Gavran's latest activity

The equation ## C=A+B ## is actually Euler's formula $$ \arctan\frac12+\arctan\frac13=\frac\pi4 $$, and there are three more formulas of...

Here is another way but it’s just a variation of @coquelicot above in post #7 but the lengths are not obvious so they must be...

a different line for ##\tan 4S## ##\tan (4S)= \tan (2S +2S) = \dfrac{\tan 2S + \tan 2S}{1 - \tan^2 (2S)}## ##\tan (2S) =...

There are four formulas of this kind. Euler's formula $$ \arctan\frac12+\arctan\frac13=\frac\pi4 $$ Hermann's formula $$...

Based on the previous posts, it can also be shown that A+B+C=D for the angles A, B, C, and D in the next picture. Or A+B+C+D=E for...

Here are some more solutions; In each case, the triangle constructed in green is similar to the triangle highlighted in brown which...

Just another approach:

I think you are right, and the correct answer is d. A copper shield can be better than a steel shield against a high-frequency magnetic...

In addition to @Steve4Physics's comments: 1. Please note that we prefer that you post your work in ##\LaTeX## which makes the equations...

Inspired by the posts by @Gavran we can show that for ##R## red balls and ##N## balls in total, the expected number of red balls at the...

I also calculated it to be 1.515 using a different but related method. What the OP is asking for is modeled by the negative...

Von Neumann's famous writing on this in his classic text on QM never really caught on (and when I first read it, it left me scratching...

## P(X=g) ## - probability that we have g green cards remaining ## P(X=b) ## - probability that we have b blue cards remaining ## P(X=r)...

Here is my solution;