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I want to find the second-order polynomial that best approximates $t^3$, with respect to the norm of the vector space $V$.

I first proved the bracketing map given in the problem was indeed an inner product for $V$.

Secondly, I converted the standard basis for $V$, $\{1, t, t^2\}$, into an orthonormal basis for $V$, using the Gram-Schmidt process and then normalizing the orthogonal set of vectors.

To answer the question of finding the second-order polynomial which best approximates $t^3$, I know that I am supposed to use this orthonormal basis, and use a sum of projection operators to find my polynomial, but I'm not sure how to proceed.

My initial attempt, if it makes any sense at all, is to project $t^3$ onto the subspace spanned by the first orthonormal basis vector, then project $t^3$ onto the subspace spanned by the second orthonormal basis vector, and then project one more time to the third subspace. If this is correct, how I can write out this polynomial explicitly?

Any help would be greatly appreciated. Thanks,

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  • $\begingroup$ which inner product are you using and around which point do you want the approximation to be? $\endgroup$ – hjhjhj57 Nov 12 '14 at 4:28
  • $\begingroup$ No specific point, I think, but rather I want the orthogonal projection of the vector, t^3, on the space V, which is spanned by the orthonormal basis that I found previously via Gram-Schmidt. Thanks, @hjhjhj57. $\endgroup$ – User001 Nov 12 '14 at 5:29
  • $\begingroup$ (and that orthogonal projection is the polynomial given by johnD below...) $\endgroup$ – User001 Nov 12 '14 at 5:30
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So you have formed an orthonormal basis $B=\{v_1,v_2,v_3\}$ for $V$ via Gram-Schmidt. You have done most of the work. The last thing you need is to find the best approximation of the function $f(t)=t^3$ in terms of $B$, which is done via $$ f(t)\approx\sum_{k=1}^3c_kv_k\quad\text{where}\quad c_k=\langle f,v_k\rangle=\int_a^b f(t)v_k(t)\,dt, $$ where in the last equality I assumed you are using the standard inner product. If not, replace that integral expression with whatever your inner product is.

Yes, your resulting polynomial $\sum_{k=1}^3c_kv_k$ will be explicit.

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  • $\begingroup$ Got it - thanks so much, @johnD. $\endgroup$ – User001 Nov 12 '14 at 5:24
  • $\begingroup$ Hi @johnD, just revisiting this problem and wanted to ask you whether I could find some orthogonal projection matrix $UU^t$ to project this polynomial $t^3$ down to the subspace of polynomials of degree at most 2 (column space of some $U$)? Or, that doesn't really work, and I should stick with the above projection formula that you gave instead? Basically, I am wondering whether both the formula above and the projection matrix $UU^t$ can be used interchangeably. Thanks, $\endgroup$ – User001 Dec 4 '15 at 7:03

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