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23

It would theoretically be possible to disprove the Riemann Hypothesis by establishing the existence of a 'too long' string of composite numbers. However, this 'too long' needs to be taken in a relative sense, that is relative to the size of the elements in the string. And, in your example the elements are huge relative to the length of the string.


22

Roughly speaking, according to the Prime Number Theorem, in the region of $x$, on average one integer out of every $\log x$ is prime. Now, Stirling's approximation says that $\log n!$ is roughly $n\log n$; so in the region of $n!$, you would expect one in every $n\log n$ numbers to be prime. This is less than one in every $n$. So to find a run of composite ...


6

$$\int_{n-1}^n\log(x) dx<\int_{n-1}^{n}\log( n) dx=\log(n)$$ using $\log(n)>\log(x)$ for $n-1\leq x<n$. Similarly: $$\int_n^{n+1}\log(x)dx>\int_n^{n+1}\log(n)dx=\log(n)$$


4

They are using the trapezoid rule over $n-1$ intervals: $(1,2), (2,3) \dots (n-1,n)$ The trapezoid rule for the first interval is $\int_1^2f(x)\;dx \approx \frac 12(f(1)+f(2))$ When you add these up over all the intervals, each internal point gets a coefficient $1$, with $\frac 12$ coming from the interval below and $\frac 12$ from the interval above but ...


3

Another case is when $k$ is a constant times $n$. Let $k = a n $ where $0 < a < 1$, so $k/n = a$. Then $\begin{align} \binom{n}{k} &=\frac{n!}{k!(n-k)!}\\\\ &\approx. \frac{\sqrt{2\pi n}(\frac{n}{e})^n} {\sqrt{2\pi k}(\frac{k}{e})^k \sqrt{2\pi (n-k)}(\frac{n-k}{e})^{n-k}}\\\\ &=\frac{1}{\sqrt{2\pi k}(k/n)^k ...


3

Throwing away small singular values gives a low rank approximation, call it $L$, to $A$. Since the Euclidean operator norm of a matrix is the largest singular value, $\| L-A \|_2$ is the largest singular value that was thrown away; call it $\sigma_k$. Hence $$\frac{\| Lx-Ax \|_2}{\| x \|_2} \leq \sigma_k.$$ If you leave behind $\ell$ singular values, then ...


2

This because on $[n-1,n]$, we have $\,\ln x<\ln n\,$ except at the end point so that by the monotonicity property of the integral: $$\int_{n-1}^n \ln x,\mathrm d\mkern1.5mux<\ln n\,[n-(n-1)]=\ln n.$$ Similarly for the other inequality.


2

Here is how to use the hint:Take $f(x)=\sqrt{x}$. Then $$\int_2^{f(x)}\frac{dt}{(\log t)^k}\leq \frac{1}{(\log 2)^k}\int_2^{f(x)}1 dt$$ due to $\log t$ being a decreasing function in $[2,\sqrt{x}]$. The last integral equals $\sqrt{x}-2$ thus showing that the contribution of the interval $[2,\sqrt{x}]$ is $O(\sqrt{x})$. Using monotonicity we see that the ...


2

Well your series should be $$\sum_{n=0}^{\infty}\frac{(-1)^n(x^{2n+1})}{(2n+1)!}$$ $$ = x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\dots$$ simply plug in $x=\frac{6\pi}{7}$ and $x=\frac{20\pi}{7}$. Then compare the difference (if the series were perfectly accurate they would be the same, because $\frac{20\pi}{7}=\frac{6\pi}{7}+2\pi$) The error is told by ...


2

I agree that (3) is correct. The reason is that it is easy to check that (3) gives the correct result if $f(x) = 1$, if $f(x) = x$, and with a bit more work if $f(x) = x^2$. Hence this also shows that (3) is indeed a solution of the linear equation system of the previous answer.


2

If you are using IEEE 754 compliant floating point numbers, it may be that square root operations are faster than you might suppose, as the square root operation is directly supported in the standard and any compliant hardware is guaranteed to round correctly (unlike for sine and cosine) and is often using a hardware implementation of the Newton-Raphson ...


2

You could try using the Newton-Raphson method, where you calculate the square root of $x$ in its entirety, using the following update (for iteration $n+1$):- $$y(n+1)=y(n)-\frac{y(n)^2-x}{2y(n)}=0.5\left(y(n)+\frac{x}{y(n)}\right)$$ and setting $y(1)=\lfloor\sqrt x\rfloor$ as a starting value. Thus, for iteration $n+1$, the fractional part $f(x)$ would be ...


2

We have \begin{align} \ln(\cos(x)) & = \dfrac{\ln(\cos^2(x))}2 = \dfrac12 \cdot \ln(1-\sin^2(x)) = - \dfrac12 \sum_{k=1}^{\infty} \dfrac{\sin^{2k}(x)}k\\ & = -\dfrac12 \sin^2(x) - \dfrac14 \sin^4(x) - \cdots\\ & = -\dfrac12 \left(x-\dfrac{x^3}{3!} + \mathcal{O}(x^5)\right)^2-\dfrac14 \left(x + \mathcal{O}(x^3)\right)^4 + \mathcal{O}(x^6)\\ & ...


2

Here is a solution: $$ \frac{1}{2}+\frac{1}{\pi}(\frac{x}{\epsilon}-\frac{x^3}{3\epsilon^{3}}+\frac{x^{5}}{5\epsilon^5}\cdots) $$


2

Yes. But you can approximate anything by anything else. The proper way how to think about it is to consider the error. In this case, it can be proved that, for example, $$ \lim_{x\to 0} \frac{\text{First expression}}{\text{Second expression}}=1. $$ But be careful! The difference, for example, $$ \lim_{x\to 0} (\text{First expression} - \text{Second ...


2

You have: $$\mathrm e^{1/(12n+1)} \leq \frac{n!}{\sqrt{2\pi n}\cdot(\frac n{\mathrm e})^n} \leq \mathrm e^{1/(12n)}$$ Let $f(n)$ be $\frac{n!}{\sqrt{2\pi n}\cdot(\frac n{\mathrm e})^n}$. You have: $$\mathrm e^{1/(12n+1)} \leq f(n) \leq \mathrm e^{1/(12n)}$$ and $$\mathrm e^{-1/(12n)} \leq \frac{1}{f(n)} \leq \mathrm e^{-1/(12n+1)}$$ Thus: $$\mathrm ...


2

Well, I'll show you where I was able to get, but I wasn't able to push through it all to the end. Maybe it will get you close enough that you could go from there. The paper by Cohen, Villegas, and Zagier (linked above) works on series of the form $$ S = \sum_{k=0}^\infty (-1)^k a_k. $$ Though they state that "$a_k$ is a reasonably well-behave function, ...


2

Let $M=A-BXC\,\,$ and $\,f=\|M\|^2_F$. Then the problem is to minimize $f$ with respect to $X$. Find the differential $$\eqalign{ df &= 2\,M:dM \cr &= 2\,(A-BXC):(-B\,dX\,C) \cr &= 2\,(BXC-A):B\,dX\,C \cr &= 2\,B^T(BXC-A)C^T:dX \cr }$$ Since $df=(\frac{\partial f}{\partial X}):dX\,$ the derivative must be $$\eqalign{ ...


1

The taylor-series $\sum_0^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n$ around $a=0$ yields $$\color{red}{\frac{1}{x}}+\frac{1}{2}+\frac{x}{12}-\frac{x^3}{720}+\frac{x^5}{30240}+O(x^6)$$ However, this is only a good approximation at around $x=0$. If you look at the graph: you can see that the further you get away from $x=0$ the "worse" your approximation ...


1

Applying the logarithm to both sides $$ y = a x^u \Rightarrow \\ \ln y = Y = \ln(a) + u \ln(x) = Y_0 + u X $$ turns the exponential curve into a linear curve $Y(X)$, so you can apply the regression techniques for linear functions to it, yielding the $Y_0$ and $u$ for the best fit linear curve. Obviously $\ln(a) = Y_0 \Rightarrow a = \exp(Y_0)$.


1

The method works for your examples : $$2\lt\sqrt 5\lt 3\Rightarrow 0\lt\frac 23\lt\frac{\sqrt 5}{3}\lt \frac 33=1$$ $$3\lt \sqrt{11}\lt 4\Rightarrow 0\lt \frac 34\lt\frac{\sqrt{11}}{4}\lt\frac 44=1$$ $$3\lt\sqrt{13}\lt 4\Rightarrow 0\lt\frac 38\lt\frac{\sqrt{13}}{8}\lt \frac 48\lt 1$$ $$3\lt\sqrt{11}\lt 4\Rightarrow 6\lt\sqrt{11}+3\lt 7\Rightarrow ...


1

$$\cos\left(\dfrac{\pi}{3}\right)=\dfrac{1}{2}$$ $$\ln\cos(\pi/3)=\ln(1/2)=\ln(1)-\ln(2)=-\ln(2)$$


1

A suitable value is a value for which $\log\cos x=\log\frac{1}{2}$, for instance $x=\frac{\pi}{3}$. That gives: $$ \log 2 \approx \frac{\pi^2}{18}+\frac{\pi^4}{972}.$$ A better choice is $x=\frac{\pi}{4}$, that leads to: $$ \log 2 \approx \frac{\pi^2}{16}+\frac{\pi^4}{1536}.$$


1

Stirling's approximation is given by $n!\approx. \sqrt{2\pi n}(n/e)^n$. Thus, for large $n$, and $n>>k$, we have $$\begin{align} \binom{n}{k}&=\frac{n!}{k!(n-k)!}\\\\ &\approx. \frac{\sqrt{2\pi n}(\frac{n}{e})^n}{k!\sqrt{2\pi (n-k)}(\frac{n-k}{e})^{n-k}}\\\\ &=\frac{n^ne^{-k}}{k!\sqrt{1-(k/n)}n^{n-k}(1-(k/n))^{n-k}}\\\\ &\approx. ...


1

The way I think about it would be to have a polynomial of degree 2 describe the three points and the second derivative would be 2 times the coefficient of $x^2$ : $$ f(x+h_1)=a\,(x+h_1)^2+b\,(x+h_1)+c\\ f(x) = a\,x^2+b\,x+c\\ f(x+h_2)=a\,(x+h_2)^2+b\,(x+h_2)+c\\ $$ this can be interpreted as a linear equation system: $$ \begin{matrix} a\\ b\\ c\\ ...


1

If $x_2-x_1 = \frac12$ then $x_2^2-x_1^2 = (x_2-x_1)(x_2+x_1) = \frac12$ is equivalent to $x_2+x_1 = 1$ The book may have wanted you to use Euler Lagrange equations?


1

First note that $${n-i \choose 2}-{n \choose 2} = \frac{(n-i)(n-i-1)-n(n-1)}{2}=\frac{-2in+i^2+i}{2}$$ Thus \begin{align} 3^{{n-i \choose 2}-{n \choose 2}}2^{{i \choose 2}} &= 3^{\frac{-2in+i^2+i}{2}}2^{\frac{i(i-1)}{2}} \\ &= 3^{-in} \cdot 3^{\frac{i^2+i}{2}} \cdot 2^{\frac{i(i-1)}{2}} \\ &< 3^{-in} \cdot 6^{\frac{i^2+i}{2}} \\ &< ...


1

You can write $$ \sum_{i=0}^{rk} \binom{rn}{i} q^i (1-q)^{rn-i} (rk-i) = \\ rk \sum_{i=0}^{rk} \binom{rn}{i} q^i (1-q)^{rn-i} - rn \sum_{i=1}^{rk} \binom{rn-1}{i-1} q^i (1-q)^{rn-i} = \\ rk \sum_{i=0}^{rk} \binom{rn}{i} q^i (1-q)^{rn-i} - rn \sum_{i=0}^{rk-1} \binom{rn-1}{i} q^{i+1} (1-q)^{rn-i-1} = \\ rk \sum_{i=0}^{rk} \binom{rn}{i} q^i (1-q)^{rn-i} - rk ...


1

As per my comment above, I'm not sure what exactly it is that you're asking. However, if it is what I said in that comment, then here's a solution. If you're given a set of points $S = \{p_1, p_2, \ldots, p_n\}$, then there are $n \choose 2$ segments made of pairs of points from $S$. If all the points in $S$, as well as $P$, are in $\mathbb{R}^2$ then you ...


1

Assuming $\max(x, y, z), \min(x,y,z), \operatorname{med}(x,y,z)$ are really $|\max(x, y, z)|, |\min(x, y, z)|, |\operatorname{med}(x, y, z)|$. Not a real proof, only a sketch. Since the formula is symmetrical under permutations of $x, y, z$, let's assume $0 \leq x \leq y \leq z$, so $$ \sqrt{x^2 + y^2 + z^2} = \alpha x + \beta y + \gamma z. $$ Since we're ...



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