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freckles
 one year ago
Does anyone know how to show \[e=2+\frac{1}{1+\frac{1}{2+\frac{2}{3+\frac{3}{4+\frac{4}{5+\cdots}}}}}\]
freckles
 one year ago
Does anyone know how to show \[e=2+\frac{1}{1+\frac{1}{2+\frac{2}{3+\frac{3}{4+\frac{4}{5+\cdots}}}}}\]

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freckles
 one year ago
Best ResponseYou've already chosen the best response.1Like showing \[\sqrt{2}=1+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\cdots }}}}\] is easy like I know that I can called that whole continued fraction thingy x but then I see that whole continued fraction thingy again inside so I know I can do this: \[x=1+\frac{1}{1+x} \\ x1=\frac{1}{x+1} \\ (x1)(x+1)1=0 \\ x^211=0 \\ x^22=0 \\ x=\sqrt{2} \text{ since } x>0\] but this one seems harder because the top number numbers are changing each time

freckles
 one year ago
Best ResponseYou've already chosen the best response.1a long with the side number

freckles
 one year ago
Best ResponseYou've already chosen the best response.1like this one is nothing like the other continued fraction I don't think I can use the same method

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2e is not algebraic, so i think the same method of solving a polynomial equation wont work for e

freckles
 one year ago
Best ResponseYou've already chosen the best response.1http://webserv.jcu.edu/math//vignettes/continued.htm in this link, it says euler used this technique I wonder exactly which technique they are referring to

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you can probably do it with pade approximants since they are connected to generalized continued fractions

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.0There's a similar identity and derivation given here http://www.math.binghamton.edu/dikran/478/Ch7.pdf ending at PDF page 14. Not exactly the same identity given by the OP, but both are very cool.

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[e^{x}=\sum_{i=0}^{\infty} \frac{(x)^i}{i!} \\ e^{x}=1x+\frac{x^2}{2}\frac{x^3}{3!}+\cdots \\ \alpha_1=1 \\ \alpha_2=\frac{1}{x} \\ \alpha_3=\frac{2}{x^2} \\ \alpha_4=\frac{3!}{x^3} \\ \cdots \\ \alpha_n=\frac{(n1)!}{x^{n1}} \\ e^{x}=\frac{1}{1}+\frac{1^2}{\frac{1}{x}1}+\frac{\frac{1}{x^2}}{\frac{2}{x^2}\frac{1}{x}}+\frac{\frac{4}{x^4}}{\frac{3!}{x^3}\frac{2}{x^2}}+\cdots +\frac{\frac{(n2)!^2}{x^{2(n2)}}}{\frac{(n1)!}{x^{n1}}\frac{(n2)!}{x^{n2}}}+\cdots\] trying to following the arctan(x) example in 7.6 ... and using some of the theorems above though this so far not making sense anyways... still playing with what they have

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2I'm reading euler's formula and it seems the super simple way to get that result : \[a_0+a_0a_1+\cdots+a_0a_1\cdots a_n = \dfrac{a_0}{1\dfrac{a_1}{1+a_1\dfrac{a_2}{1+a_2\dfrac{\vdots}{\vdots +\dfrac{a_{n1}}{1+a_{n1}\dfrac{a_n}{1+a_n}}}}}} \] Proving this is easy as induction works pretty good here

freckles
 one year ago
Best ResponseYou've already chosen the best response.1but deriving it will be killer probably :(

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Not at all, Base case, \(n=1\) : \[RHS = \dfrac{a_0}{1\dfrac{a_1}{1+a_1}} = \dfrac{a_0(1+a_1)}{1+a_1a_1}=a_0+a_0a_1=LHS\color{green}{\checkmark}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\text{ suppose } \\ a_0+a_0a_1+\cdots+a_0a_1\cdots a_k = \dfrac{a_0}{1\dfrac{a_1}{1+a_1\dfrac{a_2}{1+a_2\dfrac{\vdots}{\vdots +\dfrac{a_{k1}}{1+a_{k1}\dfrac{a_k}{1+a_k}}}}}} \\ \text{ for some integer } k \ge 0\] so we add to both sides the following: \[a_0a_1 \cdots a_k a_{k+1}\] and then we....hmmm...thinking...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2I hope you see why proving by induction is trivial here. Its just algebra manipulation. Maybe lets use the result first : consider taylor series of \(e^x\), \[e^x = 1+1*x/1 + 1*x/1*x/2+1*x/2*x/2+x^2/3 + \cdots\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2compare that with the left hand side and simply plug the values : \[\begin{align}e^x&=1+1*x/1+1*x/1*x/2+\cdots\\~\\ & = \dfrac{1}{1\dfrac{x}{1+x\dfrac{x/2}{1+x/2\vdots }}} \end{align}\] plugging in \(x=1\), we do get a continued fraction for \(e\) but oops! this is not what we're looking for hmm

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2https://en.wikipedia.org/wiki/Euler%27s_continued_fraction_formula

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.0All of these articles make appeals to "equivalence transformations." That has got to be the key to working with these...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2*typo consider taylor series of \(e^x\), \[e^x = 1+1*x/1 + 1*x/1*x/2+1*x/1*x/2*x/3 + \cdots\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[e=\frac{1}{1\frac{1}{1+1\frac{1}{2+1\frac{2(1)}{3+1\frac{3(1)}{4+1\cdots }}}}}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1and I guess we would have do another transformation to get that one continued fraction representation they had on that other site

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2oh wait, how did u get that

freckles
 one year ago
Best ResponseYou've already chosen the best response.1well I used the thingy you typed and just cleared the compound mini fraction in each subfraction if that make sense

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I assumed the thingy you were trying to type was the thing from wikepedia

freckles
 one year ago
Best ResponseYou've already chosen the best response.1and then they actually have the clearing of the fractions next

freckles
 one year ago
Best ResponseYou've already chosen the best response.1one fraction was multiplied by 2/2 another as multiplied by 3/3 ... so on...

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I think I hate continued fractions they are so hard to write

freckles
 one year ago
Best ResponseYou've already chosen the best response.1it wasn't my cleverness it was wikepedia's

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Oh I missed that from wiki, I only read the statement of theorem and got excited lol

freckles
 one year ago
Best ResponseYou've already chosen the best response.1yeah it was from that one link you gave me

freckles
 one year ago
Best ResponseYou've already chosen the best response.1but I think I need to play with fractions to see how they actually got that formula/theorem thingy

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Induction step is easy, at least for you it will be easy I'm sure...

freckles
 one year ago
Best ResponseYou've already chosen the best response.1ok thanks @ganeshie8 I will just be doing some playing of my own I will ask if I have any questions

freckles
 one year ago
Best ResponseYou've already chosen the best response.1And I'm sure I can figure it out with enough playing
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