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7

From the continuous fraction expansion, the seventh convergent is $$\gamma \approx \frac{15}{26}$$ From the limit definition $$\begin{align} \gamma &= \lim_{n \to \infty} {\left(2H_n-\frac{1}{6}H_{n^2+n-1}-\frac{5}{6}H_{n^2+n}\right)} \\ &= ...


5

There are positive integrals that relate $\log(2)$ to its first four convergents: $0,1,\frac{2}{3},\frac{7}{10}$. $$ \begin{align} \int_0^1\frac{2x}{1+x^2}dx &= \log\left(2\right) \\ \int_0^1\frac{(1-x)^2}{1+x^2}dx &= 1-\log\left(2\right) \\ \int_0^1\frac{x^2(1-x)^2}{1+x^2}dx &= \log\left(2\right)-\frac{2}{3} \\ \int_0^1\frac{x^4(1-x)^2}{1+x^2}dx ...


4

I'll try my best to get this concept through to you, so let's begin... Taylor's Formula or the Lagrange Form of the error term, denoted by $R_n(x:a)$ is given by the following equation:-$$R_n(x:a)={f^{n+1}(c)(x-a)^{n+1}\over (n+1)! }$$ Simply this gives the error involved in using $T_n(x)$ to approximate $f(x)$ about $x=a$ or $f(x)=T_n(x)+R_n(x:a)$ ...


3

The answer needs some context. Part I. One may ask why, $$\frac{1}{2\pi\sqrt{2}} - \frac{1103}{99^2} \approx 10^{-9}\tag1$$ is such a good approximation? In fact, the convergents of the continued fraction of $\displaystyle\frac{1}{2\pi\sqrt{2}}$ start as, ...


3

Let's begin by converting this to a simpler double integral: $$\begin{align}b \int_{-\infty}^{\infty} dx \, e^{i a x/b} \int_0^c du \, e^{-u \sinh^2{x}} &= b \int_0^c du \int_{-\infty}^{\infty} dx \, e^{i a x/b} \, e^{-u \sinh^2{x}}\\ &=b \int_0^c du \, e^{u/2}\,\int_0^{\infty} dx \, \cos{\left (\frac{a x}{2 b} \right )} e^{-(u/2) \cosh{x}} \\ ...


2

You can approximate the sum using an integral. If $j < x < j+1$ you have $\frac{1}{j^k} > \int_j^{j+1} \frac{1}{x^k} dx > \frac{1}{(j+1)^k}$. This will allow you to develop both an upper and a lower bound on your sum and you should find that $S_{N,k} = O(1/N^{k-1})$.


2

$$ \begin{align} &\sum_{i=0}^{n-1}\sum_{j=i+1}^{n-1}\frac{i+j+2}{(i+1)(j+1)}\cdot(i+2x)(j+2x)\tag{1}\\ &=\sum_{j=0}^{n-1}\sum_{i=j+1}^{n-1}\frac{i+j+2}{(i+1)(j+1)}\cdot(i+2x)(j+2x)\tag{2}\\ &=\sum_{i=0}^{n-1}\sum_{j=0}^{i-1}\frac{i+j+2}{(i+1)(j+1)}\cdot(i+2x)(j+2x)\tag{3}\\ ...


2

This was earlier a bit hastily closed. However, one aspect of the question might have an interesting connection to the Tribonacci constant $T$. First, let $w = \frac{\sqrt{2}}{T^{-1}+1}$, so, $$j\big(\tfrac{1+\sqrt{-11}}{2}\big) = \frac{(w^{24}-16)^3}{w^{24}} = -2^{15}$$ where $j(\tau)$ is the j-function. We then get the Ramanujan/Chudnovsky-type pi ...


2

You are right, this limit will yield $\pi$ but take a look at $\sin$. It's "degree" version is defined as $$\sin_d(x)=\sin\left(\frac{\pi}{180}x\right)$$ Since you can approximate $\sin$ using Taylor series as $$\sin(x)=\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}$$ There is no simple way to calculate $\sin_d$, as in calculations you have to use value ...


2

HINT Now $$e^{-4x} = 5x^2$$ becomes $$5x^2 \approx 8x^2-4x+1$$ which is a quadratic you can solve. The roots should approximate the rotts of the transcendental equation you started with.


2

For $\;x\;$ close enough to zero we have the good approximation you gave, so $$8x^2-4x+1=5x^2\iff3x^2-4x+1=0\implies x=\begin{cases}1\\{}\\\frac13\end{cases}$$ Take $\;\frac13\;$ as a good approx. to the wanted zero.


2

Let $\theta = \theta_0 + \psi$. Then the equation of motion beocomes $$\frac{d^{2}(\theta_0 + \psi)}{dt^2} = \sin(\theta_0+\psi) \left( \Omega^{2}\cos(\theta_0+\psi)-\frac{g}{a}\right).$$ For small $\psi$ this is easily seen to be same as $$\frac{d^{2} \psi}{dt^2} =-\Omega^{2} \sin^2\theta_0 \psi$$ after using $\cos \theta_0=\frac{g}{\Omega^2a}$ and ...


1

Let $f(x)=x^2+x$. Then you have $$I=\int_0^1 f(x) dx= \frac{5}{6}$$ while the midpoint approximation $I_n$ for $n$ subintervals is according to Wiki and the well known formulas for $\sum_k k$ and $\sum_k k^2$ $$I_n = \frac{1}{n}\sum_{k=0}^{n-1} f\left(\frac{k}{n}\right)=\\ =\frac{1}{n}\sum_{k=0}^{n-1}\left(\frac{k}{n}\right)^2 + ...


1

I believe i know the answer now if someone needs it. $M_n=h(f(\frac{1}{2n})+f(\frac{3}{2n})+f(\frac{5}{2n})+,..,+f(\frac{1-\frac{1}{n}+1}{2}))$ $M_n=\frac{1}{n}\sum_{k=0}^{n-1}f(\frac{2k+1}{2n})=\frac{1}{n}\sum_{k=0}^{n-1}(\frac{2k+1}{2n})+\frac{1}{n}\sum_{k=0}^{n-1}(\frac{2k+1}{2n})^{2}$ ...


1

The remainder term in a power series is an upper bound for the error. So your first step is to determine which is the smallest natural number $n$ such that $\left|\frac{f^{(n+1)}(0)}{(n+1)!\hspace{1ex} 9^{n+1}}\right|<10^{-4}$. Say you guess $n=3$. Then you must find $\left|\frac{f^{(4)}(0)}{4!\hspace{1ex} 9^4}\right|$. Now $f(x)=3\sqrt{1+x}$ so taking ...


1

$n\log n$ is not a very good bound for $\log \binom {n}{i}.$ Use Stirling's formula : $n!=[1+d(n)] (n/e)^n \sqrt {2 \pi n}$ where $|d(n)|<1/4 n$ for $n>0$. The largest value of $\binom {n}{i}$, which is $\binom {n}{n/2}$ for even $n$, and $\binom {n}{(n-1)/2}$ for odd $n$, is $[1+f(n)]2^n/\sqrt {\pi n}$ where $f(n)\to 0$ as $n\to \infty$.


1

The Taylor approximation $T_1(x, y) = 1 + \frac{2x}{a}$ satisfies $$k(x, y) - T_1(x, y) = \frac{2 x^2}{a^2} + O((x, y)^3),$$ but $$k(x, y) - \frac{a + x}{a - x} = O((x, y)^3) ,$$ so the latter approximation is better for small $x, y$.


1

The key is how far we step on average on a normal seek. I will do the special case where all the non-master sequences have the same density $\mu$ We want to know how many places we have to look before all the sequences have a $1$. A heuristic is that you start with $n$ sequences. After one cell you have $(1-\mu)n$ that have not seen a $1$. After two ...


1

Using Maple, I get $$ (n+1) H(n) (n+4x-2)(x-1/2) + 2 n (n+2) x + \frac{n^3-3n}{2} $$ where $$H(n) = \sum_{k=1}^n 1/k = \Psi(n+1) + \gamma$$ As $n \to \infty$, $$ H(n) \sim \ln(n) + \gamma + \dfrac{1}{2n} - \dfrac{1}{12n^2} + \dfrac{1}{120 n^4} - \dfrac{1}{252 n^6} + \dfrac{1}{240 n^8} + \ldots $$


1

One can approximate $\int_N^\infty x^{-k}dx$ in two simple ways: The sum of rectangles approximation will be a bit low, and is given by $$S_{N,k} = N^{-k}+ \sum_{i=N+1}^\infty {\frac{1}{i^k}} < N^{-k}+ \int_N^\infty x^{-k}dx = N^{-k}+\frac{1}{k-1}N^{1-k} $$ The trapezoid rule approach is a bit high (since the function is concave upward), and is given ...


1

An approximation with an accuracy similar to that of $\pi\approx3$ (error<5%) is given by the sixth root of Gelfond's constant, $$e^{\frac{\pi}{6}}\approx \phi$$ with rational term series $$e^{\frac{\pi}{6}}=\sum_{k=0}^{\infty}\frac{\left(e^{\frac{\pi}{2}} - (-1)^k ...


1

Using the 5 constants of Euler's identity $ e^{i\pi} + 1 = 0 $ it is possible to include $ \varphi $ into an equation to give an identity containing six constants as follows: $$ e^{\frac{i\pi}{1+\varphi}} + e^{-\frac{i\pi}{1+\varphi}} + e^{\frac{i\pi}{\varphi}} + e^{-\frac{i\pi}{\varphi}} = 0 $$ See article and OEIS Sequence A193537



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