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I was wondering if there were a mathematical way to note that a variable in an expression is "very small". I feel like ambiguity might be the cause of there not being one, but I'm not sure.

Example:

$\Delta A = \frac12 (r + \Delta r)^2 \Delta \theta - \frac12 r^2 \theta = r (\Delta r \Delta \theta) + \frac12(\Delta r)^2 \Delta \theta \approx r \Delta r \Delta \theta$

Error in $\Delta A $ is the term $\frac12 (\Delta r)^2 \Delta \theta$ when $\Delta r$ and $\Delta \theta$ are very small.

Is there a better way to write the bolded statement?

And on that note, is there a better way to write "Error in $\Delta A$"?

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  • $\begingroup$ Sometime the simbol $\ll$ is used as in $\Delta r \ll 1$ . $\endgroup$ Apr 8 '17 at 21:01
  • $\begingroup$ I considered that but wasn't sure if it would really be sufficient. Thanks for your input! $\endgroup$
    – Dylan
    Apr 8 '17 at 21:03
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Consider using Little o notation. This makes explicit what we mean by "a negligible error". We are ultimately interested in what happens as $\Delta r\to0$. If $f(\Delta r)=o(\Delta r)$ (i.e. if $\frac{f(\Delta r)}{\Delta r}\to0$ as $\Delta r\to0$), then $f(\Delta r)$ must be much smaller than $\Delta r$. Your final expression is a multiple of $\Delta r$, and so it is reasonable to group all lower order terms in one $o(\Delta r)$.

Another example: we know $\lim_{\Delta x\to0}\frac{\sin(x+\Delta x)-\sin(x)}{\Delta x}=\cos(x)$. We could rewrite this as $$\sin(x+\Delta x)=\sin(x)+\Delta x\cos(x)+o(\Delta x).$$ The difference between $\sin(x+\Delta x)$ and $\sin(x)+\Delta x\cos(x)$ is not zero, but on the scale of $\Delta x$ it is so small that does not affect our limit.

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  • $\begingroup$ Ha, thanks! As a CS major I had only ever heard of Big-O notation before. $\endgroup$
    – Dylan
    Apr 8 '17 at 21:01
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You can write $\Delta r, \Delta\theta < \epsilon$, if that is what you are asking.

For the error part you could say, for example:

$$ e = \lVert\Delta A - \Delta A^*\rVert $$

where $e$ is the error and $\Delta A^*$ the real value of $\Delta A$.

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