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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

  • 3
    I think you need to make more explicit what you're deriving from what -- you lost me at the point where it says "so ... and then". It's clearer if you write equations rather than just expressions that leave the reader wondering what these are supposed to be equal to. (You called the displayed expression at the very beginning an "equation", but it isn't.) From what I can tell in the current state of the question, there's a mistake: $\lim_{n\to \infty } \sqrt{n^{2} -n} = n-1/2$ doesn't make sense because the variable that's taken to a limit on the left-hand side can't occur in the result.2011-08-04
  • 0
    I second what joriki said and add that $\cos(-\pi n) = -1$ only when $n$ is odd, so unless I'm missing something, that step is wrong as well.2011-08-04
  • 0
    Like @joriki said. And the limit is not zero but $\frac14\sqrt{4+\pi^2}=0.931047945$.2011-08-04

1 Answers 1