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are polynomials dense in C(R)?

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are polynomials dense in C(R)?

Is the following statement right?
for every continuous function like f,there exist a sequence of polynomials like P_n which P_n -> f

I should have mentioned this earlier (my apologies). What you're talking about is more or less the Weierstrass Approximation Theorem, which holds for C(X), where X is a compact subset of Euclidean n-space (R^n):

This result is a special case of the Stone-Weierstrass Theorem (of which there is a real and complex version), which gives general conditions under which a sub-algebra of the continuous functions on a compact topological space is uniformly dense in the space of all continuous functions thereon:

And is what he asked true if we consider pointwise convergence?

This was discussed a few years back in the sci.math newsgroup. They have shown it to be true.. If I have time I will try to find the reference for it and post it here.

>>Prove that A is dense in C^0([a; 1];R) for all a > 0. ...

If you mean "uniformly dense," then I don't think this is true for C(R). If you replace R with a closed, bounded interval [a,b], then yes, the polynomials are uniformly dense in C([a,b]).

The continuous function O(t), which is the limit of this sequence, does not depend ... If there exists a compaetum D such that DC Pn and for every k ... and if o =f I (the restriction off on E), then q = [f/E.

(« -> oo); that is, for every continuous function /on Γ, f .... Pn,n+ k .... polynomial of degree 2«; then there exists a unique algebraic polynomial w. 2n. (z) of degree .... Under theconditions indicatedabove, the following statements aretrue: .... behaves like a polynomial in φ(ζ) of degree 2s

that one can obtain a matrix like this by eliminating every other row of a ... can extend / to an analytic function F on the annulus {z : e~l < \z\ < e}. .... Given p > 1, it is not difficult to see that there exists an s > 0 such that ... C(T), we may choose a sequence (pn of nonvanishing continuous functions on T ...

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