Meaning of Gauss' mean value theorem?

• A
• GGGGc
GGGGc
TL;DR Summary
Why does Gauss mean value theorem means f(z
0) is equal to the average of f(z) around the rim of any circle in R centre at z0? The denominator is 2pi instead of 2pi * radius which is the circumference of the circle.

GGGGc said:
TL;DR Summary: Why does Gauss mean value theorem means f(z
0) is equal to the average of f(z) around the rim of any circle in R centre at z0? The denominator is 2pi instead of 2pi * radius which is the circumference of the circle.

View attachment 335776
The short answer is because we are integrating over the angle, not over the circumference.

The slightly longer answer is that we are concerned primarily with the integral over the phases of the complex numbers surrounding z, not so much with r which, as you may note, is irrelevant due to Cauchy's integral formula.

-Dan

fresh_42
topsquark said:
The short answer is because we are integrating over the angle, not over the circumference.

The slightly longer answer is that we are concerned primarily with the integral over the phases of the complex numbers surrounding z, not so much with r which, as you may note, is irrelevant due to Cauchy's integral formula.

-Dan
Thanks for replying. Does that mean the average of f(z) is equal to f(z)/2pi since we are integrating it from 0-2pi?

GGGGc said:
Thanks for replying. Does that mean the average of f(z) is equal to f(z)/2pi since we are integrating it from 0-2pi?
No, the integral of a function f(z) (analytic over a circle of radius a about the point z) over a circle in the complex plane of radius r (r < a) about the point z is ##2 \pi f(z)##. It is an "average" in the sense that we are looking at the sum over the values of a region and that the points in the region of integration that aren't equal to z are essentially meaningless, except for a constant normalizing factor of ##2 \pi##.

Putting it a different way, to find the average of a real function:
##\displaystyle \overline{f} = \dfrac{1}{b - a} \int_a^b f(x) \, dx##

and the "average" of a complex function:
##\displaystyle \overline{f}(z) = \dfrac{1}{2 \pi} \int f(z + r e^{i \theta}) \, d \theta##

There are similarities to an average, so you can actually call it that (hey, that's what the formula is called!) but it isn't quite the same as an average in the sense of the mean value theorem.

-Dan

fresh_42
topsquark said:
No, the integral of a function f(z) (analytic over a circle of radius a about the point z) over a circle in the complex plane of radius r (r < a) about the point z is ##2 \pi f(z)##. It is an "average" in the sense that we are looking at the sum over the values of a region and that the points in the region of integration that aren't equal to z are essentially meaningless, except for a constant normalizing factor of ##2 \pi##.

Putting it a different way, to find the average of a real function:
##\displaystyle \overline{f} = \dfrac{1}{b - a} \int_a^b f(x) \, dx##

and the "average" of a complex function:
##\displaystyle \overline{f}(z) = \dfrac{1}{2 \pi} \int f(z + r e^{i \theta}) \, d \theta##

There are similarities to an average, so you can actually call it that (hey, that's what the formula is called!) but it isn't quite the same as an average in the sense of the mean value theorem.

-Dan
Many thanks for the explanation! I understand now. Thank you

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