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Note: i'm re-writing some of this to reflect some advice given below

I have reason to believe that this series:

$ d = \lim_{n\to \infty }\sqrt{{\left({\left(2 \, n^{2} - n\right)} e^{\left(2 i \, \sqrt{n^{2} - n} \pi\right)} + {\left(n e^{\left(4 i \, \sqrt{n^{2} - n} \pi\right)} + n\right)} \sqrt{n^{2} - n}\right)} e^{\left(-2 i \, \sqrt{n^{2} - n} \pi\right)}} $

definitely converges for n considered as an integer only, and I've calculated values for it up to 64K (Excel) without it going below .9

It represents a distance between 2 coordinates which approach some distance the larger n gets. I'm trying to find that limit so that I know how close the points ultimately get.

I think I've figured out how to take the limit, and doing so it looks like it goes to 0. I'm hoping someone can check my work and see if I have this right, and if not, where I've gone wrong.

First I avoid dealing with the square root and substitute: $ a = \sqrt{n^{2} -n} $

into the above so it's easier to look at (for me).

$ d = \sqrt{a n e^{\left(-2 i \, \pi a\right)} + a n e^{\left(2 i \, \pi a\right)} + 2 \, n^{2} - n} $

and then:

$ d = \sqrt{2 \, a n \cos\left(-2 \, \pi a\right) + 2 \, n^{2} - n} $

knowing that: $ \lim_{n\to \infty } \sqrt{n^{2} -n} = n-1/2 $

I substitute for a:

$ d = \sqrt{2 \,n \left( n-1/2 \right) \cos\left(-2 \, \pi \left( n-1/2 \right)\right) + 2 \, n^{2} - n} $

which reduces to:

$ d = \sqrt{ \left( 2 \, n^{2} - n \right) \cos\left(- \pi n \right) + 2 \, n^{2} - n} $

and then:

$ d = \sqrt{ \left( 2 \, n^{2} - n \right) \left( \cos\left(- \pi n \right) + 1\right)} $

and finally:

$ d = \sqrt{ \left( 2 \, n^{2} - n \right) \left( -1 + 1\right)} = 0 $

so: $ d = \lim_{n\to \infty }\sqrt{{\left({\left(2 \, n^{2} - n\right)} e^{\left(2 i \, \sqrt{n^{2} - n} \pi\right)} + {\left(n e^{\left(4 i \, \sqrt{n^{2} - n} \pi\right)} + n\right)} \sqrt{n^{2} - n}\right)} e^{\left(-2 i \, \sqrt{n^{2} - n} \pi\right)}} = 0 $

for (integer) n, which is definitely a nice answer, but have i missed something here?

Obviously I have, the answer given below of

$ \sqrt{4+\pi^2}/4 $

definitely looks much more like the values I've calculated.

Thanks in advance,

Joseph

  • 0
    Like @joriki said. And the limit is not zero but $\frac14\sqrt{4+\pi^2}=0.931047945$.2011-08-04

1 Answers 1

1

I would suggest that you introduce the auxiliary quantities $z_n:=\exp\Bigl(2i\pi n\sqrt{1-{1\over n}}\Bigr)\ .$ Using the $z_n$ your expression (without the outermost square root) looks like $Q_n:=\Bigl((2n^2-n) z_n +n^2(z_n^2+1)\sqrt{1-{1\over n}}\Bigr)\ z_n^{-1}=2n^2-n+2n^2{z_n+z_n^{-1}\over 2}\sqrt{1-{1\over n}}\ .$ In the next step we have to develop $\sqrt{1-{1\over n}}$ into powers of ${1\over n}$. Using the Taylor expansion of $\sqrt{1-x}$ at $0$ we get $\sqrt{1-{1\over n}}=1-{1\over 2n}-{1\over 8n^2}+O(n^{-3})\ .$ Now we look at $\eqalign{{z_n+z_n^{-1}\over 2}&=\cos\Bigl(2\pi n \sqrt{1-{1\over n}}\Bigr)=\cos\Bigl(2\pi n\Bigl(1-{1\over 2n}-{1\over 8n^2}+O(n^{-3})\Bigr)\Bigr)\cr &=-\cos\Bigl({\pi\over 4n}+O(n^{-2})\Bigr)=-1+{\pi^2\over 32 n^2}+O(n^{-3})\ .\cr}$ Here we have used the Taylor expansion of $\cos$ at $0$. Incidentally it has become evident that the $Q_n$ are in fact real. They can now be written as follows: $\eqalign{Q_n &= 2n^2-n +2n^2\Bigl(-1+{\pi^2\over 32 n^2}+O(n^{-4})\Bigr)\bigl(1-{1\over 2n}-{1\over 8n^2}+O(n^{-3})\bigr)\cr &= {1\over 4}+{\pi^2\over16}+O(n^{-1})\ .\cr}$ Therefore $\lim_{n\to\infty}\sqrt{Q_n}={1\over4}\sqrt{4+\pi^2}\ ,$ as stated by Didier Piau.

  • 0
    Thank you again so much, i really appreciate your help2011-08-05