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One can expand the function $(1 + x)^{\alpha}$ in a Taylor series called the binomial series. It takes on the form $$(1 + x)^{\alpha} = 1 + \alpha x + \frac{\alpha (\alpha - 1)}{2!} x^2 + \frac{\alpha (\alpha - 1)(\alpha - 2)}{3!} x^3 + \dots$$ Now if $x$ is small and $\alpha$ isn't too large relative to $x$, all the high-order terms can be neglected to ...

7

Let me show how to first construct a sequence of polynomials $p_n$ such that $p_n(z)\to1/z$ uniformly on $C=\{e^{it}\colon \theta\le t\le 2\pi-\theta\}$, for any given $\theta\in(0,\pi)$. Choosing real numbers $0 < a < b < 1$, set $$u(z)=\frac{1-az}{1-bz}.$$ For $z=e^{i\phi}$ this gives $$1-\lVert u(z)\rVert^2 = ... 7 This solution is specialized to the particular problem. As in my other solution, I am working with the arc C = \{ e^{i \theta} : \pi/4 < \theta < 7 \pi/4 \}. Let T_n(z) be the n-th Chebyshev polynomial, so it is the polynomial with leading term 2^{n-1} z^n which has |T_n(z)|\leq 1 for -1 \leq z \leq 1. Set$$g_n(z) = z^{-1} - \frac{(1 + ...

6

I'm going to put up two answers to this problem. I find it more convenient to rotate the curve $C$ to be $\{ e^{i \theta} : \pi/4 \leq \theta \leq 7 \pi /4 \}$. The first one is the standard proof of Runge's theorem, made concrete. Notice that $$\frac{1}{z} = - \sum_{k=0}^{\infty} 0.6^k \left( \frac{1}{0.6 - z} \right)^{k+1}$$ $$\frac{1}{0.6-z} = ... 4 I have not derived the approximation (2) yet, but the approximation (1) is easy: Set q(\tau)=\mathrm{e}^{2\pi\mathrm{i}\tau}. We will use the Dedekind eta function$$\eta(\tau) = q\left(\frac{\tau}{24}\right) \prod_{n=1}^\infty\left(1-q(n\tau)\right) = \frac{q\left(\frac{\tau}{24}\right)}{\sum_{n=0}^\infty P(n) q(n\tau)}$$We also know the following ... 3 The Taylor series converges for all x. If x=50, say, the terms will grow until the one \frac {50^{50}}{50!}\approx 2.9 \cdot 10^{20} The next term is -\frac {50}{51} of this and they continue decreasing. They will cancel very delicately so the final result is e^{-50}\approx 1.9\cdot 10^{-22} You are correct that the Taylor series is not a ... 2 One approach is to use Newton's Method, which provides a convergent numerical approximation (with the caveat that you choose an initial guess that is 'close enough' to the real solution; in this sense, doing so for large numbers gets more difficult). Newton's Method is based upon finding roots of a function f(x). To see how this applies to square or ... 2 The way I use to get an approximate n-th roots is to use the first terms of the binomial theorem in the form$$\sqrt[n]{a^n+b} =a \sqrt[n]{1+\frac{b}{a^n}} \sim a(1+\frac{b}{n a^n}) = a +\frac{b}{n a^{n-1}} $$For example,$$\sqrt[3]{130} =\sqrt[3]{125+5} =\sqrt[3]{5^3+5} \sim 5+\frac{5}{3\cdot 5^2} = 5+\frac{1}{15} = 5.0666... $$The correct result is ... 2 This is just a Taylor expansion of the function f(r,g) = \frac{1+r}{1+g} at the point (r,g) = (0,0). The partial derivatives of f are$$\frac{\partial f}{\partial r} = \frac{1}{1+g}$$and$$\frac{\partial f}{\partial g} = -\frac{1+r}{(1+g)^2}.$$Hence, the Taylor expansion of first order at (0,0) is 1 + r - g. 2 To answer my own question, the most accurate I can get is by using$$ \arcsin\left(x\right)=2\arctan\left(\frac{x}{1+\sqrt{1-x^{2}}}\right)\ $$so$$ \arcsin\left(\frac{a}{a+x}\right)=\frac{\pi}{2}-\frac{\sqrt{2}}{\sqrt{a}}x^{1/2}+\frac{5}{6a^{3/2}\sqrt{2}}x^{3/2}...\ $$2 Assume G is contained in L^1(\mathbb{R}). Then \lim_{x\rightarrow\infty} I_2(x) = 0 by the Riemann-Lebesgue lemma. The first integral vanishes for large x since the domain of integration shrinks; apply dominated convergence theorem to G\cdot\chi_{[0,1/n]}. For intermediate values of x, the result is false. Fix a compactly supported function G ... 2 First, by standard measure theory constructions there exists a sequence of simple functions$$\phi_n = \sum_{k=1}^{m_n} a_{k,n} \chi_{R_{k,n}}$$converging to f a.e., where R_{k,n} are open rectangles, R_{k,n} \cap R_{j,n} = \emptyset for k \ne j, and such that \bigcup_{k=1}^{m_n} \overline{R_{k,n}} = [-n,n]^2. Let$$\Omega_n := ...

2

For $m > 0$, you can substitute $t = x^m$ and get $$I_{n,m} = n \int_0^1 (1-x^m)^n x^m\,dx = \frac{n}{m} \int_0^1 (1-t)^n t^{1/m}\,dt = \frac{n}{m}B(n+1,1+1/m).$$ With the representation $$B(p,q) = \frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}$$ that becomes $$I_{n,m} = \frac{n}{m}\frac{\Gamma(n+1)\Gamma(1+1/m)}{\Gamma(n+2+1/m)}.$$ Now using Stirling's ...

1

For $\sqrt{x}$, you have already almost it: $$\begin{cases} a_1=2 & \\ a_{n+1}=\dfrac1{2}\left(a_n+\dfrac{x}{a_n}\right) \end{cases}$$ Generally speaking, given a number $x$, you can easily construct a sequence that converges to it: simply try $u_n=x+\frac1{n}$. But if you want something non trivial like this, it's absolutely not immediate in general. ...

1

$$arcsin(t)=arctan(\frac{t}{\sqrt{1-t^2}})$$further find taylor series. on evaluating you get $$arcsin(\frac{a}{a+x})=\frac{a}{(2ax+x^2)^\frac{1}{2}}-\frac{a^3}{3(2ax+x^2)\frac{3}{2}}+ \frac{a^5}{6(2ax+x^2)\frac{5}{2}}-.....$$this works for $a$ in$(0,\infty)$. example : when $x=a ,arcsin(1/2)= 0.5235$ and the formula gives the answer as $0.52389$.

1

First define the function $f(x)=(1+x)^n$ then using the Maclaurin series expansion defined as $$f(x)=P_n(x)=\sum^n_{k=0}\frac{f^{(k)}(0)}{k!}x^k$$ where $P_n(x)$ is a polynomial of degree $n$. Now we can calculate the first two derivatives of $f(x)$ \begin{align} \left.\frac{df}{dx}\right|_{x=0}&=n(1+x)^{n-1}=n \\ ...

1

Note that $d_p$ is a $1$-lipschitz function. Then theorem 1 in this article http://www.mat.ucm.es/~dazagrar/articulos/AFLRjmaa.pdf tells you that you may approximate it by a smooth function satisfying the properties you give (note that the gradient estimate is satisfied whenever the approximating function has lipschitz constant less than $2$).

1

The only function that I know has this property is essentially $$\max(x, y) \sim \frac{x e^{kx} + y e^{ky}}{e^{kx} + e^{ky}}$$ for large $k$. (@wnoise was close to the answer. In this case there is no problem if $x=y$.) This works for positive or negative numbers and tends to the maximum for $k\to\infty$. Other formulas based on powers are only valid for ...

1

Of all the $n^{th}$ degree trigonometric polynomials, the $n^{th}$ partial sum of the Fourier series of $f$ best approximates $f$. To put it mathematically, $\|f-s_n\| \leq \|f-p\| \, \, \, \ s_n, p \in T_n$ So just find the Fourier coefficients if the given functions in $[0, 2\pi]$ and you have your answer. ...

1

By definition $$\tanh x = \frac{1 - e^{-2x}}{1 + e^{-2x}},$$ so for $x \gg 1$ (something like $x > 4$) we have $$\tanh x \approx 1 - e^{-2x}.$$ Since you're interested in an approximation over a large interval we'll need $\tanh x \approx 1$ for most of it. Additionally, we at least want the approximation to be $0$ at $x = 0$. This suggests we ...

1

If you're talking about Feynman's "duel" with the abacus guy, he got lucky on that one. The number the guy chose was $1729$, which is $12^3 + 1$. Feynman knew enough to rattle that off quickly to three or four decimals by applying a series expansion. But there are algorithms for both square (cube) roots. These are straightforward to visualize in terms of ...

1

As Ross explains, this Taylor series converges for all $x$, but it's probably the case that your Taylor polynomial did not have enough terms to illustrate that yet. As for approximating, $e^{-50}=(e^{-1})^{50}$. If you want to keep using Taylor series somehow, you can use it to approximate $e^{-1}$. Only using up to the tenth degree approximant, we have ...

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