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Table 9 Absolute error bound \(\hat{r}_n\) of the estimate for the value of the power series

From: Introducing the Logarithmic finite element method: a geometrically exact planar Bernoulli beam element

n \(\varvec{\vert z \vert =0.025}\) \(\varvec{\vert z \vert =0.1}\) \(\varvec{\vert z \vert =0.25}\) \(\varvec{\vert z \vert =1}\) \(\varvec{\vert z \vert =2}\)
(a) \(m = 1\), \(J = \{ \, \}\)
0 1.27\(\cdot 10^{-2}\) 5.26\(\cdot 10^{-2}\) 1.43\(\cdot 10^{-1}\) 1 \(\infty \)
1 1.05\(\cdot 10^{-4}\) 1.72\(\cdot 10^{-3}\) 1.14\(\cdot 10^{-2}\) 2.5\(\cdot 10^{-1}\) 2
2 6.55\(\cdot 10^{-7}\) 4.27\(\cdot 10^{-5}\) 6.94\(\cdot 10^{-4}\) 5.56\(\cdot 10^{-2}\) 6.67\(\cdot 10^{-1}\)
3 3.27\(\cdot 10^{-9}\) 8.5\(\cdot 10^{-7}\) 3.43\(\cdot 10^{-5}\) 1.04\(\cdot 10^{-2}\) 2.22\(\cdot 10^{-1}\)
4 1.36\(\cdot 10^{-11}\) 1.41\(\cdot 10^{-8}\) 1.42\(\cdot 10^{-6}\) 1.67\(\cdot 10^{-3}\) 6.67\(\cdot 10^{-2}\)
(b) \(m = 5\), \(J = \{1,2,3,4\}\)
n \(\vert z \vert =0.025\) \(\vert z \vert =0.1\) \(\vert z \vert =0.25\) \(\vert z \vert =1\) \(\vert z \vert =2\)
0 4.26\(\cdot 10^{-3}\) 1.82\(\cdot 10^{-2}\) 5.26\(\cdot 10^{-2}\) 1 \(\infty \)
1 4.51\(\cdot 10^{-5}\) 7.46\(\cdot 10^{-4}\) 5\(\cdot 10^{-3}\) 1.25\(\cdot 10^{-1}\) 2
2 3.28\(\cdot 10^{-7}\) 2.15\(\cdot 10^{-5}\) 3.51\(\cdot 10^{-4}\) 2.94\(\cdot 10^{-2}\) 4\(\cdot 10^{-1}\)
3 1.82\(\cdot 10^{-9}\) 4.73\(\cdot 10^{-7}\) 1.91\(\cdot 10^{-5}\) 5.95\(\cdot 10^{-3}\) 1.33\(\cdot 10^{-1}\)
4 8.17\(\cdot 10^{-12}\) 8.49\(\cdot 10^{-9}\) 8.52\(\cdot 10^{-7}\) 1.02\(\cdot 10^{-3}\) 4.17\(\cdot 10^{-2}\)