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Uniform Convergence Implies $L^2$ Convergence and $L^2$ Convergence ...
Some of the books that discuss convergence say that uniform convergence implies $L^2$ convergence and $L^2$ convergence implies $L^1$ convergence, both while taken over a bounded interval I.
Prove that $a^n-b^n = (a-b)(a^{n-1} + a^{n-2}b + \\cdots + b^{n-1})$.
Prove that an −bn = (a − b)(an−1 +an−2b + ⋯ +bn−1) a n b n = (a b) (a n 1 + a n 2 b + + b n 1) ${a}^{n}-{b}^{n}=(a-b)({a}^{n-1}+{a}^{n-2}b+\cdots +{b}^{n-1})$. I've posted my solution below. In case someone has a more clever solution, feel free to post it! (TBH, I was surprised that there was no question on Math.SE regarding this ...
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一个最大值大于1的概率密度函数的意义是什么？ - 知乎
题主应该是把 概率质量函数 （probability mass function）和 概率密度函数 （probability density function）搞混了。 概率质量函数的取值确实无法大于1，但概率密度函数则不然。 以下内容搬运自维基百科： 在概率论中，概率质量函数是离散随机变量在各特定取值上的概率。
probability - Expectation on 1/X - Mathematics Stack Exchange
Expectation on 1/X Ask Question Asked 13 years, 11 months ago Modified 4 years, 9 months ago
real analysis - Little o notation in Taylor's formula : o (1 ...
Little o notation in Taylor's formula : o (1) Ask Question Asked 9 years, 2 months ago Modified 4 years, 11 months ago
calculus - Complex number: calculate $(1 + i)^n$. - Mathematics Stack ...
If, as your comments suggest, de Moivre's formula is "cheating", find a recurrence relation for the successive coefficients of 1 and i in (1 + i)n (1 + i) n $(1+i{)}^{n}$ i.e. real and imaginary parts, and solve that. The thing is completely equivalent. If this is part of an exercise to motivate de Moivre and prove it, that is one thing. But if you know de Moivre, why not use it?
Why is 1 raised to infinity Not defined and not "1" [duplicate]
$1$ square is $1$, so is raised $1$ to $123434234$. My maths teacher claims that $1$ raised to infinity is not $1$, but not defined. Is there any reason for this? I know that any number raised to
1 and 2 norm inequality - Mathematics Stack Exchange
While looking over my notes, my lecturer stated the following inequality;
how to prove that $\\ln(1+x)< x$ - Mathematics Stack Exchange
f′(x) = 1 1 + x − 1 = −x 1 + x <0, for x> 0 f (x) = 1 1 + x 1 = x 1 + x <0, for x> 0 ${f}^{\prime }(x)={\displaystyle \frac{1}{1+x}}-1={\displaystyle \frac{-x}{1+x}}<0,\text{ for }x>0$.
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