Even/odd composite functions

  • Thread starter roam
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  • #1
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Main Question or Discussion Point

Considering the composition of two functions ƒ · g

If g is even then does this mean that ƒ · g is even? why?

Or if g is odd and ƒ is even, then ƒ · g is even?

How can we show these statements?

Thanks.
 
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Answers and Replies

  • #2
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Consider the fact that the identity function I(x) = x is odd and the absolute value function A(x)=|x| is even.
 
  • #3
HallsofIvy
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Just like you would do any proof of this sort: use the definitions.

If g is even, then g(-x)= g(x). Now, what can you say of [itex]f\cdot g(-x)= f(g(-x))[/itex]?

If g is odd then g(-x)= -g(x). If f is even then f(-x)= f(x). Now, what can you say of [itex]f\cdot g(-x)= f(g(-x))[/itex]?
 
  • #4
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If g is even, then g(-x)= g(x). Now, what can you say of [itex]f\cdot g(-x)= f(g(-x))[/itex]?
Could you explain a little bit more on this part please? Thanks.

f(-x)= f(x)
g(-x)= g(x)

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

It is even? Because a function is even if:
f:(-a,a) -> R if for all [tex]x \in (-a,a)[/tex], f(x) = f(-x)

Please help me, I don't know if I'm right.

Regards,
 
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  • #5
tiny-tim
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Hi roam! :smile:

You need (for each part) a proof that starts "(ƒ · g)(-x) = … ", and finishes " … = (ƒ · g)(x)."

Hint: suppose g(3) = 7.

If g is even, what are (ƒ · g)(3) and (ƒ · g)(-3)?

If g is odd, what are (ƒ · g)(3) and (ƒ · g)(-3)? :smile:
 
  • #6
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If g is even, what are (ƒ · g)(3) and (ƒ · g)(-3)? even

If g is odd, what are (ƒ · g)(3) and (ƒ · g)(-3)? even

What if g is odd and f is even? would the (f · g) be even?
 
  • #7
tiny-tim
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Hi roam! :smile:

No … you're missing the point!

Follow the hint … if g(3) = 7, what is (ƒ · g)(3) (how is (ƒ · g)(3) defined? and so what is it)? And what is (ƒ · g)(-3) (same procedure)?
 
  • #8
HallsofIvy
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Could you explain a little bit more on this part please? Thanks.

f(-x)= f(x)
g(-x)= g(x)

f.g(-x) = f(-x(g(-x)))
NO! f.g(-x)= f(g(-x)) as you say below:
f.g(-x) = f(g(x))
and f(g(x))= f.g (x) doesn't it?

It is even? Because a function is even if:
f:(-a,a) -> R if for all [tex]x \in (-a,a)[/tex], f(x) = f(-x)
So you have just said, (f.g)(-x)= f.g(x), haven't you?

Please help me, I don't know if I'm right.

Regards,
 
  • #9
and f(g(x))= f.g (x) doesn't it?
Sorry for the off-topic:

Is the "=" sign typically read as "is equal to" or as "equals"...?
I thought it would be the former but according to your question tag "doesn't it" you seem to use the latter.
Hm, probably both are possible:smile:
 
  • #10
HallsofIvy
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Actually, I debated whether to say "doesn't it" or "isn't it" myself! Yes, "=" can be read as either "equals" or "is equal to".

I was thinking "f(g(x) equals f.g(x) doesn't it" but I considered "f(g(x)) is equal to f.g(x) isn't it".
 

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