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freckles

  • one year ago

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

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  1. freckles
    • one year ago
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    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
    • one year ago
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    a long with the side number

  3. freckles
    • one year ago
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    like this one is nothing like the other continued fraction I don't think I can use the same method

  4. ganeshie8
    • one year ago
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    e is not algebraic, so i think the same method of solving a polynomial equation wont work for e

  5. freckles
    • one year ago
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    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
    • one year ago
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    you can probably do it with pade approximants since they are connected to generalized continued fractions

  7. jtvatsim
    • one year ago
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    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
    • one year ago
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    \[e^{-x}=\sum_{i=0}^{\infty} \frac{(-x)^i}{i!} \\ e^{-x}=1-x+\frac{x^2}{2}-\frac{x^3}{3!}+\cdots \\ \alpha_1=1 \\ \alpha_2=\frac{1}{x} \\ \alpha_3=\frac{2}{x^2} \\ \alpha_4=\frac{3!}{x^3} \\ \cdots \\ \alpha_n=\frac{(n-1)!}{x^{n-1}} \\ e^{-x}=\frac{1}{1}+\frac{1^2}{\frac{1}{x}-1}+\frac{\frac{1}{x^2}}{\frac{2}{x^2}-\frac{1}{x}}+\frac{\frac{4}{x^4}}{\frac{3!}{x^3}-\frac{2}{x^2}}+\cdots +\frac{\frac{(n-2)!^2}{x^{2(n-2)}}}{\frac{(n-1)!}{x^{n-1}}-\frac{(n-2)!}{x^{n-2}}}+\cdots\] 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
    • one year ago
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    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
    • one year ago
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    but deriving it will be killer probably :(

  11. ganeshie8
    • one year ago
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    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
    • one year ago
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    \[\text{ suppose } \\ a_0+a_0a_1+\cdots+a_0a_1\cdots a_k = \dfrac{a_0}{1-\dfrac{a_1}{1+a_1-\dfrac{a_2}{1+a_2-\dfrac{\vdots}{\vdots +\dfrac{a_{k-1}}{1+a_{k-1}-\dfrac{a_k}{1+a_k}}}}}} \\ \text{ for some integer } k \ge 0\] so we add to both sides the following: \[a_0a_1 \cdots a_k a_{k+1}\] and then we....hmmm...thinking...

  13. ganeshie8
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    https://en.wikipedia.org/wiki/Euler%27s_continued_fraction_formula

  16. jtvatsim
    • one year ago
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    All of these articles make appeals to "equivalence transformations." That has got to be the key to working with these...

  17. ganeshie8
    • one year ago
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    *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
    • one year ago
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    \[e=\frac{1}{1-\frac{1}{1+1-\frac{1}{2+1-\frac{2(1)}{3+1-\frac{3(1)}{4+1-\cdots }}}}}\]

  19. freckles
    • one year ago
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    and I guess we would have do another transformation to get that one continued fraction representation they had on that other site

  20. ganeshie8
    • one year ago
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    oh wait, how did u get that

  21. freckles
    • one year ago
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    well I used the thingy you typed and just cleared the compound mini fraction in each sub-fraction if that make sense

  22. freckles
    • one year ago
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    I assumed the thingy you were trying to type was the thing from wikepedia

  23. freckles
    • one year ago
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    and then they actually have the clearing of the fractions next

  24. freckles
    • one year ago
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    one fraction was multiplied by 2/2 another as multiplied by 3/3 ... so on...

  25. freckles
    • one year ago
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    I think I hate continued fractions they are so hard to write

  26. ganeshie8
    • one year ago
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    that is clever!

  27. freckles
    • one year ago
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    it wasn't my cleverness it was wikepedia's

  28. ganeshie8
    • one year ago
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    Oh I missed that from wiki, I only read the statement of theorem and got excited lol

  29. freckles
    • one year ago
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    yeah it was from that one link you gave me

  30. freckles
    • one year ago
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    but I think I need to play with fractions to see how they actually got that formula/theorem thingy

  31. ganeshie8
    • one year ago
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    Induction step is easy, at least for you it will be easy I'm sure...

  32. freckles
    • one year ago
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    ok thanks @ganeshie8 I will just be doing some playing of my own I will ask if I have any questions

  33. freckles
    • one year ago
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    And I'm sure I can figure it out with enough playing

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