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#19 by N. J. A. Sloane at Thu Aug 01 00:16:16 EDT 2019
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#18 by N. J. A. Sloane at Thu Aug 01 00:16:14 EDT 2019
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| NAME
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Numerators of approximation to 2^(1/3) by Newton's method atafter n iterations.
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| STATUS
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approved
editing
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#17 by N. J. A. Sloane at Thu Aug 01 00:15:16 EDT 2019
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#16 by N. J. A. Sloane at Thu Aug 01 00:14:59 EDT 2019
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| NAME
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Numerators of approximation ofto 2^(1/3) by Newton's method at n iterations.
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| EXAMPLE
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Approximations ofto 2^(1/3):
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| STATUS
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proposed
editing
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Discussion
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| Thu Aug 01
| 00:15
| N. J. A. Sloane: approximations to sounds better
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#15 by Jon E. Schoenfield at Wed Jul 31 22:26:25 EDT 2019
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#14 by Jon E. Schoenfield at Wed Jul 31 22:26:22 EDT 2019
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| NAME
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Numerators of approximation of cuberoot(2^(1/3) by Newton's method at n iterations.
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| FORMULA
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a(n) = x(n)/GCDgcd(x(n),y(n)).))
where x(n) = 2*(a(n-1)^3*A248042(n-1)^2 + A248042(n-1)^5); y(n) = 3*a(n-1)^2*A248042(n-1)^3; x(0) = y(0) = 1.)
and y(n) = 3*a(n-1)^2*A248042(n-1)^3;
x(0) = y(0) = 1.
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| EXAMPLE
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CubeRoot(2) approximations:
Approximations of 2^(1/3):
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approved
editing
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#13 by N. J. A. Sloane at Thu Jan 15 12:36:56 EST 2015
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#12 by Kival Ngaokrajang at Sun Jan 11 19:47:44 EST 2015
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#11 by Kival Ngaokrajang at Sun Jan 11 19:45:56 EST 2015
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Discussion
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| Sun Jan 11
| 19:47
| Kival Ngaokrajang: Added x(0)=y(0)=1 & cf A002580. Thank you.
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#10 by Jon E. Schoenfield at Sun Jan 11 15:51:04 EST 2015
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Discussion
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| Sun Jan 11
| 16:42
| Robert Israel: What are x(0) and y(0)?
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