Error from central difference seems large












1












$begingroup$


for a function $f(x)=e^{2x}-cos(2x)$,



at grid points $x in {-0.3,-0.2,-0.1,0}$



I perform a central difference for the derivative at $x=-0.2$



$$frac{df}{dx}=frac{f(-0.1)-f(-0.3)}{2*(-0.1--0.3)}=0.28795$$
The derivative of this function (per mathematica) is $0.561803$
So the error is $$Abs[0.561803-0.28795]=0.273853$$



I have a formula for the upper bound of the error, and that formula is $$frac{M^*Delta x^2}{3!}=frac{8*(-0.1--0.3)}{6}=0.0533333\text{Where }M^*=text{Max}(f''(x)|xin[-0.3,0])=f(0)=8$$



The issue is the error, $0.27$, is greater than the supposid error bound, $0.05$. When I graph this function with its central difference though, it looks correct.



Graph of f(x) and finite/central differences



Any idea why my derivatives are higher than the upper bound? The error does seem pretty high.






finite-differences







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    for a function $f(x)=e^{2x}-cos(2x)$,



    at grid points $x in {-0.3,-0.2,-0.1,0}$



    I perform a central difference for the derivative at $x=-0.2$



    $$frac{df}{dx}=frac{f(-0.1)-f(-0.3)}{2*(-0.1--0.3)}=0.28795$$
    The derivative of this function (per mathematica) is $0.561803$
    So the error is $$Abs[0.561803-0.28795]=0.273853$$



    I have a formula for the upper bound of the error, and that formula is $$frac{M^*Delta x^2}{3!}=frac{8*(-0.1--0.3)}{6}=0.0533333\text{Where }M^*=text{Max}(f''(x)|xin[-0.3,0])=f(0)=8$$



    The issue is the error, $0.27$, is greater than the supposid error bound, $0.05$. When I graph this function with its central difference though, it looks correct.



    Graph of f(x) and finite/central differences



    Any idea why my derivatives are higher than the upper bound? The error does seem pretty high.






    finite-differences







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      for a function $f(x)=e^{2x}-cos(2x)$,



      at grid points $x in {-0.3,-0.2,-0.1,0}$



      I perform a central difference for the derivative at $x=-0.2$



      $$frac{df}{dx}=frac{f(-0.1)-f(-0.3)}{2*(-0.1--0.3)}=0.28795$$
      The derivative of this function (per mathematica) is $0.561803$
      So the error is $$Abs[0.561803-0.28795]=0.273853$$



      I have a formula for the upper bound of the error, and that formula is $$frac{M^*Delta x^2}{3!}=frac{8*(-0.1--0.3)}{6}=0.0533333\text{Where }M^*=text{Max}(f''(x)|xin[-0.3,0])=f(0)=8$$



      The issue is the error, $0.27$, is greater than the supposid error bound, $0.05$. When I graph this function with its central difference though, it looks correct.



      Graph of f(x) and finite/central differences



      Any idea why my derivatives are higher than the upper bound? The error does seem pretty high.






      finite-differences







      share|cite|improve this question











      $endgroup$




      for a function $f(x)=e^{2x}-cos(2x)$,



      at grid points $x in {-0.3,-0.2,-0.1,0}$



      I perform a central difference for the derivative at $x=-0.2$



      $$frac{df}{dx}=frac{f(-0.1)-f(-0.3)}{2*(-0.1--0.3)}=0.28795$$
      The derivative of this function (per mathematica) is $0.561803$
      So the error is $$Abs[0.561803-0.28795]=0.273853$$



      I have a formula for the upper bound of the error, and that formula is $$frac{M^*Delta x^2}{3!}=frac{8*(-0.1--0.3)}{6}=0.0533333\text{Where }M^*=text{Max}(f''(x)|xin[-0.3,0])=f(0)=8$$



      The issue is the error, $0.27$, is greater than the supposid error bound, $0.05$. When I graph this function with its central difference though, it looks correct.



      Graph of f(x) and finite/central differences



      Any idea why my derivatives are higher than the upper bound? The error does seem pretty high.






      finite-differences




      finite-differences






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 1 '18 at 0:48









      Key Flex

      8,18261233




      8,18261233










      asked Nov 30 '18 at 22:31









      FrankFrank

      16210




      16210






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          The problem lies in your definition of $Delta x$, remember that



          $$
          f'_{FD}(x) approx frac{f(x + Delta x) - f(x - Delta x)}{2Delta x}
          $$



          Taking $x = -0.2$ and $Delta x = 0.1$ we get



          $$
          f'_{FD}(-0.2) approx frac{f(-0.1) - f(-0.3)}{2 cdot 0.1} = 0.575941 tag{1}
          $$



          So the actual error is



          $$
          epsilon = |f'_{FD}(-0.2) - f'(-0.2)|= 0.0141374 tag{2}
          $$



          Now, the prediction for the error is



          $$
          epsilon_{max} = frac{|f^{(3)}(c)|Delta^2 x}{6} stackrel{c=-0.3}{=} 0.0148461 tag{3}
          $$



          You can indeed see that



          $$
          epsilon < epsilon_max
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks so much! But is the formula for the error bound really supposed to be the third derivative of $f$ or the second? I got the formula from someone else's notes! I couldn't find it with a google search hahaha.
            $endgroup$
            – Frank
            Dec 1 '18 at 0:04










          • $begingroup$
            @Frank It is the third derivative
            $endgroup$
            – caverac
            Dec 1 '18 at 0:40











          Your Answer





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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          The problem lies in your definition of $Delta x$, remember that



          $$
          f'_{FD}(x) approx frac{f(x + Delta x) - f(x - Delta x)}{2Delta x}
          $$



          Taking $x = -0.2$ and $Delta x = 0.1$ we get



          $$
          f'_{FD}(-0.2) approx frac{f(-0.1) - f(-0.3)}{2 cdot 0.1} = 0.575941 tag{1}
          $$



          So the actual error is



          $$
          epsilon = |f'_{FD}(-0.2) - f'(-0.2)|= 0.0141374 tag{2}
          $$



          Now, the prediction for the error is



          $$
          epsilon_{max} = frac{|f^{(3)}(c)|Delta^2 x}{6} stackrel{c=-0.3}{=} 0.0148461 tag{3}
          $$



          You can indeed see that



          $$
          epsilon < epsilon_max
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks so much! But is the formula for the error bound really supposed to be the third derivative of $f$ or the second? I got the formula from someone else's notes! I couldn't find it with a google search hahaha.
            $endgroup$
            – Frank
            Dec 1 '18 at 0:04










          • $begingroup$
            @Frank It is the third derivative
            $endgroup$
            – caverac
            Dec 1 '18 at 0:40
















          1












          $begingroup$

          The problem lies in your definition of $Delta x$, remember that



          $$
          f'_{FD}(x) approx frac{f(x + Delta x) - f(x - Delta x)}{2Delta x}
          $$



          Taking $x = -0.2$ and $Delta x = 0.1$ we get



          $$
          f'_{FD}(-0.2) approx frac{f(-0.1) - f(-0.3)}{2 cdot 0.1} = 0.575941 tag{1}
          $$



          So the actual error is



          $$
          epsilon = |f'_{FD}(-0.2) - f'(-0.2)|= 0.0141374 tag{2}
          $$



          Now, the prediction for the error is



          $$
          epsilon_{max} = frac{|f^{(3)}(c)|Delta^2 x}{6} stackrel{c=-0.3}{=} 0.0148461 tag{3}
          $$



          You can indeed see that



          $$
          epsilon < epsilon_max
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks so much! But is the formula for the error bound really supposed to be the third derivative of $f$ or the second? I got the formula from someone else's notes! I couldn't find it with a google search hahaha.
            $endgroup$
            – Frank
            Dec 1 '18 at 0:04










          • $begingroup$
            @Frank It is the third derivative
            $endgroup$
            – caverac
            Dec 1 '18 at 0:40














          1












          1








          1





          $begingroup$

          The problem lies in your definition of $Delta x$, remember that



          $$
          f'_{FD}(x) approx frac{f(x + Delta x) - f(x - Delta x)}{2Delta x}
          $$



          Taking $x = -0.2$ and $Delta x = 0.1$ we get



          $$
          f'_{FD}(-0.2) approx frac{f(-0.1) - f(-0.3)}{2 cdot 0.1} = 0.575941 tag{1}
          $$



          So the actual error is



          $$
          epsilon = |f'_{FD}(-0.2) - f'(-0.2)|= 0.0141374 tag{2}
          $$



          Now, the prediction for the error is



          $$
          epsilon_{max} = frac{|f^{(3)}(c)|Delta^2 x}{6} stackrel{c=-0.3}{=} 0.0148461 tag{3}
          $$



          You can indeed see that



          $$
          epsilon < epsilon_max
          $$






          share|cite|improve this answer









          $endgroup$



          The problem lies in your definition of $Delta x$, remember that



          $$
          f'_{FD}(x) approx frac{f(x + Delta x) - f(x - Delta x)}{2Delta x}
          $$



          Taking $x = -0.2$ and $Delta x = 0.1$ we get



          $$
          f'_{FD}(-0.2) approx frac{f(-0.1) - f(-0.3)}{2 cdot 0.1} = 0.575941 tag{1}
          $$



          So the actual error is



          $$
          epsilon = |f'_{FD}(-0.2) - f'(-0.2)|= 0.0141374 tag{2}
          $$



          Now, the prediction for the error is



          $$
          epsilon_{max} = frac{|f^{(3)}(c)|Delta^2 x}{6} stackrel{c=-0.3}{=} 0.0148461 tag{3}
          $$



          You can indeed see that



          $$
          epsilon < epsilon_max
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 30 '18 at 23:45









          caveraccaverac

          14.6k31130




          14.6k31130












          • $begingroup$
            Thanks so much! But is the formula for the error bound really supposed to be the third derivative of $f$ or the second? I got the formula from someone else's notes! I couldn't find it with a google search hahaha.
            $endgroup$
            – Frank
            Dec 1 '18 at 0:04










          • $begingroup$
            @Frank It is the third derivative
            $endgroup$
            – caverac
            Dec 1 '18 at 0:40


















          • $begingroup$
            Thanks so much! But is the formula for the error bound really supposed to be the third derivative of $f$ or the second? I got the formula from someone else's notes! I couldn't find it with a google search hahaha.
            $endgroup$
            – Frank
            Dec 1 '18 at 0:04










          • $begingroup$
            @Frank It is the third derivative
            $endgroup$
            – caverac
            Dec 1 '18 at 0:40
















          $begingroup$
          Thanks so much! But is the formula for the error bound really supposed to be the third derivative of $f$ or the second? I got the formula from someone else's notes! I couldn't find it with a google search hahaha.
          $endgroup$
          – Frank
          Dec 1 '18 at 0:04




          $begingroup$
          Thanks so much! But is the formula for the error bound really supposed to be the third derivative of $f$ or the second? I got the formula from someone else's notes! I couldn't find it with a google search hahaha.
          $endgroup$
          – Frank
          Dec 1 '18 at 0:04












          $begingroup$
          @Frank It is the third derivative
          $endgroup$
          – caverac
          Dec 1 '18 at 0:40




          $begingroup$
          @Frank It is the third derivative
          $endgroup$
          – caverac
          Dec 1 '18 at 0:40


















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