Linear algebra - best approximation

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SUMMARY

The discussion centers on the concept of best approximation in linear algebra, specifically within the context of function spaces. The user determines that the best approximation (P) to the polynomial function F (3x + 5) is itself, leading to a distance of zero (d(F, P) = 0). This outcome is attributed to the fact that both F and the approximation P are polynomials of the same degree, and P exists in the linear span (W) of the set S = {1, 2x-1}. The user seeks clarification on why this is inherently true.

PREREQUISITES
  • Understanding of function spaces, specifically V = ([0,1])
  • Knowledge of inner product definitions, particularly = ∫01 f(x)g(x)dx
  • Familiarity with linear combinations and linear spans in vector spaces
  • Basic concepts of polynomial functions and their properties
NEXT STEPS
  • Study the properties of linear spans and their implications in function approximation
  • Learn about orthogonal projections in Hilbert spaces
  • Explore the concept of polynomial approximation and the Weierstrass approximation theorem
  • Investigate the relationship between function degree and approximation accuracy
USEFUL FOR

Mathematicians, students of linear algebra, and anyone interested in function approximation techniques and polynomial analysis.

twotaileddemon
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Hi... I have a quick question. I'm given V = ([0,1], <f,g> = interval from 0 to 1 of f(x)g(x)dx, S = {1, 2x-1}, W = lin(s), and P exists in W.

I was determining the best approximation (P) to a function (F). F was some polynomial (3x + 5) and when I did the work I got P to be the same polynomial. (I used projections of both parts of S with F and added them)

The distance between them, d(F,P) was obviously 0, but I was wondering why this was true? I mean.. why it HAD to be true even before I computed P.

Is it because W and F were polynomials of the same degree and p exists in W? Thanks for your help.
Any elaboration is great :) I just want to understand why certain functions already have the best approximation to what is given.
 
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twotaileddemon said:
Hi... I have a quick question. I'm given V = ([0,1], <f,g> = interval from 0 to 1 of f(x)g(x)dx, S = {1, 2x-1}, W = lin(s), and P exists in W.

I was determining the best approximation (P) to a function (F). F was some polynomial (3x + 5) and when I did the work I got P to be the same polynomial. (I used projections of both parts of S with F and added them)

The distance between them, d(F,P) was obviously 0, but I was wondering why this was true? I mean.. why it HAD to be true even before I computed P.

Is it because W and F were polynomials of the same degree and p exists in W? Thanks for your help.
Any elaboration is great :) I just want to understand why certain functions already have the best approximation to what is given.
What you've written here is not at all clear. It would help me understand what you're trying to do if you can answer the following questions.

  1. What do you mean that V = [0, 1]? Is this just the interval 0 <= x <= 1? If so, why is it named?
  2. <f, g> is not an interval; it is an integral, namely [tex]\int_0^1 f(x)g(x)dx[/tex]
  3. What is S? It appears to be a basis with two functions in it.
  4. What is W? You show it as equal to lin(s). Do you mean that W is the set of all linear combinations of the vectors in S (not s)? It would be helpful if you said that.
  5. How did you calculate the distance from one function to another? I'm guessing that you should be using the inner product you are given.

If an arbitrary function f is in W (that is, f is a linear combination of the functions in S), then I think that an approximation using the functions in W will be exact. If, on the other hand, f is not in W (for example, f(x) = x2), then an approximation will be just that, an approximation.
 

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