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48

One simple way to derive this is to come up with a parabola approximation. Just getting the roots correct we have $$f(x)=x(\pi-x)$$ Then, we need to scale it (to get the heights correct). And we are gonna do that by dividing by another parabola $p(x)$ $$f(x)=\frac{x(\pi-x)}{p(x)}$$ Let's fix this at three points (thus defining a parabola). Easy rational ...


33

This might be more explicable if you observe that it is the same thing as $$ \csc(x) \approx -\frac{1}{4} + \frac{5 \pi}{16} \left( \frac{1}{x} + \frac{1}{\pi - x} \right) $$ The two summands in the parentheses are obvious if you want to get the poles of $\csc$ correct. If you wanted a good approximation of $\csc$ near the poles, then the coefficient out ...


13

$~\quad~$ While you're at it, try also $\cos\bigg(\dfrac\pi2x\bigg)\simeq\Big(1-x^2\Big)^\tfrac65$ and $\cos\bigg(\dfrac\pi2x\bigg)\simeq\Big(1-x^2\Big)^\tfrac76$ $~$ But since the numerical evaluation of fractional powers is significantly more time-consuming in terms of CPU, we can substantially improve this by using the binomial series for ...


5

(Cross posting my answer from the Wikipedia help desk.) Here is a way you might reverse engineer the formula, though I have no idea how Hardy derived it. Let $C(x) = \cos(πx/2)$. We know from the Taylor series that:$$C(x) = 1 - \frac{x^2}{\text{constant}} + \text{other terms}$$ Rewrite this as $C(x) = 1 - \frac{x^2}{K(x)}$ where K is to be determined. We ...


4

Consider a function of the form $$ f(x) = \sum_{j=1}^\infty c_j \cos(2 \pi j x) $$ Then the errors $ME(n)$ and $TE(n)$ in midpoint and trapezoid rules are $$\eqalign{TE(n) &= \sum_{k=1}^\infty c_{kn} \cr ME(n) &= 2 TE(2n) - TE(n) = \sum_{k=1}^\infty (-1)^{k} c_{kn} \cr}$$ For example, let $c_j = (-1)^{d(j)} 2^{-j}$ where $d(j)$ is the ...


4

You are right. In the first case, for $$\frac{f(x+h)-f(x)}{h} - f'(x)$$ you only have an $o(1)$ bound. A function like $f(x) = x\cdot\lvert x\rvert^\alpha$ for $0 < \alpha < 1$ is continuously differentiable on all of $\mathbb{R}$, but at $0$ the difference quotient converges only of the order $\lvert h\rvert^{\alpha}$ to the derivative. In the ...


3

I think your method is OK. If $||f-p||_\infty<\epsilon$ then $|\int_0^1 (f(x)-p(x))\sin(nx) \, dx|\leq \int_0^1 |f(x)-p(x)| \, dx<\epsilon$ Hence for large $n$, $|\int_0^1 f(x)\sin(nx) \, dx|\leq |\int_0^1 (f(x)-p(x))\sin(nx) \, dx| +|\int_0^1 p(x)\sin(nx) \, dx|\leq \epsilon+|\int_0^1 p(x)\sin(nx) \, dx|< 2\epsilon$ Also notice ...


2

To summarise the comments: You found that $\sqrt{16-x} \approx 4-\dfrac{x}{8}$ near $x=0$. So letting $x=1$ suggests $\sqrt{16-1} \approx 4-\dfrac{1}{8}$, i.e. $\sqrt{15} \approx \dfrac{31}{8}$. Indeed they are quite close as $\sqrt{15} \approx 3.873$ while $\frac{31}{8} = 3.875$. In fact what you have done is use the tangent at one point on a parabola ...


2

If $a\,q\ne0$, the integral does not converge. Let $s^*=-\ln(a\,q)/2$. The denominator of the integrand vanishes when $s=s^*$. Close to $s^*$ we have $$ (e^{-2s}-a\,q)^2\sim (s-s^*)^2, $$ meaning $$ \lim_{s\to s^*}\frac{(e^{-2s}-a\,q)^2}{(s-s^*)^2}=c\ne0. $$ Finally, $(s-s^*)^{-2}$ is not integrable on any neighborhood of $s^*$.


2

Are you familiar with how to approximate an error bound for alternating series? Given a sequence $\{a_n \}$ where $s = \sum_{i=1}^\infty a_n$ is a convergent alternating series and $s_n$ denotes the $n^{th}$ partial sum, then $$|s-s_n| \leq |s_n-s_{n+1}|= |a_{n+1}|$$ Hence, the quantity $|p_n-\pi|$ should be very close to $|a_{n+1}| = \frac{1}{2(n+1)+1} = ...


2

If we define $$p_n = 4 \sum_{k=1}^n \frac{(-1)^{k-1}}{2k-1},$$ then we have $$\lvert p_n - \pi\rvert = 4\sum_{k=n+1}^\infty \frac{(-1)^{k-n-1}}{2k-1}.$$ Grouping two consecutive terms in the remainder, we can write it as $$4\sum_{m = \frac{n+1}{2}}^\infty \left(\frac{1}{4m-1} - \frac{1}{4m+1}\right) = \sum_{m = \frac{n+1}{2}}^\infty \frac{8}{16m^2-1}$$ ...


2

Here is a way to solve this equation quickly that I think is more efficient that the methods proposed so far. The reason I think this is more efficient is that allows you to get about 15 digits with some multiplications (which are cheap), one square root, and at most one sin-cos evaluation (which are expensive). Your equation reminds me of the Kepler ...


2

We have to solve the equation ${\rm sinc}(x)=a$ for $x\in[0,\pi]$ in terms of the parameter $a\in[0,1]$. As Kirill has remarked, for $a\to1\!-\ $ we obtain $x\doteq\sqrt{6(1-a)}$. Therefore, if we want a single simple expression giving accurate values over the whole $a$-interval $[0,1]$ we have to take a square root at the end. For this reason we put ...


2

Every element of $B(A_0)$ is the limit of polynomials of the form $$ p(A_0)=\sum_{k=0}^na_kA_0^k,\quad n\in\mathbb N,\,\,a_k\in\mathbb C. $$ Hence the first and second bullets hold. Note that $$ \|B\|=\sup_{\|x\|=\|y\|=1}(x,By), $$ hence, if $A$ is self-adjoint, then $$ \|A^2\|=\sup_{\|x\|=\|y\|=1}(x,A^2y)=\sup_{\|x\|=\|y\|=1}(Ax,Ay)\ge ...


2

Taking the reciprocal of Stirling's Asymptotic expansion as derived in this answer: $$ n!=\frac{n^n}{e^n}\sqrt{2\pi n}\left(1+\frac{1}{12n}+\frac{1}{288n^2}-\frac{139}{51840n^3}+O\left(\frac{1}{n^4}\right)\right) $$ we get $$ \frac1{n!}=\frac{e^n}{n^n}\frac1{\sqrt{2\pi ...


1

If $x_{n+1}=g(x_n)$ is the iteration, it is sufficient that $x$ belong to a closed interval $I\subset\mathbb{R}$ such that $$ g(I)\subset I, \qquad\text{and}\qquad |g'(x)|<1 \text{ for all }x\in I. $$ This then allows you to use a fixed-point theorem to show that $g$ has a unique fixed point in $I$ and $x_n$ must converge to it. One difference between ...


1

Hint: $~\ln x=-\ln\dfrac1x~:~$ Can you take it from here ?


1

Almost as Semiclassical answered, write $x=A \times 10^{n-1}$ where $n$ is the number of digits before the decimal point, that is to say $1\leq A < 10$ and use the very fast converging expansion $$\log \left(\frac{1+y}{1-y}\right)=2\sum_{k=0}^{\infty}\frac{y^{2 k+1}}{2 k+1}$$ with $y=\frac{A-1}{A+1}$. Let us take an example : $x=123456789$; then ...


1

Take logarithms. $\alpha$ can be any negative number; if you are interested in $n\gg-\alpha$, then $\ln(n+\alpha)\approx\ln n+(\alpha/n)$, so $$ \ln C_n\approx (n-\alpha-\frac12)(\ln n+\alpha/n)-(n+\frac12)\ln n+n-\ln(2\pi)/2\\ \approx(-\alpha-1)\ln n+n+\alpha-\ln(2\pi)/2\\ C_n\approx\frac{1}{\sqrt{2\pi}}n^{-\alpha-1}e^{n+\alpha}$$


1

I don't think you can approximate it. You are performing a permutation mapping on the digits, where the indices are generated by the previous element of the sequence. So, you are mapping sequences of digits - approximations only make sense if the "importance" of numbers diminishes, but here, an early digit can already call for an accurate digit from far away ...


1

The problem with Taylor polynomials that, while they are very accurate near the center, they get worse and worse the farther away you get from it. What you want for a calculator or computer chip is to be uniformly accurate over the entire interval in question, otherwise you waste computing power with unnecessarily accurate estimates in some areas at the ...


1

More likely your calculator is using table lookup with some simple interpolation (possibly even just linear) between values in the table. Calculators (and computer math co-processor chips) can do this because you just need the answer to some known (10-15 digit most likely) precision. So as long as the error from the interpolation is below that threshold, ...


1

Because $\operatorname{sinc}$ is an even function, for $\epsilon=0$ the left side is zero. Is $\epsilon$ supposed to be small? If so, we can use $\dfrac d{dx} \operatorname{sinc}(x)=\dfrac {x \cos x-\sin x}{x^2}$ so you can evaluate the first-order Taylor expansion. The first term will contribute $3\dfrac \epsilon T (- \cos 1+\sin 1)$ as will the second. ...


1

Hint: $n~(n-1)~(n-2)~\cdots~(n-k)\approx n^{k+1}$ for large n and fixed k.


1

First of all, assuming $j,d$ are positive integers, $$ a_j = \frac{\Gamma(j+d)}{\Gamma{j+1}\Gamma{d}} = \frac{(j+d-1)!}{j!(d-1)!} = \frac{(j+d-1)\cdots(j+1)}{(d-1)!}. $$ Using $1+x \leq \exp x$, you can bound $$ \begin{align*} a_j &\leq (j+d-1)\cdots(j+1) \\ &= (j+1)^{d-1} \left(1+\frac{d-2}{j}\right) \left(1+\frac{d-3}{j}\right) \cdots ...


1

First of all, $$ \nabla f = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right\rangle = \left\langle \frac{1}{e + x + y}, \frac{1}{e + x + y} \right\rangle. $$ Evaluating both the function and its gradient at $(x, y) = (0, 0)$ gives $$ f(0, 0) = 0, \qquad \nabla f(0, 0) = \bigl\langle \tfrac{1}{e}, \tfrac{1}{e} \bigr\rangle, $$ ...


1

$$ \cos\big( \theta + \varepsilon^2 \frac{v\Omega}{L} + O(\varepsilon^{4}) \big) = \cos(\theta)\cos\big( \varepsilon^{2} \frac{v\Omega}{L} + O(\varepsilon^4) \big) - \sin(\theta)\sin\big( \varepsilon^2 \frac{v\Omega}{L} + O(\varepsilon^4) \big) $$ For $\varepsilon$ small : $$ \cos \big( \varepsilon^{2} \frac{v\Omega}{L} + O(\varepsilon^4) \big) = 1 + ...


1

I'm not sure how satisfying this answer will be, but here's something: The $n$th degree Taylor (MacLaurin) polynomial $g(x)$ of a function $f(x)$ is the unique degree $n$ polynomial such that $$ \lim_{x \to 0} \left(\frac{f(x) - g(x)}{x^n}\right)^2 = 0 $$ Is there some corresponding quantity that has been minimized? I'm not sure


1

When you round to 3 significant digits your possible results are $$ 3.45 \pm 0.005 \qquad \text{ or } \qquad 3.46 \pm 0.005 $$ The second of these intervals is the one that contain your true result, so that is what you should round to. Normally we don't write the $\pm 0.005$, but it is implicit when we write a rounded number with two digits after the ...



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