[math-fun] identity from a stranger
This just came in from a total stranger. Anyone care to answer it? NJAS
From efknightjr@hotmail.com Fri Feb 21 07:02:09 2003 Delivered-To: njas@research.att.com X-Authentication-Warning: mail-red.research.att.com: postfixfilter set sender to efknightjr@hotmail.com using -f X-Originating-IP: [64.40.53.47] From: "EFKnightJr" <EFKnightJr@hotmail.com> To: <njas@research.att.com> Subject: Arctan(Phi) Date: Fri, 21 Feb 2003 04:08:03 -0800 X-Priority: 3 X-MSMail-Priority: Normal X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2600.0000 X-OriginalArrivalTime: 21 Feb 2003 12:02:02.0197 (UTC) FILETIME=[0B472850:01C2D9A1] X-Spam-Status: No, hits=-95.8 required=5.0 tests=DEAR_SOMEBODY,DEAR_SOMETHING,HTML_WITH_BGCOLOR,MAILTO_LINK, SPAM_PHRASE_05_08,USER_AGENT_OE,USER_IN_ALL_SPAM_TO version=2.43-cvs
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Dear Sir,
Can you tell me if the following series expression is a new result?
arctan(Phi) =3D (1/2) * Sigma(k=3D0 to Infinity) ( ((-1/4)^k) * ( 1/(4k+1) + 2/(4k+2) + (1/2)/(4k+3) ) )
where Phi is the Golden Ratio and Sigma is summation notation.
I developed the expression after investigating a cone associated with the dodecahedron. The cone's base is the circle inscribed in a face of the dodecahedron, and the cone's vertex is the center of the dodecahedron. The angle between the cone's axis and side is (1/2)*arctan(2), and the tangent of that angle is the reciprocal of Phi.
That leads to an expression relating Phi, Pi, and arctan(2), namely arctan(Phi) =3D (1/2) * ( Pi - arctan(2) )
To obtain the expression in question at the start of this email, I substituted in expressions for Pi and for arctan(2) published in an article by Victor Adamchik.
The same type of reasoning also leads to a similar expression useful for base 16 computations, namely
arctan(Phi) =3D (1/2) * Sigma(k=3D0 to Infinity) ( ((1/16^k)) * ( 3/(8k+1) - (1/2)/(8k+3) - 2/(8k+4) - (3/4)/(8k+5) - 1/(8k+6) + (1/8)/(8k+7) ) )
Again, I would like to know if this is a new result.
Thank your for your time.
Sincerely,
Jack R. Bench, Jr. Email alias: EFKnightJr@Hotmail.Com
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<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <HTML><HEAD> <META http-equiv=3DContent-Type content=3D"text/html; = charset=3Diso-8859-1"> <META content=3D"MSHTML 6.00.2600.0" name=3DGENERATOR> <STYLE></STYLE> </HEAD> <BODY bgColor=3D#c0c0c0><FONT face=3DArial size=3D2> <DIV><BR>Dear Sir,</DIV> <DIV> </DIV> <DIV>Can you tell me if the following series expression<BR>is a new=20 result?</DIV> <DIV> </DIV> <DIV>arctan(Phi) =3D (1/2) * Sigma(k=3D0 to Infinity)<BR>( ((-1/4)^k) = *<BR>(=20 1/(4k+1) + 2/(4k+2) + (1/2)/(4k+3) ) )</DIV> <DIV> </DIV> <DIV>where Phi is the Golden Ratio and Sigma is summation = notation.</DIV> <DIV> </DIV> <DIV>I developed the expression after investigating a cone<BR>associated = with=20 the dodecahedron. The cone's base is the<BR>circle inscribed in a face = of the=20 dodecahedron, and the<BR>cone's vertex is the center of the = dodecahedron. The=20 angle<BR>between the cone's axis and side is (1/2)*arctan(2), and<BR>the = tangent=20 of that angle is the reciprocal of Phi.</DIV> <DIV> </DIV> <DIV>That leads to an expression relating Phi, Pi, and = arctan(2),<BR>namely=20 arctan(Phi) =3D (1/2) * ( Pi - arctan(2) )</DIV> <DIV> </DIV> <DIV>To obtain the expression in question at the start of<BR>this email, = I=20 substituted in expressions for Pi and for<BR>arctan(2) published in an = article=20 by Victor Adamchik.</DIV> <DIV> </DIV> <DIV>The same type of reasoning also leads to a similar<BR>expression = useful for=20 base 16 computations, namely</DIV> <DIV> </DIV> <DIV>arctan(Phi) =3D (1/2) * Sigma(k=3D0 to Infinity)<BR>( ((1/16^k)) = *<BR>(=20 3/(8k+1) - (1/2)/(8k+3) - 2/(8k+4) -<BR>(3/4)/(8k+5) - 1/(8k+6) + = (1/8)/(8k+7) )=20 )</DIV> <DIV> </DIV> <DIV>Again, I would like to know if this is a new result.</DIV> <DIV> </DIV> <DIV>Thank your for your time.</DIV> <DIV> </DIV> <DIV>Sincerely,</DIV> <DIV> </DIV> <DIV>Jack R. Bench, Jr.<BR>Email alias: <A=20 href=3D"mailto:EFKnightJr@Hotmail.Com">EFKnightJr@Hotmail.Com</A><BR></FO= NT></DIV></BODY></HTML>
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Assuming he(?) means inf ==== k 1 \ 1 1 2 1 - > (- -) (----------- + ------- + -------), 2 / 4 2 (4 k + 3) 4 k + 2 4 k + 1 ==== k = 0 my Macsyma series package gets atan(2)/2 + atan(1/2), as would Maple and Mma, presumably. It isn't hard to show this is atan(phi), but neither is it an obvious simplification. --rwgOC
Mma prefers Pi/8 + ArcCot[2] + ArcCot[3]/2 ----- Original Message ----- From: "R. William Gosper" <rwg@spnet.com> To: <math-fun@mailman.xmission.com> Sent: Monday, February 24, 2003 9:17 PM Subject: Re: [math-fun] identity from a stranger
Assuming he(?) means inf ==== k 1 \ 1 1 2 1 - > (- -) (----------- + ------- + -------), 2 / 4 2 (4 k + 3) 4 k + 2 4 k + 1 ==== k = 0
my Macsyma series package gets atan(2)/2 + atan(1/2), as would Maple and Mma, presumably. It isn't hard to show this is atan(phi), but neither is it an obvious simplification. --rwgOC
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Simplify the following fractions using nonstandard mathematics: 13/325, 143/242, 154/253, 26/65, 176/275, 187/286, 19/95, 124/217, 148/185, 187/748 I remember seeing fractions of this sort in a book privately published by Paul Yearout, and I know I've seen them elsewhere. Is there a definitive paper on these sorts of fractions? I didn't find a nice one where just 1's or 2's got eliminated. --Ed Pegg Jr.
This reminds me of my favorite proof of the Remainder Theorem: f( ) -------------- x - a ) f(x) f(x) - f(a) ------------- f(a) R. On Mon, 24 Feb 2003, Ed Pegg Jr wrote:
Simplify the following fractions using nonstandard mathematics:
13/325, 143/242, 154/253, 26/65, 176/275, 187/286, 19/95, 124/217, 148/185, 187/748
I remember seeing fractions of this sort in a book privately published by Paul Yearout, and I know I've seen them elsewhere. Is there a definitive paper on these sorts of fractions?
I didn't find a nice one where just 1's or 2's got eliminated.
--Ed Pegg Jr.
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participants (5)
-
Ed Pegg Jr -
N. J. A. Sloane -
R. William Gosper -
Richard Guy -
wouter meeussen