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Let $p_n(x)=x^n$ for $x\in \Bbb{R}$ and let P=span$\{p_0,p_1,p_2,p_3\dots\}$. Then-

  1. P is the vector space of all real valued continuous functions on R

  2. P is a subspace of all real valued continuous functions on R.

  3. The set $\{p_0,p_1,p_2,p_3\dots\}$ is linearly independent in the vector space of all continuous functions on R
  4. Trigonometric functions belong to P

I can easily see option 1. is false and option 2. and 3. are correct but I am not sure about option 4.

Why would any trigonometric function not belong P= span $\{p_0,p_1,p_2,p_3\dots\}$

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  • $\begingroup$ You might be interested in the concept of a Laurent decomposition. The idea is that certain functions (for example $\frac{1}{x}$) are not equal to their own taylor series (they are no 'analytic'). The trick of the Laurent decomposition is that besides positive powers of $x$, also negative powers of $x$ are allowed. Using this method of expansion every function on $\mathbb{R}$ has a laurent decomposition around every point in it's domain. There might be different compositions around different points though. $\endgroup$ – user2520938 Jul 16 '15 at 8:08
  • $\begingroup$ Same question as math.stackexchange.com/questions/1334796/… and math.stackexchange.com/questions/1334687/…, although the answers there don't fully explain your question about (4) $\endgroup$ – Chris Culter Jul 16 '15 at 8:12
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    $\begingroup$ Linear combination (and everything that follows, like: span, linear dependence etc.) is defined as a finite sum, because it's a purely algebraic concept. We can deal with infinte sums like power series, but we need additional mathematical structures and terms (mostly ones form mathematical analysis like convergence). That's why we don't say that "P spans the set of all analytical functions on $\mathbb{R}$", but rather that it forms a complete (dense) subset of the bigger space. $\endgroup$ – Lurco Jul 16 '15 at 8:25
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Your $P $ does not contain any infinite series. The span is the set of linear combinations of the $p_n $; in fact whoever wrote the question called it $P $ because it is the set of polynomials.

So the question is whether any trigonometric function is a polynomial. The answer is no, because a polynomial cannot be periodic.

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    $\begingroup$ That's the point I missed. Any element of the span would take only finitely many elements of P with non-zero coefficient . So, no series. $\endgroup$ – user118494 Jul 16 '15 at 8:15
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Some functions cannot be written as Taylor series because they have a singularity for example $f(z) = \frac{1}{1-\cos z}$.

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  • $\begingroup$ @Mitsos yeah but that is not in P $\endgroup$ – Bhaskar Vashishth Jul 16 '15 at 8:11

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