Equation for a summed series with a limit

Hi,

I'm completely new to this. I am a software developer and haven't had the calculus books open for about a quarter of a century so I do aplologize if this is not in the correct forum. Further I'm not even sure if the question is phrased correctly.

$\displaystyle \lim_{x \to \infty} $$\displaystyle \sum_{n = 0}^x $$\displaystyle (1/n)(1/(d-r \cos(2\pi n/x))^2)$

I'm basically trying to figure out the value of the summed series $\displaystyle (1/n)(1/(d-r \cos(2\pi n/x))^2)$

where there are x elements in the series as x gets large. d and r are constants.

Hoping you can help or point me in the right direction.

Cheers,

Colin.

Re: Equation for a summed series with a limit

I've taken my project to the next step using the equation you provided earlier and my next hurdle occurs with the addition of a second variable r2 and substituting the denominator of the original equation as the d variable in your solution. The latex probably explains better what I’m chasing;

$\displaystyle \lim_{x \to \infty} \sum_{n = 0}^x (1/x) \frac { d-r_{1}cos(2\pi n/x)} { ((d-r_{1} \cos(2\pi n/x))^2 – r_{2}^2)^{(3/2)}}$

Is this still a Riemann sum?

If I've got this right this should also converge to around 0.06 for d=4 and r = 1.

Re: Equation for a summed series with a limit

Quote:

Originally Posted by

**ColinJ** I've taken my project to the next step using the equation you provided earlier and my next hurdle occurs with the addition of a second variable r2 and substituting the denominator of the original equation as the d variable in your solution. The latex probably explains better what I’m chasing;

$\displaystyle \lim_{x \to \infty} \sum_{n = 0}^x (1/x) \frac { d-r_{1}cos(2\pi n/x)} { ((d-r_{1} \cos(2\pi n/x))^2 – r_{2}^2)^{(3/2)}}$

Is this still a Riemann sum?

If I've got this right this should also converge to around 0.06 for d=4 and r = 1.

Yes it is still a Riemann sum, and the limit is equal to the integral

$\displaystyle \int_0^1\!\!\frac{d-r_1\cos(2\pi x)}{\bigl((d-r_1\cos(2\pi x))^2 - r_2^2\bigr)^{3/2}}\,dx.$

That is a far more elaborate integral than the one in the previous problem, and I doubt whether it is possible to give the answer in a closed form. For particular values of d, r_1 and r_2, you can presumably get Mathematica or Wolfram to churn out an answer.

Re: Equation for a summed series with a limit

Hi,

I've spent a fair amount of time on my project and discovered that I need another term in the denominator (as below).

$\displaystyle \lim_{x \to \infty} \sum_{n = 0}^x (1/n)(1/((d-r \cos(2\pi n/x))^2 + (r \sin(2\pi n/x))^2 )) $

I'm guessing this becomes the integral;

$\displaystyle \int_0^1\frac{d \theta}{\bigl(d-r\cos(2\pi \theta))^2 + (r \sin(2 \pi \theta))^2 }$.

But, I can't for the life of me solve this thing. Any help is appreciated once again. I'm fairly sure it converges to around 0.0412468 for d = 5 and r = 1.

Regards,

Colin.

Re: Equation for a summed series with a limit

Quote:

Originally Posted by

**ColinJ** Hi,

I've spent a fair amount of time on my project and discovered that I need another term in the denominator (as below).

$\displaystyle \lim_{x \to \infty} \sum_{n = 0}^x (1/n)(1/((d-r \cos(2\pi n/x))^2 + (r \sin(2\pi n/x))^2 )) $

I'm guessing this becomes the integral;

$\displaystyle \int_0^1\frac{d \theta}{\bigl(d-r\cos(2\pi \theta))^2 + (r \sin(2 \pi \theta))^2 }$.

But, I can't for the life of me solve this thing. Any help is appreciated once again. I'm fairly sure it converges to around 0.0412468 for d = 5 and r = 1.

You are right about the integral. As it happens, this complicated-looking integral can be evaluated fairly easily by using complex numbers and the method of contour integration. If you make the substitution $\displaystyle z = e^{2\pi i\theta}$ then the integral becomes

$\displaystyle \frac i{2\pi}\oint_C\frac{dz}{rdz^2 - (r^2+d^2)z + rd},$ (**)

where the integral is taken round the unit circle C. If $\displaystyle r<d$ then Cauchy's integral formula gives the answer as $\displaystyle \boxed{\frac1{d^2-r^2}}$ (I have skipped all the details of the calculation). If d=5 and r=1, that becomes $\displaystyle 1/24\approx0.0416667,$ fairly close to the answer that you wanted.

(**) **Edit. **I have just seen that this integral looks confusing. The "dz" in the numerator is just the usual notation for an integral with respect to z, but in the denominator "d" is a constant and "$\displaystyle dz^2$" means d times z squared.

Re: Equation for a summed series with a limit

Thanks once again.

My fault on that confusing integral. Obviously d is an inappropriate variable where integrals are involved. And, I have some reading to do.

Much appreciated.

Re: Equation for a summed series with a limit

would you mind posting some of the details of the calculation? i cant follow (at least the algebra) in converting from $\displaystyle \theta $ to z.

i got to a point like this and gave up:

$\displaystyle \frac{4\,{z}^{2}}{{i}^{2}\,{r}^{2}\,{z}^{4}+{r}^{2 }\,{z}^{4}-4\,d\,r\,{z}^{3}-2\,{i}^{2}\,{r}^{2}\,{z}^{2}+2\,{r}^{2}\,{z}^{2}+4 \,{d}^{2}\,{z}^{2}-4\,d\,r\,z+{i}^{2}\,{r}^{2}+{r}^{2}}$

after making the substitution.

though i think i may be hijacking this thread.. if i am please correct me.

Quote:

Originally Posted by

**Opalg** You are right about the integral. As it happens, this complicated-looking integral can be evaluated fairly easily by using complex numbers and the method of

contour integration. If you make the substitution $\displaystyle z = e^{2\pi i\theta}$ then the integral becomes

$\displaystyle \frac i{2\pi}\oint_C\frac{dz}{rdz^2 - (r^2+d^2)z + rd},$ (**)

where the integral is taken round the unit circle C. If $\displaystyle r<d$ then Cauchy's integral formula gives the answer as $\displaystyle \boxed{\frac1{d^2-r^2}}$ (I have skipped all the details of the calculation). If d=5 and r=1, that becomes $\displaystyle 1/24\approx0.0416667,$ fairly close to the answer that you wanted.

(**)

**Edit. **I have just seen that this integral looks confusing. The "dz" in the numerator is just the usual notation for an integral with respect to z, but in the denominator "d" is a constant and "$\displaystyle dz^2$" means d times z squared.

Re: Equation for a summed series with a limit

Quote:

Originally Posted by

**phycdude** would you mind posting some of the details of the calculation? i cant follow (at least the algebra) in converting from $\displaystyle \theta $ to z.

The integral was $\displaystyle \int_0^1\frac{d \theta}{\bigl(d-r\cos(2\pi \theta))^2 + (r \sin(2 \pi \theta))^2 }$. If you make the substitution $\displaystyle z=e^{2\pi i\theta}$ then $\displaystyle \cos\theta = \tfrac12(z+z^{-1}),$ $\displaystyle \sin\theta = \tfrac1{2i}(z-z^{-1})$ and the differential $\displaystyle dz=2\pi ie^{2\pi i\theta}d\theta = 2\pi izd\theta,$ so that $\displaystyle dz = \tfrac{d\theta}{2\pi iz}.$ As θ goes from 0 to 1, z goes right round the unit circle. Thus the integral becomes

$\displaystyle \oint_C \frac1{\bigl(d-\frac r2(z+z^{-1})\bigr)^2 + \bigl(\frac r{2i}(z-z^{-1})\bigr)^2}\,\frac{dz}{2\pi iz}.$

Multiply top and bottom by 4z to get

$\displaystyle \frac1{2\pi i}\oint_C\frac{4z}{\bigl(2dz-r(z^2+1)\bigr)^2 - r^2(z^2-1)^2}dz.$

The denominator is

$\displaystyle 4d^2z^2 - 4rdz(z^2+1) + r^2(z^4+2z^2+1) - r^2(z^4-2z^2+1)$

which simplifies to $\displaystyle -4rdz(z^2+1) + 4z^2(d^2+r^2).$

Substitute that into the denominator of the integral, cancel 4z top and bottom, change the sign of the denominator and move the "i" in the constant outside the integral from the denominator to the numerator to compensate. The result is

$\displaystyle \frac i{2\pi}\oint_C\frac1{rd(z^2+1)-(d^2+r^2)z}dz.$