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

1. 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

2. freckles

a long with the side number

3. freckles

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

4. ganeshie8

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

5. 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

6. anonymous

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

7. 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.

8. 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

9. 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

10. freckles

but deriving it will be killer probably :(

11. ganeshie8

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

12. 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...

13. 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 $$e^x$$, $e^x = 1+1*x/1 + 1*x/1*x/2+1*x/2*x/2+x^2/3 + \cdots$

14. 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

15. ganeshie8
16. jtvatsim

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

17. ganeshie8

*typo consider taylor series of $$e^x$$, $e^x = 1+1*x/1 + 1*x/1*x/2+1*x/1*x/2*x/3 + \cdots$

18. freckles

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

19. freckles

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

20. ganeshie8

oh wait, how did u get that

21. freckles

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

22. freckles

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

23. freckles

and then they actually have the clearing of the fractions next

24. freckles

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

25. freckles

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

26. ganeshie8

that is clever!

27. freckles

it wasn't my cleverness it was wikepedia's

28. ganeshie8

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

29. freckles

yeah it was from that one link you gave me

30. freckles

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

31. ganeshie8

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

32. freckles

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

33. freckles

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