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I believe that I am following Hanson's proof up until this point (page 36):

$$C(n) < \dfrac{(en)^{k-1+1/(a_{k+1}+1)}w^n}{(a_1)^{(a_1-1)/a_1}(a_2)^{(a_2-1)/a_2}\dots(a_m)^{(a_k-1)/a_k}}$$

where $e$ is Euler's constant, $a_1=2$, $a_{i+1} = a_1 a_2 \dots a_i + 1$ so that:

  • $a_2 = 3$
  • $a_3 = 7$

$k$ is the value where $a_k \le n < a_{k+1}$

$w^n$ is not important for my question but it is defined as $w = a_1^{1/a_1}a_2^{1/a_2}\dots$

I am unclear at this step where:

$$C(n) < \dfrac{(en)^{k-1+1/(a_{k+1}+1)}w^n}{(a_1)^{(a_1-1)/a_1}(a_2)^{(a_2-1)/a_2}\dots(a_m)^{(a_k-1)/a_k}}< e^{k - 3/2}n^{k - 3/2}w^n$$

where $k > 2$ since $n < a_1 a_2 \dots a_k$

Here's what I understand:

  • Since each $a_i > 1$, I am clear why:

$$\dfrac{(en)^{k-1+1/(a_{k+1}+1)}w^n}{(a_1)^{(a_1-1)/a_1}(a_2)^{(a_2-1)/a_2}\dots(a_m)^{(a_k-1)/a_k}}< e^{k - 1 + 1/(a_{k+1}+1)}n^{k - 1 + 1/(a_{k+1}+1)}w^n$$

  • If he meant to use $k - \frac{3}{4}$, then I would be clear since:

$$(en)^{k-1+1/(a_{k+1}+1)}w^n < e^{k - 3/4}n^{k - 3/4}w^n$$

  • If $k - \frac{3}{2}$ is correct, then I would appreciate help in understanding how this follows.

It is not obvious to me that this is true or why it is true.

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  • $\begingroup$ @Clayton, I explored that direction but couldn't see how $a_i$ would relate to $(en)$. I think that you are right that this is where the answer should be. If I figure it out, I will update my question to show what I came up with. $\endgroup$ – Larry Freeman Oct 11 '18 at 18:58
  • $\begingroup$ Only a partial answer: the denominator simplifies to $(a_1\cdots a_k)(a_1^{-1/a_1}\cdots a_k^{-1/a_k})$. Using that $n\leq a_1\cdots a_k$ (this is by hypothesis), we can say that $(\text{right-hand side})\leq\frac{1}{n(a_1^{-1/a_1}\cdots a_k^{-1/a_k})}$. Now part of the problem that remains is the product of $a_i^{1/a_i}$ tends to a limit that is greater than $e$. With a negative exponent (in the denominator), this seems to imply that the exponent of $e$ should increase by a quantity of at least $1$. $\endgroup$ – Clayton Oct 11 '18 at 19:35
  • $\begingroup$ @Clayton, I think that this suggests that it may be a mistake. Hanson shows that $a_1 a_2 \dots a_k > n$ and $a_1^{1/a_1}a_2^{1/a_2}\dots < 2.952$ so that the denominator is roughly $n/2.952$ which is smaller than $n/e$. $\endgroup$ – Larry Freeman Oct 12 '18 at 0:35
  • $\begingroup$ Larry, that was precisely my point. @mathlove caught the mistake, though! This makes it all alright :) $\endgroup$ – Clayton Oct 12 '18 at 15:54
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First of all, from your previous question, I think (2.6) should be $$C(n) < \dfrac{(en)^{k-1+1/(a_{k+1}\color{red}{-}1)}w^n}{a_1^{(a_1-1)/a_1}a_2^{(a_2-1)/a_2}\dots a_k^{(a_k-1)/a_k}}\lt e^{k - 3/2}n^{k - 3/2}w^n$$ (see the minus sign in red in the numerator)

To prove $$\dfrac{(en)^{k-1+1/(a_{k+1}-1)}w^n}{a_1^{(a_1-1)/a_1}a_2^{(a_2-1)/a_2}\dots a_k^{(a_k-1)/a_k}}\lt e^{k - 3/2}n^{k - 3/2}w^n\tag1$$ it is sufficient to prove $$(en)^{k-1+1/(a_{k+1}-1)-(k-3/2)}\lt a_1^{(a_1-1)/a_1}a_2^{(a_2-1)/a_2}\dots a_k^{(a_k-1)/a_k},$$ i.e. $$(en)^{1/2+1/(a_{k+1}-1)}\lt a_1^{(a_1-1)/a_1}a_2^{(a_2-1)/a_2}\dots a_k^{(a_k-1)/a_k}$$ Since $n\le a_1a_2\cdots a_k$, it is sufficient to prove $$(ea_1a_2\cdots a_k)^{1/2+1/(a_{k+1}-1)}\lt a_1^{(a_1-1)/a_1}a_2^{(a_2-1)/a_2}\dots a_k^{(a_k-1)/a_k},$$ i.e. $$e^{1/2+1/(a_{k+1}-1)}\lt a_1^{1/2-1/a_1-1/(a_{k+1}-1)}a_2^{1/2-1/a_2-1/(a_{k+1}-1)}\cdots a_k^{1/2-1/a_k-1/(a_{k+1}-1)},$$ i.e. $$e^{1/2+1/(a_{k+1}-1)}2^{1/(a_{k+1}-1)}\lt a_2^{1/2-1/a_2-1/(a_{k+1}-1)}\cdots a_k^{1/2-1/a_k-1/(a_{k+1}-1)}$$ Since $a_i^{1/2-1/a_i-1/(a_{k+1}-1)}\gt 1$ for $2\le i\le k-1$, It is sufficient to prove $$e^{1/2+1/(a_{k+1}-1)}2^{1/(a_{k+1}-1)}\lt a_k^{1/2-1/a_k-1/(a_{k+1}-1)}\tag2$$ for $k\ge 3$.

Note here that seeing $(2)$ as an inequality on $k$, we see that the LHS of $(2)$ is decreasing and that the RHS of $(2)$ is increasing, so it is sufficient to prove $(2)$ for $k=3$, which is equivalent to $$2e^{22}\lt 7^{14}$$ which is indeed true.

Therefore, $(1)$ is true.

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  • $\begingroup$ Wow. Great job. I was convinced that Hanson had made another mistake. I am very glad to hear that I was wrong. $\endgroup$ – Larry Freeman Oct 12 '18 at 7:42

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