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 $x$.


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.


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 $\frac{2\epsilon}{h}$. That's how far our calculated value of the difference quotient $\frac{f(x+\epsilon)-f(x)}{\epsilon}$ can be from the true value.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.