# problem about Root and Hypergeometric2F1

See these example, why does the output is different?

expr = Root[-7 a^4 #1^2 - 2 a^2 #1^4 + #1^6 &, 4];
expr /. a -> 1 // N
ToRadicals@expr /. a -> 1 // N


1.95663668695703

0 . - 1.35219344945396 I

Table[Hypergeometric2F1[1/2-n/2,1-n/2,1-n,-4],{n,5}]
Table[Hypergeometric2F1[1/2-n/2,1-n/2,1-n,-4]//FullSimplify//Evaluate,{n,5}]//RootReduce


{1, 1, 2, 3, 5}

{1/10 (5 + Sqrt), 1/10 (5 + 3 Sqrt), 1/5 (5 + 2 Sqrt), 1/10 (15 + 7 Sqrt), 1/10 (25 + 11 Sqrt)}

• The ordering of the roots depends on a. So I think the order of operations not being commutative is probably an explanation of the first difference. (See Root, "Details".) – Michael E2 Aug 19 '15 at 16:52

Part1.

The ordering of the roots and consequently which is the fourth root depends on when a is given its value.

Table[Root[-7 a^4 #1^2 - 2 a^2 #1^4 + #1^6 &, n] // ToRadicals, {n, 6}] /.
a -> 1.


{0, 0, 0. + 1.35219 I, 0. - 1.35219 I, 1.95664, -1.95664}

Table[Root[-7 a^4 #1^2 - 2 a^2 #1^4 + #1^6 &, n] /. a -> 1. // ToRadicals, {n,
6}]


{-1.95664, 0., 0., 1.95664, 0. - 1.35219 I, 0. + 1.35219 I}

Part 2.

Clear[f]

f[n_, x_] = Hypergeometric2F1[1/2 - n/2, 1 - n/2, 1 - n, x];


From the definition of Hypergeometric2F1 you can see that the sum f[n, -4] does not converge in general

Sum[Pochhammer[1/2 - n/2, m] Pochhammer[1 - n/2, m]/
Pochhammer[1 - n, m] (-4)^m/m!, {m, 0, Infinity}]


Sum::div: Sum does not converge. >>

Sum[((-4)^m*Pochhammer[1/2 - n/2, m]*Pochhammer[1 - n/2, m])/ (m!*Pochhammer[1 - n, m]), {m, 0, Infinity}]

However, it has a "Borel" regularized value

Sum[Pochhammer[1/2 - n/2, m] Pochhammer[1 - n/2, m]/
Pochhammer[1 - n, m] (-4)^m/m!, {m, 0, Infinity},
Regularization -> "Borel"]


((1/2)*(1 + Sqrt))^n/Sqrt

This is the value used by FunctionExpand or FullSimplify

f[n, -4] // FunctionExpand // Simplify


((1/2)*(1 + Sqrt))^n/Sqrt

f[n, -4] // FullSimplify


((1/2)*(1 + Sqrt))^n/Sqrt

However, for n an explicit positive integer the Pochhammer symbols in the numerator stop the sum with a zero term and results in a polynomial

Table[{n, f[n, x]}, {n, 5}]


{{1, 1}, {2, 1}, {3, (4 - x)/4}, {4, (2 - x)/2}, {5, (1/16)*(16 - 12*x + x^2)}}

list1 = (% /. x -> -4)


{{1, 1}, {2, 1}, {3, 2}, {4, 3}, {5, 5}}

These polynomials are truncations of (hence not equal to) the Borel regularized infinite series

list2 = Transpose[{Range,
Table[f[n, -4] // FunctionExpand // Simplify // Evaluate, {n, 5}] //
RootReduce // Simplify}]


{{1, (1/10)(5 + Sqrt)}, {2, (1/10)(5 + 3*Sqrt)}, {3, 1 + 2/Sqrt}, {4, 3/2 + 7/(2*Sqrt)}, {5, 5/2 + 11/(2*Sqrt)}}

This approaches the integer values of n as a limit

list2 == Table[{m, Limit[f[n, -4] // FullSimplify // Evaluate, n -> m]}, {m,
5}]


True

Plot[f[n, -4], {n, 0, 5},
Epilog -> {AbsolutePointSize, Red, Point[list1], Blue, Point[list2]}] Note that the polynomial gets closer to the Borel regularized infinite series as n increases, i.e., as the order (number of terms) of the polynomial increases.

• More conventionally, one uses the integral representation of the Gaussian hypergeometric function as the analytic continuation off the unit disk. – J. M.'s technical difficulties Aug 19 '15 at 18:51