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Let $f \in C[a,b]$ be a function whose derivative exists on $(a,b)$. Suppose $f$ is to be evaluated at $x_0$ in $(a,b)$, but instead of computing the actual value $f(x_0)$, the approximate value, $\hat f(x_0)$, is the actual value of $f$ at $x_0 + \epsilon $, that is, $\hat f^~(x_0 ) = f(x_0 + \epsilon)$.

Use the Mean Value Theorem to estimate the absolute error $|f(x_0 )−\hat f(x_0 )|$ and the relative error $|f(x_0 )−\hat f(x_0 )|/|f(x_0)|$, assuming $f(x_0 ) \neq 0$.

Here's what I tried:

Absolute error: $|f(x_0 )−\hat f(x_0 )|=|f(x_0)-f(x_0+\epsilon)|=|f'(c)|\,\epsilon\,$ where $c \in (x_0, x_0 + \epsilon)$.

And relative error:$$\frac{|f'(c)|\epsilon}{|f(x_0)|}$$ where $c \in (x_0, x_0 + \epsilon)$. I'm not sure if this is right, and if it is how do I find $c$.

Euler_Salter
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Gjekaks
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1 Answers1

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You pretty much have the right idea. As you pointed out, $$\left|f(x_0)-f(x_0+\epsilon)\right|=\left|f^\prime(c)\right| \epsilon$$ for some $c$ in the interval $I = (x_0, x_0+\epsilon)$. Therefore, $$\left|f(x_0)-f(x_0+\epsilon)\right|\leq \epsilon \cdot \sup_{c\in I}\left|f^\prime(c)\right|.$$ Divide both sides by $|f(x_0)|$ to get a bound on the relative error.

parsiad
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