Euler's universal formula for continued fractions.

🌟Support the channel🌟
Patreon: / michaelpennmath
Channel Membership: / @michaelpennmath
Merch: teespring.com/...
My amazon shop: www.amazon.com...
🟢 Discord: / discord
🌟my other channels🌟
mathmajor: / @mathmajor
pennpav podcast: / @thepennpavpodcast7878
🌟My Links🌟
Personal Website: www.michael-pen...
Instagram: / melp2718
Twitter: / michaelpennmath
Randolph College Math: www.randolphcol...
Research Gate profile: www.researchga...
Google Scholar profile: scholar.google...
🌟How I make Thumbnails🌟
Canva: partner.canva....
Color Pallet: coolors.co/?re...
🌟Suggest a problem🌟
forms.gle/ea7P...

[@agrajyadav2951]

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?

[@Lucashallal]

Ye

[@keshavrathore5228]

Exactly

[@MarcoMate87]

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/...))))...).

[@victor1978100]

I have been waiting an searching for such a video. Thank you, Michael.

[@zakiabg845]

Is there a general formula of (a+b+c+.....+z)^n/n is an element of N what if it belongs to R?

[@victor1978100]

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.

[@aweebthatlovesmath4220]

So every analytic function can be viewed as a continued fraction?!

[@fartoxedm5638]

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

[@Lucashallal]

Yes, i think you are right

[@Lucashallal]

It should apply

[@thomashoffmann8857]

Because it's a mistake 🙂

[@fartoxedm5638]

@@hatty8071 unfortunately the number of these mistakes have been increasing for the past year and I really don't know why

[@Ahmed-Youcef1959]

@@fartoxedm5638 You don't know why !!! because Michael's video without mistakes will be boring and no one of us like to be bored .

[@DestroManiak]

Consider talking about pade approximants as well!

[@GeoffryGifari]

of course its euler...

[@fupengmou3317]

yeah I figured that mistake too, I also want a small star

[@Aerendil97]

This is great, using this method and plugging x=-1 instead of 1 yields a very natural continued fraction for e, that is much closer to the traditional form of continued fractions (no minus signs). You get e = 2 + 2/(2 + 3/(3 + 4/(4 + ...))). Really wonderful! Perhaps you can then work these continued fractions until finding the traditional continued fraction for e (which would prove that it is a transcendental number!). Since the continued fraction expansion of e is credited to Euler, this might actually be the way he derived it ?!

[@goodplacetostop2973]

13:02

[@franksaved3893]

13:04

[@wyattstevens8574]

"And... that's a *good place to stop* ."

[@kappasphere]

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

[@xizar0rg]

Substituting in "1" for "X" in EXP(x) evaluates the function at the value 1, not at zero.

[@victor1978100]

The formula at 11:13 has no mistakes. If you multiply the denominator correctly you will get a very beautiful fraction.

[@kappasphere]

​@@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

[@CHALKND]

Looks like there’s a window to talk about continued fractions now 🤓😁

[@dragonmudd]

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

[@italyball2166]

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

[@ArminVollmer]

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).

[@krisbrandenberger544]

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.

[@Happy_Abe]

How would one do this with ln(1+x)? The denominators in the sum don’t multiply into eachother

[@ericbischoff9444]

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.

[@Happy_Abe]

@@ericbischoff9444 I see, interesting. Thanks!

[@florankacaku64]

Does the inductive work to prove the infinite version of the formula?

[@diegocastaneda214]

Nice video dude. I love your channel ❤

[@ekinilseven1210]

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.

[@wyattstevens8574]

Definitely a nested product: factor the sum one term at a time to get a_0(1+a_1(1+...)).

[@aadfg0]

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.

[@wyattstevens8574]

All power series (even/odd functions, at least).

[@EconAtheist]

well now that is just cool

[@jakobthomsen1595]

Very nice :D

[@Patapom3]

Amazing!

[@BridgeBum]

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.