discrete differences, doubt about sign

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Let $Gin C(mathbb R)$ then
$lim_Nto +inftyfracG(x-frac1N)-G(x)-frac1N=G'(x)$?



I have a doubt about the sign in front of $G'$, it is a $+$ or a $-$?
Thanks to everyone




numerical-methods




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    up vote
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    down vote

    favorite












    Let $Gin C(mathbb R)$ then
    $lim_Nto +inftyfracG(x-frac1N)-G(x)-frac1N=G'(x)$?



    I have a doubt about the sign in front of $G'$, it is a $+$ or a $-$?
    Thanks to everyone




    numerical-methods




    share|cite|improve this question





















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Let $Gin C(mathbb R)$ then
      $lim_Nto +inftyfracG(x-frac1N)-G(x)-frac1N=G'(x)$?



      I have a doubt about the sign in front of $G'$, it is a $+$ or a $-$?
      Thanks to everyone




      numerical-methods




      share|cite|improve this question











      Let $Gin C(mathbb R)$ then
      $lim_Nto +inftyfracG(x-frac1N)-G(x)-frac1N=G'(x)$?



      I have a doubt about the sign in front of $G'$, it is a $+$ or a $-$?
      Thanks to everyone





      numerical-methods





      share|cite|improve this question










      share|cite|improve this question




      share|cite|improve this question









      asked Jul 20 at 15:06









      user495333

      766




      766




















          1 Answer
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          The definition of the derivative is usually stated as:



          $$
          G'(x) = lim limits_h rightarrow 0 fracG(x+h)-G(x)h
          $$



          If we replace $h$ by $-frac1N$, we get the expression in your question. So, the sign in front of $G'$ is indeed positive.






          share|cite|improve this answer





















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            1 Answer
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            up vote
            0
            down vote













            The definition of the derivative is usually stated as:



            $$
            G'(x) = lim limits_h rightarrow 0 fracG(x+h)-G(x)h
            $$



            If we replace $h$ by $-frac1N$, we get the expression in your question. So, the sign in front of $G'$ is indeed positive.






            share|cite|improve this answer

























              up vote
              0
              down vote













              The definition of the derivative is usually stated as:



              $$
              G'(x) = lim limits_h rightarrow 0 fracG(x+h)-G(x)h
              $$



              If we replace $h$ by $-frac1N$, we get the expression in your question. So, the sign in front of $G'$ is indeed positive.






              share|cite|improve this answer























                up vote
                0
                down vote










                up vote
                0
                down vote









                The definition of the derivative is usually stated as:



                $$
                G'(x) = lim limits_h rightarrow 0 fracG(x+h)-G(x)h
                $$



                If we replace $h$ by $-frac1N$, we get the expression in your question. So, the sign in front of $G'$ is indeed positive.






                share|cite|improve this answer













                The definition of the derivative is usually stated as:



                $$
                G'(x) = lim limits_h rightarrow 0 fracG(x+h)-G(x)h
                $$



                If we replace $h$ by $-frac1N$, we get the expression in your question. So, the sign in front of $G'$ is indeed positive.







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered Jul 20 at 15:12









                Sambo

                1,2561427




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