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22

Note that the expression of interest is of the form $$1-\left(\frac{n-1}{n}\right)^n=1-\left(1-\frac{1}{n}\right)^n$$ with $n=14000$. Recalling that $$\lim_{n\to \infty}\left(1-\frac{1}{n}\right)^n=e^{-1}$$ we have $$1-\left(\frac{13999}{14000}\right)^{14000}\approx 1-e^{-1}$$


12

Another way of doing it : consider $$A=\Big({13999 \over 14000}\Big )^{14000}$$ $$\log(A)=14000 \log\Big({13999 \over 14000}\Big )$$ Now, consider the very fast converging series expansion $$\log\Big({{1+x} \over {1-x}}\Big )=2 \Big(\frac{x}{1}+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\Big)$$ and make ${{1+x} \over {1-x}}={13999 \over 14000}$ which gives ...


7

\begin{align} \sum_{n=1}^{\infty}\frac{1}{2n-1}\Big(\frac13\Big)^{2n-1}&=\sum_{n=1}^{\infty}\frac{1}{n}\Big(\frac13\Big)^{n}-\sum_{n=1}^{\infty}\frac{1}{2n}\Big(\frac13\Big)^{2n}\\ &=-\log(1-\frac13)+\frac12\log(1-\frac19)\\ &=\frac12 \log2 \end{align} where we use $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}x^{n}=-\log(1-x)$.


6

LaTeX has the symbols \lessapprox ($\lessapprox$) and \gtrapprox ($\gtrapprox$). Incidentally, a long time ago I came across this awesome Short Math Guide for $\LaTeX$, a free pdf by the American Mathematical Society. Well-worth keeping it close by.


6

For a solution that gives you control over the precision of the result, you can do the following. First, write the expression as such: $$1−(\frac{13999}{14000})^{14000} = 1−(\frac{14000 - 1}{14000})^{14000} = 1−(1 - \frac{1}{14000})^{14000}$$ Then use the Binomial theorem $$(1 + x)^n = \sum_{k\,=\,0}^{n} {n \choose k}\, x^k$$ with $x = -\frac{1}{14000}$ ...


5

Using the fact that $n!\sim_{\infty }\sqrt{2\pi n}\left(\frac{n}{e}\right)^n$, $$\lim_{n\to\infty }\frac{n!}{\sqrt{3n}\left(\frac{n}{2}\right)^n}=\lim_{n\to\infty }\frac{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n}{\sqrt{3n}\left(\frac{n}{2}\right)^n}=\lim_{n\to\infty }\sqrt{\frac{2\pi}{3}}\underbrace{\left(\frac{2}{e}\right)^n}_{\to 0}=0$$ and thus ...


5

The relative error is about $8.9 \cdot 10^{-5}$. It is surprising that the relative error is somewhat small because $e$ and $\phi$ are apparently unrelated (*). It is probably a mathematical coincidence. Whether this error implies a relatively good approximation depends on why you want this approximation or where you're going to use it. (*) More ...


4

For any $x>0$, $\sqrt{1+x}\leq 1+\frac{x}{2}$ is trivial by squaring. On the other hand: $$ 1+\frac{x}{2}-\sqrt{1+x} = \frac{\frac{x^2}{4}}{1+\frac{x}{2}+\sqrt{1+x}}\leq\frac{x^2}{8+2x} $$ gives: $$ 1+\frac{x}{2}-\frac{x^2}{8+2x}\leq \sqrt{1+x} \leq 1+\frac{x}{2}-\frac{x^2}{8+4x}.$$


4

By exploiting the elliptic lambda function, we have: $$ \sqrt{2} = 2\cdot \frac{\theta_2^2(0,e^{-\pi})}{\theta_3^2(0,e^{-\pi})} $$ and by exploting the expansions of the Jacobi theta functions we have: $$ \sqrt{2} \approx 2\,\left(\frac{2-2e^{5\pi}+2e^{8\pi}-e^{9\pi}}{2+2 e^{5 \pi }+2 e^{8 \pi }+e^{9 \pi }}\right)^2$$ that looks way better and is right up to ...


3

How about this: $$\frac35+\frac{\pi}{7-\pi}-\sqrt2=10^{-6}\times6.495680826...$$ This is also interesting: $$\frac{131836323}{93222358}-\sqrt2=4\times10^{-17}$$


3

We have $$\frac{a_n}{\pi/2} = \prod_{k = n+1}^\infty \frac{(2k-1)(2k+1)}{(2k)^2} = \prod_{k = n+1}^\infty \biggl(1 - \frac{1}{4k^2}\biggr).$$ To estimate products, it is often convenient to take logarithms. Here we can get the easy upper bound $$\log \prod_{k = n+1}^\infty \biggl(1 - \frac{1}{4k^2}\biggr) = \sum_{k = n+1}^\infty \log \biggl( 1 - ...


3

By exploting the convexity of the $\log\Gamma$ function, it is not difficult to show that: $$ \log\left(\frac{\Gamma\left(\frac{x}{3}\right)}{\Gamma\left(\frac{x}{4}+1\right)\Gamma\left(\frac{x}{12}+1\right)}\right)\leq \left(\frac{\log 4}{3}-\frac{\log 3}{4}\right)x \tag{1}$$ holds for every $x$ big enough ($x\geq 3$ is fine). The RHS of $(1)$ is just the ...


3

This requires nothing more than just the definition of a derivative. The function $f(x)=\sqrt{1+x}$ has derivative $f'(x)=\frac{1}{2\sqrt{1+x}}$ for $x\geq 0$. By the fundamental definition of derivatives: $$f(x)-f(0)=f'(0)x+\epsilon(x),$$ where $\lim_{x\rightarrow 0^+}\epsilon(x)/x=0$ and $\lim_{x\rightarrow 0^+} \epsilon(x)=0$. If you're unfamiliar with ...


3

One can consider that an approximation formula is good if is achieves some information compression, i.e. it requires less bits of information than the decimal expansion of comparable accuracy. Take the case of the well-known fractional approximations of $\pi$: $$\frac{22}7=\color{green}{3.14}2857\cdots$$ $$\frac{335}{113}=\color{green}{3.141592}92\cdots$$ ...


2

You've described an NP-complete problem. Indeed, this problem is exactly the subset-sum problem with the extra restriction that the desired subset has size $N$; but if this were not NP-complete then the subset-sum problem would not be either, because it would only take $|C|$ iterations of the "change-problem" algorithm to solve subset-sum (where $C$ is the ...


2

The given series is $x+\frac {x^3} 3 + \frac {x^5} 5 + \dots \big|_{x = \frac 1 3} \\ = \frac 1 2 \log \frac {1+x} {1-x} \big|_{x = \frac 1 3} \\ = \frac 1 2 \log 2 .$


2

We have the estimates $$ 0\le\log(\Gamma(x))-\left[x\log(x)-x+\frac12\log(2\pi)-\frac12\log(x)\right]\le\frac1{12x} $$ and $$ -\frac1{360x^3}\le\log(\Gamma(x))-\left[x\log(x)-x+\frac12\log(2\pi)-\frac12\log(x)+\frac1{12x}\right]\le0 $$ Using these, along with the fact that $\log\left(\frac{\Gamma\left(\frac x3+1\right)}{\Gamma\left(\frac ...


2

Stirling's formula says the ratio of $$ \sqrt{2\pi x\,{}} \, \frac{x^x}{e^x} $$ to $x!$ approaches $1$ as $x\to\infty$. You seem to have a crude approximation of Stirling's formula. (I think Stirling's contribution to this may have been the value of the constant factor. de Moivre earlier showed that the ratio of $\sqrt{x}\ \dfrac{x^x}{e^x}$ to $x!$ ...


2

In any case, you also need to know $x_0=e^{y_0}$. Then the correction is given by $y-y_0=\ln\left(\dfrac x{x_0}\right)=\ln(1+\delta)$, where $\delta$ is small. You can evaluate it by Taylor which will converge linearly, $$\ln(1+\delta)\approx\delta-\frac{\delta^2}2+\frac{\delta^3}3\cdots$$ You can also limit yourself to the first term ...


2

Yes, there exist such functions. Let for instance $$ f(x)=\begin{cases}\sin x & 0\le x\le\pi,\\g(x) & -\pi\le x<0,\end{cases} $$ where $g\colon[-\pi,0]\to\mathbb{R}$ is to be chosen in such a way thet $f$ is continuous and piecewise $C^1$ on $[-\pi,\pi]$ and $$ \int_{-\pi}^0g(x)\cos(k\,x)\,dx=-\int_0^\pi \sin x\cos(k\,x)\,dx,\quad k=0,1,2 $$ and ...


2

$$ \sqrt {2}=-5\,\tanh \left( 1/2 \right) -{\frac {23\,\tanh \left( 3/2 \right) }{5}}+{\frac {21\,\tanh \left( 5/2 \right) }{5}}+{\frac {42\, \tanh \left( 7/2 \right) }{5}}-6\,\tanh \left( 1/3 \right) -{\frac {23 \,\tanh \left( 4/3 \right) }{5}}+\frac{4\,\tanh \left( 5/3 \right)}5 -3\, \tanh \left( 7/3 \right) -\dfrac{3\,\tanh \left( 1/4 \right)}{5} ...


2

First, replace $\operatorname{csch}$ and $\operatorname{sech}$ with $1 / \sinh$, $1 / \cosh$. Approximation in terms of hyperbolic functions ($\cosh x = \frac{e^x + e^{-x}}{2}, \sinh x = \frac{e^x - e^{-x}}{2}$), is then equivalent to approximation by powers of $e$, since \begin{align*} e^x &= \cosh x + \sinh x \\ e^{-x} &= \cosh x - \sinh x ...


2

To add emphasis, the simple continued fraction of a rational number is finite. It can be found using computer operations that exclusively use integers. For your purpose, the most important part is that, if your rational number is $r,$ and we have consecutive "convergents" $p/q$ and $s/t,$ then $$ \left| r - \frac{p}{q} \right| \leq \frac{1}{qt} < ...


2

Define $\alpha \in (0, 1)$ and $\beta \in (0, \infty)$ as $$ \alpha = p^{\gamma}(1-p)^{1-\gamma} \quad \text{and} \quad \beta = \frac{p}{1-p}. $$ Also we write $X = X_k \sim \operatorname{Binomial}(k, \gamma)$ and denote the sum by $S = S_{m,k}$. Then we can write $$ S = \Bbb{E}[ (1 - \alpha^k \beta^{X-k\gamma} )^m ]. \tag{1} $$ Since $p^X (1-p)^{k-X} ...


2

Call the number of squares $N(x)$. The statement $N(x)=\sqrt{x}+O(1)$ means $N(x)-\sqrt{x}=O(1)$, which is then defined the way that you know, i.e. that $N(x)-\sqrt{x}$ is bounded above and below as $x \to \infty$. Alternately, you could say that there exists $f$ such that $N(x)=\sqrt{x}+f(x)$ and $f(x)=O(1)$ as $x \to \infty$. This would be useful if ...


1

An alternative approach to sketching your rational function is to do some algebra to convert the problem into the equation $$x^5-x^3-\frac{1}{A}=0.$$ Note that none of $-1,0,1$ are solutions to this modified equation, so we don't need to enforce that requirement explicitly. This function is not hard to sketch. In fact, you can find its relative maximum ...


1

$\ln M$ is a rough approximation because it turns the dominant term, $e^x$, into $M$; so we are off by a factor of $\ln M$ "only": $M\ln M\ne e^{\ln M}$. What could be a good choice of $y$ to make $w=\ln M+y$ a better appoximation? Just as the first approximation dealt with the factor $M$ on the left, we may want choose $y$ so that it deals with the (now) ...


1

here is error or mistyping $$ \tan\theta=\frac{\sin \theta}{\cos\theta}\approx \frac{\theta}{1-\theta^2/2} $$ right way: $$\tan\theta=\frac{\sin \theta}{\cos\theta}=\frac{\sin \theta}{\sqrt{(1-\sin\theta^2)}}\approx \frac{\theta}{(1-\theta^2)^{1/2}} = \frac{\theta}{(1-\theta^2)^{1/2}} \approx \theta (1+\theta^2/2) \approx \theta $$ reason: linearixation of ...


1

Here are two applications: Weierstrass' theorem, from which separability of $C[a,b]$ is usually proven, is the core of most recipes for proving that quadrature rules are convergent for continuous integrands. The basic fact is that: $$|I(f)-Q_n(f)| \leq |I(f)-I(p)|+|I(p)-Q_n(p)|+|Q_n(p)-Q_n(f)|$$ where $I$ is the integration functional, $f$ is the ...


1

To get a concrete simple example, think of a function $f$ expanded in $[0,\pi]$ in a sine-only Fourier series (it is enough to think of $f$ extended to $[-\pi,0]$ as an odd function). Coefficients $b_k$ of this Fourier series are given by $b_k=\int_0^\pi f(x)\sin(kx)\,dx$ (apart a normalization factor). It is then enough to choose $f$ so that $b_1=b_2=0$. ...



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