# Tag Info

12

If you start from $\ln(2)=\int_1^2\frac{1}{x}\,dx$, then Simpson's Rule with $n=4$ gets you there fast: $$\ln(2)\approx\frac{1}{12}\left(1+4\left(\frac45\right)+2\left(\frac23\right)+4\left(\frac47\right)+\frac12\right)\approx0.693\ldots$$

10

(expanding my comments) Let's start with the fraction $\;\dfrac{355}{113}\,$ easy to remember with something like : "doubling the odds to be near the pi" (whatever this may mean...). It is easy to find starting with the continued fraction of $\pi$ and stopping just before the (relatively) large term $292$ : \begin{align} \pi&=[3; 7, 15, ...

8

If you want to stay with degree two or three but no larger, find an implementation of PSLQ and feed it the quadruple (at incredible decimal accuracy) $$\left(\pi^3, \; \pi^2, \; \pi, \; 1 \right)$$ so as to ask for integer relations, that is integers $a_3, a_2, a_1, a_0$ of not terribly large absolute value, so that $$a_3 \pi^3 + a_2 \pi^2 + a_1 \pi + a_0 ... 8 Surely not what the book's author had in mind, but still instructive: On a calculator with a \sqrt{\phantom2} button you could approximate \log c by repeatedly hitting \sqrt{\phantom2}, getting a sequence of numbers each about twice as close to 1 as the one before. After n iterations we reach c^{2^{-n}}, which for large n is about 1 + 2^{-n} ... 8 Motivated by the hint, \ln(2) is a fixed point of$$f(x)=\frac{e^x-e^{-x}-2x-\frac32}{-2}$$(This is a bit of a departure from the hint.) It happens to be an attractor. Now, I'm assuming you can exponentiate an arbitrary decimal, since the hint gives you e^x. You have$$\begin{align} a_0&=1\\ a_1&=f(a_0)=0.57\ldots\\ ...

5

What is "the" MacLaurin series you are referring to? To get $\log(2)$, for instance, we may evaluate the Taylor series of $\log(1-x)$ around $x=0$ at $x=-1$, or at $x=\frac{1}{2}$. In the second case we get the pretty fast-convergent series: $$\log(2) = \sum_{n\geq 1}\frac{1}{n 2^n} \tag{1}$$ that can be further "accelerated" through Euler's method, ...

5

A standard definition of $\log 2$ is $$\log 2:=\int_1^2{dx\over x}\ .$$ Consider the function $$h(x):=2-{4x\over3}+{8x^2\over27}\ .$$ Writing $x:={3\over2}+t$ $\>\bigl(-{1\over2}\leq t\leq{1\over2}\bigr)$, one computes $${1\over{3\over2}+t}-h\left({3\over2}+t\right)=-{16t^3\over 81+54 t}\tag{1}$$ (and this explains the choice of $h$). Using $(1)$ we ...

5

$\root 10 \of {93648}$ is marginally better than $\sqrt{10}$. But one of the comments has a much better answer, with degree of just $4$.

5

How about, $$\sqrt[3] {31}=3.14138...$$ Where, $31$ is the length of a month. If you want memorable, you could always use, $$\pi \sim \sqrt{{{69} \over {7}}}=3.139...$$ Do I really need to explain this one? You could also use, $$\sqrt{{69 \cdot 1001} \over {7 \cdot 1000}}=3.14117...$$ Where, $1001$ refers to the book 1001 Arabian Nights

5

The solution of equation $$x \log(x)=a$$ is given in terms of Lambert function $$x=\frac{a}{W(a)}$$ In the Wikipedia page, you will find quite simple formulas for the approximation of Lambert function. For example, for very large values of $a$ $$W(a) \approx L_1-L_2+\frac{L_2}{L_1}+\frac{L_2(L_2-2)}{2L_1^2}+\cdots$$ where $L_1=\log(a)$ and $L_2=\log(L_1)$. ...

5

the $2^n$ side is answered at $\text{lcm}(1,2,3,\ldots,n)\geq 2^n$ for $n\geq 7$ The governing phenomenon here is that the LCM in question is $e^{\psi(n)},$ where $\psi(n)$ is Chebyshev's second function. So, from the text comments for https://oeis.org/A003418 , we find The prime number theorem implies that LCM(1,2,...,n) = exp(n(1+o(1))) as n -> ...

4

Your proofs look fine. Another approach is to use the fact that $e^x\ge1+x$ for all $x\in\mathbb{R}$, with strict inequality for $x\ne0$. This follows from the strict convexity of $e^x-1-x$ and its minimum of $0$ at $x=0$. Then we have for $x\gt0$, $$e^{x/n}\gt1+\frac xn\implies e^x\gt\left(1+\frac xn\right)^n\tag{1}$$ and $$e^{-x/n}\gt1-\frac ... 4 Depending on your priorities, one problem with your explanation is that there are details hiding in how mathematicians found the digits of \pi in the first place. If we make those details explicit, what we get is that we have some series (many different choices will do)$$\pi = \sum_{i=0}^\infty a_i$$and a way of bounding our error, so$$\left|\pi - ...

4

Since for positive $x$ we have $\left(1+\frac{x}{2}\right)^2>1+x$, it follows that $\sqrt{1+x}<1+\frac{x}{2}$. On the other hand, $$1+\frac{x}{2}-\sqrt{1+x} = \frac{\frac{x^2}{4}}{1+\frac{x}{2}+\sqrt{1+x}} < \frac{x^2}{8\sqrt{1+x}}$$ hence for small $x$ we have that $1+\frac{x}{2}$ is an excellent approximation for $\sqrt{1+x}$, and obviously: $$... 4 If your question is When does a circle with integer radius have area that is nearly a round integer? then here are the best approximations with at least three zeros and y<10^4, found with a program: x y x/y^2 error 300000 309 3.1419863637792 0.0003937102 600000 437 3.1418711937540 ... 4 If y=\ln(x), then$$y'=\frac{1}{x}\quad\text{and}\quad y'=e^{-y}$$and you can numerically solve either differential equation to x=2, using initial condition y(1)=0. With the first differential equation, using the Runge-Kutta method with only two steps:$$ \begin{array}{rrr|rrrr} n&x_n&y_n&k_{1,n}&k_{2,n}&k_{3,n}&k_{4,n}\\ ...

3

$\ln(2)$ is a zero to $f(x)=e^x-2$. Assuming you can exponentiate an arbitrary decimal, Newton's method converges fast. \begin{align} a_0&=1\\ a_1&=a_0-\frac{e^{a_0}-2}{e^{a_0}}=a_0-1+2e^{-a_0}=0.7357\ldots\\ a_2&=a_1-1+2e^{-a_1}=0.6940\ldots\\ a_3&=a_2-1+2e^{-a_2}=0.6931\ldots\\ \end{align} and continue to the desired precision.

3

The answer you cite is using the symbols $dV$ and $dx$ in two senses. What it means is something more like $$\Delta V \approx \frac{dV}{dx} \Delta x$$ where $\Delta V$ is the change in $V$ as an ordinary real number. By definition $$\frac{dV}{dx} = \lim_{\Delta x \to 0} \frac{\Delta V}{\Delta x}$$ and hence it must be the case that we can make ...

3

The point is that $\pi$ is usually defined as a particular geometric ratio, and not as a particular numerical value (i.e., decimal expression). One can compute the numerical value of $\pi$ to some (even arbitrary) precision in many ways, but this particular integral gives a quick way to give an easy but relatively tight upper bound on its value using ...

3

Just for fun, if you don't want to use calculus, you can get it within a range of $5000$: since $\log(10000) = 4$, we see that $25000$ is too big. Since $\log(100000) = 5$, we see that $20000$ is too small.

2

Recall that the Lambert $W$ function (sometimes called the product logarithm) satisfies (but is not quite characterized by) $$W(x) \exp W(x) = x.$$ With a little work we can show that the inverse of $$L(x) := x \log x$$ is $$L^{-1}(y) = \frac{y}{W(y)} .$$ On the other hand, using that $W(e) = 1$ leads quickly to the cheap bounds $$\log y - \log \log y < ... 2 I have the following euristic arguments. It is easy to see that$$f(n)=\prod_{p\in\Bbb P} p^{\lfloor \log_p n\rfloor},$$where \Bbb P is the set of prime numbers. Subdividing the diapason, we obtain$$f(n)=\prod_{k=1}^\infty \prod_{p\in\Bbb P_k} p^k,$$where$$\Bbb P_k=\{p\in\Bbb P: p^k\le n<p^{k+1}\}.$$Then$$\prod_{k=1}^\infty n^{|\Bbb ...

2

Since $A > 0$, $$\sqrt{A^2 + \mu^2 \epsilon^2} = A \sqrt{1 + \frac{\mu^2 \epsilon^2}{A^2}}.$$ Then use the linear approximation to $\sqrt{x}$ near $1$: $$\sqrt{x} \approx 1 + \frac{1}{2} (x - 1),$$ when $x$ is close to $1$. (This comes from calculus; one of the most important reasons why the subject is so useful is the ability to use it to approximate ...

2

There is an extension of the binomial theorem that applies to a binomial raised to a non-integer power. As is mentioned in another question, the formula is $$(a+x)^n = a^n + na^{n-1}x + \frac{n(n-1)}{2!}a^{n-2}x^2 + \frac{n(n-1)(n-2)}{3!}a^{n-3}x^3 + \ldots .$$ Setting $n = \frac12$, we get \begin{align} \sqrt{a+x} = (a+x)^{1/2} &= a^{1/2} + ...

2

There are infinitely many ways to approximate pi with rational numbers, and there is a whole industry devoted to doing this to huge degrees of accuracy. Perhaps the simplest rational approximations to pi are found in its continued-fraction expansion. The first two are $3$ and $\frac{22}7$, and already the fourth, $\frac{355}{113}$, is accurate to six decimal ...

2

You can avoid Taylor's theorem with some stingy estimations. Let $s_n = \sum_{k=1}^n \frac{1}{k!}$. Then \begin{align*} 0 < e - s_n &= \frac{1}{(n+1)!} + \frac{1}{(n+2)!} + \cdots \\ &= \frac{1}{(n+1)!}\left(1 + \frac{1}{(n+2)} + \frac{1}{(n+2)(n+3)} +\cdots \right) \\ &\leq \frac{1}{(n+1)!}\left(1 + \frac{1}{(n+1)} + \frac{1}{(n+1)^2} +\cdots ...

2

Let $x' = x+0.02x$, then $V' = 2(x+0.02x)^3 = 2(1.02x)^3$. The percentage change in $V$ is \begin{align*} \frac{V' - V}{V}&= \frac{2(1.02x)^3-2x^3}{2x^3}\\ &= 1.02^3-1\\ &= 6.1208\% \end{align*} For a negative $2\%$ change in $x$, the percentage change in $V$ is similar: $$\frac{V'-V}V = 0.98^3-1 = -5.8808\%$$

2

The answer is yes. The method is detailed in the paper by S. K. Lucas. The approximate fractions are obtained from truncating the exact continued fraction of the respected numbers, so the signs are alternating. We first give a few examples. Results For $\pi$ The continued fractions are $3, 22/7, 333/106, 355/113, 103993/33102, \dots$. \begin{align} ...

2

You're using degrees where you should be using radians. $$\lim_{x\to0}\frac{\sin x} x = \text{a nonzero number, equal to 1 if radians are used.}$$ $$\frac d {dx} \sin x = C\cdot\cos x,\text{ and C=1 if radians are used.}$$ There is a reason why radians are used in calculus, and this is it.

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