Question

(Orthogonal Functions)(Complex Functions) I'm having trouble properly determining the norm of a given function; I'm getting zero, and that obviously can't be right. Workings posted below shortly.

Answer 1

\[\text{Determine the norm in set (a).} \ \ \ f(x)=e^{inx}.\]

Answer 2

\[||f(x)||^2=\int\limits_{-\pi}^{\pi}f(x)^2dx=\int\limits_{-\pi}^{\pi}\bigg(e^{inx}\cdot e^{inx}\bigg)dx=\int\limits_{-\pi}^{\pi}\bigg(e^{2inx}\bigg)dx\]

Answer 3

\[=\frac{e^{2inx}}{2 i n}\bigg]_{-\pi}^{\pi}=\frac{e^{i2\pi n}-e^{-i2\pi n}}{2 i n}\]

Answer 4

If you expand that using Euler's formula, every term in the numerator either cancels out or becomes zero upon evaluating them as trig functions, so I clearly must be doing something wrong. Where is my mistake? @wio

Answer 5

@perl

Answer 6

you dont need to expand it

Answer 7

But it's not to pi i n, it's to 2 pi i n, unless you can just separate them and leave e^2.

Answer 8

you can treat the two cases, when n is odd and when n is even. in both cases

Answer 9

e^(2 pi * i * n) = 1, for any n

Answer 10

e^(- 2 pi * i * n) = 1, for any n

Answer 11

Alright, so then you just have 1-1=0 in the numerator?

Answer 12

yes

Answer 13

But then your norm would be zero, there's no way that can be correct. The norm isn't supposed to be zero, right? That doesn't make any sense.

Answer 14

e^(2 pi * i * n) , this is the point (1,0) when n=0, that rotates 360 degrees every time you increment n by 1.

e^( pi * i * n) , this is the point (1,0) when n=0, that rotates 180 degrees every time you increment n by 1.

Answer 15

So we're saying a function is orthogonal to itself.

Answer 16

yes, `an orthogonal function` is orthogonal to itself

Answer 17

not just any function is orthogonal to itself

Answer 18

What you are saying is making little to no sense to me. What is an "orthogonal function"? If something is orthogonal, it happens to be orthogonal to something else, a set might behave orthogonally, but it's not like you have these functions and they can just be called "orthogonal". They're orthogonal under certain conditions, and it makes absolutely no sense to me, with the extension of the idea of orthogonality from vectors to functions, that a function can be orthogonal to itself.

I've never heard of a vector being orthogonal to itself, and I don't see neither how a function can just be called "orthogonal" (unless it's very explicitly used to reference a set of functions being orthogonal to one another), nor how a function can be orthogonal to itself.

Answer 19

I've just never seen or heard, to my knowledge, of a function being orthogonal to itself, so I'm having a hard time believing this.

And also the way you used orthogonal threw me off a lot. I don't understand.

Answer 20

$$\Large {
\text{In mathematics, two functions f and g}\\ \text
{ are called orthogonal}
\text{ if their inner product }\\{ \langle f,g\rangle} \text{ is zero for } f \neq g \text{ where }
\\ \langle f,g\rangle=\int_\mathbb{C} \overline {f(x)} . g(x)
}$$
http://en.wikipedia.org/wiki/Orthogonal_functions

Answer 21

Yep. I just haven't seen until now a function that has a norm of zero. To my understanding that would be like having a vector of zero length. It's trivial, it's meaningless, it's supposed to have some value, because you're supposed to use that norm to make that a unit vector, or in this instance, to make something orthonormal.

No disagreement on the definition.

Answer 22

Feeling 99% sure that a norm of zero absolutely should not be the answer here, in any case. This just doesn't make any sense.

Unless you're dealing with something that's already like, the zero vector, taking the norm of it, as a vector or otherwise, shouldn't give you zero. The length of said vector is not zero, and neither should generally be the inner product of a function with itself evaluated over some bounds.

Answer 23

"From the linearity property it is derived that x = 0 implies \langle x,x \rangle = 0, while from the positive-definiteness axiom we obtain the converse, \langle x,x \rangle = 0 implies x = 0. Combining these two, we have the property that \langle x,x \rangle = 0 if and only if x = 0."

That last sentence there.

Answer 24

page 8

Answer 25

Those are two different functions, v_j =/= v_k. That's not taking the inner product of a function with itself.

Answer 26

Yep, in agreement there entirely.

Answer 27

length and orthogonality are different concepts

Answer 28

Either way, why did it pretty overwhelmingly straightforwardly state as a property of the inner product that *unless your function itself is zero*, its norm should not be zero.

I mean, that's literally what it says. Is that or is that not true? Does it say something else?

I can't think of any examples where the norm of a nonzero functionwould be zero, and I can't find any.

Answer 29

Like, 100%, straightforward, simple yes or no, can the norm of a nonzero function be zero? Absolutely not is my current understanding, and-again-other examples would be fantastic, I am literally just failing to imagine any example, any at all, where the norm of a nonzero function is zero. Somebody has to have an example. I feel like I'm taking crazy pills.

Answer 30

lets solve
< f(x), f(x)> ^(1/2)

Answer 31

Can I get just a super straightforward, word answer to my question about the comment on the Inner Product wikipedia page.

Answer 32

It gets super infuriating in that people on here have a habit of throwing a ton of math uo-and that's fantastic, but I just need an absolutely firm, clear answer, is that just wrong?

Am I misreading it? Is it wrong? What's the issue here? Is that entry wrong, or am I misunderstanding it?

Answer 33

Can you please answer my question regarding the Wiki entry so we can clear that up and move on.

Answer 34

ok this is what i wanted
$$

\Large {{ || f(x)|| = } { \langle f(x), f(x) \rangle }^{\frac{1}{2}} }


$$

Answer 35

Alright, cool, can you answer my question on the Wiki entry, it's a really simple, worded, "yes" or "no" answer.

Answer 36

one sec

Answer 37

I don't think so, but i am checking

Answer 38

I don't understand that why, for the norm, you are integrating from zero to one, my book just has a to be, not ever zero to one.

Answer 39

*a to b as generic values of choice

Answer 40

ok i was thinking of a special case of orthogonal functions, fourier series

Answer 41

Yeah, no worries, I'm not dealing with that, though, I'm just saying, as a general statement:

If you have a nonzero function and you find its norm, the norm should not be zero.

This function that I started with, to my understanding, is a nonzero function, and I do not understand how its norm is yielding zero, then, and am assuming an algebraic error.

The only way that the function that I started with, that we've been having this whole rambling discussion over, can have a norm of zero, is if it itself is zero, according to the Wikipedia page section under elementary properties.

I do not understand how apparently e^{inx}=0.

Answer 42

im reading this
http://tutorial.math.lamar.edu/Classes/DE/PeriodicOrthogonal.aspx
it seems that we are looking at
$$
\Large {\{f_n(x)}\}
$$

Answer 43

scroll down to the definition

Answer 44

so we want to show that
$$\Large {\{ e^{\pi i n }} \} ,~ n=0,-1,1,-2,2,-3,3,...$$
is mutually orthogonal on (-pi, pi )

Answer 45

All we're trying to do is find the norm, that's all.

Answer 46

ok let me check on wolfram, be right back

Answer 47

Looks like it's nonzero on Wolfram Alpha, look at that. One hour down the drain. One hour of both of our time. I should've just gone to W|A in the first place.

Answer 48

can you give me link

Answer 49

http://www.wolframalpha.com/input/?i=integrate+from+-pi+to+pi+e%5E%7B2inx%7D

Answer 50

yes because you have to take the complex conjugate when you do dot product

Answer 51

It's 6 AM.

Answer 52

I'm going to bed, I have literally no idea whatsoever why it took nearly two hours to point out that I wasn't using the proper definition, but thanks, I guess.

At least that was figured out.

After two hours.

Answer 53

(Oh, sorry, correction, I asked two hours ago, we started talking over an hour ago. My bad. It just took two hours for that to get pointed out, but you had only been helping me out for an hour. Thanks, perl. Have a good day.)

Answer 54

$$
\Large {{ \langle f(x), f(x) \rangle } = \int\limits_{-\pi}^{\pi}f(x) ~f(x) ^{*}dx \\=\int\limits_{-\pi}^{\pi}\bigg( e^{inx}\cdot \overline {e^{inx}} ~\bigg ) dx
=\int\limits_{-\pi}^{\pi}\bigg( e^{inx}\cdot {e^{-inx}} ~\bigg ) dx\\
=\int\limits_{-\pi}^{\pi}\bigg( e^{inx-inx} ~\bigg ) dx\\
=\int\limits_{-\pi}^{\pi}\bigg( e^{0} ~\bigg ) dx\\
=\int\limits_{-\pi}^{\pi}\bigg( 1 ~\bigg ) dx\\
=\left[x \right]_{\hbox{-}\pi}^{\pi}
= \pi - (-\pi) = 2\pi
}
$$

Answer 55

the norm is the square root of this

Answer 56

so all these functions have length sqrt(2 Pi ) and these functions are mutually orthogonal , when n does not equal to m .
so dont use wolfram here, it would be inappropriate. unless you use the proper definition using the complex conjugate

Answer 57

http://www.wolframalpha.com/input/?i=evaluate+integral+e^%28nix%29*e^%28conjugate%28i%29*nx%29%2C+x+%3D+-pi..+pi

Note that if you conjugated the entire expression on the second term you would get a strange integral. wolfram does not know you are only interested in real x values on eals [0,1] and n is an integer.

strange integral (note the conjugation operator ^* ):
http://www.wolframalpha.com/input/?i=evaluate+integral+e^%28nix%29*%28e^%28i*n*x%29%29^*%2C+x+%3D+-pi..+pi

Answer 58

^I'm way more inclined to believe your result than the W|A output, do you have any idea how they got that? Also, thank you.