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HallsofIvy

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Integration is NOT simply "adding up" points.

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@lugita: yes that seems like a good way to phrase it.

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Office_Shredder

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It's because there's a hole in the domain. 1/z blows up at z = 0.

http://en.wikipedia.org/wiki/Cauchy_integral_theorem

Now a question that always bothered me is, how does the function being integrated around a circle "know" there's a hole in the middle of the circle?

That's what the OP is asking.

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Fredrik

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One way of explaining his point is that in a Riemann sum ##\sum_k \frac{1}{z_k} \Delta z_k##, it's not just the 1/z

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Office_Shredder

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The fact that the integral of z around the unit circle is zero is not because z is an odd function. All the argument says is that a function being odd is not sufficient to make it zero when going around a circle

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Fredrik

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I would say that it does. The contribution from opposite sides are the same in this case too, for precisely the same reason.But Fredrik, why doesn't this argument hold for f(z)=z?

It's precisely because it's either clockwise the whole way through or counter-clockwise the whole way through that that Δz has "opposite" values on opposite sides.And furthermore, I was under the impression that our direction of integration is not changing, it's either counter-clockwise or clockwise, and thus positive or negative accordingly.

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But then why wouldn't the integral of z over the unit circle also be non zero?

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Office_Shredder

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It's like asking why the integral from -pi to pi of cos(x) is zero even though cos(x) isn't odd... sometimes integrals are just zero because they are

But that is only perceived change in direction, you don't multiply by an extra negative just because you're down at 5pi/4.

The delta(z)s are complex numbers. The norm is whatever step size you're taking, and the direction in which these complex numbers are pointing (once you've taken many slices) is the direction of motion along the unit circle. Let's think about delta z for z

[tex]\frac{1}{100*i}*\frac{1}{1} + \frac{-1}{100*i}\frac{1}{-1} = \frac{2}{100*i}[/tex]

and the two pieces do NOT cancel

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Fredrik

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How is it "only perceived"? The Δz clearly has the same magnitude but opposite sign on the opposite side. That's not a matter of perception.But that is only perceived change in direction, you don't multiply by an extra negative just because you're down at 5pi/4.

Why would it be? Office_Shredder and I have only used this observation about contributions from opposite segments to explain why the opposite sign of 1/z on opposite sides doesn't automatically make the integral zero.But then why wouldn't the integral of z over the unit circle also be non zero?

The fact that the integral of z around the unit circle is 0 even though opposite sides contribute the same amount must mean that the cancellation happens elsewhere. Maybe the contribution from a segment is cancelled by a segment an angle π/2 away (I didn't think about it, I just made up a number), maybe it's not cancelled by any segment. All we know (unless we think it through), is that the contributions from

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Looking at the graph of cos(x) just from 0 to pi and then using the fact that it's an even function, it's quite obvious why its integral from -pi to pi is zero. I can't personally accept the fact that sometimes integrals are just zero because they are. Clearly the reason z is zero and 1/z is not is that 1/z has a singularity inside the unit circle. The reason for this is I feel for some deep reason involving the proof of Cauchy's Integral Theorem and the proof of Pullbacks for integrals.

Fredrik, could you explain to me exactly what you mean by Δz.

Edit: I guess maybe odd functions were not the right way to frame this, I'm more interested in the discrepancy between the integrals of z and 1/z.

Edit #2: Ok I see what office_shredder is saying about Δz, but I'm still curious about why this phenomenon doesn't cause the integral of z to be non zero.

Fredrik, could you explain to me exactly what you mean by Δz.

Edit: I guess maybe odd functions were not the right way to frame this, I'm more interested in the discrepancy between the integrals of z and 1/z.

Edit #2: Ok I see what office_shredder is saying about Δz, but I'm still curious about why this phenomenon doesn't cause the integral of z to be non zero.

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Office_Shredder

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Consider the Riemann sum of cos(x). [itex] cos(x_k)\delta x = cos(-x_k) \delta x[/itex] so the contributions from x and -x are equal. But the integral is still zero because of more global properties of cos(x).

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Fredrik

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The theorem we need is the one that says that if B and C are two contours with the same endpoints, and B can be continuously deformed to C without going outside an open set on which f is analytic, then ##\int_B\, f(z)dz=\int_C\, f(z)dz##. ThatClearly the reason z is zero and 1/z is not is that 1/z has a singularity inside the unit circle. The reason for this is I feel for some deep reason involving the proof of Cauchy's Integral Theorem and the proof of Pullbacks for integrals.

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lavinia

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doseem to be simply adding up the points along the curve. I'm not sure what this means..

@lugita: yes that seems like a good way to phrase it.

along the circle,

dz = e[itex]^{i\theta}[/itex]dr + ire[itex]^{i\theta}[/itex]d[itex]\theta[/itex]

but dr = 0 since the circle has constant radius.

So dz/z = ire[itex]^{i\theta}[/itex]d[itex]\theta[/itex]/re[itex]^{i\theta}[/itex] = id[itex]\theta[/itex]

So dz/z is just the differential of the angle function multiplied by i

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Fredrik

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If ##\{z_0,\dots,z_n\}## are points on the circle, thenFredrik, could you explain to me exactly what you mean by Δz.

$$\int f(z) dz\approx\sum_{k=1}^n f(z_k)(z_k-z_{k-1})=\sum_{k=1}^n f(z_k)\Delta z_k.$$

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Bacle2

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But I think we can see that the net contributions of antipodes do not cancel out

(we can ignore the contribution of r, since it cancels out). While it is true that:

1/z= 1/e^{iθ} =e^{-iθ} =cos(-θ)+isin(-θ)=cosθ-isinθ

and, for the antipode e^{i(θ+∏)}= -cosθ+isinθ ,

The contributions of θ=0 must cancel out that of both θ=∏. But the contribution of ∏

must also cancel out that of 2∏.

(we can ignore the contribution of r, since it cancels out). While it is true that:

1/z= 1/e

and, for the antipode e

The contributions of θ=0 must cancel out that of both θ=∏. But the contribution of ∏

must also cancel out that of 2∏.

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Office_Shredder

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But I think we can see that the net contributions of antipodes do not cancel out

(we can ignore the contribution of r, since it cancels out). While it is true that:

1/z= 1/e^{iθ}=e^{-iθ}=cos(-θ)+isin(-θ)=cosθ-isinθ

and, for the antipode e^{i(θ+∏)}= -cosθ+isinθ ,

The contributions of θ=0 must cancel out that of both θ=∏. But the contribution of ∏

must also cancel out that of 2∏.

This is incorrect. When integrating from 0 to 2pi the angles 0, pi and 2pi are the only angles for which you get this happening. And assuming your Riemann sum is a left or right hand sum the angles 0 and 2pi won't both show up

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Bacle2

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This is incorrect. When integrating from 0 to 2pi the angles 0, pi and 2pi are the only angles for which you get this happening. And assuming your Riemann sum is a left or right hand sum the angles 0 and 2pi won't both show up

Of course; I was referring to the talk about odd real-valued functions canceling out,

that some where talking about ,and it is an informal argument. I should have been more

clear on that.

A rigorous argument would have required , as you said, Riemann sums, choices of

elements in partitions, etc.

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