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

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- freckles

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

- freckles

a long with the side number

- freckles

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- ganeshie8

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

- freckles

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

- anonymous

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

- jtvatsim

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.

- freckles

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

- ganeshie8

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

- freckles

but deriving it will be killer probably :(

- ganeshie8

Not at all,
Base case, \(n=1\) :
\[RHS = \dfrac{a_0}{1-\dfrac{a_1}{1+a_1}} = \dfrac{a_0(1+a_1)}{1+a_1-a_1}=a_0+a_0a_1=LHS\color{green}{\checkmark}\]

- freckles

\[\text{ suppose } \\ a_0+a_0a_1+\cdots+a_0a_1\cdots a_k = \dfrac{a_0}{1-\dfrac{a_1}{1+a_1-\dfrac{a_2}{1+a_2-\dfrac{\vdots}{\vdots +\dfrac{a_{k-1}}{1+a_{k-1}-\dfrac{a_k}{1+a_k}}}}}} \\ \text{ for some integer } k \ge 0\]
so we add to both sides the following:
\[a_0a_1 \cdots a_k a_{k+1}\]
and then we....hmmm...thinking...

- ganeshie8

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

- ganeshie8

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

- ganeshie8

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

- jtvatsim

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

- ganeshie8

*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

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

- freckles

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

- ganeshie8

oh wait, how did u get that

- freckles

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

- freckles

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

- freckles

and then they actually have the clearing of the fractions next

- freckles

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

- freckles

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

- ganeshie8

that is clever!

- freckles

it wasn't my cleverness
it was wikepedia's

- ganeshie8

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

- freckles

yeah it was from that one link you gave me

- freckles

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

- ganeshie8

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

- freckles

ok thanks @ganeshie8
I will just be doing some playing of my own
I will ask if I have any questions

- freckles

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.