# Tag Info

12

This is what I'm trying to show on a diagram indicating explicitly all quantities and the approximation is quite rough. Here is a circle with center $O$. Quantities $OA=OB=OC=OD=AB=1$ and $OC\perp OA, OD\parallel OA$. Line segments $AC$ and $BD$ have an intersection $E$. We can easily deduce the following quantity. $$AC^2=OA^2+OC^2=2\qquad ... 7 Riffing on @Shuchang's answer ... Starting with unit circle \bigcirc O, one easily constructs A, B, C, D with |\overline{AB}| = \sqrt{2} and |\overline{CD}| = \sqrt{3}. Quadrisecting \overline{AB} and \overline{CD} one draws \bigcirc A and \bigcirc{B} to provide chords of length \sqrt{2}/4 and \sqrt{3}/4. Chains of congruent ... 5 It depends on what is meant by "polynomial". If only \sum c_n z^n, then every function that is uniformly approximable by polynomials must be holomorphic on the interior of J. Although that condition is trivially satisfied if J has empty interior, that doesn't mean that for such J every continuous function is the uniform limit of polynomials. For ... 4 Hint. If f: [1,\infty)\to\mathbb R is continuous, and g(x)=f(1/x), then g is continuous in (0,1]. If f also has a (finite) limit, as x\to\infty, then g is continuous at 0, and altogether it is continuous in the whole closed interval [0,1]. If now p is a polynomial, such that$$ \max_{x\in [0,1]}\lvert g(x)-p(x)\rvert<\varepsilon, ...

3

Hint Any mathematical function can be, at least locally, approximated by so called Taylor or Mc Laurin expansions. To make it as simple as possible, the tangent to a curve is, at the point where it is defined, a local approximation of the curve. So, write the equation of the tangent to the curve $y=sin(x)$ at $x=0$ and you will obtain, for the tangent ...

3

There aren't any specific distinct "roots". The set of points $(x,y)$ where $y^3+y +xy-x^3-2=0$ is an entire curve in the $xy$ plane. Here's what the curve looks like: You have to decide which point of that curve you want, by providing a second equation, for example. Then you can use Newton's method to solve for $x$ and $y$. If you don't care which point ...

3

By the way, $$\sum_{n=1}^\infty \frac{1}{n^n} \approx 1+\frac{\sin \left(\frac{\pi }{e}\right)}{\pi }$$ seems to be a better approximation. According to RIES, $$\sum_{n=1}^\infty \frac{1}{n^n} \approx\left(\frac{\phi +\pi }{2}\right)^{\frac{1}{\pi+\frac{1}{4} }}$$ is still better but not as good as the one proposed by JJacquelin. Next one, please !

2

Seems to me that you are getting ready for Taylor series of trig functions. I would suggest to google this and you are getting lots of answers http://en.wikipedia.org/wiki/Taylor_series would do but there are many many other great sites. As far as usefullness, that can't be even described in one sentence. I appreciate you being inquisitive. That approach ...

2

It seems the following. The answer is negative. Let $I=[0;1]$ be the unit segment endowed with the standard topology. Put $X=I^{\omega_1}$. By Tychonoff Theorem $X$ is compact. By Hewitt-Marczewski-Pondiczeri Theorem [Eng, 2.3.15], $X$ is separable. For each subset $S$ of the set $\omega_1$ as $\pi_S$ we denote the projection from $X=\prod\{I_\alpha\colon ... 2 If a cubic and a quadratic agree at the endpoints and midpoint of an interval$[a, b]$, their difference is a cubic vanishing at the endpoints and midpoint, hence is a multiple of$p(x) = (x - a)(x - b)\bigl(2x - (a + b)\bigr)$. But$p$is "odd with respect to the midpoint" in the sense that$p(b - x) = -p(a + x)$for$a \leq x \leq b$, so the integral of ... 2 Well, it's rather simple $$d = \sqrt{a^2 + b^2 - 2ab \ cos(\theta)}$$ $$d^2 = a^2 + b^2 - 2ab \ cos(\theta)$$ $$d^2 - a^2 - b^2=2ab \ cos(\theta)$$ $$\frac{d^2 - a^2 - b^2}{2ab} = \ cos(\theta)$$ $$\frac{d^2}{2ab} - \frac{a^2 + b^2}{2ab} = \ cos(\theta)$$ and since$a$and$b$are fixed,$k = \frac{a^2 + b^2}{2ab}$$$\ cos(\theta) = \frac{d^2}{2ab} -k$$ ... 2 I can only see a relation with trigonometric functions, i.e.something like$\sqrt{3}=2-\tan (\pi/12)$and $$\sqrt{2}=\frac{4\cos(\pi/12)}{3-\tan(\pi/12)}.$$ If we consider the continued fraction of$\sqrt{2}$,$\sqrt{3}$and$4/\pi$we see some similarities, too. However, there are also arguments indicating that both numbers are only accidentally close. ... 1 You could use $$\cos(\frac{\pi}2) = 0$$ $$\sin(\frac{\pi}{2}) = 1$$ and the half-angle formulas (for$0 \le \alpha \le \frac{\pi}2$) $$\cos(\frac{\alpha}{2}) = \sqrt{\frac12(1 + \cos{\alpha})}$$ $$\sin(\frac{\alpha}{2}) = \sqrt{\frac12 (1 - \cos \alpha)}$$ and the angle-sum formulas $$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha ... 1 Sounds like an approach to your problem is to let a=b for simplicity so$$ d = \sqrt{a^2+b^2-2ab \cos x} = a \sqrt{2-2\cos x} $$and then just plot the function f(x) = \sqrt{2-2\cos x}, getting which is clearly a wave, but not sure how sinusoidal it is. It is, however, similar to a similar transformation of the sine, just off by a horizontal ... 1 The function that describes the distance of a fixed point from a point travelling on a circle cannot be perfectly "sinusoidal", since it exhibits discontinuities for the derivatives, as shown by @gt6989b, for instance. However we can state that the coefficients of the Fourier cosine series of$$ f(\theta) = \sqrt{1-\lambda \cos\theta}, $$where we set ... 1 Indeed, there a way to angle this useful behavior without such series. In analysis involving real numbers \sin x will always be appreciably different from x. However, in a framework involving an infinitesimal-enriched continuum such as the hyperreals the richer syntax allows one to formulate ideas such as the replaceability of \sin x by x for ... 1 The correct way to expand \dfrac{x-y}{x+y} is using the binomial theorem.$$ \frac{1}{x+y} = \frac1x\left(1+\frac{y}{x}\right)^{-1} = \frac1x\left(1-\frac{y}{x} + \ldots\right)$$Hence$$ \frac{x-y}{x+y} = \frac{x-y}{x}\left(1-\frac{y}{x} + \ldots\right) \approx 1 -2\frac{y}{x}$$1 Since you admit that x>>y, let is define z=\frac{y}{x} and rewrite$$\log \left(\frac{x-y}{x+y}\right)=\log \left(\frac{1-z}{1+z}\right)$$and use Taylor expansion for both numerator and denominator. So, as the beginning of the expansion, you have$$\log \left(\frac{1-z}{1+z}\right)=-2 z-\frac{2 z^3}{3}-\frac{2 z^5}{5}-\frac{2 z^7}{7}-\frac{2 ... 1 Very accurate approximations can be computed thanks to series expansions such as the example given by Claude Leibovici (13 exact digits in case of$S(10)$) Other methods of numerical calculs leads to a lot of numerical appoximations of various kind. Some examples are compared below. Many surprising formulas are very easily obtained with the method of ... 1 Squaring is the best method to prove the inequality. However, there is a detour, which involves squaring and nastier computation (and somewhat magical factorization), that may be what you're looking for. Substitute$x = \frac{29}{\sqrt 2}$in$t(x) = x^2 - 41x + 420to get \begin{align} t\left(\frac{29}{\sqrt 2}\right) = \frac{29^2}{2} - 41 \cdot ... 1 Initially setr_i=nw_i$and round to the closest number. Then add up the$r_i$'s and check against$n$. If they are equal, you are done. If not let$\Delta n=\sum_i r_i-n$be the number of seats of excess. If$\Delta n$is positive, rank the$r_i$'s by how much they were rounded up and decrease the$\Delta n$largest round ups. If$\Delta n\$ is ...

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