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After I use Simplify on an expression I get$ \dfrac{1}{2}\sqrt{-\dfrac{\sqrt{(-b^2+16|c|^2)(4|c^2|+b\Im(c))^2}}{4a(4|c|^2+b\Im(c)])}}$ . This expression can clearly be simplified further by noticing that the square bracket term in the numerator cancels the other bracket term in the denominator so $ \dfrac{1}{2}\sqrt{-\dfrac{\sqrt{(-b^2+16|c|^2)}}{4a}}$ . This is clearly a much simpler form since it includes less terms, so my questionRead more

A calculation yields the following result: $ \text{kh}\to \frac{2 \sqrt{3} \sqrt{\nu +n-1} \sqrt{\text{R}^2 n^3+3 \nu +12 \nu n+15 n-3}+6 \nu -6 (2 \nu +1) n-6}{\text{R} (\nu +1) n^2}$ Knowing that $ n \gg 1$ , I can approximate this solution (by hand) to: $ \text{kh}\to \frac{2 \left(\sqrt{3 \left(\text{R}^2 \text{n}^2+12 \nu +15\right)}-3 (2 \nu +1)\right)}{\text{R} (\nuRead more

Question: How can I tell Mathematica to replace a sub-expression (in a simplified expression) with a symbol if the symbol is defined to be that sub-expression? Say you have defined: f[x_, y_] := some_complicated_expr_using_x_and_y_1 g[x_, y_] := some_complicated_expr_using_x_and_y_2 h := some_complicated_expr_using_f_and_g then you do D[h,x] // Simplify, you would get an expression that has sub-expressionsRead more

Expression $ ((A \oplus B) \land \lnot C) \lor (\lnot(A \oplus B) \land C)$ simplifies to $ A \oplus B \oplus C$ . This is my attempt at simplification: 1. $ ((A \oplus B) \land \lnot C) \lor (\lnot(A \oplus B) \land C)$ 2. $ (((A \land \lnot B) \lor (\lnot A \land B)) \landRead more