Approximaths

Month: September 2026

  • Stieltjes integral II

    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

  • Stieltjes integral I

    In this post we would like to illustrate the calculation of an integral using perturbation techniques. We would like to calculate the following Stieltjes integral:

    \displaystyle I = \int_{0}^{\infty} \frac{e^{-t}}{1+t}dt

    This integral can be evaluated exactly (I = 0.5963), but we’ll use perturbation techniques to compare the results obtained using these approximation techniques with the exactly calculated value. In order to do so, we will insert the parameter \epsilon as follows:

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

    For \epsilon = 0 we have I(\epsilon) = 1. This is the solution to the unperturbed problem. We observe that:

    \displaystyle \frac{1}{1-x} = 1 + x + x^2 + ...
    \displaystyle = 1 + x + x^2 + ... + x^n + x^{n+1}(1 + x + x^2 + ...)
    \displaystyle = 1 + x + x^2 + ... + x^n + \frac{x^{n+1}}{1-x}

    Defining x := -\epsilon^2 t and using \int_{0}^{\infty} e^{-t} t^n dt = n! in the formula above we obtain:

    \displaystyle \frac{1}{1 +\epsilon^2t} = 1 -\epsilon^2 t + \epsilon^4 t^2 + ... + (-1)^n \epsilon^{2n} t^{n} + \frac{(-1)^{n+1} \epsilon^{2n+2} t^{n+1}}{1+\epsilon^2 t}
    \displaystyle I(\epsilon) = \int_{0}^{\infty} e^{-t}(1 -\epsilon^2 t + \epsilon^4 t^2 + ... + (-1)^n \epsilon^{2n} t^{n} + \frac{(-1)^{n+1} \epsilon^{2n+2} t^{n+1}}{1 + \epsilon^2 t}) dt
    \displaystyle = \int_{0}^{\infty} (e^{-t} -\epsilon^2 e^{-t} t + \epsilon^4 e^{-t} t^2 + ... + (-1)^n \epsilon^{2n} e^{-t} t^{n} + \frac{(-1)^{n+1} \epsilon^{2n+2} e^{-t} t^{n+1}}{1 + \epsilon^2 t}) dt
    \displaystyle = 1 - \epsilon^2 + \epsilon^4 2! + ... + (-1)^{n}\epsilon^{2n} n! + (-1)^{n+1} \epsilon^{2n+2} I_{n+1}(\epsilon)

    where I_{n+1}(\epsilon) := \int_{0}^{\infty} \frac{e^{-t} t^{n+1}}{1 + \epsilon^2 t} dt. Finally we have:

    \displaystyle I(\epsilon) = \sum_{k=0}^{n} (-1)^{k} \epsilon^{2k} k! + (-1)^{n+1} \epsilon^{2n+2} I_{n+1}(\epsilon)

  • Quantum Mechanical Interpretation

    The differential equation presented in the previous posts

    \displaystyle y'' + (1 - x)y = 0

    appears in quantum mechanics when a particle is subjected to a linear potential. The time-independent Schrödinger equation is

    \displaystyle -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V(x)\psi(x) = E\psi(x)

    Consider a linear potential of the form

    \displaystyle V(x) = V_0 - Fx

    where F is a constant force. The Schrödinger equation then becomes

    \displaystyle -\frac{\hbar^2}{2m}\psi''(x) + \bigl(V_0 - Fx\bigr)\psi(x) = E\psi(x)

    Rearranging the terms gives

    \displaystyle \psi''(x) + \frac{2m}{\hbar^2}\bigl(E - V_0 + Fx\bigr)\psi(x) = 0

    We introduce the turning point x_0 (in quantum mechanics, it marks the transition between an oscillatory (classically allowed) region and an exponentially decaying (classically forbidden) region) defined by

    \displaystyle E - V_0 + Fx_0 = 0
    \displaystyle x_0 = \frac{V_0 - E}{F}

    The equation then becomes

    \displaystyle \psi''(x) + \frac{2mF}{\hbar^2}(x - x_0)\psi(x) = 0

    Introducing the dimensionless variable

    \displaystyle z = -\left(\frac{2mF}{\hbar^2}\right)^{1/3}(x - x_0)

    we finally obtain

    \displaystyle \frac{d^2\psi}{dz^2} - z\psi = 0

    Thus, the equation

    \displaystyle y'' + (1 - x)y = 0

    is, after a suitable rescaling and translation of the independent variable, an Airy equation. It describes, in particular, the behavior of a quantum wavefunction near a turning point, i.e. a position where

    \displaystyle E = V(x)

    In a region where E > V(x), the solution is oscillatory, whereas in a region where E < V(x), it exhibits exponential behavior. The Airy function provides the transition between these two regimes (see fig. below).

  • y” + (1-x) y= 0 (Part II)

    The first terms of the equation presented in the previous post are:

    \displaystyle a_0''(x) + a_0(x) - x a_0(x)\epsilon
    \displaystyle + a_1''(x)\epsilon + a_1(x)\epsilon - x a_1(x)\epsilon^2
    \displaystyle + a_2''(x)\epsilon^2 + a_2(x)\epsilon^2 - x a_2(x)\epsilon^3 + \cdots

    the differential equation can be written as:

    \displaystyle a_0''(x) + a_0(x) + (a_1''(x) + a_1(x) - x a_0(x))\epsilon + (a_2''(x) + a_2(x) - x a_1(x))\epsilon^2 + \cdots = 0
    \displaystyle a_0''(x) + a_0(x) = 0
    \displaystyle a_1''(x) + a_1(x) - x a_0(x) = 0
    \displaystyle a_2''(x) + a_2(x) - x a_1(x) = 0

    This implies

    \displaystyle a_0(x) = \cos(x)
    \displaystyle a_1(x) = \frac{1}{4}x^2 \sin(x) + \frac{1}{4} x \cos(x) - \frac{1}{4} \sin(x)

    We can check that a_1(x) is correct:

    \displaystyle a_1(x) = \frac{1}{4}x^2 \sin(x) + \frac{1}{4} x \cos(x) - \frac{1}{4} \sin(x)
    \displaystyle a_1'(x) = \frac{1}{4}x^2 \cos(x) + \frac{1}{2} x \sin(x) + \frac{1}{4} \cos(x) - \frac{1}{4} x \sin(x) - \frac{1}{4} \cos(x)
    \displaystyle a_1'(x) = \frac{1}{4}x^2 \cos(x) + \frac{1}{4} x \sin(x)
    \displaystyle a_1''(x) = \frac{1}{2}x \cos(x) - \frac{1}{4} x^2 \sin(x) + \frac{1}{4} \sin(x) + \frac{1}{4} x \cos(x)

    We can check that the condition:

    \displaystyle a_1''(x) + a_1(x) - x a_0(x) = 0

    is fulfilled. Therefore:

    \displaystyle y_{\epsilon}(x) = \cos(x) + \left(\frac{1}{4}x^2 \sin(x) + \frac{1}{4} x \cos(x) - \frac{1}{4} \sin(x)\right) \epsilon + \cdots

    Setting \epsilon = 1 gives the first-order perturbative approximation to the corresponding solution of the Airy equation.

    \displaystyle y(x) = \cos(x) + \frac{1}{4}x^2 \sin(x) + \frac{1}{4} x \cos(x) - \frac{1}{4} \sin(x) + \cdots