_0|}{2},frac{epsilon|y_0|^2}{2})=frac{|y_0|}{2}$$



          $$frac{|y_0|}{2}lefrac{epsilon|y_0|^2}{2}$$



          $$frac1{|y_0|}leepsilontag{1}$$



          In this particular case:



          $$|y-y_0|lt frac{|y_0|}{2}$$



          $$-frac{|y_0|}{2}<y-y_0lt frac{|y_0|}{2}$$



          $$y_0-frac{|y_0|}{2}<y< y_0+frac{|y_0|}{2}tag{2}$$



          If $y_0>0$, (2) becomes:



          $$frac{y_0}{2}<y< frac{3y_0}{2}$$



          $$frac{2}{y_0}>frac1y> frac{2}{3y_0}$$



          $$frac{2}{y_0}-frac{1}{y_0}>frac1y - frac{1}{y_0}> frac{2}{3y_0}-frac{1}{y_0}$$



          $$frac{1}{y_0}>frac1y - frac{1}{y_0}> -frac{1}{3y_0}$$



          $$frac{1}{y_0}>|frac1y - frac{1}{y_0}|tag{3}$$



          Because of (1) and (3) we have:



          $$epsilon>|frac1y - frac{1}{y_0}|$$



          If $y_0<0$, (2) becomes:



          $$y_0+frac{y_0}{2}<y< y_0-frac{y_0}{2}$$



          $$frac{3y_0}{2}<y< frac{y_0}{2}$$



          Notice that $y$ is squeezed between negative values so we can write:



          $$frac2{3y_0}>frac1y> frac2{y_0}$$



          $$frac2{3y_0}-frac{1}{y_0}>frac1y-frac1{y_0}> frac2{y_0}-frac1{y_0}$$



          $$-frac{1}{3y_0}>frac1y-frac1{y_0}> frac1{y_0}$$



          $$|frac1y-frac1{y_0}|< frac1{|y_0|}tag{4}$$



          Because of (1) and (4) we have:



          $$|frac1y-frac1{y_0}|< epsilon$$



          Case 2:



          $$text{min}(frac{|y_0|}{2},frac{epsilon|y_0|^2}{2})=frac{epsilon|y_0|^2}{2}$$



          $$frac{|y_0|}{2}gefrac{epsilon|y_0|^2}{2})$$



          $$1geepsilon|y_0|tag{5}$$



          In this particular case:



          $$|y-y_0|lt frac{epsilon|y_0|^2}{2}$$



          $$-frac{epsilon|y_0|^2}{2}lt y-y_0lt frac{epsilon|y_0|^2}{2}$$



          $$y_0-frac{epsilon|y_0|^2}{2}lt y lt y_0+frac{epsilon|y_0|^2}{2}tag{6}$$



          For $y_0>0$, (6) becomes:



          $$y_0-frac{epsilon y_0^2}{2}lt y lt y_0+frac{epsilon y_0^2}{2}$$



          Notice that because of (5) $y$ is squeezed between positive values so we can write:



          $$frac{1}{y_0-frac{epsilon y_0^2}{2}}gt frac1y gt frac1{y_0+frac{epsilon y_0^2}{2}}$$



          $$frac{1}{y_0-frac{epsilon y_0^2}{2}}-frac 1{y_0} gt frac1y-frac 1{y_0} gt frac1{y_0+frac{epsilon y_0^2}{2}}-frac 1{y_0}$$



          This evolves into:



          $$frac{epsilon}{2-epsilon y_0}>frac1y-frac 1{y_0}>-frac{epsilon}{2+epsilon y_0}tag{7}$$



          Notice that because of (5):



          $$2-epsilon y_0 ge 1 implies epsilon ge frac{epsilon}{2-epsilon y_0}$$



          $$2+epsilon y_0 ge 2 implies -frac{epsilon}{2+epsilon y_0}>-frac{epsilon}{2}$$



          With this in mind, (7) now becomes:



          $$epsilon >frac1y-frac 1{y_0}>-frac{epsilon}{2}$$



          $$epsilon >|frac1y-frac 1{y_0}|$$



          I'm leaving the last case ($y_0<0$) to you as an exercise.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for your answer. The equation $|y-y_0|<frac{epsilon|y_0|^2}{2}$ can be multiplied with $frac{1}{|y y_0|}$ , because $frac{1}{|yy_0|}>0$ . Thus $|frac{1}{y_0}-frac{1}{y}|<frac{epsilon|y_0|}{2|y|}$. It is $|frac{1}{y_0}-frac{1}{y}|=|frac{1}{y}-frac{1}{y_0}|$, That's why $|frac{1}{y}-frac{1}{y_0}|<frac{epsilon|y_0|}{2|y|}$ is true. $||y|-|y_0||leq |y-y_0|<frac{|y_0|}{2} Rightarrow |frac{|y|}{|y_0|}-1|<frac{1}{2} Rightarrow 2>frac{|y_0|}{|y|}>frac{2}{3}$. This means $|frac{1}{y}-frac{1}{y_0}|<frac{epsilon|y_0|}{2|y|}<epsilon$, because $2>frac{|y_0|}{|y|}$.
            $endgroup$
            – Ahmed Hossam
            Dec 10 '18 at 19:20


















          0












          $begingroup$

          Case 1:



          $$text{min}(frac{|y_0|}{2},frac{epsilon|y_0|^2}{2})=frac{|y_0|}{2}$$



          $$frac{|y_0|}{2}lefrac{epsilon|y_0|^2}{2}$$



          $$frac1{|y_0|}leepsilontag{1}$$



          In this particular case:



          $$|y-y_0|lt frac{|y_0|}{2}$$



          $$-frac{|y_0|}{2}<y-y_0lt frac{|y_0|}{2}$$



          $$y_0-frac{|y_0|}{2}<y< y_0+frac{|y_0|}{2}tag{2}$$



          If $y_0>0$, (2) becomes:



          $$frac{y_0}{2}<y< frac{3y_0}{2}$$



          $$frac{2}{y_0}>frac1y> frac{2}{3y_0}$$



          $$frac{2}{y_0}-frac{1}{y_0}>frac1y - frac{1}{y_0}> frac{2}{3y_0}-frac{1}{y_0}$$



          $$frac{1}{y_0}>frac1y - frac{1}{y_0}> -frac{1}{3y_0}$$



          $$frac{1}{y_0}>|frac1y - frac{1}{y_0}|tag{3}$$



          Because of (1) and (3) we have:



          $$epsilon>|frac1y - frac{1}{y_0}|$$



          If $y_0<0$, (2) becomes:



          $$y_0+frac{y_0}{2}<y< y_0-frac{y_0}{2}$$



          $$frac{3y_0}{2}<y< frac{y_0}{2}$$



          Notice that $y$ is squeezed between negative values so we can write:



          $$frac2{3y_0}>frac1y> frac2{y_0}$$



          $$frac2{3y_0}-frac{1}{y_0}>frac1y-frac1{y_0}> frac2{y_0}-frac1{y_0}$$



          $$-frac{1}{3y_0}>frac1y-frac1{y_0}> frac1{y_0}$$



          $$|frac1y-frac1{y_0}|< frac1{|y_0|}tag{4}$$



          Because of (1) and (4) we have:



          $$|frac1y-frac1{y_0}|< epsilon$$



          Case 2:



          $$text{min}(frac{|y_0|}{2},frac{epsilon|y_0|^2}{2})=frac{epsilon|y_0|^2}{2}$$



          $$frac{|y_0|}{2}gefrac{epsilon|y_0|^2}{2})$$



          $$1geepsilon|y_0|tag{5}$$



          In this particular case:



          $$|y-y_0|lt frac{epsilon|y_0|^2}{2}$$



          $$-frac{epsilon|y_0|^2}{2}lt y-y_0lt frac{epsilon|y_0|^2}{2}$$



          $$y_0-frac{epsilon|y_0|^2}{2}lt y lt y_0+frac{epsilon|y_0|^2}{2}tag{6}$$



          For $y_0>0$, (6) becomes:



          $$y_0-frac{epsilon y_0^2}{2}lt y lt y_0+frac{epsilon y_0^2}{2}$$



          Notice that because of (5) $y$ is squeezed between positive values so we can write:



          $$frac{1}{y_0-frac{epsilon y_0^2}{2}}gt frac1y gt frac1{y_0+frac{epsilon y_0^2}{2}}$$



          $$frac{1}{y_0-frac{epsilon y_0^2}{2}}-frac 1{y_0} gt frac1y-frac 1{y_0} gt frac1{y_0+frac{epsilon y_0^2}{2}}-frac 1{y_0}$$



          This evolves into:



          $$frac{epsilon}{2-epsilon y_0}>frac1y-frac 1{y_0}>-frac{epsilon}{2+epsilon y_0}tag{7}$$



          Notice that because of (5):



          $$2-epsilon y_0 ge 1 implies epsilon ge frac{epsilon}{2-epsilon y_0}$$



          $$2+epsilon y_0 ge 2 implies -frac{epsilon}{2+epsilon y_0}>-frac{epsilon}{2}$$



          With this in mind, (7) now becomes:



          $$epsilon >frac1y-frac 1{y_0}>-frac{epsilon}{2}$$



          $$epsilon >|frac1y-frac 1{y_0}|$$



          I'm leaving the last case ($y_0<0$) to you as an exercise.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for your answer. The equation $|y-y_0|<frac{epsilon|y_0|^2}{2}$ can be multiplied with $frac{1}{|y y_0|}$ , because $frac{1}{|yy_0|}>0$ . Thus $|frac{1}{y_0}-frac{1}{y}|<frac{epsilon|y_0|}{2|y|}$. It is $|frac{1}{y_0}-frac{1}{y}|=|frac{1}{y}-frac{1}{y_0}|$, That's why $|frac{1}{y}-frac{1}{y_0}|<frac{epsilon|y_0|}{2|y|}$ is true. $||y|-|y_0||leq |y-y_0|<frac{|y_0|}{2} Rightarrow |frac{|y|}{|y_0|}-1|<frac{1}{2} Rightarrow 2>frac{|y_0|}{|y|}>frac{2}{3}$. This means $|frac{1}{y}-frac{1}{y_0}|<frac{epsilon|y_0|}{2|y|}<epsilon$, because $2>frac{|y_0|}{|y|}$.
            $endgroup$
            – Ahmed Hossam
            Dec 10 '18 at 19:20
















          0












          0








          0





          $begingroup$

          Case 1:



          $$text{min}(frac{|y_0|}{2},frac{epsilon|y_0|^2}{2})=frac{|y_0|}{2}$$



          $$frac{|y_0|}{2}lefrac{epsilon|y_0|^2}{2}$$



          $$frac1{|y_0|}leepsilontag{1}$$



          In this particular case:



          $$|y-y_0|lt frac{|y_0|}{2}$$



          $$-frac{|y_0|}{2}<y-y_0lt frac{|y_0|}{2}$$



          $$y_0-frac{|y_0|}{2}<y< y_0+frac{|y_0|}{2}tag{2}$$



          If $y_0>0$, (2) becomes:



          $$frac{y_0}{2}<y< frac{3y_0}{2}$$



          $$frac{2}{y_0}>frac1y> frac{2}{3y_0}$$



          $$frac{2}{y_0}-frac{1}{y_0}>frac1y - frac{1}{y_0}> frac{2}{3y_0}-frac{1}{y_0}$$



          $$frac{1}{y_0}>frac1y - frac{1}{y_0}> -frac{1}{3y_0}$$



          $$frac{1}{y_0}>|frac1y - frac{1}{y_0}|tag{3}$$



          Because of (1) and (3) we have:



          $$epsilon>|frac1y - frac{1}{y_0}|$$



          If $y_0<0$, (2) becomes:



          $$y_0+frac{y_0}{2}<y< y_0-frac{y_0}{2}$$



          $$frac{3y_0}{2}<y< frac{y_0}{2}$$



          Notice that $y$ is squeezed between negative values so we can write:



          $$frac2{3y_0}>frac1y> frac2{y_0}$$



          $$frac2{3y_0}-frac{1}{y_0}>frac1y-frac1{y_0}> frac2{y_0}-frac1{y_0}$$



          $$-frac{1}{3y_0}>frac1y-frac1{y_0}> frac1{y_0}$$



          $$|frac1y-frac1{y_0}|< frac1{|y_0|}tag{4}$$



          Because of (1) and (4) we have:



          $$|frac1y-frac1{y_0}|< epsilon$$



          Case 2:



          $$text{min}(frac{|y_0|}{2},frac{epsilon|y_0|^2}{2})=frac{epsilon|y_0|^2}{2}$$



          $$frac{|y_0|}{2}gefrac{epsilon|y_0|^2}{2})$$



          $$1geepsilon|y_0|tag{5}$$



          In this particular case:



          $$|y-y_0|lt frac{epsilon|y_0|^2}{2}$$



          $$-frac{epsilon|y_0|^2}{2}lt y-y_0lt frac{epsilon|y_0|^2}{2}$$



          $$y_0-frac{epsilon|y_0|^2}{2}lt y lt y_0+frac{epsilon|y_0|^2}{2}tag{6}$$



          For $y_0>0$, (6) becomes:



          $$y_0-frac{epsilon y_0^2}{2}lt y lt y_0+frac{epsilon y_0^2}{2}$$



          Notice that because of (5) $y$ is squeezed between positive values so we can write:



          $$frac{1}{y_0-frac{epsilon y_0^2}{2}}gt frac1y gt frac1{y_0+frac{epsilon y_0^2}{2}}$$



          $$frac{1}{y_0-frac{epsilon y_0^2}{2}}-frac 1{y_0} gt frac1y-frac 1{y_0} gt frac1{y_0+frac{epsilon y_0^2}{2}}-frac 1{y_0}$$



          This evolves into:



          $$frac{epsilon}{2-epsilon y_0}>frac1y-frac 1{y_0}>-frac{epsilon}{2+epsilon y_0}tag{7}$$



          Notice that because of (5):



          $$2-epsilon y_0 ge 1 implies epsilon ge frac{epsilon}{2-epsilon y_0}$$



          $$2+epsilon y_0 ge 2 implies -frac{epsilon}{2+epsilon y_0}>-frac{epsilon}{2}$$



          With this in mind, (7) now becomes:



          $$epsilon >frac1y-frac 1{y_0}>-frac{epsilon}{2}$$



          $$epsilon >|frac1y-frac 1{y_0}|$$



          I'm leaving the last case ($y_0<0$) to you as an exercise.






          share|cite|improve this answer











          $endgroup$



          Case 1:



          $$text{min}(frac{|y_0|}{2},frac{epsilon|y_0|^2}{2})=frac{|y_0|}{2}$$



          $$frac{|y_0|}{2}lefrac{epsilon|y_0|^2}{2}$$



          $$frac1{|y_0|}leepsilontag{1}$$



          In this particular case:



          $$|y-y_0|lt frac{|y_0|}{2}$$



          $$-frac{|y_0|}{2}<y-y_0lt frac{|y_0|}{2}$$



          $$y_0-frac{|y_0|}{2}<y< y_0+frac{|y_0|}{2}tag{2}$$



          If $y_0>0$, (2) becomes:



          $$frac{y_0}{2}<y< frac{3y_0}{2}$$



          $$frac{2}{y_0}>frac1y> frac{2}{3y_0}$$



          $$frac{2}{y_0}-frac{1}{y_0}>frac1y - frac{1}{y_0}> frac{2}{3y_0}-frac{1}{y_0}$$



          $$frac{1}{y_0}>frac1y - frac{1}{y_0}> -frac{1}{3y_0}$$



          $$frac{1}{y_0}>|frac1y - frac{1}{y_0}|tag{3}$$



          Because of (1) and (3) we have:



          $$epsilon>|frac1y - frac{1}{y_0}|$$



          If $y_0<0$, (2) becomes:



          $$y_0+frac{y_0}{2}<y< y_0-frac{y_0}{2}$$



          $$frac{3y_0}{2}<y< frac{y_0}{2}$$



          Notice that $y$ is squeezed between negative values so we can write:



          $$frac2{3y_0}>frac1y> frac2{y_0}$$



          $$frac2{3y_0}-frac{1}{y_0}>frac1y-frac1{y_0}> frac2{y_0}-frac1{y_0}$$



          $$-frac{1}{3y_0}>frac1y-frac1{y_0}> frac1{y_0}$$



          $$|frac1y-frac1{y_0}|< frac1{|y_0|}tag{4}$$



          Because of (1) and (4) we have:



          $$|frac1y-frac1{y_0}|< epsilon$$



          Case 2:



          $$text{min}(frac{|y_0|}{2},frac{epsilon|y_0|^2}{2})=frac{epsilon|y_0|^2}{2}$$



          $$frac{|y_0|}{2}gefrac{epsilon|y_0|^2}{2})$$



          $$1geepsilon|y_0|tag{5}$$



          In this particular case:



          $$|y-y_0|lt frac{epsilon|y_0|^2}{2}$$



          $$-frac{epsilon|y_0|^2}{2}lt y-y_0lt frac{epsilon|y_0|^2}{2}$$



          $$y_0-frac{epsilon|y_0|^2}{2}lt y lt y_0+frac{epsilon|y_0|^2}{2}tag{6}$$



          For $y_0>0$, (6) becomes:



          $$y_0-frac{epsilon y_0^2}{2}lt y lt y_0+frac{epsilon y_0^2}{2}$$



          Notice that because of (5) $y$ is squeezed between positive values so we can write:



          $$frac{1}{y_0-frac{epsilon y_0^2}{2}}gt frac1y gt frac1{y_0+frac{epsilon y_0^2}{2}}$$



          $$frac{1}{y_0-frac{epsilon y_0^2}{2}}-frac 1{y_0} gt frac1y-frac 1{y_0} gt frac1{y_0+frac{epsilon y_0^2}{2}}-frac 1{y_0}$$



          This evolves into:



          $$frac{epsilon}{2-epsilon y_0}>frac1y-frac 1{y_0}>-frac{epsilon}{2+epsilon y_0}tag{7}$$



          Notice that because of (5):



          $$2-epsilon y_0 ge 1 implies epsilon ge frac{epsilon}{2-epsilon y_0}$$



          $$2+epsilon y_0 ge 2 implies -frac{epsilon}{2+epsilon y_0}>-frac{epsilon}{2}$$



          With this in mind, (7) now becomes:



          $$epsilon >frac1y-frac 1{y_0}>-frac{epsilon}{2}$$



          $$epsilon >|frac1y-frac 1{y_0}|$$



          I'm leaving the last case ($y_0<0$) to you as an exercise.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 10 '18 at 10:51

























          answered Dec 10 '18 at 10:42









          OldboyOldboy

          7,8401935




          7,8401935












          • $begingroup$
            Thanks for your answer. The equation $|y-y_0|<frac{epsilon|y_0|^2}{2}$ can be multiplied with $frac{1}{|y y_0|}$ , because $frac{1}{|yy_0|}>0$ . Thus $|frac{1}{y_0}-frac{1}{y}|<frac{epsilon|y_0|}{2|y|}$. It is $|frac{1}{y_0}-frac{1}{y}|=|frac{1}{y}-frac{1}{y_0}|$, That's why $|frac{1}{y}-frac{1}{y_0}|<frac{epsilon|y_0|}{2|y|}$ is true. $||y|-|y_0||leq |y-y_0|<frac{|y_0|}{2} Rightarrow |frac{|y|}{|y_0|}-1|<frac{1}{2} Rightarrow 2>frac{|y_0|}{|y|}>frac{2}{3}$. This means $|frac{1}{y}-frac{1}{y_0}|<frac{epsilon|y_0|}{2|y|}<epsilon$, because $2>frac{|y_0|}{|y|}$.
            $endgroup$
            – Ahmed Hossam
            Dec 10 '18 at 19:20




















          • $begingroup$
            Thanks for your answer. The equation $|y-y_0|<frac{epsilon|y_0|^2}{2}$ can be multiplied with $frac{1}{|y y_0|}$ , because $frac{1}{|yy_0|}>0$ . Thus $|frac{1}{y_0}-frac{1}{y}|<frac{epsilon|y_0|}{2|y|}$. It is $|frac{1}{y_0}-frac{1}{y}|=|frac{1}{y}-frac{1}{y_0}|$, That's why $|frac{1}{y}-frac{1}{y_0}|<frac{epsilon|y_0|}{2|y|}$ is true. $||y|-|y_0||leq |y-y_0|<frac{|y_0|}{2} Rightarrow |frac{|y|}{|y_0|}-1|<frac{1}{2} Rightarrow 2>frac{|y_0|}{|y|}>frac{2}{3}$. This means $|frac{1}{y}-frac{1}{y_0}|<frac{epsilon|y_0|}{2|y|}<epsilon$, because $2>frac{|y_0|}{|y|}$.
            $endgroup$
            – Ahmed Hossam
            Dec 10 '18 at 19:20


















          $begingroup$
          Thanks for your answer. The equation $|y-y_0|<frac{epsilon|y_0|^2}{2}$ can be multiplied with $frac{1}{|y y_0|}$ , because $frac{1}{|yy_0|}>0$ . Thus $|frac{1}{y_0}-frac{1}{y}|<frac{epsilon|y_0|}{2|y|}$. It is $|frac{1}{y_0}-frac{1}{y}|=|frac{1}{y}-frac{1}{y_0}|$, That's why $|frac{1}{y}-frac{1}{y_0}|<frac{epsilon|y_0|}{2|y|}$ is true. $||y|-|y_0||leq |y-y_0|<frac{|y_0|}{2} Rightarrow |frac{|y|}{|y_0|}-1|<frac{1}{2} Rightarrow 2>frac{|y_0|}{|y|}>frac{2}{3}$. This means $|frac{1}{y}-frac{1}{y_0}|<frac{epsilon|y_0|}{2|y|}<epsilon$, because $2>frac{|y_0|}{|y|}$.
          $endgroup$
          – Ahmed Hossam
          Dec 10 '18 at 19:20






          $begingroup$
          Thanks for your answer. The equation $|y-y_0|<frac{epsilon|y_0|^2}{2}$ can be multiplied with $frac{1}{|y y_0|}$ , because $frac{1}{|yy_0|}>0$ . Thus $|frac{1}{y_0}-frac{1}{y}|<frac{epsilon|y_0|}{2|y|}$. It is $|frac{1}{y_0}-frac{1}{y}|=|frac{1}{y}-frac{1}{y_0}|$, That's why $|frac{1}{y}-frac{1}{y_0}|<frac{epsilon|y_0|}{2|y|}$ is true. $||y|-|y_0||leq |y-y_0|<frac{|y_0|}{2} Rightarrow |frac{|y|}{|y_0|}-1|<frac{1}{2} Rightarrow 2>frac{|y_0|}{|y|}>frac{2}{3}$. This means $|frac{1}{y}-frac{1}{y_0}|<frac{epsilon|y_0|}{2|y|}<epsilon$, because $2>frac{|y_0|}{|y|}$.
          $endgroup$
          – Ahmed Hossam
          Dec 10 '18 at 19:20




















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