Inspired by Patrick Corn and Myself

Let a n + 1 = 2 a n a_{n+1} = 2^{a_n} and b n + 1 = 3 b n b_{n+1} = 3^{b_n} both for n 1. n \ge 1.

If a 1 = 2 a_1 = 2 and b 1 = 3 b_1 = 3 , then find the minimum l l and maximum k k such that

a n + k < b n < a n + l a_{n+k}<b_{n}<a_{n+l} for all n 2 n \geq 2

Input your answer as k + l k+l .


Inspiration


The answer is 3.

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1 solution

Zk Lin
Feb 12, 2016

This proof is actually lifted verbatim from my longer proof to Two Looming Towers of Power by Patrick Corn. I rely on this lemma heavily while constructing a proof to that question.

We claim that k = 1 , l = 2 k=1,l=2 .

To prove that a n + 1 < b n a_{n+1} < b_{n} , we proceed with induction.

Base case n = 2 n=2 :

16 = a 3 < b 2 = 27 16= a_{3} <b_{2}=27 is obviously true.

Assuming a n + 1 < b n a_{n+1} < b_{n} is true, we derive

a n + 1 log 2 < a n + 1 log 3 < b n log 3 a_{n+1} \log 2 < a_{n+1} \log 3 < b_{n} \log 3

log 2 a n + 1 < log 3 b n \log 2^{a_{n+1}} <\log 3^{b_{n}}

2 a n + 1 < 3 b n 2^{a_{n+1}}<3^{b_{n}}

a n + 2 < b n + 1 a_{n+2}<b_{n+1} as required.

To prove that b n < a n + 2 b_{n} < a_{n+2} , we proceed by induction again.

We will actually prove a stronger condition, that is 4 b n < a n + 2 4b_{n}<a_{n+2}

Base case n = 2 n=2 :

108 = 4 b 2 < a 4 = 2 16 108= 4b_{2}< a_{4}=2^{16} is obviously true.

Assuming 4 b n < a n + 2 4b_{n}<a_{n+2} is true,

4 b n + 1 = 4. 3 b n < 4 b n . 3 b n < 1 6 b n = 2 4 b n < 2 a n + 2 = a n + 3 4b_{n+1}=4.3^{b_{n}}<4^{b_{n}}.3^{b_{n}}<16^{b_{n}}=2^{4b_{n}}<2^{a_{n+2}}=a_{n+3} as required.

The second part of the inequality is proven by Deedlit from MathStackExchange . I could never have applied mathematical induction so skilfully myself!

Moderator note:

I call the induction in the second half " stronger induction ", in which it is surprisingly easier to prove a stronger result instead.

I call the induction in the second half " stronger induction ", in which it is surprisingly easier to prove a stronger result instead.

Calvin Lin Staff - 5 years, 3 months ago

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