Special techniques - Numerical Analysis

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from pylab import *

8.11.1Change of variable¶

Many times, a change of variable can improve the regularity of the integrand.

Example 3 (Infinite interval of integration)

I = \int_1^\infty \frac{f(x)}{x^p} \ud x, \qquad p > 1

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Assume $\lim_{x \to \infty} f(x)$ exists and $f(x)$ is smooth on $[1,\infty)$ . Change of variable

\begin{gather} x = \frac{1}{u^\alpha}, \qquad \ud x = -\frac{\alpha}{u^{1+\alpha}} \ud u, \quad \alpha > 0 \\ \Downarrow \\ I = \alpha \int_0^1 u^{(p-1)\alpha-1} f(1/u^\alpha) \ud u \end{gather}

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Integrand can be singular at $u=0$ , so pick α so that the exponent $(p-1)\alpha-1$ is large $(\ge 1)$ .

As a specific example consider

\begin{gather} I = \int_1^\infty \frac{f(x)}{x\sqrt{x}} \ud x \\ x = \frac{1}{u^4} \quad\implies\quad I = 4 \int_0^1 u f(1/u^4) \ud u \end{gather}

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For $x \to \infty$ , $f(x)$ behaves like

\begin{gather} f(x) = c_0 + \frac{c_1}{x} + \frac{c_2}{x^2} + \ldots \\ \Downarrow \\ uf(1/u^4) = c_0 u + c_1 u^5 + c_2 u^9 + \ldots \end{gather}

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and the integrand behaves nicely as $u \to 0$ .

Example 4 (Chebyshev quadrature)

I = \int_{-1}^1 \frac{f(x)}{\sqrt{1-x^2}} \ud x

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For this weight function, the orthogonal polynomials are Chebyshev polynomials. The integration nodes are roots of $T_n(x) = \cos(n \cos^{-1}x)$ (Chebyshev points of I kind)

x_j = \cos\left( \frac{2j-1}{2n}\pi \right), \qquad j=1,2,\ldots,n

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and weights

w_j = \frac{\pi}{n}

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so that

I \approx I_n = \frac{\pi}{n} \sum_{j=1}^n f(x_j)

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This corresponds to doing the change of variable $x = \cos\theta$ and applying the mid-point rule in the θ variable

I = \int_0^\pi f(\cos\theta) \ud\theta \approx \frac{\pi}{n} \sum_{j=1}^n f(\cos\theta_j), \qquad \theta_j = \frac{(2j-1)\pi}{2n}

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8.11.2Analytic treatment of singularity¶

Consider an integral with a singular term

I = \int_0^b f(x) \log(x) \ud x

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Since $\log(x)$ is singular at $x=0$ , we have to compute this with some care. Divide the interval into two parts

I = \int_0^\epsilon f(x) \log(x) \ud x + \int_{\epsilon}^b f(x) \log(x) \ud x =: I_1 + I_2

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$I_2$ can be computed by some standard technique and $I_1$ is computed semi-analytically. Assume $f(x)$ has convergent Taylor series on $[0,\epsilon]$

f(x) = \sum_{j=0}^\infty a_j x^j, \qquad x \in [0,\epsilon]

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Then

I_1 = \int_0^\epsilon \left( \sum_j a_j x^j \right) \log(x) \ud x = \sum_{j=0}^\infty \frac{a_j \epsilon^{j+1}}{j+1} \left[ \log(\epsilon) - \frac{1}{j+1} \right]

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8.11.3Product integration¶

Consider an integral with a singular term

I(f) = \int_a^b w(x) f(x) \ud x

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where $w(x)$ is singular and $f(x)$ is a nice function. Let $\{ f_n \}$ be a sequence of approximations to $f$ such that

$|w|$ is integrable.
$\lim_{n\to\infty} \norm{f - f_n}_\infty = 0$
The integrals
$I_n(f) = \int_a^b w(x) f_n(x) \ud x$
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can be analytically computed.

Then

|I(f) - I_n(f)| \le \norm{f-f_n}_\infty \int_a^b |w(x)| \ud x \to 0

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Example 6 (Singular integrand)

I(f) = \int_0^b f(x) \log(x) \ud x

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Let $f_n$ be piecewise linear approximation on a uniform partition

x_j = jh, \quad j=0,1,\ldots,n \qquad\textrm{where}\qquad h = \frac{b}{n}

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and

f_n(x) = \frac{1}{h}[ (x_j - x) f_{j-1} + (x-x_{j-1}) f_j], \qquad x_{j-1} \le x \le x_j

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for which we know the error estimate from Section 6.9.1

\norm{f-f_n}_\infty \le \frac{h^2}{8} \norm{f''}_\infty

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Then, with $I_n(f) = I(f_n)$

|I(f) - I_n(f)| \le \frac{h^2}{8} \norm{f''}_\infty \int_0^b |\log(x)| \ud x \to 0

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$I_n(f)$ can be computed analytically. In particular, in the first sub-interval $[x_0,x_1]=[0,h]$

f_n(x) = \frac{h-x}{h} f_0 + \frac{x}{h} f_1

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we have

\int_{x_0}^{x_1} \log(x) f_n(x) \ud x = \frac{f_0}{h} \int_0^h (h-x)\log(x) \ud x + \frac{f_1}{h} \int_0^h x \log(x) \ud x

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and these are finite since

\int_0^h \log(x) \ud x = h[\log(h)-1], \qquad \int_0^h x\log(x)\ud x = \frac{h^2}{4}[ 2\log(h) - 1 ]

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The integrals in other intervals can also be computed analytically.

Numerical Analysis

Gauss quadrature - II