Intereting Posts

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Local and global logarithms

With a computer or calculator, it is easy to show that

$$

2^{2^\sqrt{3}} = 10.000478 \ldots > 10.

$$

How can we prove that $2^{2^{\sqrt3}}>10$ without a calculator?

- Euler's Approximation of pi.
- Approximation symbol: Is π ≈ 3.14.. equal to π ≒ 3.14..?
- Intuitively, why is the Euler-Mascheroni constant near sqrt(1/3)?
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- Maximum of Polynomials in the Unit Circle
- Very accurate approximations for $\sum\limits_{n=0}^\infty \frac{n}{a^n-1}$ and $\sum\limits_{n=0}^\infty \frac{n^{2m+1}}{e^n-1}$

- When is $2^x+3^y$ a perfect square?
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- Why would the reflections of the orthocentre lie on the circumcircle?
- Square matrices satisfying certain relations must have dimension divisible by $3$
- 2011 AIME Problem 12, probability round table
- Let $k$ be a natural number . Then $3k+1$ , $4k+1$ and $6k+1$ cannot all be square numbers.
- Exploring Properties of Pascal's Triangle $\pmod 2$
- Integral $ \int_0^\infty \frac{\ln(1+\sigma x)\ln(1+\omega x^2)}{x^3}dx$
- $g(x+y)+g(x)g(y)=g(xy)+g(x)+g(y)$ for all $x,y$.
- Does there exist a sequence $\{a_n\}_{n\ge1}$ with $a_n < a_{n+1}+a_{n^2}$ such that $\sum_{n=1}^{\infty}a_n$ converges?

I suspect any adequate answer to this question is going to be very computation-heavy. Here’s one.

First we claim that

$$

\sqrt{3} > \frac{3691}{2131} \tag{1}

$$

This fraction is not pulled out of a hat; rather, it is found by taking the continued fraction $\sqrt{3} = [1; 1, 2, 1, 2, 1, 2, \ldots]$ and carrying it out a while. Anyway, the above is proved directly by

$$

3 \cdot 2131^2 = 3 \cdot 4541161 = 13623483 > 13623481 = 3691^2.

$$

We next claim that

$$

2^{{3691}/{2131}} > \frac{1465}{441} \tag{2}

$$

To prove this, one needs to show $2^{3691} 441^{2131} > 1465^{2131}$.

With repeated squaring, each term requires squaring about $12$ times, so this is doable$^{a}$ with a pencil and paper.

The last step is

$$

2^{1465/441} > 10 \tag{3}

$$

Again one uses repeated squaring, and if one is working in base 10 one only needs to count the number of digits in $2^{1465}$. This should not take nearly as long as the previous computation.$^{b}$

From (1), (2), and (3),

$$

2^{2^{\sqrt{3}}} > 2^{2^{3691/2131}} > 2^{1465 / 441} > 10. \quad \square

$$

$^{a}$ Honestly, this is quite a stretch. I can’t guarantee it won’t take like a year of work.

$^{b}$ As a rough estimate, perhaps a day or so.

Here is another mythical answer:

We want to prove $(\log_2 \log_2 10)^2 < 3$.

One way is to use the procedure outlined here to compute $\log_2$. The only mild complication is knowing when to stop. Since $\log_2$ is non-decreasing, it suffices to find $x\ge\log_2 10 $ and $y \ge \log_2 x$ such that $y^2 <3$.

The procedure is straightforward, I am just repeating the parts necessary to see how an upper bound is found. Given $x>0$, we first compute the integer part of $\log_2 x$ by finding the smallest $n$ such that ${x \over 2^n} \in [1,2)$, then $\log_2 x = n + \log_2 {x \over 2^n}$. Then, supposing $x \in (1,2)$, we repeatedly square $x$ until $x^{2^n} \in [2,4)$. Then we have

$\log_2 x = {1 \over 2^n}(1+ \log_2 {x^{2^n} \over 2})$, where ${x^{2^n} \over 2} \in [1,2)$, and so we can repeat ad nauseam.

For the purposes of this problem, we note that in the latter step, we always have $\log_2 x \le {1 \over 2^n}(1+ 1)$, since $\log_2 {x^{2^n} \over 2} \le 1$. So by replacing $\log_2 {x^{2^n} \over 2}$ by $1$ at any stage we obtain an upper bound. The error will be $\le {1 \over 2^{n_1}} \cdots {1 \over {2^{n_k}}}$, where the $n_1,…,n_k$ are the number of ‘squarings’ at each step.

Then it is a matter of trial and error to find suitable $x,y$:

$x = 3+{1 \over 4} + {1 \over 16} + {1 \over 128} + {1 \over 1024} + {1 \over 2048}+{1 \over 8192}+ {1 \over 65536}+ {1 \over 65536} = {217706 \over 65536} \ge \log_2 10$.

$y = 1+{1 \over 2} + {1 \over 8} +{1 \over 32} + {1 \over 128} + {1 \over 256} + {1 \over 1024} + {1 \over 2048} + {1 \over 16384} + {1 \over 65536}+ {1 \over 65536}= { 113510 \over 65536 } \ge\log_2 x$.

We have $y^2=({ 113510 \over 65536 })^2 < 3$.

No calculators or computers were harmed during this computation.

There is nothing magical about a calculator: any computation performed on a calculator can be done by hand (and perhaps with extra rigor). (*)

Most special functions have series approximations that are known to converge particularly quickly; but Taylor’s theorem with remainder can be applied to almost all important special functions even without knowing these specialized series. If $f(x)$ is smooth on $(a-\epsilon, b+\epsilon)$ and all derivatives $f^{(n)}$ are bounded on that interval:

$$|f^{(n)}(x)| \leq K^n\qquad \forall x\in (a-\epsilon, b+\epsilon)$$

then Taylor’s theorem guarantees that

$$|f(b) – F^i(b)| \leq \frac{K^i(b-a)^{i+1}}{(i+1)!}$$

where $F^i$ is the Taylor expansion to $i$th order

$$F^i(b) = \sum_{j=0}^i f^{(j)}(a) \frac{(b-a)^j}{j!}.$$

In particular if $K^n$ can be chosen to decay sufficiently quickly in $n$, and $f^{(n)}(a)$ is easy to evaluate exactly at a special value $a$, the above can be used to approximate $f(b)$ to arbitrary precision by hand.

For this particular problem, we want to show that

$$\sqrt{3}\log 2 \leq \log\log 10 – \log \log 2.$$

$\sqrt{x}$ and $\log x$ can be trivially estimated using the above. $\log \log x$ can be computed by composition (this will require computing $\log x$ to high precision) or directly using Taylor expansion about $x=e$ (NB the formulas for the higher-order derivatives are not particularly pleasant).

(*) However even when guaranteed accuracy is required, the advantages of hand calculation over computer algebra packages like Mathematica with arbitrary-precision support are rather dubious.

First take logs

$$2^{\sqrt3}>\log_2 10$$

Then again

$$\sqrt{3}>\log_2 \log_2 10$$

Now, $$\log_2(10)=\frac{\log 10}{\log 2}$$

With $\log 10\approx 2.302$ and $\log 2\approx 0.6931$, We find $$\log_2(10)\approx3.32131$$

And the $\log_2$ of that is $(\approx1.731)$ Now, $\sqrt 3\approx 1.732,$ and therefore we conclude that $$2^{2^{\sqrt3}}>10$$

Note that these values can be found with Newton’s method and taylor series, though tedious to find and calculate.

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