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I am currently reading about Fourier series and Orthogonality of functions and Complete Sets of functions. Below are two extracts from the book I’m reading for which I simply do not understand:

`Extract 1:`

$\sin {nx}$ is a complete set on $(0, \pi)$; we used this fact when

we started with a function given on $(0, \pi)$, defined it on $(−\pi, 0)$ to make it odd, and then expanded it in a sine series.

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`Extract 2:`

A function given on $(0, l)$ can be expanded in a sine series by

defining it on $(−l, 0)$ to make it odd, or in a cosine series by defining it on $(−l, 0)$ to make it even (where $l$ is the period of the function).

For the first extract I don’t understand why *‘defining $\sin{nx}$ on $(-\pi,0)$’* makes it an odd function as it was my understanding that $\sin{nx}$ **is** odd on $[0,\pi]$.

For the second extract I am starting to wonder if there is a typo; since it says that we can define it on $[-l,0]$ to make a sine series **or** a cosine series. This doesn’t make any sense to me, as I don’t see how that interval can represent $\fbox{both}$ a sine series *and* a cosine series.

Most *importantly*; Why is $\sin{nx}$ a complete set of orthogonal functions on $[0,\pi]$?

Any ideas?

Many thanks.

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I think this question touches an important and essential aspect of a function. Namely, a function is *more* than the mapping $$y=f(x)$$

It is an object which also crucially depends on the domain and the codomain where it is defined.

\begin{align*}

&f:X\rightarrow Y\\

&y=f(x)

\end{align*}

It is important to know, that the

definitionof domain and codomain have fundamental influence to the properties of a function.Question: Is $\sin x$ odd? Answer: It depends! It depends on the definition of domain and codomain.

\begin{align*}

f:(0,\pi)\rightarrow\mathbb{R}\\

f(x)=\sin(x)

\end{align*}

This is not an odd function (at least in a non-void sense), since there is no $x\in(0,\pi)$ with $\sin(x)=-\sin(-x)$. Negative $x$ are not evendefinedfor it. But, if weextendthe domain to $X=[-\pi,\pi]$, then $f(x)=\sin(x)$ becomes an odd function.

With respect to your second question, let us assume a function $f$ is defined at $(0,l)$

\begin{align*}

&f:(0,l)\rightarrow\mathbb{R}\\

&y=f(x)

\end{align*}

Let us consider a function $g$ with

\begin{align*}

&g:(-l,l)\rightarrow\mathbb{R}\\

&g(x)=\begin{cases}

f(x)\qquad\quad &x\geq 0\\

f(-x)\qquad &x<0

\end{cases}

\end{align*}

Here $g$ isdefinedas even function, since $g(x)=g(-x)$ for all $x$ in the domain of $g$.Let us consider a function $h$ with

\begin{align*}

&h:(-l,l)\rightarrow\mathbb{R}\\

&h(x)=\begin{cases}

f(x)\qquad\quad\quad &x \geq 0\\

-f(-x)\qquad &x <0

\end{cases}

\end{align*}Here $h$ is defined as

oddfunction, since $h(x)=-h(-x)$ for all $x$ in the domain of $h$.We observe we can

extendthe function $f$ either to an odd or even function, so that approximation via a family of odd functions like $(\sin nx)$ or a family of even functions $(\cos nx)$ becomes feasible.

Note:When we talk about $\sin$ as odd function, we implicitely consider an appropriate domain.

*Epiloque:* To say it less formally, it is essential *where* an object *lives*.

Question: Is $f(x)=|x|$ differentiable? Answer: It depends! Question: Is $f(x)=\text{sign}(x)$ continuous? Answer: It depends! The term *continuous* touches another essential aspect. It does not merely depend on the domain and codomain as a whole, but also on the fine grained structure, namely what are the open sets, what are neighborhoods of points. This is a main theme in point set topology.

Add-on [2016-04-10]according to a comment of OP.It is stated in the question, that $(\sin nx)$ is a

completeset on $(0,\pi)$, which may sound somewhat peculiar, since the statement that

\begin{align*}

\{1,\cos x,\sin x,\cos 2x, \sin 2x, \ldots\}\tag{1}

\end{align*}

is acompleteset of orthogonal functions in $(-\pi,\pi)$ is much more common. In fact it’s again thedomainwhich is crucial for the correctness of OPs statement.Recall, a set of orthogonal functions is complete, if the function $0$ is the only function which is orthogonal to all elements of the set. This means, the set cannot be extended by another orthogonal function. In a more general context we talk about $L^2$ functions and call a set of orthogonal functions complete, if the only function orthogonal to all members of the set is $0$

almost everywhere.A proof showing this claim on $(0,\pi)$ based on the completeness of (1) on $(-\pi,\pi)$ is given in

this paper.

To show that the infinite set of functions $\sin(nx)$ or $$\{\sin{x}, \sin(2x), \sin(3x),…..,\sin(mx), \sin(nx)\}$$ is a complete set of orthogonal functions on $[0,\pi]$, where $m,n \in \mathbb{Z}$

We need to show that $$\int_{x=0}^{\pi}\sin(mx)\cdot\sin(nx)\,\mathrm{d}x=0 \quad\ \text{for}\quad m\ne n$$

To do this we make use of the identity: $$\sin(mx)\cdot\sin(nx)=\frac{\cos\Big((m-n)x\Big) -\cos\Big((m+n)x\Big)}{2}$$

$\Large\therefore$$$ \begin{align}\int_{x=0}^{\pi}\sin(mx)\cdot\sin(nx)\,\mathrm{d}x & = \int_{x=0}^{\pi}\frac{\cos\Big((m-n)x\Big) -\cos\Big((m+n)x\Big)}{2}\,\mathrm{d}x \\&= \frac12\left[\frac{\sin\Big((m-n)x\Big)}{(m-n)}-\frac{\sin\Big((m+n)x\Big)}{(m+n)}\right]_{x=0}^{\pi}\\&=0 \end{align}$$

Since the sine of zero or any multiple of $\pi$ is zero.

$\large{\fbox{}}$

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