It is possible to find the number of complex roots of a function $f(z)$ in a contour $\gamma$ using complex contour integration. I was wondering if you can do the same thing on the real line: take a real-valued function $f(x)$ and find the number of real roots in a contour (usually a circle with a center $(a,b)$ and radius $r$) using "real contour integration" (I don't know if it exists or does it have another name)?
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You can use the argument principle to find the number of real roots of a function in an interval; but you're using complex contour integration, not real contour integration (which does exist, by the way). If you're thinking of a function of two variables $f(x,y),$ I'm not quite sure of the generalization. – Adrian Keister Sep 30 '19 at 18:48
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@AdrianKeister I thought the argument principle only applied to complex roots. How would you use it to find number of real roots in an interval? Also, can you give me an example of "real contour integration"? Any links I can use? – MathPowers Sep 30 '19 at 18:51
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Sure, but this will be an answer, as it's too long for a comment. – Adrian Keister Sep 30 '19 at 18:56
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@AdrianKeister Ok, thank you. – MathPowers Sep 30 '19 at 18:56
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@AdrianKeister And for the record, $f(x)$ is univariate. – MathPowers Sep 30 '19 at 19:04
1 Answers
You can use the Argument Principle to find the number of roots of a function in an interval (technically, the number of roots minus the number of poles). It works like this. Suppose you have a function $f(x)$ on an interval $(a,b).$ Then you switch to a complex variable $f(z),$ and you parametrize a circle $C$ centered at $c=(a+b)/2,$ with radius $r=(b-1)/2.$ It's centered at the center of the interval, and its radius is half the width of the interval. You could do other contours, but this is perhaps the simplest. The Argument Principle, then, says that the integral $$\frac{1}{2\pi i}\oint_C\frac{f'(z)}{f(z)}\,dz=Z-P, $$ where $Z$ is the number of zeros inside $C,$ and $P$ the number of poles inside $C.$ To carry this a bit further, the parametrization would look like this: $$z=c+re^{i\theta},\; 0\le \theta\le 2\pi. $$ Then $dz=ie^{i\theta}\,d\theta,$ and you would write $$\frac{1}{2\pi i}\int_0^{2\pi}\frac{f'\big(c+re^{i\theta}\big)}{f\big(c+re^{i\theta}\big)}\,ie^{i\theta}\,d\theta=Z-P. $$ Important note: if you're concerned about picking up zeros and poles not on the real line, then you can choose a rectangular contour as follows: \begin{array}{|c|c|c|c|} \hline &z &t\;\text{interval} &dz\\ \hline \gamma_1 &a+i\varepsilon(1-2t) &[0,1] &-2i\varepsilon\,dt \\ \hline \gamma_2 &a+t(b-a)-i\varepsilon &[0,1] &(b-a)\,dt \\ \hline \gamma_3 &b+i\varepsilon(-1+2t) &[0,1] &2i\varepsilon\,dt \\ \hline \gamma_4 &b+t(a-b)+i\varepsilon &[0,1] &(a-b)\,dt \\ \hline \end{array} For many functions you can choose $\varepsilon>0$ small enough not to pick up any zeros or poles inside this rectangle, traversed counter-clockwise. The final contour $C$ would then be $C=\gamma_1\cup\gamma_2\cup\gamma_3\cup\gamma_4.$
So much for the Argument Principle.
An example of real contour integration is the work function calculation: $$W=\int_C\mathbf{F}\cdot d\mathbf{r}. $$ This is the work done by the vector force $\mathbf{F}$ along $C.$ Notice that, at any point on $C,$ only the component of $\mathbf{F}$ parallel to the tangent differential $d\mathbf{r}$ contributes to the work done.
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Comments are not for extended discussion; this conversation has been moved to chat. – quid Oct 05 '19 at 01:07