Proving the Odd Function Property: f of f(x) as an Odd Function

[t_n_p]

Homework Statement

Show cos²(x) is periodic

The Attempt at a Solution

I don't know how , could somebody please show me step by step?
Thanks in advance!

edit: I've got another show/justify question.
If f is an odd function, then show f of f is an odd function.

Whilst I know odd of odd results in an odd function, it seems such an explanation is not suffucient.
I have been shown this example, but can't really make any sense of it
f o f(-x) = f(f(-x))
= f(-f(x))
= -f(f(x))
= -f o f(x)

 

[D H]

A function f(x) is periodic if there exists some P>0 such that f(x+P)=f(x) over the domain of f. Try rewriting cos²(x) in terms of cos(2x).

 

[Mute]

For the first one, a function f(x) is periodic, with a period of T if f(x) = f(x + T). You can use this fact to solve your first problem:

$$\cos^2(x) = \cos^2(x + T)$$

A useful identity is $$\cos^2x = \frac{1}{2}\left(1 + \cos{2x})$$. Use that identity on both sides of the above equation, then group terms in x. Since T is a constant, it must be independent of x, so something must kill all of the x terms. Find the condition on T such that this happens. If you can find such a T, then f(x) is periodic with a period of T.

For the second one, the properties of odd functions should be enough.

An odd function is a function such that $$f(-x) = -f(x)$$, so

$$f(f(-x))=f(-f(x)) = f(-y) = -f(y) = -f(f(x))$$

Using $f(x) = y$ to show a little more explicitly how it works. Is that not a sufficient demonstration of what the question asks for?

 

[t_n_p]

in regards to the 2nd question, I don't understand how the negative can move positions from inside the brackets to outside and then outside again.

 

[D H]

You are told f(x) is an odd function: f(-x)=-f(x). You are asked to show that f(f(x)) is an odd function. Your "example" is a lot more than just an example. It is a proof of this conjecture.

Define y=f(x). Then f(f(x)) = f(y). You want to evaluate f(f(-x)). What is f(-x)? It is -f(x)=-y, since f is an odd function. (Note: This is not true for all functions, but it is true for all odd functions by definition.) Thus f(f(-x))=f(-f(x))=f(-y). Once again using that f is odd, f(-y)=-f(y). Using y=f(x), -f(y)=-f(f(x)). Thus f(f(x))=-f(f(x)), so f(f(x)) is odd.

 

[Dick]

A function f(x) is periodic if there exists some P>0 such that f(x+P)=f(x) over the domain of f. Try rewriting cos²(x) in terms of cos(2x).

Why would you need an identity? It just says to show that it is periodic, not to find the smallest period. cos(x+2pi)=cos(x) -> cos^2(x+2pi)=cos^(x).

 

[t_n_p]

I'm really quite confused.
I have been shown this alternative method

cos²(x+2pi)
=[cos(x+2pi)]²
=[cos(x)+cos(2pi)]²
=[cos(x)]²

However I don't see how the cos(2pi) part dissapears as cos(2pi)=1

 

[Gib Z]

Umm Cos (a+b) doesn't equal cos A + cos B by the way, who ever showed you that method. But the general method was correct, show that cos x is periodic to show cos^2 x is.

 

[t_n_p]

My bad, it should be
=[cos(x+2pi)]²
=[cos(x)]²

without that step, I thought something was fishy...

But my question still remains, how do you get from [cos(x+2pi)]² to [cos(x)]²?

 

[Gib Z]

umm...Think of a unit circle?

 

[t_n_p]

umm...Think of a unit circle?

silly me...

still struggling to comprehend how the negative can move through the brackets in the 2nd question though..

 

[Gib Z]

Ok f(x)= - f(x) is given to us.

We want to show f( f(x) ) = - f( f(x) )

Ok so on the LHS replace the f(x) inside, with -f(x) since we know we can.

Now we can f( f(x) ) = f ( -f(x) )

Now let's replace with the f(x) with u for a second,to make things less confusing. So it becomes f (-u). f is an odd function though, so f(-u)=-f(u)

Hence f( f(x) ) = - f( f(x) ). Done!

 

[t_n_p]

Ok f(x)= - f(x) is given to us.

We want to show f( f(x) ) = - f( f(x) )

Ok so on the LHS replace the f(x) inside, with -f(x) since we know we can.

Now we can f( f(x) ) = f ( -f(x) )

Now let's replace with the f(x) with u for a second,to make things less confusing. So it becomes f (-u). f is an odd function though, so f(-u)=-f(u)

Hence f( f(x) ) = - f( f(x) ). Done!

I don't get this part. Why do we replace with -f(x)?

 

[Gib Z]

f(x) = - f(x) is given to us?

 

[NateTG]

Ok f(x)= - f(x) is given to us.
...

Your notation...could use some help...or something.

f(x)=-f(x)
so
2f(x)=0
so
f(x)=0
...

 

[VietDao29]

edit: I've got another show/justify question.
If f is an odd function, then show f of f is an odd function.

Whilst I know odd of odd results in an odd function, it seems such an explanation is not suffucient.
I have been shown this example, but can't really make any sense of it
f o f(-x) = f(f(-x))
= f(-f(x))
= -f(f(x))
= -f o f(x)

Ok, if f is an odd function, then the following property should hold: f(-x) = -f(x), or, some also may write f(x) = -(f(-x)).

Now, f o f(-x) = f(f(-x)), you can get this step, right?
Since f(x) is odd, so f(-x) = -f(x). Right? So we have:
f o f(-x) = f(f(-x)) = f(-f(x))

Now, again, since f(x) is odd, we have: f(-f(x)) = -f(f(x)) (Notice the movement of the minus sign)

If you still cannot see why this is true, let k = f(x), we have:
f o f(-x) = f(f(-x)) = f(-f(x)) = f(-k) = -f(k) = -f(f(x))

Now, the final step -f(f(x)) is actually, -f o f(x).

So, we have shown that:
f o f(-x) = - f o f(x), hence, f o f(x) is an odd function.

Can you get it?

Ok, here's some other similar problems, you can try to see if you can do it.

---------------------------

Problem 1:
f(x) is an odd function, and g(x) is an even function.
a. Is f o g(x) odd, or even?
b. Is g o f(x) odd, or even?

Problem 2:
If f o g o h(x) is an even function, and we know that f(x), and g(x) are all odd functions, what can we say about h(x)?

 

[Gib Z]

Your notation...could use some help...or something.

f(x)=-f(x)
so
2f(x)=0
so
f(x)=0
...

I have a severe mental retardation, i can see that now >.< ahh i can't believe i didn't notice that...its f(-x) = -f(x) btw :(

 

[t_n_p]

After many minutes thinking over it, I finally got it. Thanks to all!