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laborious
laborious, a. (ləˈbɔərɪəs) Also 6 -yous(e. [ad. F. laborieux (12–13th c. in Hatz.-Darm.) or ad. L. labōriōs-us, f. labor labour: see -ious.] 1. Given to labour or toil; doing much work; assiduous in work, hard-working.1390 Gower Conf. II. 90 If thou wolt here Of hem that whilom vertuous Were and the...
Oxford English Dictionary
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Abraham the Laborious
Abraham the Laborious (fl. 14th century) was a monk of Kiev.
He is regarded as a saint, with a feast day of 21 August at Kiev.
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laborious
laborious/ləˈbɔ:rɪəs; lə`bɔrɪəs/ adj1 (of work, etc) needing much effort (指工作等)艰苦的, 费力的 a laborious task 艰苦的工作.2 showing signs of greateffort; not fluent or natural 吃力的; 不流畅的; 不自然的 a laborious style of writing 艰涩的文体. Cf 参看 laboured (labour2).
牛津英汉双解词典
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Laborious Ukraine Electoral Bloc
The Laborious Ukraine Electoral Bloc or Labour Ukraine () is a former political alliance in Ukraine that participated in the 1998 Ukrainian parliamentary
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A more structural proof using homomorphisms and similar tools that every ideal of $M_n(R)$ is of the form $M_n(I)$ Let $R$ be a ring with unity , we know that if $J$ is an ideal of $M_n(R)$ then for some ideal $I$ of ...
The "very tedious... laborious" version is more accessible, by comparison, and probably can be streamlined a bit to be tolerable.
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Convergence of the series $\sum\limits_{n = 4}^\infty {\frac{{n + 1}}{{(n + 5)(n + 4)(n - 3)}}}$? To analyze the convergence of the $$\sum\limits_{n = 4}^\infty {\frac{{n + 1}}{{(n + 5)(n + 4)(n - 3)}}}$$ series I use...
Assume that the series begins at $n=4$. Then, we have $$n+5\ge n$$ $$n+4\ge n$$ $$n-3\ge \frac14 n$$ $$n+1\le 2n$$ Therefore, have $$\frac{n+1}{(n+5)(n+4)(n-3)}\le \frac{2n}{\frac14 n^3}=8\frac1{n^2}$$ Finally, using the result $\sum_{n=1}^\infty \frac1{n^2}=\frac{\pi^2}{6}$ reveals $$\sum_{n=4}^{\i...
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Derivation of the Exponential Nature of $e^x$ Presumably, the transcendental number $e$ was first found by taking the power series solution to the (arguably most fundamental) differential equation $f'(x)=f(x)$, with t...
I believe that $e$ was first found as $$ e=\lim_{n\to\infty}\left(1+\frac1n\right)^n\tag{1} $$ perhaps arising from compound interest problems and the like. * * * However, if we start with $f'(x)=f(x)$, then we have for any fixed $y$ $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\frac{f(x+y)}{f(x)}...
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Simpler way to compute a definite integral without resorting to partial fractions? I found the method of partial fractions very laborious to solve this definite integral : $$\int_0^\infty \frac{\sqrt[3]{x}}{1 + x^2}\,...
Perhaps this is simpler. Make the substitution $\displaystyle x^{2/3} = t$. Giving us $\displaystyle \frac{2 x^{1/3}}{3 x^{2/3}} dx = dt$, i.e $\displaystyle x^{1/3} dx = \frac{3}{2} t dt$ This gives us that the integral is $$I = \frac{3}{2} \int_{0}^{\infty} \frac{t}{1 + t^3} \ \text{d}t$$ Now make...
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The remainder of $n^{11}$ when divided by $23$. Let $n$ be a natural number such that $n$ is not divisible by $23$. Then the remainder when $n^{11}$ is divided by $23$ is $±1(\mod 23)$. I have solve it by a laborious...
By Fermat's little theorem: $$n^{22} \equiv 1 \pmod {23}$$ Since $23$ is a prime, $\Bbb Z/23\Bbb Z$ is a field, so $x^2-1$ only has two roots over $\Bbb Z/23\Bbb Z$, but we know that those two roots are $\pm1$, so $n^{11} \equiv \pm1 \pmod {23}$ since $n^{11}$ is a solution to $x^2-1=0$. Informally,...
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Relative trace and algebraic integers In a number field, the trace of an element over $\mathbb{Q}$ gives necessary conditions on the algebraic integers--the trace of an algebraic integer over $\mathbb{Q}$ is an intege...
The corresponding thing is true: if $\alpha$ in $K$ is an algebraic integer, then its trace down to any subfield $k$ is an integer of $k$.
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How to generate complicated looking identities such as $\sqrt [3] {2 + \sqrt 5} - \sqrt [3] {2 - \sqrt 5}=1$ easily? How to generate complicated looking identities, or even more complicated looking identies such as $\...
!Image Here's a nice "generator" of the result I commented in the comments.
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Multiple Mysql databases backup tool? My OS is Ubuntu 10.10. I have several small sites (all on various shared hosts) and I'm thinking that it'd be probably good idea to regularly backup mysql databases from these rem...
Specifying the password that way is just insecure on your machine. Over the network it's just the same way that MySQL clients connect to your database. As far as I read, the username and password are hashed, so you are not sending your password in plain text. In your situation I would write a simple...
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Find $\lim_{x\to 0}\frac{\sin x-x\cos x}{x^3}$ $$\lim_{x\to 0}\frac{\sin x-x\cos x}{x^3}$$ How do I go about doing this? I can see no simple way of doing this. Application of l'Hopital's rule would be very laborious....
Taylor expanding we find that: \begin{equation} \begin{aligned} \lim_{x\to 0}\frac{\sin x-x\cos x}{x^3}&=\lim_{x\to 0}\frac{x-\frac{x^3}{6}+\mathcal{O}\left(x^4\right)-x\left[1-\frac{x^2}{2}+\mathcal{O}\left(x^4\right)\right]}{x^3}\\\ &=\lim_{x\to 0}\frac{x^3\left(\frac{1}{2}-\frac{1}{6}\right)+\mat...
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Linear relation inside of a triangle Let $d(P,l)$ be distance between point $P$ and line $l$. Inside triangle $ABC$ there are points $D$ and $E$, for which $$ d(D,AC)+d(D,BC)=d(D,AB), $$ $$ d(E,AC)+d(E,BC)=d(E,AB). $$...
The proof is as simple as it gets: $$ $$x_2=d_2+\frac{u}{DE}(e_2-d_2)$$ $$x_1+x_2=d_1+\frac{u}{DE}(e_1-d_1)+d_2+\frac{u}{DE}(e_2-d_2)=$$ $$x_1+...
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Is the symmetric group $S_4$ cyclic > **Is the symmetric group $S_4$ cyclic?** By writing all $24$ elements we can write the tabular form of $S_4$. Then choosing each element of $S_4$, we can find its order and thus,...
Another approach, if $S^4$ is abelian, then every subgroup is abelian, but $S^3\leq S^4$ and $S^3$ is the very first and unique (up to isomorphism) non-abelian group. $S^3$ in $S^4$ is $\langle(1234),(12)\rangle$.
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