函数逼近求最佳a值。

楼上代码运行后得：最小化最大误差的最佳 a=46.4826……
可实际上a=46.449……
另外楼上代码运行有点卡，不知道怎么改代码使之运行快一点？
以上情况还望前辈们指导一二。

用Mathematica检验所求的最大误差值是不是最小值：
NumberForm[N[NMaximize[{Abs[Hypergeometric2F1[-(1/2), -(1/2), 1, x^2] -
(1 + (3 x^2)/(
10 + Sqrt[
4 - 3 x^2])) (1 + (22/(7 \[Pi]) - 1) ((2 x)/(1 + x))^
46.4483021270782)], 0 < x < 1}, x]], 10]

特别的，当a≈46.4484 误差=0.00001872138763348019
NumberForm[N[NMaximize[{Abs[Hypergeometric2F1[-(1/2), -(1/2), 1, x^2] -
(1 + (3 x^2)/(
10 + Sqrt[
4 - 3 x^2])) (1 + (22/(7 \[Pi]) - 1) ((2 x)/(1 + x))^
46.4484)], 0 < x < 1}, x]], 10]

从图片上显示误差
a=46.4826，误差=0.000018792507331166775`
a=46.4484，误差=0.00001872138763348019

a=46.4483021270782……
代码：
(*目标函数 G(\[Lambda])*)
G[\[Lambda]_] := HypergeometricPFQ[{-1/2, -1/2}, {1}, \[Lambda]^2];
(*逼近函数 F(\[Lambda],a)*)
F[\[Lambda]_,
a_] := (1 + (3 \[Lambda]^2)/(10 +
Sqrt[4 - 3 \[Lambda]^2]))*(1 + (22/(7 \[Pi]) -
1) (2 \[Lambda]/(1 + \[Lambda]))^a);
(*定义绝对误差函数*)
error[\[Lambda]_, a_] := Abs[F[\[Lambda], a] - G[\[Lambda]]];
(*计算给定 a 值时的最大绝对误差*)
maxError[a_?NumericQ] :=
NMaximize[{error[\[Lambda], a], 0 < \[Lambda] < 1}, \[Lambda]][[1]];
(*寻找最优 a 值*)
optimalA =
NMinimize[{maxError[a], a > 0}, a,
Method -> {"SimulatedAnnealing", "PerturbationScale" -> 3},
PrecisionGoal -> 3];
(*输出最优 a 值*)
aOpt = a /. optimalA[[2]];
\[Lambda]Opt = \[Lambda] /.
Last[NMaximize[{error[\[Lambda], aOpt],
0 < \[Lambda] < 1}, \[Lambda]]];
Print["最优 a 值为: ", NumberForm[aOpt, 20]];
Print["最大误差为: ", NumberForm[optimalA[[1]], 20]];
Print["最优 \[Lambda] 值为: ", NumberForm[\[Lambda]Opt, 20]];
(*误差图像*)
Plot[error[\[Lambda], aOpt], {\[Lambda], 0, 1}, PlotRange -> All,
PlotLabel -> "最大误差 = " <> ToString[NumberForm[optimalA[[1]], 10]]]

可惜的是13楼的代码运行时有点卡，但能运行。

ClearAll["Global`*"]
G[\[Lambda]_] := Hypergeometric2F1[-1/2, -1/2, 1, \[Lambda]^2]
c = 22/(7 \[Pi]) - 1;
a = 46.4483021270782`50;
F[\[Lambda]_,
a_] := (1 + (3 \[Lambda]^2)/(10 + Sqrt[4 - 3 \[Lambda]^2]))*(1 +
c*(2 \[Lambda]/(1 + \[Lambda]))^a)
error[\[Lambda]_] := Abs[G[\[Lambda]] - F[\[Lambda], a]];
obj[\[Lambda]_] = (G[\[Lambda]] - F[\[Lambda], a])^2;
crit = \[Lambda] /.
NSolve[{D[obj[\[Lambda]], \[Lambda]] == 0,
0 <= \[Lambda] <= 1}, \[Lambda], Reals, WorkingPrecision -> 30];
errorAtCrit = error /@ crit;
Transpose[{crit, errorAtCrit}] // Select[#, #[[2]] > 10^-6 &] & //
Grid[#, Frame -> All] &
Plot[error[\[Lambda]], {\[Lambda], 0, 1}, PlotRange -> All]