lims
$ \lim\limits_{x -> 0} \frac{sinx} {x} = $
$ \lim\limits_{x -> \infty} (1 + \frac{1} {x})^x = $
Derivative
$ (x^\alpha)^{(n)} = $
$ (x^n)^{(n)} = $
$ (\frac{1}{x})^{(n)} = $
$ (sinx)^{(n)} $
$ (cosx)^{(n)} $
$ (\frac{1}{x + a})^{(n)} = $
$ (a^x)^{(n)} = $
series
Fourier
$ a_n = $
$ b_n = $
$ f(x) ~ $
Dirichlet
$ S(x) = $ans
lims
1
e
Derivative
$ \alpha(\alpha - 1)(\alpha - 2) … (\alpha - n + 1)x^{\alpha - n} $
n!
$\frac{(-1)^n n!}{x^{n+1}} $
$ sin(x + \frac{n}{2} \pi) $
$ cos(x + \frac{n}{2} \pi) $
$ \frac{(-1)^n n!}{(x + a)^{n+1}} $
$ a^x ln^n a ,(a > 0)$
series
Fourier
$ \frac{1}{l} \int^l_{-l} f(x) cos \frac{n \pi x}{l} dx , n = 0,1,2,3…$
$ \frac{1}{l} \int^l_{-l} f(x) sin \frac{n \pi x}{l} dx, n = 1,2,3…$
$ \frac{a0}{2} + \sum\limits{n = 1}^{\infty}(a_n cos\frac{n \pi x }{l} + b_n sin\frac{n \pi x}{l}) $
当x是f(x)的连续点时，f(x)
当x是f(x)的第一类间断点时, $ \frac{f(x-0) + f(x + 0)}{2} $
English
what if 如果。。。将会。。。有没有这样一种可能……
how 如何，多么