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A306648
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Decimal expansion of 3 + 36/(5*sqrt(3)*Pi) + 48/(5*Pi^2).
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2
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5, 2, 9, 5, 8, 7, 2, 7, 1, 1, 9, 7, 8, 7, 4, 3, 5, 2, 4, 6, 8, 8, 4, 2, 6, 4, 4, 2, 7, 6, 4, 5, 1, 1, 9, 9, 2, 8, 3, 5, 5, 7, 1, 7, 8, 9, 7, 3, 3, 5, 0, 4, 4, 3, 5, 6, 3, 6, 3, 2, 5, 1, 5, 4, 8, 8, 2, 5, 3, 4, 6, 8, 9, 5, 6, 3, 7, 0, 7, 7, 4, 3, 0, 5, 2, 5, 2, 1, 0, 9, 8, 8, 6, 1, 5, 7, 9, 7, 7, 5
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OFFSET
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1,1
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COMMENTS
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This is the mean square end-to-end distance of the 3-step self-avoiding walk with full excluded volume in the 2-dimensional continuum.
Take 4 touching circles of diameter 1 which are joined as a chain and each is free to move around its neighbors' perimeters, but no circle can overlap another. This value is the average of the square of the distance from the middle of the first circle to the middle of the fourth circle, averaged over all possible configurations the chain of 4 non-overlapping circles can take.
To derive the exact value consider the 4-circle chain where each circle has diameter 1. Let t1 be the angle between the first and third circles centered at the second circle, and t2 be the angle between the second and fourth circles centered at the third circle. Using the law of cosines and basic geometry one can show the square of the distance between the centers of the first and fourth circles, r2_4, can be written as r2_4 = 4*sin^2(t1/2) - 4*sin(t1/2)*sin(t1/2+t2) + 1. When 2*Pi/3 <= t1 <= Pi the fourth circle cannot overlap the first, so t2 is free to vary over its full range of Pi/3 to 5*Pi/3. When Pi/3 <= t1 < 2*Pi/3 it can overlap, so the lower bound of t2 is restricted to the angle at which the fourth circle touches the first -- this corresponds to t2 = Pi - t1. To average over all the t1 and t2 angles we must therefore calculate the two t1 ranges separately. The required integral for the mean square distance thus becomes <r2_4> = ( Integrate(r2_4,{t2,Pi/3,5*Pi/3},{t1,2*Pi/3,Pi}) + Integrate(r2_4,{t2,Pi-t1,5*Pi/3},{t1,Pi/3,2*Pi/3}) ) / ( Integrate(1,{t2,Pi/3,5*Pi/3},{t1,2*Pi/3,Pi}) + Integrate(1,{t2,Pi-t1,5*Pi/3},{t1,Pi/3,2*Pi/3}) ), which includes division by the required normalization integrals. Solving this definite integral gives the exact value for <r2_4> as 3 + 36/(5*sqrt(3)*Pi) + 48/(5*Pi^2). Note that to find the mean distance, and also the mean square distance for the 5-circle chain, requires integration of sqrt(r2_4) which is a non-elementary integral so only numerical approximations are possible -- these values are approximately 2.25134... and 8.27291... respectively.
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REFERENCES
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Scott R. Shannon, The Two Dimensional Polymer Chain - Statics and Dynamics, PhD Thesis, Monash University, Melbourne, Australia, (1997).
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LINKS
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EXAMPLE
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5.295872711978743524688426442764511992835571789733504435636...
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MAPLE
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evalf[120](3+36/(5*Pi*sqrt(3))+48/(5*Pi^2)); # Muniru A Asiru, Mar 05 2019
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MATHEMATICA
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First@ RealDigits[3 + 36/(5 Sqrt[3] Pi) + 48/(5 Pi^2), 10, 105] (* Michael De Vlieger, Mar 11 2019 *)
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PROG
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(PARI) 3+36/(5*sqrt(3)*Pi)+48/(5*Pi^2) \\ Michel Marcus, Mar 04 2019
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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