Finding error in a an approximation












0












$begingroup$



We want to see the total error in approximating



$$ f'(x) approx frac{ f(x+h)-f(x) }{h} $$



where $f: R to R$ is differentiable. We can find $theta in [x,x+h]$
by Taylor's to that



$$ f(x+h) = f(x) + f'(x) h + f''( theta ) h^2 /2 $$



If the error in function values is bounded by $epsilon$, prove that
the rounding error is bounded by $2 epsilon /h$ and the truncation
error is bounded by $Mh/2$ where $M$ is a bound for $|f''(t)|$ for $t$
near $x$.




Try



We know truncation error is difference between true result and the result that would be produced by algorithm. By the result given above, we see that



$$ frac{ f(x+h) - f(x) }{h} = f'(x) + f''(theta)h/2 $$



So that



$$ underbrace{ frac{ f(x+h) - f(x) }{h} }_{approx} - underbrace{f'(x)}_{true ; result} = f''(theta)h/2 $$



So that trucantion error $E_T$ is absolute value of the above:



$$ E_T = |f''(theta) | h/2 leq Mh/2 $$



So we have our first result. However, for the rounding error I dont see how it is $2 epsilon / h $. Can someone explain what they really mean? Perhaps am I misunderstanding this part.










share|cite|improve this question









$endgroup$

















    0












    $begingroup$



    We want to see the total error in approximating



    $$ f'(x) approx frac{ f(x+h)-f(x) }{h} $$



    where $f: R to R$ is differentiable. We can find $theta in [x,x+h]$
    by Taylor's to that



    $$ f(x+h) = f(x) + f'(x) h + f''( theta ) h^2 /2 $$



    If the error in function values is bounded by $epsilon$, prove that
    the rounding error is bounded by $2 epsilon /h$ and the truncation
    error is bounded by $Mh/2$ where $M$ is a bound for $|f''(t)|$ for $t$
    near $x$.




    Try



    We know truncation error is difference between true result and the result that would be produced by algorithm. By the result given above, we see that



    $$ frac{ f(x+h) - f(x) }{h} = f'(x) + f''(theta)h/2 $$



    So that



    $$ underbrace{ frac{ f(x+h) - f(x) }{h} }_{approx} - underbrace{f'(x)}_{true ; result} = f''(theta)h/2 $$



    So that trucantion error $E_T$ is absolute value of the above:



    $$ E_T = |f''(theta) | h/2 leq Mh/2 $$



    So we have our first result. However, for the rounding error I dont see how it is $2 epsilon / h $. Can someone explain what they really mean? Perhaps am I misunderstanding this part.










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$



      We want to see the total error in approximating



      $$ f'(x) approx frac{ f(x+h)-f(x) }{h} $$



      where $f: R to R$ is differentiable. We can find $theta in [x,x+h]$
      by Taylor's to that



      $$ f(x+h) = f(x) + f'(x) h + f''( theta ) h^2 /2 $$



      If the error in function values is bounded by $epsilon$, prove that
      the rounding error is bounded by $2 epsilon /h$ and the truncation
      error is bounded by $Mh/2$ where $M$ is a bound for $|f''(t)|$ for $t$
      near $x$.




      Try



      We know truncation error is difference between true result and the result that would be produced by algorithm. By the result given above, we see that



      $$ frac{ f(x+h) - f(x) }{h} = f'(x) + f''(theta)h/2 $$



      So that



      $$ underbrace{ frac{ f(x+h) - f(x) }{h} }_{approx} - underbrace{f'(x)}_{true ; result} = f''(theta)h/2 $$



      So that trucantion error $E_T$ is absolute value of the above:



      $$ E_T = |f''(theta) | h/2 leq Mh/2 $$



      So we have our first result. However, for the rounding error I dont see how it is $2 epsilon / h $. Can someone explain what they really mean? Perhaps am I misunderstanding this part.










      share|cite|improve this question









      $endgroup$





      We want to see the total error in approximating



      $$ f'(x) approx frac{ f(x+h)-f(x) }{h} $$



      where $f: R to R$ is differentiable. We can find $theta in [x,x+h]$
      by Taylor's to that



      $$ f(x+h) = f(x) + f'(x) h + f''( theta ) h^2 /2 $$



      If the error in function values is bounded by $epsilon$, prove that
      the rounding error is bounded by $2 epsilon /h$ and the truncation
      error is bounded by $Mh/2$ where $M$ is a bound for $|f''(t)|$ for $t$
      near $x$.




      Try



      We know truncation error is difference between true result and the result that would be produced by algorithm. By the result given above, we see that



      $$ frac{ f(x+h) - f(x) }{h} = f'(x) + f''(theta)h/2 $$



      So that



      $$ underbrace{ frac{ f(x+h) - f(x) }{h} }_{approx} - underbrace{f'(x)}_{true ; result} = f''(theta)h/2 $$



      So that trucantion error $E_T$ is absolute value of the above:



      $$ E_T = |f''(theta) | h/2 leq Mh/2 $$



      So we have our first result. However, for the rounding error I dont see how it is $2 epsilon / h $. Can someone explain what they really mean? Perhaps am I misunderstanding this part.







      numerical-methods rounding-error truncation-error






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      asked Jan 2 at 7:02









      JamesJames

      2,636425




      2,636425






















          1 Answer
          1






          active

          oldest

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          3












          $begingroup$

          The rounding error? Take that upper bound $epsilon$ for the error in function values. We're subtracting two instances of it, which doubles that (at most $epsilon - (-epsilon)$). Then we divide by $h$, for $frac{2epsilon}{h}$. That's how far our calculated value of the difference quotient $frac{f(x+epsilon)-f(x)}{epsilon}$ can be from the true value.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            what do they mean by function values? Are we trying to estimate f(x) exact with $f(x')$ approx.? That is we are given that $|f(x) - f(x') | leq epsilon $? and we want to estimate $|f'(x) - f'(x')|$?
            $endgroup$
            – James
            Jan 2 at 7:28






          • 1




            $begingroup$
            Say we're trying to calculate some reasonably nice function value - say, $exp(1.234)$. But this is a computer system - we're looking for an explicit number, and our numbers are represented with finite blocks of memory (likely 64 or 128 bits on a modern system). There's only so many numbers that can be represented exactly that way, and $e^{1.234}$ isn't one of them - so, to get our function value, we'll have to round it off. We'll estimate it as best we can, but we can't possibly do better than the closest number that has a floating-point representation. That's roundoff error.
            $endgroup$
            – jmerry
            Jan 2 at 7:55












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          1 Answer
          1






          active

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          active

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          3












          $begingroup$

          The rounding error? Take that upper bound $epsilon$ for the error in function values. We're subtracting two instances of it, which doubles that (at most $epsilon - (-epsilon)$). Then we divide by $h$, for $frac{2epsilon}{h}$. That's how far our calculated value of the difference quotient $frac{f(x+epsilon)-f(x)}{epsilon}$ can be from the true value.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            what do they mean by function values? Are we trying to estimate f(x) exact with $f(x')$ approx.? That is we are given that $|f(x) - f(x') | leq epsilon $? and we want to estimate $|f'(x) - f'(x')|$?
            $endgroup$
            – James
            Jan 2 at 7:28






          • 1




            $begingroup$
            Say we're trying to calculate some reasonably nice function value - say, $exp(1.234)$. But this is a computer system - we're looking for an explicit number, and our numbers are represented with finite blocks of memory (likely 64 or 128 bits on a modern system). There's only so many numbers that can be represented exactly that way, and $e^{1.234}$ isn't one of them - so, to get our function value, we'll have to round it off. We'll estimate it as best we can, but we can't possibly do better than the closest number that has a floating-point representation. That's roundoff error.
            $endgroup$
            – jmerry
            Jan 2 at 7:55
















          3












          $begingroup$

          The rounding error? Take that upper bound $epsilon$ for the error in function values. We're subtracting two instances of it, which doubles that (at most $epsilon - (-epsilon)$). Then we divide by $h$, for $frac{2epsilon}{h}$. That's how far our calculated value of the difference quotient $frac{f(x+epsilon)-f(x)}{epsilon}$ can be from the true value.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            what do they mean by function values? Are we trying to estimate f(x) exact with $f(x')$ approx.? That is we are given that $|f(x) - f(x') | leq epsilon $? and we want to estimate $|f'(x) - f'(x')|$?
            $endgroup$
            – James
            Jan 2 at 7:28






          • 1




            $begingroup$
            Say we're trying to calculate some reasonably nice function value - say, $exp(1.234)$. But this is a computer system - we're looking for an explicit number, and our numbers are represented with finite blocks of memory (likely 64 or 128 bits on a modern system). There's only so many numbers that can be represented exactly that way, and $e^{1.234}$ isn't one of them - so, to get our function value, we'll have to round it off. We'll estimate it as best we can, but we can't possibly do better than the closest number that has a floating-point representation. That's roundoff error.
            $endgroup$
            – jmerry
            Jan 2 at 7:55














          3












          3








          3





          $begingroup$

          The rounding error? Take that upper bound $epsilon$ for the error in function values. We're subtracting two instances of it, which doubles that (at most $epsilon - (-epsilon)$). Then we divide by $h$, for $frac{2epsilon}{h}$. That's how far our calculated value of the difference quotient $frac{f(x+epsilon)-f(x)}{epsilon}$ can be from the true value.






          share|cite|improve this answer









          $endgroup$



          The rounding error? Take that upper bound $epsilon$ for the error in function values. We're subtracting two instances of it, which doubles that (at most $epsilon - (-epsilon)$). Then we divide by $h$, for $frac{2epsilon}{h}$. That's how far our calculated value of the difference quotient $frac{f(x+epsilon)-f(x)}{epsilon}$ can be from the true value.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 2 at 7:14









          jmerryjmerry

          17.1k11633




          17.1k11633












          • $begingroup$
            what do they mean by function values? Are we trying to estimate f(x) exact with $f(x')$ approx.? That is we are given that $|f(x) - f(x') | leq epsilon $? and we want to estimate $|f'(x) - f'(x')|$?
            $endgroup$
            – James
            Jan 2 at 7:28






          • 1




            $begingroup$
            Say we're trying to calculate some reasonably nice function value - say, $exp(1.234)$. But this is a computer system - we're looking for an explicit number, and our numbers are represented with finite blocks of memory (likely 64 or 128 bits on a modern system). There's only so many numbers that can be represented exactly that way, and $e^{1.234}$ isn't one of them - so, to get our function value, we'll have to round it off. We'll estimate it as best we can, but we can't possibly do better than the closest number that has a floating-point representation. That's roundoff error.
            $endgroup$
            – jmerry
            Jan 2 at 7:55


















          • $begingroup$
            what do they mean by function values? Are we trying to estimate f(x) exact with $f(x')$ approx.? That is we are given that $|f(x) - f(x') | leq epsilon $? and we want to estimate $|f'(x) - f'(x')|$?
            $endgroup$
            – James
            Jan 2 at 7:28






          • 1




            $begingroup$
            Say we're trying to calculate some reasonably nice function value - say, $exp(1.234)$. But this is a computer system - we're looking for an explicit number, and our numbers are represented with finite blocks of memory (likely 64 or 128 bits on a modern system). There's only so many numbers that can be represented exactly that way, and $e^{1.234}$ isn't one of them - so, to get our function value, we'll have to round it off. We'll estimate it as best we can, but we can't possibly do better than the closest number that has a floating-point representation. That's roundoff error.
            $endgroup$
            – jmerry
            Jan 2 at 7:55
















          $begingroup$
          what do they mean by function values? Are we trying to estimate f(x) exact with $f(x')$ approx.? That is we are given that $|f(x) - f(x') | leq epsilon $? and we want to estimate $|f'(x) - f'(x')|$?
          $endgroup$
          – James
          Jan 2 at 7:28




          $begingroup$
          what do they mean by function values? Are we trying to estimate f(x) exact with $f(x')$ approx.? That is we are given that $|f(x) - f(x') | leq epsilon $? and we want to estimate $|f'(x) - f'(x')|$?
          $endgroup$
          – James
          Jan 2 at 7:28




          1




          1




          $begingroup$
          Say we're trying to calculate some reasonably nice function value - say, $exp(1.234)$. But this is a computer system - we're looking for an explicit number, and our numbers are represented with finite blocks of memory (likely 64 or 128 bits on a modern system). There's only so many numbers that can be represented exactly that way, and $e^{1.234}$ isn't one of them - so, to get our function value, we'll have to round it off. We'll estimate it as best we can, but we can't possibly do better than the closest number that has a floating-point representation. That's roundoff error.
          $endgroup$
          – jmerry
          Jan 2 at 7:55




          $begingroup$
          Say we're trying to calculate some reasonably nice function value - say, $exp(1.234)$. But this is a computer system - we're looking for an explicit number, and our numbers are represented with finite blocks of memory (likely 64 or 128 bits on a modern system). There's only so many numbers that can be represented exactly that way, and $e^{1.234}$ isn't one of them - so, to get our function value, we'll have to round it off. We'll estimate it as best we can, but we can't possibly do better than the closest number that has a floating-point representation. That's roundoff error.
          $endgroup$
          – jmerry
          Jan 2 at 7:55


















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