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# How to handle UM using Wolfram Language?

Posted 3 months ago
 Hi, I'm trying to understand how WM handles UM.That is I expect that a calculation should give some UM, but I don't succeed. Let's take as an example (314.159 m Sqrt[((7.34102*10^-6 - 0.0000178801 I) s^2)/m^2])/s  The results should be a scalar that I can calculate using coth[] but I'm unable to "convince" WM to give me that,Thanks for any help
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Posted 3 months ago
 Applying your second "trick" and crashing my head against the wall I've got the desired result. However plotting a graph with the resulting function is much faster (0.235151) when I disregard the UM's than when I use them (2.97654). Doing physics calculations with WM can be a nightmare.
Posted 3 months ago
 \$Assumptions = {m > 0, s > 0, kg > 0} Z[x_] := Sqrt[R[x] K[x]] // Simplify; Cleaner Thanks again
Posted 3 months ago
 Look up Quantity in the documentation: Quantity[314.159, "Meters"] * Sqrt[Quantity[7.34102*10^-6 - 0.0000178801 I, "Seconds"^2/"Meters"^2] ]/Quantity[1, "Seconds"] % // Coth You can also use your notation, but adding the information that s and m are positive: (314.159 m * Sqrt[((7.34102*10^-6 - 0.0000178801 I) s^2)/m^2])/s Simplify[%, s > 0 && m > 0]