# Tag Info

8

By my calculator this expression only agrees with $\pi$ to about 13 decimal digits. Since there are 22 separate digits in the expression, we should expect that very many different expressions of that shape approximate $\pi$ with a similar precision -- there's nothing particular remarkable about one of them, nor any "explanation" other than it just happens to ...

4

Let $f(x)=\dfrac{x(x-1)}{2}$. We note first that $$\forall\,x,y\ge0,\quad f(x+y)-f(x)-f(y)=xy\ge0$$ Thus, by induction, for nonnegative numbers $a_1,\ldots,a_n$ we have $$\sum_{i=1}^n f(a_i)\le f\left(\sum_{i=1}^na_i\right)$$ Thus, $$A(n)\le \frac{B(n)(B(n)-1)}{2}$$ With equality if $a_1=\cdots=a_{n-1}=0$ for example. So, this is an optimal inequality.

3

Let $u = 2t$ (so: $t = \frac{u}{2}$), substitute into the first approximation, square root both sides, and then flip the sides.

3

With this initial value and the "normal" look of the differential equation, one can expect that the solution staying small means that it is $O(ϵ)$ for all times $x\in[0,\infty)$. Setting $y=ϵz$ one expects $z$ to stay bounded. $z$ now satisfies the scaled differential equation $$z''-z+ϵ^2z^3=0\text{ with } z(0)=1,$$ which now has a clear perturbation ...

3

Hint. As integration and the right hand side are linear in $f$, the quadrature formula is exact for all polynomials of degree $\le 4$ iff it integrates the five functions $x \mapsto x^i$, $i = 0,\ldots, 4$ exactly. Just compute the right hand side values for those five $f$ and you are done.

2

$$\lesssim\text{or} \gtrsim$$

2

We're going to substitute $y=y_0 + y_1$ into the differential equation then throw away most (but not too many) of the smallest terms. This last part is a bit tricky but it will start to make more sense as you become more familiar with it. So, suppose that $y = y_0 + y_1$ where $y_1 \ll y_0$ (and $y_0$ is small as well). Substituting this into the equation ...

2

$0.99^9 = (1 - 0.01)^9 = \sum_{i=0}^9 \binom{9}{i}(-1)^i(0.01)^i$, by the binomial theorem. Terms for $i \ge 2$ are below your accuracy.

2

The usual parametrization of a torus knot is \left. \begin{aligned} x(t) &= \bigl(R + r\cos(pt)\bigr) \cos(qt), \\ y(t) &= \bigl(R + r\cos(pt)\bigr) \sin(qt), \\ z(t) &= r\sin(pt), \end{aligned} \right\} \qquad 0 \leq t \leq 2\pi. The arc length is $$\ell = \int_{0}^{2\pi} \sqrt{p^{2} r^{2} + q^{2}\bigl(R + r\cos(pt)\bigr)^{2}}\, dt.$$ ...

2

If you want to do your calculation with natural logarithms, the expression is $p^k(1-p)^{n-k} = \exp( k \log_e(p)+(n-k)\log_e(1-p) )$ which is the first part of your expression. This should lead you to $0.05^{250}0.95^{250} \approx 1.49 \times 10^{-331}$.

1

For the set of equations, we have one parameter, $\lambda$, that we can adjust. If one is looking for a "best fit" in the sense of mean-squared error, then one seeks the value of $\lambda$ that minimizes the function $$f(\lambda)=\left(e^{-0.05\lambda}-0.5469\right)^2+\left(e^{-0.1\lambda}-0.3229\right)^2+\cdots +\left(e^{-0.2\lambda}-0.1226\right)^2 \tag ... 1 The integral \int\int_Ae^{x^2+y^2}dA gives the volume of a solid. It's easy to see that this solid is contained in the cylinder centered at the origin of radius 3 with height e^9. The Monte Carlo method challenges us to find a way to sample points uniformly from the interior of this cylinder. The fraction of these points which also happen to fall ... 1 Let's use \epsilon instead of \Delta^2, and write it as$$\sqrt{1+\epsilon}$$Then,$$\sqrt{1+\epsilon}=1+\frac{1}{2}\epsilon$$Squaring both sides, we find that$$1+\epsilon=1+\epsilon+\frac{1}{4}\epsilon^2$$This is true if \epsilon is small enough that \epsilon^2 can be neglected. The first two terms of the MacLaurin series for \sqrt{1+x} are ... 1 Assume for induction that the backward divided difference of f satisfy$$f[x_0,x_1,\ldots,x_k] = \frac{1}{x_0x_1\ldots x_k}$$for k=n. We then find that$$f[x_0,x_1,\ldots,x_n,x_{n+1}] \equiv \frac{f[x_0,x_1,\ldots,x_n]-f[x_1,\ldots,x_n,x_{n+1}]}{x_{n+1}-x_0} \\= \frac{\frac{1}{x_0x_1\ldots x_n} - \frac{1}{x_1\ldots x_nx_{n+1}} }{x_{n+1}-x_0} = ...

1

$$\int_0^t K(t-u) e^{-ku} \text{d}u = \int_0^\varepsilon K(t-u) e^{-ku} \text{d}u + O(e^{k\varepsilon})$$ where $1/k\ll \varepsilon \ll 1$. Letting $z=ku$ gives $$\frac{1}{k}\int_0^{\varepsilon k} K\left(t-\frac{z}{k}\right)e^{-z}\text{d}z \sim \frac{K(t)}{k}\int_0^{\infty}e^{-z}\text{d}z$$ as $\varepsilon k \gg 1$ and by making the approximation ...

1

If $a_i =0$ then $a_i (a_i -1 ) \leq a_i^2.$ If $a_i \geq 1$ then $a_i (a_i -1 ) \leq \left(\frac{2a_i -1}{2}\right)^2 =\left( a_i -\frac{1}{2}\right)^2\leq a_i^2$ hence $$2A(n) \leq \sum_i a_i^2 \leq \left(\sum_i a_i \right)^2 =B(n) ^2$$ so $$A(n)\leqslant \frac{B(n)^2}{2}.$$

1

Let me write $P_k=P(X=x_k)$. The plan is to take the trivial estimate $1\ge \sum\limits_{k=m}^{m+d} P_k$ with some $d\approx \sqrt{n}$, and replace $P_k$ by $c\cdot P_m$. So, we need a suitable upper bound on $\frac{P_m}{P_k}$. For $1\le j\le d\le\sqrt{n}$, $$\frac{P_m}{P_{m+j}} = \frac{\binom{n}{m}p^{m}q^{n-m}}{\binom{n}{m+j}p^{m+j}q^{n-m-j}} = ... 1 I think z_n and 1/(z-z_n) do not really help here. You want a function that is greater at z_0 than anywhere on K. Among many choices,$$ f(z) = \begin{cases} 0,\quad z\in K \\ 1,\quad z=z_0\end{cases} $$is one of simplest. Note that f can be extended to a holomorphic function in a neighborhood of K\cup \{z_0\}. As you said, K\cup \{z_0\} has ... 1 I suppose some minor mistakes (probably a sign error in the development of the square). If$$f(x)=x^3 \quad , \quad p(x)=\frac{3 }{2}x^2-\frac{1}{2}x$$then$$f(x)-p(x)=x^3-\frac{3 }{2}x^2+\frac{1}{2}x$$Now square carefuly and group terms for same powers of x to get$$\big(f(x)-p(x)\big)^2=x^6-3 x^5+\frac{13 }{4}x^4-\frac{3 }{2}x^3+\frac{1}{4}x^2$$... 1$$\int_0^1[ae^x+b\cos(\pi x/2)]{\rm d}x=a(e-1)+\frac{2b}{\pi}=\underbrace{1}_{A_1}\underbrace{(ae)}_{f(1)}+\underbrace{\frac{2b}{(a+b)\pi}}_{A_0}\underbrace{(a+b)}_{f(0)}$$1 One thing you may be aware that, if X_0=Y_0, this equation does not have unique solution (the Jacobian is singular). Therefore, you may instead consider choosing the initial start point carefully if the Newton method is used. If X_0>Y_0, then choose \alpha(0) > \beta(0), and vice versa. Another way to avoid simple Newton update is to consider ... 1 We have inequality$$\frac{1}{x+1} \leq \ln \left(1 +\frac{1}{x} \right) \leq \frac{1}{x} $$Denote s_j =\sum_{k=1}^j a_k  then$$\sum_{t=2}^n \frac{a_t }{s_{t-1}} =\sum_{t=2}^n \frac{1 }{\frac{s_{t-1}}{a_t}}\geq \sum_{t=2}^n \ln \left( 1 +\frac{1 }{\frac{s_{t-1}}{a_t}}\right)=\sum_{t=2}^n ...

1

Go through the following PDFs: http://www.ams.org/bookstore/pspdf/mbk-48-prev.pdf & https://www.dropbox.com/s/7ahnld1wtvqiurz/Continued%20fraction%20expansion.pdf?dl=0 This will clarify your issue.

1

As usual please double check. You have essentially solved it except for your change of variable: You have $_1F_1\Big(1+n;2+m+n;a (x-T)\Big)=e^{a(x-T)} \, _1F_1\Big(1+m;2+m+n;a(T-x)\Big)$ $\int_0^T (T-x)^{1+m+n} x^k \, _1F_1\Big(1+m;2+m+n;a(T-x) \Big)dx$ Now we apply $y=a\left(T-x\right)$ , $x=T-\frac{y}{a}=\frac{a\cdot T-y}{a}$, $dx=-\frac{y}{a}$ ...

1

I don't know how to do it in LaTeX, but yes - just replace the underbar in $\leq$ with a squiggle and that's it. It is used when you have bounds, but the bound is more conveniently expressed as an approximation. We used it when we did numerical error analysis using Taylor approximations, and also when something is bound by a random variable that is ...

1

The initial value does matter: for $x_0=1$ the method converges to $\pi/2$ but for $x_0=4$ the method converges to $3\pi/2$. As the theory predicts, for $x_0$ close enough to each root $(2k+1)\pi/2$, the method converges to that root. The basins of attractions for each root are likely to be complicated fractal sets.

1

Yes, the initial seed matters quite a lot. If you're "close" to one of the roots, you'll converge to that root. Exactly how close is required is complicated, though. The image below shows the regions of attraction for the cosine function in a neighborhood of the origin in the complex plane. Initial seeds chosen from green region on the left converge to ...

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