Even and Odd functions

darfmore · 2012-03-16 21:12:43

Hello again guys,

I have been given a problem to solve and I'm not too sure how to go about it.

The question reads:

"A function f: R --> R is said to be even if f(-x) = f(x) for all x, and odd if f(-x) = -f(x) for all x.

a) Show that any function f: R --> R can be written as f(x) = E(x) + O(x) where E is even and O is odd.

b) Prove that there is only one way of writing f in this way"

Now so far I have not even found a way to start writing the proof of either of these things. I understand part a fully in that even and odd functions make sense to me, so I know that any even power of x must be even and any odd power must be odd, so an even/odd function can only contain even/odd terms. It also makes sense that if you collect all of the even terms together and the odd terms together than you can write the expression in a.

How could I go about proving either of these things to be true?
Any help understanding the thought process behind solving these kinds of problems would be greatly appreciated!

Thanks in advance,
Darfmore

Bob · 2012-03-16 21:30:49

hi darfmore,

I'll make a quick post as you're on line too. Then I'll try to answer your questions.

What is odd / even coming up.

Bob

Bob · 2012-03-16 21:38:03

hi again,

Yes even means things like

But also

which, you may know, can be written as a power series with only even powers.

and odd like

One way to recognise them is to look at the graph.

Even functions have line of symmetry along the 'y' axis and odd functions have rotational symmetry around (0,0).

Now to think about that question.

I'll have to get a piece of paper but I seem to remember a way to do this. (Long time ago so I've got to trawl back through years of memories.

Bob

Bob · 2012-03-16 21:51:03

Right. Here is one way. There are probably others. ( LATER EDIT: but see post #6 )

It is possible with any function to make an even function. Also to make a odd function.

eg.

For f(x).

Consider g(x) = f(x) + f(-x)

g(-x) = f(-x) + f(--x) = f(-x) + f(x) = f(x) + f(-x) = g(x).

So g is an even function.

and consider h(x) = f(x) - f(-x)

h(-x) = f(-x) - f(--x) = f(-x) - f(x) = - ( f(x) - f(-x) ) = - h(x)

So h is an odd function.

So we have f is the sum of an even and an odd function.

Bob

Bob · 2012-03-16 22:00:47

Here's an example.

exp(x) is neither odd nor even.

Bob

ps. I see you've got to show this is the only way. Didn't know that was the case. hhmmmm, working on it. B

Bob · 2012-03-16 22:08:31

OK. Got it.

One way to prove something is unique is to assume it isn't and get a contradiction.

Suppose f(x) = E1(x) + O1(x) = E2(x) + O2(x). ie there are two ways of doing it.

Then E1(x) - E2(x) = O2(x) - O1(x)

So an even function = an odd function =><=

So there aren't two ways.

Hopefully that wraps it up, but post back if you need clarification

Bob

Sylvia104 · 2012-03-23 02:16:22

bob bundy wrote:

So an even function = an odd function =><=

It isn't exactly a contradiction. There is exactly one function that is both odd and even and that's .

Bob · 2012-03-23 04:52:16

Ok fair enough.

I had my next post ready in my head with a proof of this and of the odd + odd is odd etc rules, but he never responded.

Why does that happen sometimes?

Bob

anonimnystefy · 2012-03-23 07:32:21

Sylvia104 wrote:

bob bundy wrote:
So an even function = an odd function =><=

It isn't exactly a contradiction. There is exactly one function that is both odd and even and that's
.

then E1(x) must equal E2(x) and O1(x) must equal O2(x) so it's not unique either way!

Bob · 2012-03-23 10:14:15

hi Stefy,

If f(x) = 0 for all x then I suppose you could say f(x) = 0 + 0 where the first 0 is even and the second is odd. That is a unique representation.

But Sylvia104 is correct; my proof lacks rigor. I realised this after the post and was ready to tighten it up if asked.

Bob

anonimnystefy · 2012-03-23 10:21:49

but your proof is correct.i don't see what's wrong with it.

Bob · 2012-03-23 16:02:38

The proof is OK if you make certain assumptions.

(i) the sum of two odd functions is also odd.

(ii) the sum of two even functions is also even.

(iii) a multiple of an even (odd) function is also even (odd).

(iv) the function is not zero for all x.

I should have made my proof more rigorous, but it might have made it harder to follow so I left steps out.

The contradiction is OK subject to all those assumptions.

Bob

anonimnystefy · 2012-03-23 16:29:48

The first three are facts not assumptions.

The forth one does not have to apply as you have shown.

Bob · 2012-03-23 21:00:33

hi Stefy,

In mathematics you start with axioms and prove theorems from them (big simplification, I know).

What I'm saying is this. We have definitions for odd and even functions. There may be other properties of these functions which turn out to be true (which I guess makes them facts) but, in a properly argued thesis these extras should be proved, otherwise they might not be true. This is not trivial.

For example, you might expect that an odd function multiplied by an odd function will also be odd. But this is not so. (counter example: x^3 times x^3) So to be properly rigorous, prove everything.

eg. odd added to odd is also odd

proof.

suppose f(x) and g(x) are odd functions

Consider h(x) = f(x) + g(x)

h(-x) = f(-x) + g(-x) = -f(x) + -g(x) = - {f(x) + g(x)} = - h(x)

so h is also an odd function.

My proof (leading to contradiction) uses this fact, so now I've fully justified that part of the proof. I'll leave the other properties as an exercise to the reader.

Bob

anonimnystefy · 2012-03-23 21:08:09

I love that axiomatic part of math, although I never tried doing axiomatic aspirants.But I love axiomatic geometry.

I didn't expect that.My intuition is very good when it comes to that stuff.I would have noticed there was a counter-example.

Oh, how I hate does words.As an exercise to the reader.

Btw, how well do you know probability?

Bob · 2012-03-23 21:23:48

hi Stefy,

Before joining the forum I would have said my understanding of probabliity was fairly good. Then I came across some of bobbym's posts and realised there's a whole world of formulas I've never even heard of. So, my answer depends on who I am being compared with.

Mostly, I don't bother to learn formulas at all, I just work stuff out from scratch when I need to. For example, when I was doing A level, I couldn't be bothered to learn the quadratic formula so I used to do all quadratics that wouldn't factorise by 'completing the square'. Slower, but I think it helped my understanding tremendously. And if a question required a proof then I was very happy because that's what I did all the time anyway.

So, if you have a probability question I'll have a look. And we can always admit defeat and let 'the afore-mentioned' help us out.

Bob

anonimnystefy · 2012-03-23 21:29:10

A random person would notice that you are a professor from a mile.You only concentrated to the question in my post.

I don't have any definite problem.I just want to learn probability a little.

Math Is Fun Forum

#1 2012-03-16 21:12:43

Even and Odd functions

#2 2012-03-16 21:30:49

Re: Even and Odd functions

#3 2012-03-16 21:38:03

Re: Even and Odd functions

#4 2012-03-16 21:51:03

Re: Even and Odd functions

#5 2012-03-16 22:00:47

Re: Even and Odd functions

#6 2012-03-16 22:08:31

Re: Even and Odd functions

#7 2012-03-23 02:16:22

Re: Even and Odd functions

#8 2012-03-23 04:52:16

Re: Even and Odd functions

#9 2012-03-23 07:32:21

Re: Even and Odd functions

#10 2012-03-23 10:14:15

Re: Even and Odd functions

#11 2012-03-23 10:21:49

Re: Even and Odd functions

#12 2012-03-23 16:02:38

Re: Even and Odd functions

#13 2012-03-23 16:29:48

Re: Even and Odd functions

#14 2012-03-23 21:00:33

Re: Even and Odd functions

#15 2012-03-23 21:08:09

Re: Even and Odd functions

#16 2012-03-23 21:23:48

Re: Even and Odd functions

#17 2012-03-23 21:29:10

Re: Even and Odd functions

Board footer