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2

Even more is true: the leading digits of $p^n$ can be any sequence you want. Note how $$\text{the leading digits of }p^n\text{ are }\overline m$$ is the same as saying $$\overline m\cdot 10^k\leq p^n<(\overline m+1)\cdot 10^k\text{ for some }k\geq0.$$ Equivalently, $$\log_{10}(\overline m)\leq n\log_{10}(p)-k<\log_{10}(\overline m+1).$$ Indeed there ...


2

You get this because the golden ratio is a quadratic algebraic integer. Let $\tau=(1+\sqrt5)/2$ and $\overline{\tau}=(1-\sqrt5)/2$. These are the zeros of the polynomial $$ p(x)=(x-\tau)(x-\overline{\tau})=x^2-x-1. $$ Let then $a/b$ be a rational number (in reduced form). Because $\tau$ is irrational we get that $p(a/b)\neq0$. But $$ p(a/b)=(\frac ...


2

Fleshing out my comment into an answer: we know that $e^9\approx 8100$, so that the integer part of $e^{e^9}$ is roughly 12,000 bits. This means that we can get a value to sufficient accuracy by: Computing $e$ to 'sufficient' accuracy (being conservative, $2^{16}$ bits should be enough here; it's not hard to do a more accurate error analysis, but the ...


4

We could prove it is not an integer by providing explicit bounds. If you really wanted to, you could note the following common inequality, holding for all positive $n$: $$\left(1+\frac{1}n\right)^n<e<\left(1+\frac{1}n\right)^{n+1}.$$ Which yields a series of rational bounds for $e$, where both sides converge to $e$ as $n$ heads to infinity. The point ...


1

$((x-a)(x-b)(a-b))^2 = -23$ which is not a square in $\Bbb Q$. And so $\sqrt{-23} \in \Bbb Q(x,a)$. This extension must have even degree over $\Bbb Q$. If $\Bbb Q(x,a)$ were equal to $\Bbb Q(x)$ it would have degree $3$, which is impossible. Hence those fields are different, and so $a \notin \Bbb Q(x)$. $d$ cannot be a square in $\Bbb Q(x)$.


1

Presumably you're not allowing $a_n = b_n = c_n = 0$. Clearly yes, in some sense: if you restrict the possible $(a, b, c)$ in some way (e.g. a bound on $\max(|a|,|b|,|c|)$) that leaves finitely many possible choices, one of them must have the smallest value of $|at^2 + b t + c |$. If, for example, you impose $a = b$, you're essentially approximating $t^2 + ...


0

the following series is used for $\sqrt{3}$ $$\sum_{n=0}^{\infty }\frac{(2n-1)!!}{n!3^n}=1+1/3+1/6+5/54+35/684+..$$


1

I outlined a really easy to use method originally created by the Babylonians to approximate the square root of a number $N$ in this answer. Here's the idea: make an initial guess as to the square root, and let this initial guess be given by $r_1$. Now, continue improving your guesses by using $$ r_{n+1}=\frac{1}{2}\left(r_n+\frac{N}{r_n}\right),\tag{1} $$ ...


1

Brahmagupta's equation gives such approximations. An integer solution for $X^2-dY^2=1$ leads to the rational number $X/Y$ an approximation for $\surd d$. $$49-3\times16=1$$ $$7^2-(4\surd3 )^2=1$$ $$(7-4\surd3)(7+4\surd3)=1$$ Now raising to an arbitrary power $$ (7-4\surd3)^n(7+4\surd3)^n=1$$ Expressing $(7+4\surd3)^n$ in the form $a_n+b_n\surd3$ you can ...


2

To test if $x>\sqrt3$, it suffices to check if $x^2>3$, since these are equivalent when $x$ is positive. With that in mind, I then see if $1.00>\sqrt3$, if $1.01>\sqrt3$, if $1.02>\sqrt3$, if $1.03>\sqrt3$, in order, until I find the smallest one that's bigger than $\sqrt3$. I then take the one immediately before that. Hey, nobody said ...


4

May be, we could consider solving for $x$ $$f(x)=x^2-3=0$$ and use Newton method which will give $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=\frac{x_n^2+3}{2 x_n}$$ Starting with $x_0=1$, we should get as iterates $2$, $\frac{7}{4}$, $\frac{97}{56}$, $\frac{18817}{10864}$, $\frac{708158977}{408855776}$. Using Halley method instead, which will give ...


8

It is an overkill for sure, but since the continued fraction of $\sqrt{3}$ is given by: $$ \sqrt{3}=[1;\overline{1,2}],\tag{1}$$ we have that: $$ [1,1,2,1,2,1,2] = \frac{71}{41} \tag{2} $$ is an accurate approximation, and: $$\left|\sqrt{3}-\frac{71}{41}\right|\leq\frac{1}{41^2}<\frac{1}{1000}.\tag{3}$$


2

There are $100$ distinct rational numbers between $1$ and $2$ of the form $1.XY5$. Square them and find where the square changes from below $3$ to above $3$. Take the average of the largest number whose square is below $3$ and the smallest number whose square is above $3$.


5

One way is: use decimal search to find the first 2 decimal places, then multiply by 100, truncate to an integer, and use this value over 100. Decimal search is: Find each digit by seeing if a certain value is two high or two low. Eg: calculate $1.5^2$. If this is < 3, try $1.6^2$, $1.7^2$ until we find a value which squares to more than 3. Then do the ...



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