User:AchatesAVC/math

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$\sqrt{-1}=i$

Er...what's up with this:

$\int \frac{du}{\sqrt{a^2 -u^2}} = \sinh ^{-1} \frac{u}{a} + C$

Now we assume that $\sinh ^{-1} \frac{u}{a}$ is undefined for $a = 0$ But if we take $a = 0$ in the integral, we get the following:

$\int \frac{du}{\sqrt{-u^2}} = \int \frac{du}{ui} = i \int \frac{du}{u}$

Then integrating.

$i \int \frac{du}{u} = i \ln u + C$

Now does this mean that:

$\sinh ^{-1} \frac{u}{0} + C_0 = \sinh ^{-1} \infty + C_0 = i \ln u + C_1$

If so, that is strange...

${dy \over dx} = \lim _{\Delta x \to \infty} {{f(x+\Delta x)-f(x)} \over \Delta x}$

$\ln x = \int _{1} ^{x} {1 \over u} du = \lim _{h \to 0} {{x^h - 1} \over h}$

$e 2π i + 1 = 0$

$e 2π i = cosθ + i sinθ$

$\mathfrak{z} ^n _k = \sqrt[n] |{\mathfrak{z}}|(\cos {{\theta + {2 \pi k}} \over n} + i \sin {{\theta + {2 \pi k}} \over n})$

Special Case: ${1 \over a} - {1 \over {a+1} } = {1 \over {a(a+1)} }$

General Case: ${n \over a} - {n \over b} = {{n(b-a)} \over ab}$

${ {d(e^u)} \over dx} = e^u {du \over dx}$

For initial acceleration, velocity, and displacement:
$a_0,\ v_0,\ s_0$

$\frac{d^3 s}{dt^3} = j$

$a = \frac{d^2 s}{dt^2} = \int j\ dt= jt + a_0$

$v = \frac{ds}{dt} = \int a\ dt = \int (jt + a_0)dt = \frac{1}{2} jt^2 + a_0 t + v_0$

$s =\int v\ dt= \int (\frac{1}{2} jt^2 + a_0t + v_0)\ dt = \frac{1}{6}jt^3 + \frac{1}{2}a_0 t^2 +v_0 t + s_0$

$X_1 = X_2 - \frac{dm_2}{m_1+m_2}$

$\mathcal{F}_g = \frac{\mathcal{G} m_1 m_2}{|r|^2}\ \frac{|r|}{\mathbf{r}}$

$X_{(1,2)} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Cauchy-Riemann:

$\frac{\partial u}{\partial x}(x_0,y_0) = \frac{\partial v}{\partial y}(x_0,y_0)$

$\frac{\partial v}{\partial x}(x_0,y_0)= - \frac{\partial u}{\partial y}(x_0,y_0)$

$\operatorname{If}\ r<1:$

$a_\infty = \sum ^\infty _1 ar^{n-1} = \frac{a}{1-r}$

$P(X,N) = \bigg( \begin{matrix} N \\ X \end{matrix} \bigg) p^x (1-p)^{n-x}$

$P (X = N) = p N - 1 (1 - p)$

$\frac{1}{P} = \frac{1}{n-1} \bigg( {\frac{1}{R_1} + \frac{1}{R_2}} \bigg)$

$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$

$\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2} dt.$

$\operatorname{NormalCDF} (X, \mu, \sigma) = \frac{1}{2} \bigg( 1 + \frac{2}{\sqrt{\pi}} \int _0 ^{\frac{x-\mu}{\sigma \sqrt{2}}} e^{-t^2} dt \bigg)$

Random Integrals:

$\int x \sin x dx$

$u=x\quad du=dx\quad dv=\sin x\quad v=-\cos x$

$uv - \int v du= -x\cos x +\sin x$

$\int \frac{3}{x \ln(x)}\ dx= 3\ln|\ln(x)|$

$\int \frac{1}{x^2-4}\ dx\ =\ \frac{1}{4}\ln\bigg|\frac{\ln|x-2|}{\ln|x+2|}\bigg| + C$

$i 2 = - 1$

$P = I 2 R$

$\operatorname{cis} (\theta) = \cos (\theta) + i\sin (\theta)$

$re^{i\theta}=|r|\operatorname{cis}(\theta)$

$|\mathfrak{z}|=\mathfrak{z}\bar{\mathfrak{z}}$

$x = r cosθ$

$y = r sinθ$

$r = x 2 + y 2$

$\sinh (x) = \frac{e^x -e^{-x}}{2}$

$\cosh(x) = \frac{e^x + e^{-x}}{2}$

$\int \frac{du}{u^2 + a^2} = \frac{1}{a} \tan ^{-1} (\frac{u}{a})$

$\int \frac{du}{\sqrt{a^2-u^2}} = \sin ^{-1} (\frac{u}{a})$

$\int \frac{du}{\sqrt{a^2+u^2}} = \sinh ^{-1} (\frac{u}{a})$

$V = \int _a ^b 2 \pi x f(x) dx$

Prove:

$\lim _{k \to \infty} A \bigg( 1 + \frac{r}{k}\bigg)^{kt}= Ae^{rt}$

This holds if:

$\lim _{k \to \infty} \bigg( 1 + \frac{r}{k} \bigg) ^k = e^r$

So:

$\bigg( 1 + \frac{r}{k} \bigg) ^k = e^{\ln \big( 1 + \frac{r}{k} \big) ^k}$

Which is equivalent to:

$e^{k \ln \big( 1 + \frac{r}{k} \big)}$

Then our supposition holds true if:

$\lim_{k \to \infty} k \ln \bigg(1 + \frac{r}{k} \bigg) = r$

Now we substitute a new variable $n = \frac{1}{k}$ into the equation. This gives:

$\lim_{n \to 0} \frac{\ln (1 + rn)}{n}$

Then by L'Hôpital's rule the above expression is equivalent to:

$\lim_{n \to 0} \frac{ \frac{r}{1 + rn}}{1} = lim_{n \to 0} \frac{r}{1+rn}=\frac{r}{1}=r$

Therefore:

$\lim _{k \to \infty} \bigg( 1 + \frac{r}{k} \bigg) ^k = e^r$

$\lim_{x \to 0} \frac{\sin x}{x} = 1$

Proofs:

By L'Hôpital's Rule:

$\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\cos x}{1}= \cos 0 = 1$

By Sandwich Theorem:

$sin x < x$

$\frac{\sin x}{x} < 1$

$\cos x < \frac{\sin x}{x}$

$\lim_{x \to 0} \cos x =1$

$\lim_{x \to 0} 1 = 1$

Therefore:

$\lim_{x \to 0} \frac{\sin x}{x} = 1$

$\int \frac{du}{\sqrt{a^2-u^2}} = \sinh ^{-1} \frac{u}{a}$

$cos 2 x + sin 2 x = 1$

$sec 2 x = 1 + tan 2 x$

$csc 2 x = cot 2 x + 1$

$\operatorname{Fish} (x) = \iint _2 ^43 \ln \sinh x + 7x^4 -exi$

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