Help with proving a function can be written f = E + O

[pyrosilver]

Homework Statement

a. Prove that any function f with domain R can be written f = E + O, where E is even and O is odd.

b. Prove that this way of writing f is unique.

Homework Equations

The Attempt at a Solution

Well, I didn't know where to start and I don't know where I'm trying to go, so I don't know if I'm even on the right track.

since even: f(x) = f(-x) and odd: f(x) = -f(-x) -- as well as f(-x) = -f(x)...

E(x) + O(x) = E (x) + O(x)
E(-x) + O(x) = E(x) + O(x)
E(-x) + O(-x) = E(x) - O(x)
(E+O)(-x) = (E-O)(x)

not sure where to go from here. so if someone could help me figure out what I'm trying to end up with? or where to go from here?I would very much appreciate it. thanks!

 

[LCKurtz]

Assume you could write f(x) = E(x) + O(x). See if you can figure out what E and O would have to be. Then figure out if that works.

 

[pyrosilver]

I'm still a little confused. So here's what I did:

for finding out what E is:
f(x) = E(x) + O(x)
f(x) - O(x) = E(x)
(f-O)(x) = E(x)

then for O:
f(x) = E(x) + O(x)
f(x) - E(x) = O(x)
(f-E)(x) = O(x)

Is that right?

so then f(x) = (f-O)(x) + (f-E)(x)? I'm a little confused on what I'm supposed to do and where I should end up

 

[lurflurf]

almost
f(x) = E(x) + O(x)
so what is
f(-x) ?
show E and O exist and are unique

 

[LCKurtz]

"f(-x), does it equal E(x) - O(x)?

Yes, so you have f(x) = E(x) + O(x) and f(-x) = E(x) - O(x). Solve those two equations for E and O and what do you get?

 

[pyrosilver]

for the first, both E(x) and O(x) are E(x) + O(x). For the second, both E(x) and O(x) are E(x) - O(x). I'm still a bit confused though because i started with what I wanted to show?

 

[LCKurtz]

for the first, both E(x) and O(x) are E(x) + O(x). For the second, both E(x) and O(x) are E(x) - O(x). I'm still a bit confused though because i started with what I wanted to show?

What you are doing right now is show that if f can be expressed as the sum of an even and odd function, what must they be? Treat E and O as the two unknowns in the two equations and solve for them in terms of f.

Once you know what they must be if they are to work, then you can check to see if they actually do work.

 

[pyrosilver]

so for f(-x) = E(x) - 0(x):

E(x) = f(-x) + O(x)
O(x) = E(x) - f(-x)

and for f(x) = E(x) + O(x):

E(x) = f(x) - O(x)
O(x) = f(x) - E(x)

is that right?

"Once you know what they must be if they are to work, then you can check to see if they actually do work. "

how do I prove that there isn't a function that is not one of these, though? sorry I'm still a little confused

 

[LCKurtz]

so for f(-x) = E(x) - 0(x):

E(x) = f(-x) + O(x)
O(x) = E(x) - f(-x)

and for f(x) = E(x) + O(x):

E(x) = f(x) - O(x)
O(x) = f(x) - E(x)

is that right?

"Once you know what they must be if they are to work, then you can check to see if they actually do work. "

how do I prove that there isn't a function that is not one of these, though? sorry I'm still a little confused

No, that isn't what you want. You have two equations:

f(x) = E(x) + O(x)
f(-x) = E(x) - O(x)

Two equations in two unknowns. The unknowns are O and E. Solve for the two unknowns in terms of f. You don't solve two equations in two unknowns by having the unknowns in your formula for the unknowns.

Once you show that if can be written as E + O, then E and O must have a certain form, you know E and O couldn't have some other form.

 

[pyrosilver]

ok so in f(-x), E and O both equal f(-x) -f(-x), and in f(x), E and O are f(x) + f(x)?

 

[LCKurtz]

Do you know how to solve two simultaneous equations in two unknowns like:

2x + 3y = 5
-x + 4y = 3

for x and y?

 

[pyrosilver]

No, I've never done that before

 

[LCKurtz]

Well, no wonder you are having trouble with this. What level mathematics are you taking? Where did you get this problem?

 

[pyrosilver]

i'm taking a calc theory class but I'm in 8th grade and my teachers made me skip a couple levels of math. I'm trying to get out of this class because I'm really struggling but they won't let me

 

[LCKurtz]

So you haven't had, for example, Algebra I and II and trigonometry??

 

[pyrosilver]

I don't know what I've had. back in 5th grade I was ahead of my class so my teacher had me do other stuff, and that's kind of what I've been doing for the past couple years. I've never taken an official course like geometry or algebra II, I've just kind of done what they've told me to do :( and now they put me in a high school junior-senior level calc theory class and they won't let me drop down because i still have a decent grade in it somehow

 

[LCKurtz]

What do your parents have to say about your situation? I don't know what I can do to help you with it. Does your school have a school counselor? There should be someone locally that you could get through to.

 

[pyrosilver]

Okay i just googled how to solve for two equations simultaneously, I don't know if i didi this right. But I went:

f(-x) = E(x) - O(x)
O(x) = E(x) - f(-x)

so then

f(x) = E(x) + E(x) - f(-x)
-2E(x) = -f(-x) + f(x)

then can I divide by -2? I'm so sorry I have a feeling I'm being really slow and I don't mean to waste your time I'm just still confused on what my end product needs to be.

edit: my parents want me to stay in the class. I won't be at this school for much longer, so i guess I'm just going to stick it out and then drop down in high school. my parents mostly want me to do this because they think me taking this class makes my resume look good for high school. I just got to stick it out a little while longer :(

 

[LCKurtz]

Okay i just googled how to solve for two equations simultaneously, I don't know if i didi this right. But I went:

f(-x) = E(x) - O(x)
O(x) = E(x) - f(-x)

so then

f(x) = E(x) + E(x) - f(-x)
-2E(x) = -f(-x) + f(x)

then can I divide by -2? I'm so sorry I have a feeling I'm being really slow and I don't mean to waste your time I'm just still confused on what my end product needs to be.

edit: my parents want me to stay in the class. I won't be at this school for much longer, so i guess I'm just going to stick it out and then drop down in high school. my parents mostly want me to do this because they think me taking this class makes my resume look good for high school. I just got to stick it out a little while longer :(

It's easiest to add the two equations and divide by 2 to solve for E(x), then subtract the two equations and solve for O(x). Then verify that your E(x) that you get really is even and the O(x) you get is really odd, and that they add up to f(x)

I have to run now. Good luck.

 

[lurflurf]

That sound like fun what is calc theory. This is a common type of proof called an existence uniqueness proof. We desire to show that there is at least one answer, and at most one answer. Thus exactly one answer.
We have here the equations
f(x)=E(x)+O(x)
f(-x)=E(x)-O(-x)
This is a common type of equation called a system of simultaneous linear equations. One normally learns to solve them in prealgebra or the first (elementary) algebra class. If you did not read up on it now. The idea is we will be able to write the equations switched around so
E(x)=a*f(x)+b*f(-x)
O(x)=c*f(x)+d*f(-x)
for properly chosen numbers a,b,c,d. Try it.