- **MattcAnderson**
(*https://www.mersenneforum.org/forumdisplay.php?f=146*)

- - **√2 as a fraction**
(*https://www.mersenneforum.org/showthread.php?t=27320*)

√2 as a fraction2 Attachment(s)
Hi all,
One of the scanned pages is upside down, but you can print it out if you want. Regards, Matt |

@Matt - Here's an easy construction for square roots approximations of any arbitrary numbers. No need for matrices.
Use [URL="https://en.wikipedia.org/wiki/Newton%27s_method"]Newton's method[/URL] for solving f(x)=x[SUP]2[/SUP]-a=0. You know f'(x). It is 2x. x[SUB]new[/SUB] = x - f(x)/f'(x) = x - (x^2-a)/(2x) = (x^2+a)/2x [SPOILER]...or (x+a/x)/2 as frequently taught in schools[/SPOILER] For [$]\sqrt 2[/$]: use a=2 and apply this repeatedly: [CODE]a=2; x=1; x=(x+a/x)/2 3/2 x=(x+a/x)/2 17/12 x=(x+a/x)/2 577/408 x=(x+a/x)/2 665857/470832 x=(x+a/x)/2 886731088897/627013566048 x=(x+a/x)/2 1572584048032918633353217/1111984844349868137938112[/CODE] For [$]\sqrt 10[/$]: use a=10 and apply this repeatedly: [CODE]a=10; x=3; x=(x+a/x)/2 19/6 x=(x+a/x)/2 721/228 x=(x+a/x)/2 1039681/328776 x=(x+a/x)/2 2161873163521/683644320912 x=(x+a/x)/2 9347391150304592810234881/2955904621546382351702304 ...[/CODE] Now, try the same to get fast approximation of a cubic root of 2: x[SUB]new[/SUB] = x - f(x)/f'(x) = x - (x[SUP]3[/SUP]-a)/(3x[SUP]2[/SUP]) = (2x^3+a)/(3x^2) ... |

[QUOTE=MattcAnderson;593086]Hi all,
One of the scanned pages is upside down, but you can print it out if you want. Regards, Matt[/QUOTE]Continued fraction for sqrt 2 is 1;2. |

If n is a positive integer and d is a divisor of n, the simple continued fraction for [tex]\sqrt{n^{2}+d}[/tex] is
n, 2n/d, 2n, 2n/d, 2n, 2n/d,... |

Thanks Batalov and others, Some of us are 'into' math and computers. I appreciate the effort.
AS a next step. Look at a fraction for square root of 3. I have not memorized that the square root of 3 is shown to be sqrt(3) = 1.732050808. minus some error due to the fact that the square root of 3 is an irrational number. I am not ashamed to share this with you all. Matt |

1 Attachment(s)
I did a little copying of the definition of continued fraction from Wikipedia. Thank you for showing that to me.
Regards, Matt I assume that the infinite continued fraction for the square root of 2 is 1+1/(2 + 1/(2 + ...)). |

[QUOTE=MattcAnderson;593129]I did a little copying of the definition of continued fraction from Wikipedia. Thank you for showing that to me.
Regards, Matt I assume that the infinite continued fraction for the square root of 2 is 1+1/(2 + 1/(2 + ...)).[/QUOTE]You assume correctly. |

[QUOTE=Dr Sardonicus;593108]If n is a positive integer and d is a divisor of n, the simple continued fraction for [tex]\sqrt{n^{2}+d}[/tex] is
n, 2n/d, 2n, 2n/d, 2n, 2n/d,...[/QUOTE]That should read n; 2n/d, 2n, ... in conventional notation. The ; is the continued fraction equivalent to the decimal point. |

Thank you for that typing and effort @Batalov
I know that requires some effort and learning and typing. As a lifetime member of The Mathematics Association of America, I just thought I would share. Again thanks. For what it's worth, *griz* |

some more data1 Attachment(s)
look.
Cheers Matt |

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