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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.

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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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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