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2

Let $a = \displaystyle{ \frac{11 + \sqrt{106}}{2}}$ and $b = \displaystyle{ \frac{21 + 2 \sqrt{106}}{2}}.$ Then $$x = (a + \sqrt{a^2 - 1}) (b + \sqrt{b^2 + 1}).$$ As requested, this exhibits $x$ as a product of two quartic units. (For the purists, note that $a$ and $b$ are only half algebraic integers, but the expressions above are genuinely units. I wrote ...


0

The fastest known algorithms are based on the Bailey-Borwein-Plouffe formula. A particular variation later developed by Plouffe (see here) can be used to calculate the base-$10$ digits of $\pi$ by using the formula $$ \pi + 3 = \sum_{n=1}^{\infty} \frac{n 2^n n!^2}{(2n)!} $$ Plouffe's method calculates the $n^{\text{th}}$ digit of $\pi$ in ...


1

At the time of my writing, there are 7 lengthy, beautifully written answers. They all, in one way or another, take as a launching point the definition of $e^x$ as an infinite series, or as a representation of a complex number in polar form on the complex plane. What an eight-grader might find puzzling is why one would approach e in this manner. There is a ...


1

[I see that Brian Tung has already posted a similar answer to this one. Although this is shorter, it's saying essentially the same thing, so I don't know if it will be easier to understand. Perhaps harder!] If: (1) you're familiar with the operation of differentiation of real-valued functions of real numbers; (2) you know that the function $x \mapsto e^x$ ...


2

As in my other answer, I won't be trying to prove anything to you -- power series, differential equations, etc. will be over your head for the moment. Instead I'll offer a different interpretation of the exponential function that hopefully you will find enlightening. Instead of seeing $e^{i\theta}$ as a number -- let's interpret it as an operator on ...


4

Just to be clear on what it would take to decide this based on rational approximation: we want to compare the ratio $$ f(x)=\frac{x^x}{(x-1)^{x+1}} $$ to unity at $x=(\pi + 1)$. Its logarithm is $$ \begin{eqnarray} g(x)=x\log x - (x+1)\log(x-1) &=& x\log x - (x+1)\log x - (x+1)\log(1-1/x) \\ &=& -\log x - (x+1)\log(1-1/x) \\ &=& ...


7

(The following is not meant to be serious mathematics.) In decimal there are $1000$ three place numbers. The probability that at the $N^{\rm th}$ decimal place of $\pi$ the last three figures enounce exactly the number $N$ therefore is ${1\over1000}$, and the probability that this does not happen is ${999\over1000}$. Assuming independence of the involved ...


2

As commented, the question is missing an essential piece of information, the ground field. To get a somewhat non-trivial question, the ground field should probably be $\mathbb Q(\pi)$. Now the following reasoning works: Any $a\in\mathbb Q(\pi)$ has the form $a = \frac{f(\pi)}{g(\pi)}$ with $f,g\in \mathbb Q[x]$, $g\neq 0$. If $a^3 = \pi$, then $f(\pi)^3 - ...


2

Let $x=(\pi+1)^{\pi+1}$ and $y=\pi^{\pi+2}$ Since $\ln x=(\pi+1)ln(\pi+1)$ and $\ln y=(\pi+2)\ln\pi$ and $\ln x$ is increasing , compare $\ln x$ and $\ln y$ So I depend on a internet calculating, $\ln x-\ln y<0$ and so $x<y$ But $\ln x-\ln y=-0.00019...$ tell us that it is difficult to clear for manually. (I tried to prove for " ...


14

In my attempts to solve the problem of the OP, I found it useful to choose a slightly broader perspective. I started by defining a function $f(x)$ in which an adjustable parameter $a$ appears: $$f(x) = (x + a)log(x+a) - (x+2a)log(x)$$ In terms of parameter $a$, we seek the value of $x$ for which $f(x)$ equals zero. Later on we will focus on the case $a=1$. ...


0

(This answer is taken from the corresponding question on MO.) This is known since 2010 at least for $n\leq 11$ -- see this entry in the OEIS or F. Bellards's page about digits of $\pi$. In fact, every sequence of length $11$ occurs once in the first $2\ 512\ 258\ 603\ 207$ digits of $\pi$.


2

Consider the function f(x) =($\frac{x+1}{x}$)^(x+1); the sign of the derivative on [3, 4] depends of the expression xln($\frac{x+1}{x})$ - 1 and both f ‘(3) and f ‘(4) are negative and not null on the interval. Hence f is decreasing over [3, 4]. We have f(3) = 3.160493827 > $\pi$ and f(4) = 3.051757813 < $\pi$ . We note that f(3) is nearer of $\pi$ than ...


6

Unfortunately, needles are needed. This work is for $(\pi + 1)/\pi$, not $(\pi+2)/(\pi + 1)$, if you are going to make it more precise, third order approximation is needed. I think needles are not necessary. $a = \pi$, We try to prove $\ln(a + 1)/\ln(a) < (\frac{a+1}{a})$. well, $a \sim 355/113\sim 3$, the error on estimate of $\pi$ is very tiny, will ...


4

If the number is positive, then raising it to the power of an irrational number is well defined. This is because an irrational number can be defined as a converging sequence of rational numbers (like 3, 3.1, 3.14, 3.141, 3.1415 etc), and as these approach the irrational number then the power of these rational numbers also converges to a fixed value, which is ...


1

Let $$x=a.a_1a_2a_3... and$$ $$y=b.b_1b_2d_3...>0$$. You can approximate x^y as a limit of the sequence $$a^b,a.a_1^{(b.b_1)},a,a_1a_2^{(b.b_1b_2)},....$$ As for physics application suppose you are dealing with an experiment which is governed by the differential equation $$yy'-(y')^2=(y^2)/x$$ One of the solutions is $$x^x$$ and if $$x=2^.5$$ then you get ...


2

Observe that $$ a^r=e^{r\ln a} $$ and use the Taylor series for $e^x$.


7

Formally, we have $a^b = e^{b \ln(a)}$ and $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$$ $$\ln x = \int_1^x \frac{dt}{t}$$ And for integer $n$, we define $x^n$ as $$\prod^n_{i=1} x$$ This is needed because we don't want to define the powers in $e^x$ circulary. Also note that since we use $\ln a$ in this definition, we ...



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