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I have the following expression:

$$\sum_{i=0}^{n-1}[\rho^i-\frac{1}{n}(\frac{\rho^{i+1}-1}{\rho-1})]^2 $$

I would like to reduce this to a closed form expression. What I have tried so far:

  1. Remove the square by multiplying out. This give an inner expression of: $$\rho^{2i}-2\frac{\rho^i}{n}(\frac{\rho^{i+1}-1}{\rho-1})+\frac{1}{n^2}\frac{(\rho^{i+1}-1)^2}{(\rho-1)^2}$$

  2. Split out the sum on each term $$\sum_{i=0}^{n-1}\rho^{2i}-\sum_{i=0}^{n-1}2\frac{\rho^i}{n}(\frac{\rho^{i+1}-1}{\rho-1})+\sum_{i=0}^{n-1}\frac{1}{n^2}\frac{(\rho^{i+1}-1)^2}{(\rho-1)^2}$$

  3. Hope for some nice expressions to arise e.g. using geometric series.

I have only done some simple manipulations with geometric series before and can't get further than the above. All help is appreciated!

RVA92
  • 115

1 Answers1

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There is further expansion possible:

$$\sum_{i=0}^{n-1} \rho^i (\rho^{i+1} - 1) = \rho \sum_{i=0}^{n-1} (\rho^2)^i - \sum_{i=0}^{n-1} \rho^i,$$

and

$$\sum_{i=0}^{n-1} (\rho^{i+1}-1)^2 = \rho^2 \sum_{i=0}^{n-1} (\rho^2)^i - 2\rho \sum_{i=0}^{n-1} \rho^i + \sum_{i=0}^{n-1} 1.$$

So if we let $$\sum_{i=0}^{n-1} \rho^i = S_1, \quad \sum_{i=0}^{n-1} (\rho^2)^i = S_2,$$ then the given sum equals

$$S_2 - \frac{2}{n(\rho-1)} \left(\rho S_2 - S_1\right) + \frac{1}{n^2(\rho-1)^2} \left(\rho^2 S_2 - 2\rho S_1 + n\right).$$ Now put everything over a common denominator and simplify:

$$\frac{n^2(\rho-1)^2 S_2 - 2n(\rho-1)(\rho S_2 - S_1) + \rho^2 S_2 - 2\rho S_1 + n}{n^2 (\rho-1)^2} = \frac{(n+\rho-n\rho)^2 S_2 - 2(n + \rho - n\rho)S_1 + n}{n^2 (\rho-1)^2}.$$

Now because $$S_1 = \frac{\rho^n - 1}{\rho - 1}, \quad S_2 = \frac{\rho^{2n}-1}{\rho^2-1}, $$ we can substitute these to get the final result, but since it is a rather lengthy expression, I have omitted it.

heropup
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