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Table 3 shows numerical errors of this example in analogy to errors for this example in [28].

Numerical solutions in the form of errors of the example are given in Table 1.

Tables 3, 4 show the maximum absolute errors of the Example 2 for different values of (epsilon) and (mu).

Table 2 contains the errors for both this example of 40 events and the example of 4,000 events from the introduction.

Numerical results can be seen in Table 1 and also Table 2 illustrates the absolute values of the errors obtained here and the absolute errors of [18] for this example.

Figure 4 The comparison between F 2 ( t ) and F 2, 4 ( t ) for Example 2. The comparison between solution of Example 2. Trapezoidal rule Suppose that the region of integration is subdivided into 5 equal intervals of width h = 0.4, 6 integration nodes t i = 0.4 i for i = 0, …, 5. Table 4 illustrates the absolute values of the errors obtained here and the absolute errors of [18] for this example.

Note that the bias of the retrieval error in this example is small relative to the error variance, as the mean of the training set is very close to the mean of the validation and test sets.

In Table 1 we list the maximum pointwise error of Example 1, while in Table 2, we compare the best errors of Example 1.

Open image in new window Fig. 2 Absolute errors of Example 3 for (S=48).

A comparison between the absolute errors of Example 3 for different values of α, β is explained at x = 1 in Table 3.

end{aligned}Using the hybrid method with (Q=5, P=5,10,20) for (sin (0,1)), the maximum absolute errors of Example 6 are reported in Table 5.