aΦ/ax⋍±mΦ⋍±√(2k+1)Φ

p⋍(１/2)∓⒤√(2k+1)
⓪□F　(Fは証明できる)
①(□+m＾2)φ⋍0			ﾊｲｾﾞﾝﾍﾞﾙｸﾞp138
②	aΦ/ax⋍±mΦ⋍±√(2k+1)Φ
∵	Φ⋍Φ₀･e^{±mx}⋍Φ₀･e^{±√(2k+1)x}
∴	x＾2+ｘ+Q,25⋍(x+１/2)^2+Q⋍0
k	文献値	文/11	+(s/11)	√Q⋍√2k+1
1	14.13472	1.284975	0.454545	1.73952
2	21.02203	1.911094	0.363636	2.27473
3	25.01085	2.273714	0.363636	2.63735
4				3
5	30.42487	2.765897	0.727273	3.49317
6	32.93506	2.994096	0.636364	3.63046
7	37.58617	3.416925	0.454545	3.87147
8	40.91871	3.719883	0.272727	3.99261
9	43.32707	3.938825	0.454545	4.39337
10				4.58257569
11	48.00515	4.364105	0.454545	4.81865
12	49.77383	4.524894	0.363636	4.88853
13	52.97	4.815455	0.454545	5.27
14	56.446	5.131455	0.454545	5.586
15	59.347	5.395182	0.181818	5.577
16	60.831	5.530091	0.090909	5.621
17				5.91607978
18	65.112	5.919273	0.272727	6.192
29				7.61577311
30	75.8	6.890909	0.909091	7.8

F{-11-1}ﾐｶｴﾘｽ･ﾒﾝﾃﾝ､ﾓｼﾞｭﾗｰ⋍-1

F[-1-1-1[]ﾓｼﾞｭﾗｰ∵k^2⋍1.133^2

F[-s,s-1,-s] F[2,-3,2]∴ 逆数が有効

超幾何関数 F[αβγ] f(x)⇒1/f(x)

	超幾何関数	F[αβγ]	f(x)⇒1/f(x)
⓪	F[1/t,-t,1/t]	F[1,-1,1]	1-x⇒1/(1-x)
	F[-1/t,t,-1/t]	F[-1,1,-1]	1+x⇒1/(1+x)
①	F[-s,s-1,-s]	F[2,-3,2]∴	逆数が有効
		F[4,-5,4]
	F[s-1,-s,s-1]	F[-3,2,-3]
		F[-5,4,-5]
②	ｺﾗｰｹﾞﾝ	F[3,1,6]