Linear Algebra and Its Applications, Exercise 3.4.24

Exercise 3.4.24. As discussed on page 178, the first three Legendre polynomials are 1, $x$ , and $x^2 - \frac{1}{3}$ . Find the next Legendre polynomial; it will be a cubic polynomial defined for $-1 \le x \le 1$ and will be orthogonal to the first three Legendre polynomials.

Answer: The process of finding the fourth Legendre poloynomial is essentially an application of Gram-Schmidt orthogonalization. The first three polynomials are

$v_1 = 1 \qquad v_2 = x \qquad v_3 = x^2 - \frac{1}{3}$

We can find the fourth Legendre polynomial by starting with $x^3$ and subtracting off the projections of $x_3$ on the first three polynomials:

$v_4 = x^3 - \frac{(v_1, x^3)}{(v_1, v_1)}v_1 - \frac{(v_2, x^3)}{(v_2, v_2)}v_2 - \frac{(v_3, x^3)}{(v_3, v_3)}v_3$

$= \frac{(1, x^3)}{(1, 1)}\cdot 1 - \frac{(x, x^3)}{(x, x)}x - \frac{(x^2-\frac{1}{3}, x^3)}{(x^2-\frac{1}{3}, x^2-\frac{1}{3})}(x^2-\frac{1}{3})$

For the first term we have

$(1, x^3) = \int_{-1}^1 1 \cdot x^3 \;\mathrm{d}x = \int_{-1}^1 x^3 \;\mathrm{d}x = 0$

so that the first term $\frac{(v_1, x^3)}{(v_1, v_1)}v_1$ does not appear in the expression for $v_4$ .

The third term $\frac{(v_3, x^3)}{(v_3, v_3)}v_3$ drops out for the same reason: its numerator is

$(x^2-\frac{1}{3}, x^3) = \int_{-1}^1 (x^2 - \frac{1}{3}) x^3 \;\mathrm{d}x$

$= \int_{-1}^1 x^5 \;\mathrm{d}x - \frac{1}{3} \int_{-1}^1 x^3 \;\mathrm{d}x = 0 - \frac{1}{3} \cdot 0 = 0$

That leaves the second term $\frac{(v_2, x^3)}{(v_2, v_2)}v_2$ with numerator of

$(x, x^3) = \int_{-1}^1 x \cdot x^3 \;\mathrm{d}x = \int_{-1}^1 x^4 \;\mathrm{d}x$

$= \left( \frac{1}{5} x^5 \right) \;\big|_{-1}^1 = \frac{1}{5} \cdot 1^5 - \frac{1}{5} \cdot (-1)^5 = \frac{1}{5} - (-\frac{1}{5}) = \frac{2}{5}$

and denominator

$(x, x) = \int_{-1}^1 x^2 \;\mathrm{d}x = \left( \frac{1}{3}x^3 \right) \;\big|_{-1}^1 = \frac{1}{3} \cdot 1^3 - \frac{1}{3} \cdot (-1)^3 = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$

We then have

$v_4 = x^3 - \left[ \frac{2}{5}/\frac{2}{3} \right] x = x^3 - \frac{3}{5}x$

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fifth Edition and the accompanying free online course, and Dr Strang’s other books.

Linear Algebra and Its Applications, Exercise 3.4.24

Leave a Reply Cancel reply

Archives

Meta

Linear Algebra and Its Applications, Exercise 3.4.24

Share this:

Like this:

Related

Leave a Reply Cancel reply

Archives

Meta