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Math Attack's user avatar
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Math Attack's user avatar
Math Attack
  • Member for 1 year, 1 month
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  • Italy
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About

I calculate formulas when I'm bored to pass the time.
(Ergo I may not be very linear in the type of questions)

My favorite formulas:

  • $\displaystyle z!_{(\alpha)}=\alpha^{\frac{z}{\alpha}}\Gamma\left(1+\frac{z}{\alpha}\right)\prod_{j=1}^{\alpha-1}\left(\frac{\alpha^{\frac{\alpha-j}{\alpha}}}{\Gamma\left(\frac{j}{\alpha}\right)}\right)^{C_{\alpha}(z-j)}\qquad \text{where }C_{\alpha}(z)=\frac{1}{\alpha}\sum_{k=1}^{\alpha}\cos\left(\cos^{-1}\left(\cos\left(\frac{2k\pi}{\alpha}\right)\right)z\right)$
  • $\displaystyle\Gamma(z)=\frac{(-1)^m}{m!(z+m)}\sum_{n=0}^{\infty}\frac{B_n(a_1,...,a_n)}{n!}(z+m)^{n}\qquad\text{where }\begin{cases}a_n=(n-1)!\left(H_m^{(n)}+(-1)^n\zeta(n)\right)\\\zeta(1)\overset{\mathcal{R}}{=}\gamma\end{cases}$
  • $\displaystyle\sum_{n=1}^{x}n^{s-1}\ln(n)^m\propto(-1)^m\left[\zeta^{(m)}(1-s)+\sum_{k=0}^{m}\frac{m!}{k!}\sum_{j=0}^{s}\binom{s}{j}\frac{B_{s-j}^{+}x^j}{s^{m+1-k}}B_k\left(\left\{l!\left(H_j^{(l+1)}-H_s^{(l+1)}\right)-\delta_l\ln(x)\right\}_{l=0}^{k-1}\right)\right]$
  • $\displaystyle\sum_{n=0}^{\infty}n!_{(\alpha)}\overset{\mathcal{R}}{=}\frac{1}{\alpha\sqrt[\alpha]{e}}\text{Ei}\left(\frac{1}{\alpha}\right)+\sum_{k=1}^{\alpha-1}\left[\frac{\cos\left(\frac{k\pi}{\alpha}\right)}{\sqrt[\alpha]{e\alpha^k}}\Gamma\left(1-\frac{k}{\alpha}\right)+\frac{1}{k}{}_1F_1\left(\left.{1 \atop 1+\frac{k}{\alpha}}\right|-\frac{1}{\alpha}\right)-1\right]$
  • $\displaystyle z!_{(\infty)}=\exp\left(\sum_{k=1}^{\infty}\ln(k)\cdot\text{sinc}(z-k)\right)$
  • $\displaystyle\int\operatorname{tan^{-1}}(\alpha\cdot\cos(x))x^n\mathrm{d}x=2n!\sum_{k=0}^{n}\frac{x^{n-k}}{(n-k)!}\Im\left[i^k\cdot\text{Ti}_{k+2}\left(\frac{\sqrt{\alpha^2+1}-1}{\alpha}\cdot e^{ix}\right)\right]$
  • $\displaystyle\int_0^x \frac{t^{m-1}}{(1\pm t^n)^p}\mathrm{d}t=\frac{x^{m}}{n}\sum_{j=1}^{p-1}\frac{\displaystyle\left[\prod_{k=j+1}^{p-1}\frac{nk-m}{nk}\right]}{j\left(1\pm x^{n}\right)^{j}}+(I(x)-J(x))\left[\prod_{k=1}^{p-1}\frac{nk-m}{nk}\right]$
    $\displaystyle I(x)=\frac{1}{n}\sum_{k=1}^{n}\Re\left[e^{m\theta}\ln(1-xe^{-i\theta})\right]\quad \begin{cases}\text{When }+:&J(x)=\displaystyle\sum_{k=1}^{\left\lfloor{\frac{m-1}{n}}\right\rfloor}(-1)^{k}\frac{x^{m-nk}}{m-nk}\qquad \theta=\frac{2k-1}{n}\pi\\ \text{When }-:&\displaystyle J(x)=\sum_{k=1}^{\left\lfloor{\frac{m-1}{n}}\right\rfloor}\frac{x^{m-nk}}{m-nk}\quad\qquad\qquad \theta=\frac{2k}{n}\pi \end{cases}$
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