Created
November 19, 2019 13:13
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diff --git a/construction.tex b/construction.tex | |
index 79aa218..babee01 100644 | |
--- a/construction.tex | |
+++ b/construction.tex | |
@@ -144,11 +144,18 @@ We will now show that the savings from the interlink set technique are significa | |
We can split it up at some arbitrary index $k \geq 0$: | |
- \begin{align*} | |
- \sum_{j=0}^\infty (2^{-i})^j &= \sum_{j=0}^k 2^{-ij} + \sum_{j=k+1}^\infty (2^{-i})^j\\ | |
- &= \sum_{j=0}^k 2^{-ij} + \frac{2^{-i(k + 1)}}{1 - 2^{-i}} | |
+ \[ | |
+ \sum_{j=0}^\infty (2^{-i})^j = \sum_{j=0}^k 2^{-ij} + \sum_{j=k+1}^\infty (2^{-i})^j | |
+ \] | |
+ | |
+ Applying the formula for geometric series sum from $j=k+1$ to $\infty$ | |
+ on the second term, we obtain: | |
+ | |
+ \[ | |
+ \sum_{j=0}^k 2^{-ij} + \sum_{j=k+1}^\infty (2^{-i})^j | |
+ = \sum_{j=0}^k 2^{-ij} + \frac{2^{-i(k + 1)}}{1 - 2^{-i}} | |
= \sum_{j=0}^k 2^{-ij} + \frac{1}{2^{i(k + 1)} - 2^{ik}} | |
- \end{align*} | |
+ \] | |
\end{proof} |
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