I Evaluated An Infinite Fraction

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[@francis6888]

Fun fact: this is the formula for the silver ratio

[@bobbyheffley4955]

The area of a regular octagon is twice the squared side length times the silver ratio.

[@SyberMath]

Wow! Thanks ❤️

[@SyberMath]

😍

[@MrLikeG]

I think you can simplify the answer to (1+√2)

[@vladimirkaplun5774]

The most educative part -convergence of the continuous fraction - is still missing.

[@SyberMath]

Can you fill in the gaps? 😊

[@Qermaq]

@@SyberMath I don't know how to put it rigorously, but in my manner of thinking I would look at the partial results as you did mid-video and see how they progress as compared to the value (root2+1) we expect through the previous calculation.
2/1, 5/2, 12/5, 29,12, 70/29, ... b/c, (2b+c)/b, ...
They zig-zag above and below the predicted value, and each is closer. 2 is too small, 2.5 is too big but closer, 2.4 is too small but closer, 2.41667 is too big but closer, and the pattern continues.
If we plotted the absolute value of the partial result from the expected value, we'd find it is continuous and tends to 0. Thus the continuing fraction must continue to converge toward the expected value.
(I think it's like an integral in calculus, but I never took a class so I might have the term wrong.)

[@vladimirkaplun5774]

@@SyberMath Even the phenomena of oscillation of the approximants around the limit deserves to be discussed. As well as very fast convergence in comparison with other approximation.

[@Qermaq]

The end bit is stuff I've wondered about but didn't even understand well enough to ask.
The next "nice" conjugate pair I see that works similarly is root5+2 and root5-2. This gives us the same as here except all the 2s are 4s. In fact any radical in the form root(n^2+1) +/- n will give us this setup but where we have 2 here we'd have 2n.
3 + 1/(3 + 1/(3 + ... - this one is different. The initial methods you used to reach a close value all will work, but that last bit where you backwards-engineer it won't quite work. Or will it?

[@SyberMath]

3 + 1/(3 + 1/(3 + ... = x gives us (3 + sqrt(13))/2. It's easier to work with 2x = 3 + sqrt(13), though
Notice that (sqrt(13) + 3)(sqrt(13) - 3) = 4. This implies
sqrt(13) - 3 = 4/((sqrt(13) + 3)) ***
Start with 3 + sqrt(13)
3 + sqrt(13)
= 6 + sqrt(13) - 3 ...use ***
= 6 + 4/((sqrt(13) + 3))
= 6 + 4/(6 + sqrt(13) - 3) ...use ***
= 6 + 4/(6 + 4/(sqrt(13) + 3)) ...
= 6 + 4/(6 + 4/(6 + 4/...))
Cut this in half and it should be the original expression
hard to divide an infinite expression by 2 but finite cases might help with the pattern.
Consider
3 + 1/(3 + 1/3). Now double this
2[3 + 1/(3 + 1/3)] = 6 + 2/(3 + 1/3) = 6 + 4/(6 + 2/3) ...
The 2 in the last expression will turn into a 4 when 2/(3 + 1/3) is multiplied by 2/2 and this goes on forever...
I hope this helps.
We could probably proceed with (3 + sqrt(13))/2, too but I don't think it'll be this easy.

[@scottleung9587]

Got it!

[@yakupbuyankara5903]

2^(1/2)+1.

[@honestadministrator]

z = k + 1/( k + 1/..) = k + 1/z
z = ( k/2 + √ ( 1 + k^2/4)) is the general solution to these continued fraction

[@yoav613]

You did it before so i will say it again,there is a nive way to solve a(n+1)=2+1/a(n),a(1)=2 by a(n)=b(n+1)/b(n),b(1)=1,b(2)=2 so you will get b(n+2)=2b(n+1)+b(n) and by assumming b(n)=r^n you can find aclosed form for a(n) take the limit to inf and you will find it goes to sqrt2+1😊

[@SyberMath]

Wow! 😍

[@giuseppemalaguti435]

t=2+1/t...t^2-2t-1=0...t=1+√2

[@bobbyheffley4955]

Silver ratio

[@vikramadityakodavalla3795]

first

[@roger7341]

If I enter 2 followed by (1/x, +, 2, =)^n on my calculator the result eventually reaches 2.414213562373095..., which matches 1+√2 to the best of the calculator's ability.