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Mar 18 at 18:38 comment added Robert Bryant @George: You are correct. That minus sign was an error on my part (caused by my miscopying the formula). I have fixed it now.
Mar 18 at 18:33 history edited Robert Bryant CC BY-SA 4.0
Fixed a sign mistake in the formula for Y_1
Mar 17 at 21:07 comment added George Should we choose $Y_1=e^{-u}\cos(v)$?
Mar 17 at 20:37 comment added George I think $X$ and $Y$ aren't linearly independent where $sin(v)=cos(v)$.
Mar 9 at 20:58 history answered Robert Bryant CC BY-SA 4.0