Question

(Orthogonal Functions) More info below, finding the orthogonality of a complex function in the Hermitian sense.

Answer 1

\[\text{Show that the set $e^{inx}, -\pi \leq x \leq \pi, n \in Z,$ is orthogonal} \]

Answer 2

(Orthogonal in the Hermitian sense)

Answer 3

My idea of this would be that if \[w(x)=u(x)+iv(x)\]is the general form of a complex function, then in this situation \[iv(x)=e^{inx}\]\[u(x)=e^0\]

(Need to find the complex conjugate to compute the Hermitian)

That just sounds fishy to me, though. Would that really just mean that its complex conjugate is 1? @SithsAndGiggles

Answer 4

The way I'm currently imagining this is that\[u(x)=0, \ \ \ iv(x)=inx, \ \ \ \overline{w}(x)=u(x)-iv(x)=e^{0-inx}=e^{-inx}\]

Answer 5

Is this right? @perl

Answer 6

@Zarkon

Answer 7

(Misread originally the definition of a complex function, put things as I understand them with the correct definition in the fourth from current post)

Answer 8

demonstrate that
$$
\Large
{ \int_{-\pi}^{\pi}e^{inx}\cdot \overline { e^{inx} }=0



}
$$

Answer 9

I'm not even sure about the definition; what I want to understand before I do anything else is whether the complex conjugate of \[e^{inx}\]
is\[e^{-inx}\]

Answer 10

Also, how did you scale up the font? I'm not sure how to do that in the OS LaTeX editor.

Answer 11

Alright, so that answers my question, cool. Then you just have the definite integral of zero, which is c, but since you're evaluating at those bounds, it's just c-c=0.

Answer 12

oh wait, its not so easy

Answer 13

(?)

Answer 14

But yeah, how did you scale up the font in the OS LaTeX editor, I need to know how to do that.

Answer 15

type \Large , or \large

Answer 16

you can use brackets around multiple sentences
\Large { formulas and stuff }

Answer 17

Alright, awesome, I wasn't aware that would work in this environment.

But yeah, why is it not so easy/what is wrong with what I said?

Answer 18

$$
\Large
{ \Large
{ \int_{-\pi}^{\pi}e^{inx}\cdot \overline { e^{imx} }=0
\\\text { } \\ \text{however}\\\text { } \\
\overline { e^{imx} } = e^{-imx}

}


}

$$

Answer 19

for any m not equal to n

Answer 20

scroll up one page to get the definition of Hermitian orthogonal

Answer 21

I get an HTTP 500 error, but I'll try again.

Answer 22

one sec, i see

Answer 23

Yeah, no, I agree with you on that, heh, I just am not sure how to do it from here.

Answer 24

click the first search link
https://www.google.com/search?tbm=bks&hl=en&q=show+function+is+orthogonal+hermitian+sense

Answer 25

then scroll up a page to get the definition of hermitian

Answer 26

Nah, it worked the second time for whatever reason, and I'm in agreement/knew that, just thinking about how to now demonstrate that with m =/= n.

Answer 27

we have to solve that integral

Answer 28

\[\int\limits_{}^{}e^{inx}e^{-imx}=e^{n/m}dx\]

Answer 29

\[\large{ \int\limits_{-\pi}^{\pi}e^{n/m}dx=xe^{n/m}\bigg]^{\pi}_{-\pi} }\]

Answer 30

Ah, nevermind, I see what you're doing, my bad

Answer 31

$$

\Large

{ \int_{-\pi}^{\pi}e^{inx}\cdot \overline { e^{imx}} ~dx
\\=
\int_{-\pi}^{\pi}e^{inx}\cdot { e^{-imx} }~dx\\
=\int_{-\pi}^{\pi}e^{inx-imx}~dx\\
=\int_{-\pi}^{\pi}e^{i(n-m)x}~dx\\

}





$$

Answer 32

\[\large{\int\limits_{-\pi}^{\pi}e^{(ix)(n-m)}dx }\]

Answer 33

if you write click on my latex code, you can see 'show text as ' and choose TEx

Answer 34

(Lol for whatever reason didn't read your last line, will try to integrate/move forward from here)

Alright

Answer 35

choose "show math as" choose Tex command.
then you can copy paste
(begin and end the latex code with double dollar signs)

Answer 36

i never use the equation editor. all the latex commands are online anyway

Answer 37

ok now we integrate this

Answer 38

\[\Large{\int\limits_{-\pi}^{\pi}e^{(m-n)ix}dx= \left[ \frac{e^{(m-n)ix}}{(m-n)i} \right] }\]

Answer 39

I'm not sure why, as opposed to using TeX Studio, where that doesn't happen, the left bracket isn't disappearing when I go "\left.", but in any case

Answer 40

$$
\Large

{ \int_{-\pi}^{\pi}e^{inx}\cdot \overline { e^{imx}} ~dx
\\=
\int_{-\pi}^{\pi}e^{inx}\cdot { e^{-imx} }~dx\\
=\int_{-\pi}^{\pi}e^{inx-imx}~dx\\
=\int_{-\pi}^{\pi}e^{i(n-m)x}~dx\\
u = i(n-m)x\\
du = i(n-m)~dx\\
\frac{du}{i(n-m)}= dx \\
\int_{-\pi}^{\pi}e^{i(n-m)x}~dx = \left [\frac{e^{(n-m)ix} } {(n-m)i} \right]_{-\pi}^{\pi}


}

$$

Answer 41

i tried the same, to delete the left bracket. wont allow it

Answer 42

Sorry, had to go to class. But alright, looks like this makes sense.