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a.) Find a basis for the orthogonal complement to the following subspaces of $\mathbb{R^4}$. The subspace spanned by $(1,2,-1,3)^T$,$(-2,0,1,-2)^T$,$(-1,2,0,1)^T$.

b.) Use the Legendre polynomials to find the best quadractic and cupic approxiamtion to $t^4$, based on the $L^2$ norm on [-1,1].

For a, I found the orthogonal matrix from the spanned vectors given which gave me

$K=\pmatrix{1&\frac{-7}{5}&\frac{-7}{5}\\2&\frac{6}{5}&\frac{6}{5}\\-1&\frac{2}{5}&\frac{2}{5}\\3&\frac{-1}{5}&\frac{-1}{5}}$ but how can I find the basis of the spanned vectors, becuase I am confused on what they are exactly trying to get me to find.

For b, I found the quadratic $\frac{1}{5}+ \frac{4}{7}(\frac{-1}{2}+\frac{3}{2}t^2) = \frac{-3}{35} + \frac{6}{7}t^2$ but how can I find the cubic using the same process?

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    $\begingroup$ I found a link to math.tamu.edu/~yvorobet/MATH304-504/Lect4-04web.pdf in one of the linked questions. That might be useful for gaining better intuition about the Legendre polynomials, note my definition has different normalization than those notes. I'm not sure your context, but there is some variation on what precisely these polynomials are taken to be...beware. $\endgroup$ Nov 13 '12 at 5:09
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The Legendre polynomials can be generated by neat formulas, or looked up in tables... I happen to know that relative the inner product $<f,g> = \int_{-1}^{1}f(t)g(t)\, dt$ we can calculate that $$ <1,1>=2, <t,t>=2/3, <t^2,t^2>=2/5 $$ $$ <1,t>=0, <1,t^2>=2/3, <t,t^2>=0 $$ We apply Gram-Schmidt on $\{ 1,t,t^2 \}$ to produce an orthonormal set:

Let $u_1=1/\sqrt{2}$ then clearly $<u_1,u_1>=1$. Next, Let $v_2 = t-<t,u_1>u_1 =t$ then normalize to $u_2=t\sqrt{3/2}$. Next, Let $v_3 = t^2-<t^2,u_1>u_1-<t^2,u_2>u_2$ and once again normalize to $u_3= \sqrt{\frac{8}{45}}(t^2-\frac{1}{3})$. Next, $$v_4 = t^3-<t^3,u_1>u_1-<t^3,u_2>u_2-<t^3,u_3>u_3$$ and set $u_4 = \frac{1}{\sqrt{<v_4,v_4>}}v_4$. Next, $$v_5 = t^4-<t^4,u_1>u_1-<t^4,u_2>u_2-<t^4,u_3>u_3-<t^4,u_4>u_4$$ and normalize by settting $u_5 = \frac{1}{\sqrt{<v_5,v_5>}}v_5$. I hope you have these in a table since your question just says to use these Legendre polynomials. Very well, what's the point of these particular polynomials $S=\{ u_1,u_2,u_3,u_4,u_5 \}$? The point is that $S$ is $<,>$-orthonormal: $$ <u_i,u_j> = \delta_{ij}$$ and projections onto spaces spanned by vectors in $S$ is a simple matter: setting $W = span(S)$ we write $$ Proj_W(f) = \sum_{i=1}^5 <u_i,f>u_i $$ For the quadratic case set $W_2 = span\{u_1,u_2,u_3\}=P_2$ and you could calculate, $$ Proj_{W_2}(f) = <f,u_1>u_1+<f,u_2>u_2+<f,u_3>u_3 $$ You need to first find the Legendre polynomials upto the appropriate order and second form the natural projections as I indicate above. I think this is what is intended. I'll leave part (a.) to someone else for now.

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  • $\begingroup$ Thank you a lot!! This is very helpful! Do you think you can give me a hint on how to do part a? $\endgroup$
    – Q.matin
    Nov 13 '12 at 5:13
  • $\begingroup$ Sure, to find space orthogonal to column vectors $v_1,v_2, \dots , v_k$ just form $A=[v_1|v_2|\cdots |v_k]$ and calculate the basis for $Null(A^T)$. This is same as setting $v_j \cdot x=0$ for $j=1,2, \dots, k$ and solving for the set of all such $x$. I assume you've found the basis for the nullspace of a matrix before... $\endgroup$ Nov 13 '12 at 5:50
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    $\begingroup$ there is a mistake in my 2nd-order term. Hope to fix later. $\endgroup$ Nov 13 '12 at 13:58

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