Question

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

Answer 1

Like 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:
\[x=1+\frac{1}{1+x} \\ x-1=\frac{1}{x+1} \\ (x-1)(x+1)-1=0 \\ x^2-1-1=0 \\ x^2-2=0 \\ x=\sqrt{2} \text{ since } x>0\]
but this one seems harder because the top number numbers are changing each time

Answer 2

a long with the side number

Answer 3

like this one is nothing like the other continued fraction
I don't think I can use the same method

Answer 4

e is not algebraic, so i think the same method of solving a polynomial equation wont work for e

Answer 5

http://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

Answer 6

you can probably do it with pade approximants since they are connected to generalized continued fractions

Answer 7

There'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.

Answer 8

\[e^{-x}=\sum_{i=0}^{\infty} \frac{(-x)^i}{i!} \\ e^{-x}=1-x+\frac{x^2}{2}-\frac{x^3}{3!}+\cdots \\ \alpha_1=1 \\ \alpha_2=\frac{1}{x} \\ \alpha_3=\frac{2}{x^2} \\ \alpha_4=\frac{3!}{x^3} \\ \cdots \\ \alpha_n=\frac{(n-1)!}{x^{n-1}} \\ e^{-x}=\frac{1}{1}+\frac{1^2}{\frac{1}{x}-1}+\frac{\frac{1}{x^2}}{\frac{2}{x^2}-\frac{1}{x}}+\frac{\frac{4}{x^4}}{\frac{3!}{x^3}-\frac{2}{x^2}}+\cdots +\frac{\frac{(n-2)!^2}{x^{2(n-2)}}}{\frac{(n-1)!}{x^{n-1}}-\frac{(n-2)!}{x^{n-2}}}+\cdots\]

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

Answer 9

I'm reading euler's formula and it seems the super simple way to get that result :

\[a_0+a_0a_1+\cdots+a_0a_1\cdots a_n = \dfrac{a_0}{1-\dfrac{a_1}{1+a_1-\dfrac{a_2}{1+a_2-\dfrac{\vdots}{\vdots +\dfrac{a_{n-1}}{1+a_{n-1}-\dfrac{a_n}{1+a_n}}}}}} \]

Proving this is easy as induction works pretty good here

Answer 10

but deriving it will be killer probably :(

Answer 11

Not at all,

Base case, \(n=1\) :
\[RHS = \dfrac{a_0}{1-\dfrac{a_1}{1+a_1}} = \dfrac{a_0(1+a_1)}{1+a_1-a_1}=a_0+a_0a_1=LHS\color{green}{\checkmark}\]

Answer 12

\[\text{ suppose } \\ a_0+a_0a_1+\cdots+a_0a_1\cdots a_k = \dfrac{a_0}{1-\dfrac{a_1}{1+a_1-\dfrac{a_2}{1+a_2-\dfrac{\vdots}{\vdots +\dfrac{a_{k-1}}{1+a_{k-1}-\dfrac{a_k}{1+a_k}}}}}} \\ \text{ for some integer } k \ge 0\]

so we add to both sides the following:
\[a_0a_1 \cdots a_k a_{k+1}\]


and then we....hmmm...thinking...

Answer 13

I hope you see why proving by induction is trivial here. Its just algebra manipulation. Maybe lets use the result first :

consider taylor series of \(e^x\),
\[e^x = 1+1*x/1 + 1*x/1*x/2+1*x/2*x/2+x^2/3 + \cdots\]

Answer 14

compare 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

Answer 15

https://en.wikipedia.org/wiki/Euler%27s_continued_fraction_formula

Answer 16

All of these articles make appeals to "equivalence transformations." That has got to be the key to working with these...

Answer 17

*typo

consider taylor series of \(e^x\),
\[e^x = 1+1*x/1 + 1*x/1*x/2+1*x/1*x/2*x/3 + \cdots\]

Answer 18

\[e=\frac{1}{1-\frac{1}{1+1-\frac{1}{2+1-\frac{2(1)}{3+1-\frac{3(1)}{4+1-\cdots }}}}}\]

Answer 19

and I guess we would have do another transformation to get that one continued fraction representation they had on that other site

Answer 20

oh wait, how did u get that

Answer 21

well I used the thingy you typed
and just cleared the compound mini fraction in each sub-fraction if that make sense

Answer 22

I assumed the thingy you were trying to type was the thing from wikepedia

Answer 23

and then they actually have the clearing of the fractions next

Answer 24

one fraction was multiplied by 2/2
another as multiplied by 3/3
... so on...

Answer 25

I think I hate continued fractions
they are so hard to write

Answer 26

that is clever!

Answer 27

it wasn't my cleverness
it was wikepedia's

Answer 28

Oh I missed that from wiki, I only read the statement of theorem and got excited lol

Answer 29

yeah it was from that one link you gave me

Answer 30

but I think I need to play with fractions to see how they actually got that formula/theorem thingy

Answer 31

Induction step is easy, at least for you it will be easy I'm sure...

Answer 32

ok thanks @ganeshie8
I will just be doing some playing of my own

I will ask if I have any questions

Answer 33

And I'm sure I can figure it out with enough playing