rigour

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rigour
rigour (ˈrɪgə(r)) Forms: 4–6 rygour(e, 6 r(e)ygur, 5–7 rygor (5 -ore); 5–6 rigoure (5 rigur), 4– rigour, 4–9 (now U.S.) rigor. [a. OF. rigor, rigour (13th c.; mod.F. rigueur), = Sp. and Pg. rigor, It. rigore, ad. L. rigor rigor.] I. 1. Severity in dealing with a person or persons; extreme strictness... Oxford English Dictionary
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rigour
rigour(US rigor), / ˈrɪgə(r); `rɪɡɚ/ n (fml 文)1 [U] severity; strictness; (esp mental) discipline 严格; 严厉; (尤指思想的)严谨, 严密 the utmost rigour of the law 法律的苛严 intellectual rigour 思想的缜密.2 [C often pl 常作复数]harshness (of weather, conditions, etc) (气候、 条件等的)严酷; 艰苦 the rigour(s) of an Arctic winter, of priso... 牛津英汉双解词典
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Rigour in mathematics Mathematics is very rigorous and everything must be proven properly even things that may seem true and obvious. Can you give me examples of conjectures/theories that seemed true but through rigo...
Finding the roots of a linear polynomial is trivial. Already the Babylonians could find roots of quadratic polynomials. Methods to solve cubic polynomials and fourth-degree polynomials were discovered in the sixteenth century, all using radicals (i.e. $n$th roots for some $n$). Isn't it obvious that...
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Implication of mean value or intermediate value theorem Let $f$ be a continuously differentiable function on $[a,b]$ with $f(a)=f(b)$ and $f'(a)=f'(b)$. Then, do there exist $x_1,x_2\in(a,b)$ such that $f'(x_1)=f'(x_2...
By the mean value theorem, there exists $c \in (a,b)$ s.t. : $$0=\frac {f(b)-f(a)}{b-a}=f'(c).$$ Knowing that $f'$ is continuous, by IVT, there exists $x_1 \in (a,c)$ such that $f'(x_1)=\dfrac{f'(a)}2$. By the same reasonning, there exists $x_2 \in (c,b)$ such that $f'(x_2)=\dfrac{f'(b)}2$. Since $f...
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Translation of "rigour, pragmatism" in Chinese - Reverso Context
Translations in context of "rigour, pragmatism" in English-Chinese from Reverso Context: Trust is not decreed. It is earned through actions in the field, through listening skills, anticipation, rigour, pragmatism and the ability to adapt.
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Show that if $f$ is continuous on $[0, \infty)$ and $\lim_{x \to \infty} f(x) = a$ that $\lim_{x \to \infty} 1/x \int_0^x f(t)dt = a$. Intuitively, I see this statement as saying that if a continuous function has a fi...
Suppose that $f(x) \to 0$. Then for any $\varepsilon>0$ there exists $A>0$ fixed such that $|f(x)|A$. Therefore $\frac{1}{x}|\int_0^xf(t)dt| \le \frac{1}{x}\int_0^A|f(t)|dt + \frac{1}{x}\int_A^x|f(t)|dt \le \frac{A}{x}\sup_{t \in [0,A]}|f(t)|+\varepsilon\frac{x-A}{x}$ The first term can be made $<\v...
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Borel-Cantelli Lemma Proof. !enter image description here **I struggle to understand the transition between the steps in the red box.** Especially why the limit there, i get the intuition behind the limit, its becau...
The sequence $(A_n)_{n \in \mathbb{N}}$ is not necessarily decreasing. However, if we define $$B_n := \bigcup_{k \geq n} A_k$$ then $B_n$ is decreasing since $$B_{n+1} = \bigcup_{k \geq n+1} A_k \subset \bigcup_{k \geq n} A_k = B_n.$$ Since $B_n \downarrow B$ for $B:= \bigcap_{n \geq 1} B_k$, the co...
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What does the continuum hypothesis imply? Are there any fundamental/interesting results that are a consequence of assuming the continuum hypothesis as an additional axiom? I'm sorry if this question was already asked...
There are many cardinals which have aleph values undetermined in ZFC, but for which we can say that they are either infinite and less than continuum, or uncountable and no greater than continuum, both of which would immediately resolve their (aleph) value if we had CH. Many of those can be resolved ...
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Find lim: $x_n=\dfrac{1+\frac12+...+\frac1{2^n}}{1+\frac14+...+\frac1{4^n}}$ Find limit: $x_n=\dfrac{1+\frac12+...+\frac1{2^n}}{1+\frac14+...+\frac1{4^n}}$ as $n \rightarrow \infty$ My "intuition" says that it shoul...
By induction you can prove the following formula for the sum of terms of a geometric progression: $$a + ar + ar^2 + ... + ar^n = {a - ar^{n+1} \over 1 - r}$$ So you can apply this in the numerator with $a = 1, r = {1 \over 2}$, and in the denominator with $a = 1, r = {1 \over 4}$. You obtain $$x_n =...
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Prove that for rational $x$,$y$ and $\epsilon$ if $|x-y| \leq \epsilon$ , $ \forall \epsilon > 0$ then $x=y$ I know that this is a repeated question, but I wanted to show my attempt. Suppose $x \neq y$ and (wlog) $x...
> _...But since $\delta 0, \epsilon \in \mathbb Q $..._ This part is not right. Make sure to use the _right_ argument with the _right_ inequality! > Since $\delta >0$, there **exists** $\epsilon \in \Bbb Q$ such that $0 < \epsilon < \delta$ and therefore $\epsilon < |x-y| = \delta$, which is absurd.
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Ring map-integers proof Prove that there is a ring map from $\mathbb{Z}/n\mathbb{Z}$ to $\mathbb{Z}/m\mathbb{Z}$ iff $m|n$. I am not able to prove this with the appropriate rigour, could someone show the steps for a...
$\def\Z{\mathbb Z}$ First suppose that there is a ring morphism $\phi\colon\Z/n\Z \to \Z/m\Z$. Then $0= \phi(0) = \phi(n \cdot 1) = n \cdot \phi(1) = n$, hence $n + m\Z = 0$, which gives $m \mid n$. If on the other hand $m \mid n$, define $\phi\colon \Z/n\Z \to \Z/m\Z$ by $\phi(a + n\Z) = a + m\Z$, ...
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Functional Relationship Between Epsilon and Delta In my textbook on vector calculus, I was studying up on the epsilon-delta definition of limits (new to me at the time). I believe I've understood what it's all about i...
What you _have_ to do is to show that for any $\varepsilon>0$, there is at least one $\delta>0$ which works. Theoretically, there is no need to supply one, only show that one exists. However, in practice, given $\varepsilon$, actually supplying a $\delta$ that works turns out to be the most practica...
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Integral equals zero implies function equals $0$ a.e. Let $f$ be a nonnegative bounded measurable function on a set of finite measure $E$. Assume $\int_E f = 0$. Show that $f=0$ a.e. on $E$ So I feel like I understan...
Its easier to show the contrapositive statement. If $f$ is not zero almost everywhere then $A:=\\{x\in E:f(x)>0\\}$ have positive measure. Now setting $A_n:=\\{x\in E:f(x)>1/n\\}$ its easy to check that $$ A=\bigcup_{n\in\Bbb N}A_n\tag1 $$ and because $\\{A_n\\}_{n\in\Bbb N}$ is an increasing sequen...
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