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Topic: ***Spoiler*** inequality (Read 894 times) |
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anonymous
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\[
\begin{split}
\sum_{n=1}^{\infty} (a_n-a_{n+1})a_{n+2} \\
= \sum_{n=1}^{\infty} (a_1^{2^{n-1}}-a_1^{2^n})a_1^{2^{n+1}} \\
\leq \sum_{i=1}^{\infty} (a_1^i-a_1^{i+1})a_1^{2i+2} \\
= (1-a_1)a_1^2\sum_{i=1}^{\infty} a_1^{3i} \\
= (1-a_1)a_1^5/(1-a_1^3) \\
= a_1^5/(1+a_1+a_1^2) \\
< 1/3
\end{split}
\]
The second to third line uses telescopic sum, and there is a geometric series in the fourth line.
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wowbagger
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Re: ***Spoiler*** inequality
« Reply #1 on: Jun 24th, 2003, 5:17am » |
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For those who aren't used to reading LaTeX:
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| « Last Edit: Jun 24th, 2003, 5:17am by wowbagger » |
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James Fingas
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Re: ***Spoiler*** inequality
« Reply #2 on: Jun 24th, 2003, 9:26am » |
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How do we justify the third line? It's not true that every term in the new sum is larger than the corresponding term in the old sum. For instance, with i=n=2 and a1=0.99, the terms are 0.017997 for the old and 0.009415 for the new.
I am assuming that you have a good justification ... maybe I don't understand telescopic sums like I thought I did?
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anonymous
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Re: ***Spoiler*** inequality
« Reply #3 on: Jun 24th, 2003, 10:27am » |
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I'll take n=3 as an example
when n=3,
\[
\begin{split}
(a^{2^{n-1}}-a^{2^n})a^{2^{n+1}} \\
=(a^4-a^8)a^{16} \\
=(a^4-a^5)a^{16}+(a^5-a^6)a^{16}+(a^6-a^7)a^{16}+(a^7-a^8)a^{16} \\
<(a^4-a^5)a^{10}+(a^5-a^6)a^{12}+(a^6-a^7)a^{14}+(a^7-a^8)a^{16} \\
=\sum_{i=2^{n-1}}^{2^n-1}(a^i-a^{i+1})a^{2i+2}
\end{split}
\]
When n is summed from n=1 to infinity, then i is also summed from i=1 to infinity.
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| « Last Edit: Sep 7th, 2003, 2:12pm by Icarus » |
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James Fingas
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Re: ***Spoiler*** inequality
« Reply #4 on: Jun 24th, 2003, 11:24am » |
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Okay, I see where you're coming from now. Too non-obvious for me though...
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towr
wu::riddles Moderator
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Re: ***Spoiler*** inequality
« Reply #5 on: Jun 24th, 2003, 11:29am » |
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for those who'd rather not read latex, even though/if they can/could
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| « Last Edit: Jun 24th, 2003, 11:30am by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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