Math Tricks & Simplifications
\(\ds \begin{align}
\bf[1] & \quad \frac{d}{dt}\left[\Ln{\frac{A(t)}{B(t)}}\right] = \overbrace{\frac{d}{dt} \Ln{A(t)}}^{D_1} - \overbrace{\frac{d}{dt}\Ln{B(t)}}^{D_2};\ \ D_1=\frac{1}{A(t)}\frac{dA(t)}{dt},\ D_2=\frac{1}{B(t)}\frac{dB(t)}{dt}\\
\bf[2] & \quad \sum_{j=0\t{ and even}}^{\infty}\frac{\vplup{\lambda}^j}{j!}=\frac{1}{2}\sum_{j=0}^{\infty}\frac{{\lambda}^j}{j!}+\frac{1}{2}\sum_{j=0}^{\infty}\frac{(-\lambda )^j}{j!} = \frac{1}{2}e^{\lambda} +\frac{1}{2}e^{-\lambda} \\
\bf[3] & \quad \frac{d}{dy}\intlim{a}{g(y)}f(x)dx=p(g(y))\frac{dg(y)}{dy} \\
\bf[4] & \quad k_1\sin{\theta}+k_2\cos{\theta} = A\Cos{ \theta -\alpha },\quad A=\sqrt{k_1^2+k_2^2},\ \alpha=\atan{k_1/k_2} \\
\bf[5] & \quad \bn{M}\Frac{x}{1+x^2}=x\cdot\bn{M}(f(u))\vert_{u=-x^2}=x\cdot(1+(-x^2)+(-x^2)^2+...)=x-x^3+x^5-... \\
& \quad \bn{M}(f)=\t{Maclaurin}(f), f(u)=\frac{1}{1-u} \\
\bf[6] & \quad \abs{A+Be^{j\alpha}+Ce^{jk\alpha}}=\sqrt{\abs{(A+Bz+Cz^k)(A+Bz^{-1}+Cz^{-k})}},\quad z= e^{j\alpha} \\
\end{align} \)
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