Trigonometric Polynomial vs Fourier Polynomial

[laser1]

Homework Statement

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Relevant Equations

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what is the difference? It seems like in T, you choose the RHS first, but in f, you choose the LHS first. Is this the only difference?

Because the Fourier coefficients of f is derived in a standard way, right? As in, couldn't I derive the coefficients for T in the same way as I did for f? In which case, isn't T and f the same?

Thank you

 

[Mark44]

what is the difference?

I don't see any difference other than a Fourier polynomial is a trig polynomial for a specific function.

 

[laser1]

I don't see any difference other than a Fourier polynomial is a trig polynomial for a specific function.

Later on in the notes it says this so they seem different but idk really.

 

[erobz]

Later on in the notes it says this so they seem different but idk really.


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That $f(x)$ in the integrand has no subscript $N$. Notice in your OP its $f_N(x)$ that is the Fourier Polynomial. I don't claim to know what the fuss is about with saying $T_n(x)$ is different from $f_N(x)$ but in general with the method of least squares you are looking for the function that minimizes error between your chosen function $f(x)$ and the data. My guess is that in the first definition you present it is supposed to be $F_N(x)$ not $f_N(x)$, a typo. If I'm talking out my backside I apologize, but I think in the very least it's worth noting this discrepancy.

 

[FactChecker]

Is any trigonometric polynomial its own Fourier polynomial? I think it would be. And clearly any Fourier polynomial is a trigonometric polynomial.

 

[erobz]

Is any trigonometric polynomial its own Fourier polynomial? I think it would be. And clearly any Fourier polynomial is a trigonometric polynomial.

They seem to be saying in Theorem 8 that the coefficients of the Fourier Polynomial $a_n,b_n$ of $F_N(x)$ are special; they would be found by minimizing $Err(N)$ w.r.t. $\alpha_n,\beta_n$?

 

[DaveE]

I suspect the difference is subtle and mostly semantic. In the trigonometric polynomial $T_n(x)$ is defined by the choice of $\alpha_n$ and $\beta_n$. In the Fourier polynomial $a_n$ and $b_n$ are determined by the definition of $f(x)$. The Fourier polynomials would then be a subset of all trigonometric polynomials defined by the function $f(x)$.

 

[FactChecker]

They seem to be saying in Theorem 8 that the coefficients of the Fourier Polynomial $a_n,b_n$ of $F_N(x)$ are special; they would be found by minimizing $Err(N)$ w.r.t. $\alpha_n,\beta_n$?

Yes. The OP says that the Fourier polynomial coefficients for a function are the same as the usual Fourier series coefficients as far as the Fourier polynomial goes. So the Fourier polynomial is just a truncated Fourier series. My question is whether the truncated Fourier series coefficients of a trigonometric polynomial are identical to the coefficients of the trigonometric polynomial.

 

[erobz]

Yes. The OP says that the Fourier polynomial coefficients for a function are the same as the usual Fourier series coefficients as far as the Fourier polynomial goes. So the Fourier polynomial is just a truncated Fourier series. My question is whether the truncated Fourier series coefficients of a trigonometric polynomial are identical to the coefficients of the trigonometric polynomial.

Ooooh…I’m behind in the ideas here; too big for my britches.

 

[FactChecker]

Ooooh…I’m behind in the ideas here; too big for my britches.

Ha! Me too. But I think the polynomials are identical. I don't have what it takes right now to prove it.

 

[laser1]

Ha! Me too. But I think the polynomials are identical. I don't have what it takes right now to prove it.

I think they are identical too, but I don't know why they are talked about differently in my notes. Maybe because, as mentioned above, for the T you choose the coefficients first, whereas for the f, you choose the function first.

 

[vela]

what is the difference? It seems like in T, you choose the RHS first, but in f, you choose the LHS first. Is this the only difference?

Perhaps this analogous example might help:

We know that any finite sum of the kind​

$$p_N(x) = \alpha_0 + \alpha_1 x + \cdots + \alpha_N x^N,$$ where $\alpha_i$ are real constants, is called a polynomial of degree $N$. If the polynomial is associated with a function $f(x)$, then​

$$f_N(x) = a_0 + a_1 x + \cdots + a_N x^N$$ is called the Taylor polynomial of $f(x)$ of order $N$, where $a_i = f^{(i)}(0)/n!$ are the usual Taylor series coefficients.​

The difference in the text is that $T_N(x)$ is a generic trigonometric polynomial and the Fourier polynomial $f_N(x)$ is a specific trigonometric polynomial where the coefficients are the usual Fourier coefficients from the Fourier series.

It's not obvious that the Fourier polynomial of order $N$ obtained by truncating the Fourier series is the trigonometric polynomial of degree $N$ that best approximates the function $f(x)$ in the least-squares sense. Theorem 8 says it is.

 

[pasmith]

My question is whether the truncated Fourier series coefficients of a trigonometric polynomial are identical to the coefficients of the trigonometric polynomial.

Yes.

Given a family of smooth functions \phi_n : [a.b] \to \mathbb{R} for n = 0, 1, \dots, which are orthogonal with respect to the inner product <br /> \langle f,g \rangle = \int_a^b w(x)f(x)g(x)\,dx, \quad w(x) &gt; 0\,\forall x \in (a,b) with induced norm \|f\| = \langle f,f \rangle^{1/2}, we can interpolate any smooth function f: [a,b] \to \mathbb{R} as I_N[f] = \sum_{n=0}^N \frac{\langle f, \phi_n \rangle}{\|\phi_n\|^2} \phi_n. Here the error f - I_N[f] is orthogonal to the space spanned by the \phi_n, but this also minimizes the square error over all approximations of the form \sum_{n=0}^N a_n \phi_n, since <br /> \frac{\partial}{\partial a_m} \left\| f - \sum_{n=0}^N a_n \phi_n \right\|^2 = -2\langle f, \phi_m\rangle + 2 a_m \|\phi_m\|^2. One can easily verify that <br /> I_N\left[ \sum_{n=0}^N b_n \phi_n \right] = \sum_{n=0}^N b_n\phi_n and indeed that for M &gt; N,<br /> I_N\left[ \sum_{n=0}^M b_n \phi_n \right] = \sum_{n=0}^N b_n\phi_n. This is what we are doing when we obtain truncated fourier series coefficients.

 

[laser1]

Perhaps this analogous example might help:

We know that any finite sum of the kind​

$$p_N(x) = \alpha_0 + \alpha_1 x + \cdots + \alpha_N x^N,$$ where $\alpha_i$ are real constants, is called a polynomial of degree $N$. If the polynomial is associated with a function $f(x)$, then​

$$f_N(x) = a_0 + a_1 x + \cdots + a_N x^N$$ is called the Taylor polynomial of $f(x)$ of order $N$, where $a_i = f^{(i)}(0)/n!$ are the usual Taylor series coefficients.​

The difference in the text is that $T_N(x)$ is a generic trigonometric polynomial and the Fourier polynomial $f_N(x)$ is a specific trigonometric polynomial where the coefficients are the usual Fourier coefficients from the Fourier series.

It's not obvious that the Fourier polynomial of order $N$ obtained by truncating the Fourier series is the trigonometric polynomial of degree $N$ that best approximates the function $f(x)$ in the least-squares sense. Theorem 8 says it is.

Nice analogy. But, can't I just define $f_N(x)$ to be a $p_N(x)$? Therefore they are the same. When you say that "it is not obvious that..." but if the trigonometric polynomial is the Fourier polynomial, then why wouldn't it be obvious?

 

[vela]

Nice analogy. But, can't I just define $f_N(x)$ to be a $p_N(x)$? Therefore they are the same.

I'm not sure how to respond other than to repeat myself. $p_N$ defines what a polynomial is, generally. $f_N$ defines what a Taylor polynomial is. Similarly, the text defines what a trigonometric polynomial is, generally, and what the Fourier polynomial of $f(x)$, which is a specific polynomial, is.

It's like defining what a dog is and then identifying what a German shepherd is. I doubt you'd claim that a German shepherd is the same as every dog.

When you say that "it is not obvious that..." but if the trigonometric polynomial is the Fourier polynomial, then why wouldn't it be obvious?

It may seem to be true intuitively, but that doesn't cut it in math. The statement that the Fourier series with those coefficients converges to $f$ is a different claim than saying the Fourier polynomial best approximates $f$ in the least-squares sense.

 

[WWGD]

My understanding is that , given a basis for $S:=L^2[-\pi, \pi]$ ( as an example); here the basis being $B:=\{\pi, cos(nx), sin(nx); n=1,2,3,...\}$, the orthogonal projection of any $f$ onto the space spanned, generated by B minimizes the ($L^2$) , aka, mean-squared, distance between f and the series $\Sigma_{n=1}^{\infty} a_0 +a_ncos(nx) + b_n sin(nx)$ (i.e., the vector space generated by $S$). And the Fourier coefficients are precisely the result of that projection. As such, the Fourier coefficients provide the best approximation. Maybe @mathwonk can comment on it.

 

[laser1]

I'm not sure how to respond other than to repeat myself. pN defines what a polynomial is, generally. fN defines what a Taylor polynomial is. Similarly, the text defines what a trigonometric polynomial is, generally, and what the Fourier polynomial of f(x), which is a specific polynomial, is.

What I think: All polynomials are taylor polynomials.
What I also think: All taylor polynomials are polynomials.

All german shepherds are dogs. All dogs are not german shepherds.

So I don't think this analogy holds unless one of my statements above is incorrect.

 

[laser1]

the Fourier coefficients provide the best approximation.

sorry if I am behind in logic, but I would like to clarify something. Are the fourier coefficients equivalent to the trigonometric coefficients? I believe they are.

However, if they are, then I don't understand the logic.

 

[WWGD]

sorry if I am behind in logic, but I would like to clarify something. Are the fourier coefficients equivalent to the trigonometric coefficients? I believe they are.

However, if they are, then I don't understand the logic.

Hi, the Fourier Coefficients guarantee the best approximation in the mean-square sense, ($L^2$) to your choice of $f$ under the basis $\{ cos(nx), sin(nx), \pi\}$. Thus they are the subset of Trigonometric coefficients $a_n, b_n$ that minimize $|| f - [\Sigma_{n=1}^{\infty} a_0 + a_nCos(nx) +b_n Sin(nx) ] ||$. It follows from the result that the orthogonal projection of a vector (function, here) minimizes the distance from the vector onto a subspace.

 

[mathwonk]

"Because the Fourier coefficients of f is derived in a standard way, right? As in, couldn't I derive the coefficients for T in the same way as I did for f? In which case, isn't T and f the same?"

It is a little hard for me to understand what is being asked. It is true that if we start from T_N and define f to be the function defined by T_N, then indeed f_N does equal T_N. But then all higher fourier coefficients of this f would be zero.

So perhaps the confusion is in looking at only one trig polynomial at a time. I.e. the interest of trig polynomials is in using them, as has been said, to approximate functions more general than those given by a single trig polynomial. Thus if we start from a general (integrable) function f, we get not one, but a whole infinite sequence of trig polynomials, namely the fourier polynomials of f. Then one might ask, if we are given a sequence of trig polynomials, when does there exist a function f such that this sequence is the sequence of Fourier polynomials for f.

One necessary condition would be that given any two polynomials T_N and T_M in the sequence with N<M, the polynomial T_M must have the same coefficients as T_N up through degree N.

Even if you are given such a sequence of compatible trig polynomials, one could still ask how to recover a function f for which the given sequence might be its Fourier polynomials.

Another analogy might be to define a finite decimal, and then note that any real number yields an infinite sequence of finite decimal approximations. It seems then rather odd, to me, to ask what Is the difference between a finite decimal and the finite approximation 3.1416 to π. Isn't 3.1416 just the same as a finite decimal? Well yes, it is true that if we start from 3.1416 and define the real number r to be r=3.1416, then 3.1416 will indeed be the best finite approximation to r, but this does not tell us how to recover π.

Similarly, given any trig polynomial T_N, there are infinitely many functions f for which this T_N is the Fourier polynomial of degree N, but the more challenging question seems to be how to recover f from the full sequence of Fourier polynomials for f.

Sorry if I have not identified the point of confusion.

 

[FactChecker]

I think it is more accurate to say "Fourier polynomial of order N of the function $f(x)$".
Every trigonometric polynomial is the Fourier polynomial of itself. Clearly, every Fourier polynomial is a trigonometric polynomial.
So the only difference is when a specific function, $f(x)$, is introduced. Then a given trigonometric polynomial of degree N may or may not be the Fourier polynomial of order N of the function $f(x)$.

 

[vela]

What I think: All polynomials are taylor polynomials.
What I also think: All taylor polynomials are polynomials.

You've kind of moved the goalposts here. There's the general definition of a polynomial and a specific Taylor polynomial of a function $f$. You're essentially saying that any polynomial is the Taylor polynomial of $f$.

 

[DaveE]

What I think: All polynomials are taylor polynomials.

"Taylor" polynomials make no sense unless you have a function associated with them. Of course you have the trivial case, where your function definition is a polynomial, and then you derive the Taylor expansion, which, like magic, is exactly the same thing!

edit: Perhaps I wasn't clear. My point is that there aren't "Taylor" polynomials. There are Taylor expansions OF A FUNCTION. Otherwise it's just another polynomial.

 

[WWGD]

"Taylor" polynomials make no sense unless you have a function associated with them. Of course you have the trivial case, where your function definition is a polynomial, and then you derive the Taylor expansion, which, like magic, is exactly the same thing!

And in the Real case, the Taylor Polynomial may not even converge to the function it was derived from.

 

[pasmith]

What I think: All polynomials are taylor polynomials.
What I also think: All taylor polynomials are polynomials.

Certainly every polynomial is a taylor polynomial of some function (itself, for example), but is every polynomial a taylor polynomial of a specific given function?

 

[FactChecker]

Certainly every polynomial is a taylor polynomial of some function (itself, for example), but is every polynomial a taylor polynomial of a specific given function?

A general Taylor series does not have to be expanded around $x=0$. Maybe we should say Maclaurin.