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I was going through some example question in my lecture notes and didn't quite understand how my lecturer got to the final formula.

it goes like this.

what is the condition number when we evaluate a function $y = f(x)$ at an approximation of $x$, $\hat{x} = x + \Delta x$, rather than at the true value $x$?

$cond = \left |\frac{{\frac{\Delta y}{y}}}{\frac{\Delta x}{x}} \right | = \left |\frac{{\frac{f(x+\Delta x) - f(x)}{f(x)}}}{\frac{\Delta x}{x}} \right | = \left |\frac{f(x + \Delta x) - f(x)}{\Delta x} \frac {x}{f(x))}\right | \approx \left | \frac{x f'(x)}{f(x)} \right |$

I could see how he got $cond = \left |\frac{{\frac{\Delta y}{y}}}{\frac{\Delta x}{x}} \right | = \left |\frac{{\frac{f(x+\Delta x) - f(x)}{f(x)}}}{\frac{\Delta x}{x}} \right | = \left |\frac{f(x + \Delta x) - f(x)}{\Delta x} \frac {x}{f(x))}\right |$ this part

but I could not get $\approx \left | \frac{x f'(x)}{f(x)} \right |$ this part.

How did he derived $\left | \frac{x f'(x)}{f(x)} \right |$ from

$\left |\frac{f(x + \Delta x) - f(x)}{\Delta x} \frac {x}{f(x))}\right |$ this?

Thank you in advance.

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  • $\begingroup$ Well, $\frac{f(x+\Delta x)}{\Delta x}\rightarrow f'(x)$ as $\Delta x\rightarrow 0$ so you can approximate this quotient by $f'(x)$ $\endgroup$ – shdp Mar 11 '17 at 11:15
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You only have to use the definition of the derivative. See: https://en.wikipedia.org/wiki/Derivative

$$f'(x)=\lim_\limits{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}.$$

Here, you have $h=\Delta x$.

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