Mathematica says the exact value of y that maximizes the distance between arctan(1.1y) and arctan(0.9y) is 1/sqrt(0.99) which is approximately 1.00504 (its negative also works). Feynman should still choose 1, because this isn’t a calculation to do in your head, and 1 is a good guess.
His chances of winning, however, are about 3.1%: the range of x-values that work is roughly 0.1, and the period of tangent is pi.
Excellent solution. Looking back on my tinkering with excel, I made an error in how I determined whether or not Feynman would have been right.
Unsurprisingly, that method would have still given the wrong answer (a 3% chance still isn’t very good), but it’s an example of using your knowledge of the general behavior of a function to make an educated guess.
Mathematica says the exact value of y that maximizes the distance between arctan(1.1y) and arctan(0.9y) is 1/sqrt(0.99) which is approximately 1.00504 (its negative also works). Feynman should still choose 1, because this isn’t a calculation to do in your head, and 1 is a good guess.
His chances of winning, however, are about 3.1%: the range of x-values that work is roughly 0.1, and the period of tangent is pi.
Excellent solution. Looking back on my tinkering with excel, I made an error in how I determined whether or not Feynman would have been right.
Unsurprisingly, that method would have still given the wrong answer (a 3% chance still isn’t very good), but it’s an example of using your knowledge of the general behavior of a function to make an educated guess.