$\int_0^\infty1/(x^n+1)\mathrm{d}x$ 的值是?

Someone

利用留数定理计算 $$\int_0^\infty\dfrac{1}{x^n+1}\mathrm{d}x$$ 其中 $n$ 是正整数。

若 $a_{n-1}+2a_n \rightarrow 0$,则 $a_n$ 收敛还是发散?

Someone

设 $\{a_n\}$ 满足 $$\lim_{n\rightarrow\infty} a_{n-1}+2a_n = 0$$,试讨论 $\{a_n\}$ 的敛散性。

若函数在实轴上连续,且在正负无穷都有极限,求证函数在实轴上一致连续?

Someone

设 $f(x)$ 在 $\mathbb{R}$ 上连续,且 $f(+\infty), f(-\infty)$都存在(极限),求证 $f(x)$ 在 $\mathbb{R}$ 上一致连续

Injective, surjective既分別??

Someone

可唔可以用中文解釋一下injective, surjective同 bijective既分別呀???唔該晒!!

What does it mean if we disable K-rule in Agda?

Someone

TL;DR: Can I say, "K-rule in Agda enables people to match $ \forall a.a \equiv a $ with $ refl $"? In https://agda.readthedocs.io/en/v2.5.4.1/language/without-k.html#without-k, K-rule is introduced as an implicit rule and it's defaultly enabled. If I understand it correctly, it means parameter of type $ \forall a.a \equiv a $ can be matched with $ refl $. If we disable K-rule, what will happen? What kind of codes is it going to prevent me writing? Because we can always construct $ \forall a . a \equiv a $, even without K, we can always get an instance of $ T $ by passing $ refl $ to any functions with type $ \forall a.a \equiv a \rightarrow T $. Agda's doc has given me an example which indeed shows a circumstance that can only work with K: ```agda K : {A : Set} {x : A} (P : x ≡ x → Set) → P refl → (x≡x : x ≡ x) → P x≡x K P p refl = p ``` In this code, if we can pattern match `x≡x` with `refl`, `P refl` can be trivially equivalent to `P x≡x` (but without K, we can't). So does that mean: "K enables people to match $ \forall a.a \equiv a $ with $ refl $"? I didn't find the answer on the Agda doc. If we disable K, will the semantic of Agda's equality type (CH-ISO) change? Source: https://cs.stackexchange.com/q/98339

What's "this", really?

Someone

Generally speaking, in object oriented programming languages, "this" is a keyword for referencing the object "itself". however, if I write in this way struct DDF { struct { int fantansy } dark; } deep; void makeFantansy(DDF *this, int value) { this->dark.fantansy = value; } What's the difference between this and "Classes" grammer~ga

如何在一堆复数中选出一部分,使得这部分和的模不小于所有数的模的和的某个常数倍?

Someone

设 $z_1,\cdots,z_n$ 是任意$n$个复数,证明必有$\{1,2,\cdots,n\}$的子集$E$,使得 $$\left|\sum_{j\in E}z_j\right|\ge\dfrac{1}{\pi}\sum_{j=1}^{n}|z_j|$$

在过原点的直线一侧的复数们,取倒数之后,是否仍在这条直线的某一侧?

Someone

如果 $z_1,\cdots, z_n$ 都位于过原点的直线的一侧,证明 $\dfrac{1}{z_1},\cdots,\dfrac{1}{z_n}$也必位于该直线的某一侧,而且满足 $$z_1+\cdots+z_n\ne 0$$ $$\dfrac{1}{z_1}+\cdots+\dfrac{1}{z_n}\ne 0$$

证明 $\sum \dfrac{a_n}{1+na_n}$ 发散?

Someone

假定 $a_n>0$,而 $\sum a_n$ 发散。 证明 $\sum \dfrac{a_n}{1+na_n}$ 发散

Result<Success, Failure> 究竟比 checked exception 好在哪儿?

Someone

为什么大家都说前者好?这两者不是明明是一样的吗?

Flutter 是不是就是移动版的 Duilib?

Someone

或者说就是移动版的 DirectUI 的实现?即它的本质竞争力就是自己通过 GPU 绘图,绕过了系统控件?

「设计直接产出视图代码」的根本难点在哪儿?

Someone

为什么历史上视图这么做的人都失败了?比如 storyboard,沦为笑柄,Dreamweaver 更是大家唾弃生成的代码无法维护。 是自定义视图很难渲染吗?如果我们限制视图的多样性呢? 这里面有本质的困难吗?

????????????????????

Someone

提问不见了

为什么水区也有字数限制?

Someone

> 错误: 回复至少要 10 个字符。 > 错误: 问题必须以问号结尾 # 水区应该随便水。

$f=u+iv$ 是全纯函数,$xu+yv = (x^2+y^2)e^x \cos y$,问 $f$ 是什么?

Someone

我试图把 $xu+yv$ 解释为某个新函数的一部分,比如 $\overline{z}f$ 的实部。但这个函数不是全纯的。所以我不知道怎么办了……(也许 $\dfrac{\partial}{\partial\overline{z}}(\overline{z}f)=1$?但怎么利用这一点呢?)

如何计算 $\sum_{i = 0}^{n}\ln\binom{n}{i}\Big/n^2$?

Someone

如何计算 $$\sum_{i = 0}^{n}\ln\binom{n}{i}\Big/n^2$$? 我只能这么估计: $$\binom{n}{i}<\left(\dfrac{n\cdot e}{k}\right)^k$$ $$\sum_{i = 0}^{n}\ln\binom{n}{i}\Big/n^2<\dfrac{1}{n}\dfrac{k}{n}\left(\ln(\dfrac{n}{k}\cdot e)\right)\rightarrow\int_0^1t\ln\dfrac{e}{t}dt=\dfrac{3}{4}$$ 但数值计算的结果似乎是 $\dfrac{1}{2}$。 怎么才能证明这点呢?

如何在不引入相关模块的情况下获取类型?

Someone

比如说,我想在前端代码中引入模型,但是模型的代码其实是 `extends Sequelize.Model` 的,那么我应该怎么获取到模型的类型,却又避免引入 `sequelize` 呢?(实际上前端也是无法引入 `sequelize` 的,因为它依赖很多 C++ 的包)

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