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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\(\cdot 10^{-2}\)