I have a question and I have a definition $\{\varphi_n(t)=\frac{1}{\sqrt{2\pi}}e^{int}\}$.I know that $n\in \mathbb Z$ for this question. does int means integer?
I couldn't find any reference with a web search.
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The exponent $int$ is indeed a product of the imaginery number $i$, an integer $n$ and a variable $t$. By Euler's formula, we have $$e^{int} = \cos(nt) + i\sin(nt).$$
What is important about this sequence of functions $\{\phi_n(t)\}_{n\in\mathbb{Z}}$ is that they form a orthonormal Schauder basis for the inner product space $L^2(0,2\pi)$, i.e.
$$\int_0^{2\pi} \phi_m(t) \overline{\phi_n(t)} \, dt = \begin{cases} 1 & \text{ if } m = n,\\ 0 & \text{ otherwise.}\end{cases}$$
and for any $f \in L^2(0,2\pi)$, there exists a sequence $(f_j)_{j=1}^\infty \subset \langle \phi_n \rangle$ such that $$\int_0^{2\pi} \Big( f(t) - f_j(t) \Big)^2 \, dt \to 0 \text{ as } j \to \infty. $$
Empiricist
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My question contains $\int_{-\pi}^\pi$ whithin $L_2$. Can you kindly expand on the meaning of $\langle \phi_i ,\phi_j \rangle = \delta_{i,j}$? Does it means $\langle \phi_i ,\phi_j \rangle =1$ for $i=j$ and $\langle \phi_i ,\phi_j \rangle =0$ for $i \neq j$? – havakok Jun 11 '16 at 07:17
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1Since the trigonometric functions are $2\pi$ periodic, the same happens when you domain is $(-\pi, \pi)$. For the second question, yes, I have clarified in my answer. – Empiricist Jun 11 '16 at 07:20
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I believe from context $\varphi_n(t)=\frac{1}{\sqrt{2\pi}}e^{int}=\frac{1}{\sqrt{2\pi}}(\cos(nt)+i\sin(nt))$
$i^2=-1$
$n \in \mathbb Z$
with argument $t$.
JasonM
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$\text{int}$ here $i$ is iota and $e^{int}$ =$ \cos nt +i\sin nt $
Edit: if $i$ does not refer to iota then it can represent any number, although usually $i^2=-1$.
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6The $i$ doesn't carry through into the argument of the cosine or sine functions. – okrzysik Jun 11 '16 at 07:04
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