√2 as a fraction
 

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

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 data
 

look.

Cheers

Matt