Decomposition of a function into even and odd parts

[PainterGuy]

TL;DR Summary

A function could be decomposed into even and odd parts but do those parts equally contribute to form the resultant vector?

Hi,

I understand how any function could be decomposed into even and odd parts assuming the function isn't a purely even or odd to start with.

It's just like saying that any vector in x-y plane could be decomposed into its x- and y-component assuming it doesn't lie parallel to x- or y-axis. Its x-component could be 'stronger' than its y-component, or vice versa. Or, both components could be equal.

It it possible to make a similar statement about even and odd parts of a function? Is it possible that even part is dominant compared to its odd part, and so on?

Thank you!

 

[fresh_42]

What do you think?

Even and odd in this context is very misleading, because even and odd functions are something else.
What you do is a split into a sort of conjugates over the reals. Now simply try some lines of different signed slope, parabolas etc. and see what you get.

 

[mfb]

There are (multiple) ways to assign a magnitude to functions. How useful that is depends on the application.

Even and odd in this context is very misleading, because even and odd functions are something else.

No?
https://en.wikipedia.org/wiki/Even_and_odd_functionsf(x)=f(-x) for all x is even, f(x)=-f(-x) for all x is odd, everything else is neither even nor odd but can be written as the sum of the two.

 

[fresh_42]

There are (multiple) ways to assign a magnitude to functions. How useful that is depends on the application.No?
https://en.wikipedia.org/wiki/Even_and_odd_functionsf(x)=f(-x) for all x is even, f(x)=-f(-x) for all x is odd, everything else is neither even nor odd but can be written as the sum of the two.

Indeed. I overread the minus sign in the second term.

 

[PainterGuy]

Thank you!

There are (multiple) ways to assign a magnitude to functions. How useful that is depends on the application.

I'm not sure if I was able to put forward my question clearly.

f(x) = (x-1)^2
f_e(x) = x^2+1
f_o(x) = -2x

For the given f(x), I believe that it could be said that its even part is dominant over its odd part in terms of magnitude. Thanks.

 

[WWGD]

Thank you!
I'm not sure if I was able to put forward my question clearly.

f(x) = (x-1)^2
f_e(x) = x^2+1
f_o(x) = -2x

For the given f(x), I believe that it could be said that its even part is dominant over its odd part in terms of magnitude. Thanks.

Remember, as Mfb stated, it depends on your choice for magnitude function. If you use modulus/absolute value, then, yes. Other norms are , e.g., when working on a compact subset ( in the Real line, i.e., a closed, finite interval) , the Sup/Max over the interval ( for continuous functions).

 

[PainterGuy]

Thank you.

In case of f(x), we could use the absolute value of a fraction. If the even part is larger then minimum value of fraction |f_e(x)/f_o(x)|would be "1", or we can write it as |f_o(x)/f_e(x)| and the maximum value would be "1".

f(x) = (x-1)^2
f_e(x) = x^2+1
f_o(x) = -2x

 

[mfb]

This is a very special case where one function is always larger in magnitude. if you have $f_e = x^2-1$ and $f_o=2x$ then things get more complicated.

 

[PainterGuy]

This is a very special case where one function is always larger in magnitude. if you have $f_e = x^2-1$ and $f_o=2x$ then things get more complicated.

Thank you.

I agree with you. So, how we handle such cases? Should we use finite intervals?

 

[mfb]

Just give up the idea that "one function is dominant over the other" would be a universal concept.

$f_e(x)=|x| cos(x)$, $f_o(x)=|x| sin(x)$ if you need a more wild example. Their ratio is a well-known function.

 

[PainterGuy]

Just give up the idea that "one function is dominant over the other" would be a universal concept.

Thank you.

Okay. I give it up. But I believe that in many cases, this would be true and one part would be dominant over the other, and this is also true as you suggest that generalizing this result is difficult if not impossible.

 

[Stephen Tashi]

Okay. I give it up. But I believe that in many cases, this would be true and one part would be dominant over the other, and this is also true as you suggest that generalizing this result is difficult if not impossible.

You needn't give up. You should exert yourself to define "dominant over the other" precisely. What do you want that to mean? If we are given two functions f(x), g(x), how do we decide if f(x) is "dominant over" g(x)?

Restrict the type of functions that are considered. You picked a polynomial as an example. Are polynomials a subset of the "many cases" where your idea is true?

 

[PainterGuy]

Thank you!

What do you want that to mean? If we are given two functions f(x), g(x), how do we decide if f(x) is "dominant over" g(x)?

I'm going to repeat the content from post #7. You can see that the red curve,

, gradually goes toward zero and in this case, roughly speaking, the integral from 0 → ∞ would have a finite value. It was easier to decide if x²+1 is dominant over -2x.

But this definition doesn't hold as was pointed out by @mfb. For example, see the plot below for $f_e(x)=|x| cos(x)$, $f_o(x)=|x| sin(x)$. So, to be honest, I don't know how to define when one function is dominant over the other. Perhaps, we could define 'dominant' over a closed interval.

Thanks a lot.

 

[mfb]

Some things you can look at:
$$\lim_{x \to \infty} \left|\frac{f_e(x)}{f_o(x)}\right|$$
This will always be $\infty$ or 0 if both functions are polynomials, and it only depends on which function has the largest power of x. It doesn't exist in general, however, as the sine/cosine example shows.
$$\frac{max(|f_e(x)|)}{max(|f_o(x)|)}$$
...if these maxima exist.
$$\frac{\int f^2_e(x)dx}{\int f^2_o(x)dx}$$
where the integral runs from x=0 to infinity for square integrable functions or over one period for periodic functions.

 

[Stephen Tashi]

Perhaps, we could define 'dominant' over a closed interval.

Good idea.

A well known technique in mathematics is treat certain sets of functions as infinite dimensional vector spaces. In your original post, you make an analogy to 2 dimensional vectors.

Look at the definition of a vector space - the sophisticated definition, not the secondary school definition. You will see that the secondary school definition of vectors (vectors as arrows in finite dimensions) adds to the properties of an abstract vector space because arrows in finite dimensions have properties that abstract vectors need not have. In particular, arrows in finite dimensions have the concept of "length". So arrows in a finite dimension can be regarded as a normed vector space.

If you look at the abstract definition of a norm, it is more general that the particular example of computing the length of an arrow from its cartesian coordinates. On sets of functions, there are various ways of defining norms. A commonly used norm on a sets of (real valued) functions is defined by $||f(x)|| =\sqrt{ \int_a^b{ (f(x))^2 dx}}$. Using that concept of "length", you can test whether $||f(x)|| > ||g(x)||$ It needn't involve considering a ratio.

Before you go too far into refining the concept of dominance, you should consider whether decomposing a given function into an even and odd function can be done in more than one way. If so, could one decomposition have the even function dominant and the other decomposition have the odd function dominant?

 

[PainterGuy]

I really appreciate your help.

So, there are different ways to decide which function of the two is dominant and it depends upon the type of functions we are dealing with. For example, in case of polynomial function, one way of deciding the dominance was suggested @mfb .

Using that concept of "length", you can test whether $||f(x)|| > ||g(x)||$ It needn't involve considering a ratio.

So, this is another way and it might be more general than others but still not universal.

Before you go too far into refining the concept of dominance, you should consider whether decomposing a given function into an even and odd function can be done in more than one way. If so, could one decomposition have the even function dominant and the other decomposition have the odd function dominant?

I'm not sure if there are more than one way to decompose a function into its even and odd parts. I have always come across the following two formulas used for decomposition but I'm not very knowledgeable and couldn't research it much at the moment because of some other commitments so perhaps I should research it more and come back later.

Thank you for your time and help!

 

[SSequence]

Before you go too far into refining the concept of dominance, you should consider whether decomposing a given function into an even and odd function can be done in more than one way. If so, could one decomposition have the even function dominant and the other decomposition have the odd function dominant?

Let's take the example of constant function $0$. So we want:
$Even(x)+Odd(x)=f(x)$ ----for all $x \in \mathbb{R}$

Now since $f$ is the constant function $0$, in our case we get:
$Even(x)+Odd(x)=0$ ----for all $x \in \mathbb{R}$

In this specific case, it can be shown that the decomposition is entirely unique.

============

First consider the case of $Even(0)$ and $Odd(0)$. First observe that we must have $Even(0)=0$ because:
$Odd(0)=-Odd(-0)$ ------ definition of Odd function
$Odd(0)=-Odd(0)$
$2 \cdot Odd(0)=0$
$Odd(0)=0$

Now using:
$Even(0)+Odd(0)=0$
$Even(0)+0=0$
$Even(0)=0$

============

Now consider two arbitrary values $x_0,a \in \mathbb{R}$ ... where we have $x_0 \neq 0$ and $a \neq 0$. Suppose we had:
Case-1: $Even(x_0)=a$

From $Even(x_0)=a$
we get:
$Even(-x_0)=a$ ------ definition of Even function

But we also must have:
$Even(x_0)+Odd(x_0)=0$
$a+Odd(x_0)=0$
$Odd(x_0)=-a$

And now we can conclude:
$Odd(-x_0)=-Odd(x_0)$ ------ definition of Odd function
$Odd(-x_0)=-(-a)$
$Odd(-x_0)=a$

But now we get:
$Even(-x_0)+Odd(-x_0)=a+a=2a \neq 0$
which is a contradiction.

Case-2: $Odd(x_0)=a$

From $Odd(x_0)=a$
we get:
$Odd(-x_0)=-a$ ------ definition of Odd function

But we also must have:
$Even(x_0)+Odd(x_0)=0$
$Even(x_0)+a=0$
$Even(x_0)=-a$

And now we can conclude:
$Even(-x_0)=Even(x_0)$ ------ definition of Odd function
$Even(-x_0)=-a$

But now we get:
$Even(-x_0)+Odd(-x_0)=-a-a=-2a \neq 0$
which is a contradiction.

 

[SSequence]

One can genralise the above too. Briefly, here is how. We want:
$Even(x)+Odd(x)=f(x)$ ----for all $x \in \mathbb{R}$

$Even(0)+Odd(0)=0$ ----for all $x \in \mathbb{R}$
Using the property of odd function we can show that:
$Odd(0)=0$
and hence from the equality written above we can also conclude that $Even(0)=f(0)$.

For the more general case, once consider an arbitrary value $x_0 \in \mathbb{R}$ ... where we have $x_0 \neq 0$. Now let:
$Even(x_0)=e$
$Odd(x_0)=o$

We have:
$Even(x_0)+Odd(x_0)=f(x_0)$
$e+o=f(x_0)$ eq.[1]
The above equation is the first main equation.

To derive the second equation:
$Even(-x_0)=Even(x_0)$ ----property of even function
$Even(-x_0)=e$

$Odd(-x_0)=-Odd(x_0)$ ----property of even function
$Odd(-x_0)=-o$

And now we have:
$Even(-x_0)+Odd(-x_0)=f(-x_0)$
$e+(-o)=f(-x_0)$
$e-o=f(-x_0)$ eq.[2]

Writing eq.[1] and eq.[2] above we get:
$e+o=f(x_0)$
$e-o=f(-x_0)$I think the solution has to be unique because we have linear equations in two variables $e$, $o$. Also, re-call that we set (with the choice of $x_0 \neq 0$ being arbitrary):
$Even(x_0)=e$
$Odd(x_0)=o$

 

[mfb]

@SSequence: The decomposition is always unique and the proof is very short.
Let x be non-negative, then e(x)+o(x)=f(x) and e(-x)+o(-x)=f(-x), which can be rewritten as e(x)-o(x)=f(-x).
Two equations, two unknowns, the solution is e(x)=1/2 (f(x)+f(-x)) and o(x)=1/2 (f(x)-f(-x)), no ambiguity. This applies to all non-negative x, the extension to negative x follows from symmetry. qed.

 

[SSequence]

Yes, I did the same thing (more or less) in post#18.

==========

Also one small thing: "eventual dominance" (defined in obvious way) does have some important properties. So I think that it will def. form some kind of ordering. But I don't know exactly which one, as I am not quite familiar with all the varieties of orderings.

Probably more can be said than what I wrote below (but I will need to think more carefully for that).

For example, writing $f<g$ for "$f$ eventually dominates $g$".
---- As mentioned above there will be functions $f,g$ so that $f<g$ and $g<f$ are both false.

---- We do have the following property:
whenever $f<g$ and $g<h$ are true, we will always have $f<h$

---- We also have the following property:
whenever we have $f<g$ as true, then $g<f$ is false

 

[jbriggs444]

So I think that it will def. form some kind of ordering.

A partial ordering. As has been pointed out, there are pairs of unequal functions where neither "eventually dominates" the other.

 

[SSequence]

I think you are probably right as I remember reading something similar before.

 

[Stephen Tashi]

One can genralise the above the above too.

We can be more bold about generalizing. A general viewpoint is to consider situations where decomposition of a function into sums of functions is equivalent to solving a set of simultaneous equations and a relation among values of component functions can be used to reduce the number of variables. My guess is that this has been studied in some branch of higher mathematics. A moderately general situation would be where values of a component function are related to other values of that function by an invertible linear transformation. (One might call the linear transformation a "symmetry" of the component function.)

 

[PainterGuy]

Hi again,

I'm sorry that I took a break from this discussion so if you don't mind I would like to conclude it.

The following is a follow-up to my post #16 above.

$$\frac{max(|f_e(x)|)}{max(|f_o(x)|)}$$
...if these maxima exist.
$$\frac{\int f^2_e(x)dx}{\int f^2_o(x)dx}$$
where the integral runs from x=0 to infinity for square integrable functions or over one period for periodic functions.

I had assumed that you were referring to global maxima and that method could only be used when both even and odd functions have global maxima. Please correct me if I'm wrong. Thanks.I believe that each function, f(x), is made up of a unique pair of even and odd functions but there could be more than one way/method to find that pair. The function in itself shouldn't be even or odd. I'm saying it in reference to quoted part of @Stephen Tashi 's post in my positing #16.

The decomposition is always unique and the proof is very short.
Let x be non-negative, then e(x)+o(x)=f(x) and e(-x)+o(-x)=f(-x), which can be rewritten as e(x)-o(x)=f(-x).

Once I was trying to find out that who first came with the idea of decomposing a function into its even and odd parts so this might interest you: https://hsm.stackexchange.com/q/9942/9341

Thanks a lot for your help!

 

[nuuskur]

What do you assume about your functions? Are they necessarily smooth? Is one necessarily dominating the other? Studying some finite collection of examples is not very useful outside of potentially giving us some (mis-leading) intuition.

 

[PainterGuy]

What do you assume about your functions? Are they necessarily smooth? Is one necessarily dominating the other? Studying some finite collection of examples is not very useful outside of potentially giving us some (mis-leading) intuition.

Thank you but I have already been told that and agree with those replies. In my last posting I was only trying to clarify a point from @mfb 's post.

 

[olgerm]

If function f is real:
$f_{even}(x)=\int_{-\infty}^{\infty}(d\omega*e^{i*2\pi*\omega*x}* Re(\int_{-\infty}^{\infty}(dx_1*e^{-i*2\pi*\omega*x_1}*f(x_1))))$
$f_{odd}(x)=\int_{-\infty}^{\infty}(d\omega*e^{i*2\pi*\omega*x}* Im(\int_{-\infty}^{\infty}(dx_1*e^{-i*2\pi*\omega*x_1}*f(x_1))))$
or
$f_{even}(x)=IFT( Re(FT(f)))(x)$
$f_{odd}(x)=IFT( Im(FT(f)))(x)$
FT fouruer transform. IFT is inverst Fourier transform.i am not sure if it also always works if f is complex function.

 

[mfb]

Let $g(x)=i f(x)$. Then $\hat g(\omega) = i \hat f(\omega)$ due to the linearity of the Fourier transform. What was the real part is now the imaginary part and vice versa (+sign), so these formulas don't work any more.

 

[olgerm]

For complex functions it should be:
$f_{even}(x)=\int_{-\infty}^{\infty}(d\omega*e^{i*2\pi*\omega*x}* Re(\int_{-\infty}^{\infty}(dx_1*e^{-i*2\pi*\omega*x_1}*Re(f(x_1)))))+\sqrt{-1}*\int_{-\infty}^{\infty}(d\omega*e^{i*2\pi*\omega*x}* Re(\int_{-\infty}^{\infty}(dx_1*e^{-i*2\pi*\omega*x_1}*Im(f(x_1)))))$
$f_{odd}(x)=\int_{-\infty}^{\infty}(d\omega*e^{i*2\pi*\omega*x}* Im(\int_{-\infty}^{\infty}(dx_1*e^{-i*2\pi*\omega*x_1}*Re(f(x_1)))))+\sqrt{-1}*\int_{-\infty}^{\infty}(d\omega*e^{i*2\pi*\omega*x}* Re(\int_{-\infty}^{\infty}(dx_1*e^{-i*2\pi*\omega*x_1}*Re(f(x_1)))))$
or
$f_{even}(x)=IFT( Re(FT(Re(f))))(x)+\sqrt{-1}*IFT( Re(FT(Im(f))))(x)$
$f_{odd} (x)=IFT( Im(FT(Re(f))))(x)+\sqrt{-1}*IFT( Im(FT(Im(f))))(x)$
FT fouruer transform. IFT is inverst Fourier transform.