Euler's universal formula for continued fractions.
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- Опубліковано 5 лип 2023
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I think the series of e^x should be 1/(1-x/((1+x)-2x/((2+x)-3x/((3+x)-4x/...))))...). The denominators from the terms should multiply to the numerators of the next dividing thing, right?
Ye
Exactly
I think Michael wrote wrong and you wrote wrong too. The correct formula should be 1/(1-x/((1+x)-x/((2+x)-2x/((3+x)-3x/...))))...).
11:26 Shouldn't we also multiply (1/3)x by 2 and then multiply the fraction by 3? So, we will have 2x in the numerator.
11:18 why didn't 2 apply to the x/3 term(which is then turned into x)? The same question with 3, 4, etc
Yes, i think you are right
It should apply
Because it's a mistake 🙂
@@hatty8071 unfortunately the number of these mistakes have been increasing for the past year and I really don't know why
@@fartoxedm5638
You don't know why !!!
because Michael's video without mistakes will be boring and no one of us like to be bored .
I have been waiting an searching for such a video. Thank you, Michael.
So every analytic function can be viewed as a continued fraction?!
Consider talking about pade approximants as well!
Nice video dude. I love your channel ❤
yeah I figured that mistake too, I also want a small star
Timely! You just gave me a new tool to play around with. I thought of a problem the other day that needs fast convergence for tan^2 (or sec^2) for programming reasons, i may need to see if this approach helps.
Amazing!
Hey, Michael! So an easy way to find out what 1/x is (where you wrote the x in blue) is to use Euler's Continued Function Formula, where the role of a_0 is being played by 1 and the roles of the other a_i's are being played by
a_(i+1)'s.
well now that is just cool
Very nice :D
12:15 According to the formula of e^x that was just derived, wouldn't that expression evaluate to e^0 (i.e. 1) instead of e? It kind of makes sense considering that the derived formula has a mistake, but I'm still wondering how the jump to the formula for e was made
Substituting in "1" for "X" in EXP(x) evaluates the function at the value 1, not at zero.
The formula at 11:13 has no mistakes. If you multiply the denominator correctly you will get a very beautiful fraction.
@@xizar0rgYou're right, I didn't see that the +1 was already collapsed into the 1 2 3... pattern to shift it up by one level so I thought that he added 0 instead. That explains how he got there, thanks
Finite fraction formula looked trivial but then the application to infinite fractions and continued fraction for e vindicated the video. I wonder how many other continued fractions can be derived from such a humble method.
All power series (even/odd functions, at least).
Is there a general formula of (a+b+c+.....+z)^n/n is an element of N what if it belongs to R?
13:02
13:04
"And... that's a *good place to stop* ."
Does the inductive work to prove the infinite version of the formula?
Looks like there’s a window to talk about continued fractions now 🤓😁
Nice! What Michael is actually showing here, expressed in Wolfram Language, is the equivalence of sumOfProducts[k_] :=Sum[Product[a[i], {i, 0, n}], {n, 0, k}] (LHS) and contFraction[k_] :=
a[0]/(1 - Fold[#2/(1 + #2 - #1) &, 0, Reverse@Array[a, k]]) (RHS).
How would one do this with ln(1+x)?
The denominators in the sum don’t multiply into eachother
you multiply preceding term with (-1)(n-1)/n .... that's not as clean as with a factorial, but maybe this leads somewhere? I did not try, to be honest.
@@ericbischoff9444 I see, interesting. Thanks!
Is there a way to write the LHS which is a sum of products as product of sums? As in, sum_(n=0)^N prod_(i=0)^n a_i = prod_(n=0)^N sum_(i=0)^n a'_i ?? Potentially the indices, limits, and the series might change.
Definitely a nested product: factor the sum one term at a time to get a_0(1+a_1(1+...)).
of course its euler...
Was hoping we could use this to show that 1+2+3+4+... = -1/12 but alas I don't think we can pull it off
I doubt you can do it. First of all, that result is gotten through some compromises (saying that the sequence 1,0,1,0,1... converges to 1/2 for example), also you'd need to find a way to write every natural number as a recursive product