Approximaths

Stieltjes integral II

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If we give \epsilon the value 1 for the series from the previous post and draw up a table of the values of the integral as a function of n, we can see that the values first converge towards the exact value and then move away from it (see table below). This reminds us of the “astronomers’ series” described by H. Poincaré. We also note that we cannot approach the exact value with arbitrary precision using this method.

n Approximation Error
0 1 0.4036
1 0 -0.5963
2 2 1.4037
3 -4 -4.5963
4 20 19.4037
5 -100 -100.5963

Given such a series, a natural question is: Up to what value of n should we sum in order to get as close as possible to the exact value ? In order to determine the ideal value of n, we must first note the following facts concerning the remainder of the series above:

\displaystyle I_{n+1}(\epsilon) := \int_{0}^{\infty} \frac{e^{-t} t^{n+1}}{1 + \epsilon^2 t} dt
\displaystyle |I_{n+1}(\epsilon)| > \int_{1}^{\infty} \frac{e^{-t} t^{n+1}}{1 + \epsilon^2 t} dt > \int_{1}^{\infty} \frac{e^{-t} t^{n+1}}{t + \epsilon^2 t} dt

Defining t := 1+s (\implies dt = ds), we have:

\displaystyle \int_{1}^{\infty} \frac{e^{-t} t^{n+1}}{t + \epsilon^2 t} dt = \int_{0}^{\infty} \frac{e^{-1-s} (1+s)^{n+1}}{(1+s) + \epsilon^2 (1+s)} ds
\displaystyle = \frac{e^{-1}}{1 + \epsilon^2} \int_{0}^{\infty} e^{-s} (1+s)^n ds

We observe that:

\displaystyle \frac{e^{-1}}{1 + \epsilon^2} \int_{0}^{\infty} e^{-s} (1+s)^n ds > \frac{e^{-1}}{1 + \epsilon^2} \int_{0}^{\infty} e^{-s} s^n ds
\displaystyle = \frac{e^{-1}}{1 + \epsilon^2}n!

Therefore:

\displaystyle |I_{n+1}(\epsilon)| > \frac{e^{-1}}{1 + \epsilon^2}n!
\displaystyle \epsilon^{2n+2}|I_{n+1}(\epsilon)| > \epsilon^{2n+2}\frac{e^{-1}}{1 + \epsilon^2}n!

Based on the above calculations, we can now define a lower bound (L_{n+1}(\epsilon)) and an upper bound (U_{n+1}(\epsilon)) for the error made when evaluating the Stieltjes integral using the sum derived above.

\displaystyle L_{n+1}(\epsilon) := \epsilon^{2n+2}\frac{e^{-1}}{1 + \epsilon^2}n!
\displaystyle < \left|I_{\epsilon} - \sum_{k=0}^{n} (-1)^k \epsilon^{2k} k!\right|
\displaystyle < \epsilon^{2n+2} (n+1)! =: U_{n+1}(\epsilon)

For \epsilon = 1, we have L_{n+1}(\epsilon) = \frac{e^{-1}}{2}n! and U_{n+1}(\epsilon) = (n+1)!. The approximation, error, and corresponding upper and lower bounds for the error are shown for \epsilon = 1 in the table below. In this table, we can clearly see that the approximations are getting worse and worse, while the lower and upper bounds are getting larger and larger. This makes the perturbative method, as implemented in this example, unattractive.

n Approximation |Error| L_{n+1}(\epsilon) U_{n+1}(\epsilon)
0 1 0.4036 0.18393 1
1 0 0.5963 0.18393 2
2 2 1.4037 0.36788 6
3 -4 4.5963 1.10364 24
4 20 19.4037 4.41455 120
5 -100 100.5963 22.0728 720

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