Question

What is f=? https://youtu.be/jcKRGpMiVTw?t=555 (He doesn't say the answer in the video, but this is where I found the question)

Answer 1

\[\huge f= 1 + \frac{3 +\frac{1+\frac{3+\cdots}{2+\cdots}}{3+\frac{1+\cdots}{3+\cdots}}}{2+\frac{2+\frac{2+\cdots}{1+\cdots}}{1+\frac{3+\cdots}{2+\cdots}}}\]

Answer 2

focus on the numerator, denominator independently:
$$f=\frac{x}y\\x=3+\frac{1+\frac{3+\dots}{2+\dots}}{3+\frac{1+\dots}{3+\dots}}=3+\frac{f}{x}\\y=2+\frac{2+\frac{2+\dots}{1+\dots}}{1+\frac{3+\dots}{2+\dots}}=2+\frac{y}f$$
so it follows $$x=3+\frac{f}x\implies x^2-3x-f=0\implies x=\frac12(3\pm\sqrt{9+4f})\\y=2+\frac{y}f\implies y=\frac2{1-1/f}$$
presumably \(x,y,f>0\) so we pick \(x=\frac12(3+\sqrt{9+4f})\)

Answer 3

I tried a few simple means of solving it analytically but I'm not sure it worked. I don't believe a solution exists according to the above. the paradoxical thing is that the solution I did find numerically has x, y < 0 (!)

Answer 4

Aha ok I see. I think you made a typo at the start so I don't know if that affects any of the rest of the work or not \(f=1+\frac{x}{y}\) but I see what you're doing here.

Answer 5

oh oops, that fixes a lot of things

Answer 6

$$f=1+\frac{x}y\\f-1=\frac14 (3+\sqrt{9+4f})(1-1/f)\\4f=3+\sqrt{9+4f}\\4f-3=\sqrt{9+4f}\\16f^2-24f+9=9+4f\\16f^2-28f=0\\f(4f-7)=0\implies f=7/4$$

Answer 7

Hmm I might have made a mistake, I just worked it out and got \(f=\frac{5}{4}\).

Answer 8

Wait, I think this should be a minus sign under the square root \(\sqrt{9\color{red}{-}4 f}\) or I did it wrong, oh well either way I get the trick and that's what matters haha.