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6

Consider $$f(x)=x-(1+\ln(1+\ln(1+\ln(x)))).$$ It is $$f'(x)=1-\frac{1}{1+\ln(1+\ln(x))}\cdot \frac{1}{1+\ln (x)}\cdot \frac 1x=\frac{(1+\ln(1+\ln(x)))(1+\ln x)x-1}{(1+\ln(1+\ln(x)))(1+\ln x)x}.$$ If $x\in(e^{1/e-1},1)$ then $f'(x)<0$ (that is, $f$ is strictly decreasing in $(e^{1/e-1},1)$) and if $x\in(1,\infty)$ it is $f'(x)>0$ (that is, $f$ is ...


6

That is known as a "Borwein Integral", named after one of the authors of the paper you linked. http://en.wikipedia.org/wiki/Borwein_integral Wikipedia has some references which explain what is going on. Here is one of them. http://schmid-werren.ch/hanspeter/publications/2014elemath.pdf I would have left this as a comment but this site requires 50 ...


3

This is not a complete answer, but I followed their advice "see [16, chap. 2] for additional details" (that's the book Experimentation in Mathematics). In section 2.5.2 they show that the cosine product (without the factor $\cos 2x$) equals $\prod_{k=0}^{\infty} \operatorname{sinc}\left(\frac{2x}{2k+1}\right)$, whose integral can be computed using the ...


3

The following should at least partly answer your question. Let $X_1,\ldots,X_k$ be random variables i.i.d according to $B(n,\frac{1}{2})$. Let $$Y_k=min\{X_1,\ldots,X_k\}$$ The following is valid: \begin{align*} E(Y_k) &= \frac{1}{2^{nk}}\sum_{t=1}^{n}\left(\sum_{j=t}^{n}\binom{n}{j}\right)^k\tag{1}\\ \\ E(Y_k) &\sim ...


2

In general the differential equation is solved by $$ y(x) = \exp\left(\frac{1}{\epsilon} \int_0^x Q(\xi)\,d\xi\right)\left[y(0) + \frac{1}{\epsilon}\int_0^x \exp\left(-\frac{1}{\epsilon} \int_0^\zeta Q(\xi)\,d\xi\right) R(\zeta)\,d\zeta \right]. \tag{1} $$ Unless we require $y(0) = 0$ then we will always have a term of the form ...


2

I am not sure of the purpose of the exercise as none of the three items approximate the sequence, but here goes. This is a just a straight plug in values and take differences for each $n$ for the four quantities. Following this, we can setup the table and arrive at: $$\left( \begin{array}{cccc} \text{n} & \text{$|$x$\_$n - r$\_$n$|$} & ...


2

These rather unexpected and large coefficients arise, in the calculation of the integral and of the successive minimization, only if we search an expression for $a$ where the numerator has no fractional terms and no "collected" factors: in fact, in this case we necessarily have to perform some multiplications that lead to relatively large coefficients. ...


1

$$\log\frac{r+l_1}{r-l_2}=\log\frac{1+(l_1/r)}{1-l_2/r}\\=\log(1+l_1/r)-\log(1-l_2/r)\\ \approx l_1/r-(-l_2/r) = (l_1+l_2)/r$$


1

Hint $$\frac{r+l_2}{r-l_1}=\frac{r-l_1+l_1+l_2}{r-l_1}=1+\frac{l_1+l_2}{r-l_1}\approx 1+\frac{l_1+l_2}{r}$$ Now, use, for small $x$, $\log(1+x)\approx x$.


1

This answer is to address a question in the comments of the other answer on whether the exponentially diverging terms drop out in regions where $\operatorname{Re} Q(x) < 0$. We'll study the particular case of $Q(x) = -\sin x$ and $R(x) = 1$ over the interval $2\pi < x < 3\pi$. Note that we have $Q(x) < 0$ for all $x$ in question. The ...


1

The PSLQ algorithm you can determine a,b,c to an arbitrary tolerance. I implemented this in python as follows from sympy.mpmath import pslq x = 0.5**.25 def f(a,b,c): return a*x**3 + b*x**2 + c*x - round(a*x**3 + b*x**2 + c*x) def nextabc(a,b,c,i): return pslq([x**3,x**2,x,1], maxcoeff=i+1, tol=i**-3, maxsteps=10000) a=1 b=1 c=1 pa = 0 pb = 0 pc ...


1

You can make an approximation by the normal distribution: $\large{1-P(X\leq 59)\approx 1-\Phi \left( \frac{59+0.5-50}{\sqrt{100\cdot 0.5 \cdot 0.5}} \right)}$ $\Phi(z)$ is the cummulative function of the standard normal distribution. 0.5 is the continuous correction factor. Formula for approximation the cdf of the standard normal distribution: ...


1

Don you want the following: $$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots$$ $$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots $$


1

Near to zero it is usual to use the following approximation: $$\sin x \approx x $$ $$\cos x \approx 1-\displaystyle \frac{x^2}{2}$$ $$\tan x \approx x $$ you can convince yourself, for example, using the relationships that @paul gives.


1

There is a general construction that can be used here. Consider any nonnegative measurable function $f$. Now define $$A_{n,m} = \left \{ x : \frac{m-1}{n} \leq f(x) \leq \frac{m}{n} \right \}, \quad m = 1,2,\dots,n^2, \\ A_{n,n^2+1} = \{ x : f(x) \geq n \}.$$ Then the sequence of simple functions $$s_n(x) = \sum_{m=1}^{n^2} \frac{m-1}{n} 1_{A_{n,m}}(x) + ...


1

You can establish quite nice approximations using Pade expansions of the functions. For example $$\sin(x)\approx \frac{x-\frac{7 x^3}{60}}{1+\frac{x^2}{20}}$$ $$\cos(x)\approx\frac{1-\frac{5 x^2}{12}}{1+\frac{x^2}{12}}$$ $$\tan(x)\approx\frac{x-\frac{x^3}{15}}{1-\frac{2 x^2}{5}}$$ are quite good. For sure, if you increase the degrees of numerator and ...


1

You can mimic the way it is done in 2D, by randomly sampling points in the unit cube and calcultate the fraction that lie in the unit sphere inside. The theoretical result is $\dfrac{\frac43\pi}{8} = \dfrac{\pi}{6}$. However, it don't see it being more efficient than the 2D version, because you need to make $1.5$ times more random samples, and the calculus ...


1

Assuming the improper integral exists, your function can be written as $$F(y) = C \exp\left(\int_0^y f(x)\; dx\right)$$ where $C = \exp(-\int_0^\infty f(x)\; dx)$. Assuming the required derivatives exist, the Taylor series of $F(y)$ around $y=0$ starts $$ F(y) = C + C f(0)\; y + \dfrac{C}{2} \left( f'(0) + f(0)^2\right) y^2 + \dfrac{C}{6} \left( f''(0) + ...


1

I would say either of those inferences would be extremely dicey due to selection biases. That is, the set of voters is unlikely to be particularly representative of the set of viewers. My thinking is that you could argue that the set of voters is a representative sample of the set of viewers if the voters were randomly drawn from the set of viewers, but ...


1

Given $\alpha$ a positive integer and $0<\delta\le1$, set $$ \epsilon=(1+\delta)^{1/\alpha}-1. $$ We claim that if $|z-\omega|<\epsilon$, then $|z^\alpha-\omega^\alpha| < \delta$; in particular, the real and imaginary parts of that latter difference are both less than $\delta$. To see this, consider the function $$ f(z) = z^{\alpha-1} + ...



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