Fifth-order AGM-formula for the period of a large-angle pendulum

Big-Alabo, Akuro

doi:10.1590/1806-9126-RBEF-2023-0014

Akuro Big-Alabo ^** Correspondence email address: akuro.big-alabo@uniport.edu.ng

University of Port Harcourt, Faculty of Engineering, Department of Mechanical Engineering, Applied Mechanics & Design Research Group, Port Harcourt, Nigeria.

https://orcid.org/0000-0003-2821-0897

* Correspondence email address: akuro.big-alabo@uniport.edu.ng

n	a _n	b _n	Number of exact digits
0	1.0000000000000000000000000000000	$\sqrt{2} = 1$ .4142135623730950488016887242097	n/a
1	1.2071067811865475244008443621048	1.1892071150027210667174999705605	1
2	1.1981569480946342955591721663327	1.1981235214931201226065855718201	5
3	1.1981402347938772090828788690764	1.1981402346773072057983837881898	10
4	1.1981402347355922074406313286331	1.1981402347355922074392136559275	20

	Approximate solutions					Exact
A (◦)	T₁/T₀	T₂/T₀	T₃/T₀	T₄/T₀	T₅/T₀	T_ex/T₀
010	1.00178	1.00197	1.00178	1.00191	1.00191	1.00191
015	1.00402	1.00443	1.00419	1.0043	1.0043	1.0043
020	1.00718	1.0079	1.00757	1.00767	1.00767	1.00767
025	1.01127	1.01237	1.01195	1.01203	1.01203	1.01203
030	1.01632	1.01787	1.01736	1.01741	1.01741	1.01741
035	1.02236	1.02442	1.02381	1.02383	1.02383	1.02383
040	1.02944	1.03204	1.03134	1.03134	1.03134	1.03134
045	1.0376	1.04076	1.04	1.03997	1.03997	1.03997
050	1.04691	1.05064	1.04984	1.04978	1.04978	1.04978
055	1.05742	1.06172	1.06092	1.06083	1.06083	1.06083
060	1.06922	1.07407	1.0733	1.07318	1.07318	1.07318
065	1.0824	1.08776	1.08706	1.08692	1.08692	1.08692
070	1.09707	1.10289	1.10231	1.10214	1.10214	1.10214
075	1.11335	1.11956	1.11914	1.11896	1.11896	1.11896
080	1.13139	1.13791	1.13768	1.13749	1.13749	1.13749
085	1.15137	1.1581	1.15809	1.15789	1.15789	1.15789
090	1.17347	1.18031	1.18053	1.18034	1.18034	1.18034
095	1.19795	1.20476	1.20522	1.20504	1.20504	1.20504
100	1.22507	1.23171	1.2324	1.23223	1.23223	1.23223
105	1.25519	1.26147	1.26236	1.26221	1.26221	1.26221
110	1.28868	1.29441	1.29546	1.29534	1.29534	1.29534
115	1.32605	1.33099	1.33214	1.33205	1.33205	1.33205
120	1.36786	1.37176	1.37293	1.37288	1.37288	1.37288
125	1.41486	1.4174	1.41852	1.41851	1.41851	1.41851
130	1.46793	1.4688	1.46978	1.46982	1.46982	1.46982
135	1.52822	1.52711	1.52787	1.52795	1.52795	1.52795
140	1.59723	1.59386	1.59433	1.59445	1.59445	1.59445
145	1.67699	1.6712	1.67133	1.67148	1.67148	1.67148
150	1.7704	1.76225	1.76203	1.7622	1.7622	1.7622
155	1.88196	1.87187	1.87132	1.8715	1.8715	1.8715
160	2.01924	2.00813	2.00733	2.00751	2.00751	2.00751
165	2.19684	2.18619	2.18529	2.18544	2.18544	2.18544
170	2.44827	2.44004	2.43926	2.43936	2.43936	2.43936
175	2.88197	2.87804	2.87761	2.87766	2.87766	2.87766
176	3.02255	3.0196	3.01927	3.01931	3.01931	3.01931
177	3.20431	3.20231	3.20208	3.20211	3.20211	3.20211
178	3.4612	3.46009	3.45996	3.45996	3.45997	3.45997
179	3.90148	3.90111	3.90106	3.90103	3.90107	3.90107
179.9	5.36688	5.36687	5.36687	5.36503	5.36687	5.36687

\frac{T}{T_{0}} = \frac{1}{π} [\ln 2 + \sqrt{\frac{2}{1 + \sin^{2} (A / 2)}} \cdot \ln (\frac{\sqrt{2} + \sqrt{1 + \sin^{2} (A / 2)}}{\sqrt{2} - \sqrt{1 + \sin^{2} (A / 2)}})] - \frac{{(\cos (A / 2))}^{1.6}}{70}

\frac{T}{T_{0}} = \frac{16}{{\begin{matrix} 1 + \cos (A / 2) + 2 \sqrt{\cos (A / 2)} \\ + 2^{3 / 2} {(\cos (A / 2))}^{1 / 4} {[1 + \cos (A / 2)]}^{1 / 4} \\ + 2^{7 / 4} {(\cos (A / 2))}^{1 / 8} {[1 + \cos (A / 2)]}^{1 / 4} \\ \cdot {[1 + \cos (A / 2) + 2 \sqrt{\cos (A / 2)}]}^{1 / 2} \end{matrix}}}

a_{n} = \frac{1}{16} {[\sqrt{a_{n - 4}} + \sqrt{b_{n - 4}} + 2 {(\frac{a_{n - 4} + b_{n - 4}}{2})}^{1 / 4} \cdot {(a_{n - 4} b_{n - 4})}^{1 / 8}]}^{2}

a_{5} = \frac{1}{16} [\sqrt{\frac{a_{0} + b_{0}}{2}} + {(a_{0} b_{0})}^{1 / 4} + {(2 (\sqrt{a_{0}} + \sqrt{b_{0}}) {(a_{0} b_{0})}^{1 / 8} {(\frac{a_{0} + b_{0}}{2})}^{1 / 4})}^{1 / 2}]^{2}

K_{a} (m) = \frac{8 π}{{[\begin{matrix} \sqrt{\frac{1 + β}{2}} + β^{1 / 4} \\ + {(2 (1 + \sqrt{β}) β^{1 / 8} {(\frac{1 + β}{2})}^{1 / 4})}^{1 / 2} \end{matrix}]}^{2}}

\frac{T}{T_{0}} = \frac{16}{{[\begin{matrix} \sqrt{\frac{1 + \cos (\frac{A}{2})}{2}} + \cos^{1 / 4} (\frac{A}{2}) \\ + (2 (1 + \sqrt{\cos (\frac{A}{2})}) \cos^{1 / 8} (\frac{A}{2}) \\ \cdot {(\frac{1 + \cos (\frac{A}{2})}{2})}^{1 / 4})^{1 / 2} \end{matrix}]}^{2}}

\frac{T}{T_{0}} = \frac{16}{{[\begin{matrix} \cos (\frac{A}{4}) + \cos^{1 / 4} (\frac{A}{2}) \\ + \sqrt{2 (1 + \cos^{1 / 2} (\frac{A}{2})) \cos^{1 / 8} (\frac{A}{2}) \cos^{1 / 2} (\frac{A}{4})} \end{matrix}]}^{2}}

\frac{T_{1}}{T_{0}} = \frac{\begin{matrix} \frac{2}{π} \tan^{2} (A / 2) \ln (\frac{4}{\cos (A / 2)}) \\ + \frac{7.17}{(1 - \cos (A / 2))} \ln (\frac{1}{\cos (A / 2)}) \end{matrix}}{7.17 + \tan^{2} (A / 2)}

\frac{T_{2}}{T_{0}} = \frac{\begin{matrix} \frac{\cos^{2} (A / 2)}{(1 - \cos (A / 2))} \ln (\frac{1}{\cos (A / 2)}) \\ + \frac{2 \sin^{2} (A / 2)}{π} \ln (\frac{4}{\cos (A / 2)}) \end{matrix}}{1 - (π / 25) \cos^{2} (A / 2) \sin^{2} (A / 2)}

\frac{T_{3}}{T_{0}} = \frac{1}{π} [\ln 2 + \sqrt{\frac{2}{1 + \sin^{2} (A / 2)}} \cdot \ln (\frac{\sqrt{2} + \sqrt{1 + \sin^{2} (A / 2)}}{\sqrt{2} - \sqrt{1 + \sin^{2} (A / 2)}})] - \frac{{(\cos (A / 2))}^{1.6}}{70}

\frac{T_{4}}{T_{0}} = \frac{16}{{\begin{matrix} 1 + \cos (A / 2) + 2 \sqrt{\cos (A / 2)} \\ + 2^{3 / 2} {(\cos (A / 2))}^{1 / 4} {[1 + \cos (A / 2)]}^{1 / 4} \\ + 2^{7 / 4} {(\cos (A / 2))}^{1 / 8} {[1 + \cos (A / 2)]}^{1 / 4} \\ \cdot {[1 + \cos (A / 2) + 2 \sqrt{\cos (A / 2)}]}^{1 / 2} \end{matrix}}}

= \frac{16}{{[1 + \sqrt{\cos (A / 2)} + 2 \sqrt{\cos (A / 4) {(\cos (A / 2))}^{1 / 4}}]}^{2}}

\frac{T_{5}}{T_{0}} = \frac{16}{{[\begin{matrix} \cos (\frac{A}{4}) + \cos^{1 / 4} (\frac{A}{2}) \\ + \sqrt{2 (1 + \cos^{1 / 2} (\frac{A}{2})) \cos^{1 / 8} (\frac{A}{2}) \cos^{1 / 2} (\frac{A}{4})} \end{matrix}]}^{2}}

[1] * Correspondence email address: akuro.big-alabo@uniport.edu.ng

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Fifth-order AGM-formula for the period of a large-angle pendulum