If I mis-understood Benny, then let him say so Hall. His comment above most certainly looked like he did not to any objective person reading it. I wished only to be helpful, not subversive like you.
Benny, t is "time". The populations are functions of time, that is, x(t) and y(t). You don't see time in the ODEs since they'er "homogeneous", that is, the differential equation is not a function of time. You could plot x(t) and y(t) independently but usually the "phase portrait" (y as a...
Alright, how about this one:
\int_0^{2\pi}\frac{1}{(Cos(\theta)+2)^n}d\theta
Can it be done using residue integration for any integer n? Can it be done any other way?
What about just for n=1, 2,3, maybe 4, then Mathematical Induction? I don't know. Might work on it though.:smile:
Hello Loom. So you did the z-substitution and got:
\frac{1}{i}\oint \frac{z^2+1}{z(z^2+6z+1)}dz
right?
So we're integrating around a circle of radius 1. Now, when you factor the denom, only 0 and one other root are poles inside this circle. Calculating the residues of both I get...
I yield to you. Sorry guys for poisioning your thread. I do very much believe in staying focused on the topic of discussion and failed here. Will try to do a better job in the future.
In the interest of some closure in this matter I'll summarize Riemann's evaluation of the integral. He relies on the two expressions for the auxiliary function Xi:
\xi(s)=\pi^{-s/2}\Gamma(s/2+1)(s-1)\zeta(s)
and:
\xi(s)=\xi(0)\mathop\Pi\limits_{\rho}(1-\frac{s}{p})
Now, taking...
Yea, I'm taking a class however I don't feel comfortable asking my professor anything about this . . . he'll get the wrong impression and I'd just as soon remain unnoticed.
You forgot the factoral coefficient in the calculation of a residue of higher order, you know, the :
\frac{1}{(m-1)!}
part. That would have one-halfed it down to 1/2 or . . . like Shmoe said, it's already in a Laurent series form with coefficient 1/2.
That makes sense. Thank you. I see the problem now, the whole problem indeed: need to crawl before I can walk. I'm impatient though. It's all very interesting to me.
We're so anthropocentric aren't we: purposes in life, goals, direction, and meaning. That's all good I think from a Darwinist perspective. Hey, whatever works in the interest of survival and reproductive success. Religion works too. Stripped however of it's humane trappings lies a dark...
Thanks Lokofer. I'm stuborn though.:wink:
Suppose I wanted to calculate the residue at the first complex zero of Zeta for the integrand above. That then would be:
\mathop\text{Res}\limits_{z=\rho_1}\left\{\frac{ln(\zeta(z))x^z}{z}\right\}=
\frac{1}{2\pi i}\oint\frac{ln(\zeta(z))x^z}{z}dz...
Can anyone tell me if Riemann's Prime Counting function can be solved by residue integration?
Here it is:
J(x)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{ln(\zeta(s))x^s}{s}ds
which has the solution:
J(x)=li(x)-\sum_{\rho}li(x^\rho)-ln(2)+
\int_x^{\infty}\frac{dt}{t(t^2-1)ln(t)}
I...
Very well Hall. I obviously don't have it then. Thanks.
And atoms are made of quarks and those of perhaps strings. But I do not in the least believe that is the end of it nor are super-clusters the end at the other extreme. Rather I suspect we encounter singularities which change the...
Thanks Shmoe. I think the differential of z caused the problem for Loom. It's just:
dz=ie^{i\theta}d\theta
so then:
d\theta=\frac{1}{iz}dz
all the rest should fall into place. Also, I got most of the examples I gave above from the quintessential engineering math reference book...
Very well Matt. I am struck by the similarities between the properties of non-linear systems and the geometry of math itself. Not non-linear geometry but the very geometry of mathematics itself: nested, fractal, and ergodic (the last property explaning why we can get to the same result from...