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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 ...

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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 ...

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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 ...

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[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 $x \mapsto e^x$ ...

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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 ...

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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... andy=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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