Re: even/odd functions

On Jun 10, 2:26 am, Gary Wessle <phd...@xxxxxxxxx> wrote:
José Carlos Santos <jcsan...@xxxxxxxx> writes:



On 09-06-2007 19:43, Gary Wessle wrote:

testing with odd functions y=x and even functions y=x^2
is it always the case that

odd-function / even-function = even-function
even-function / odd-function = even-function
odd-function / odd-function = even-function
even-function / even-function = even-function

thanks
Say f is a function of a single argument.

1) f(-x) = f(x) for all x means f is even
2) f(-x) = -f(x) for all x means f is odd

Use these definitions to prove the results (as Arturo Magadin
suggested).
interesting results I arrived to, can some one second this!
odd-function / even-function = odd-function
even-function / odd-function = odd-function
odd-function / odd-function = neither
even-function / even-function = neither
Nobody can, since the third and the fourth one are false.

could some one give me examples of non-trig odd-function and
even-function so that I can apply the (1) and (2) test on them?
Sure. f(x) = x is odd and f(x) = x^2 is even.

I have a feeling that the results depend on the function and is not
generalized, we'll see.
Your feeling is wrong.
well, if I divide the odd function f(x) = x by itself I get f(x) = 1
and the same goes for the even function f(x) = x^2.
so y=1 is an even-function. does that settles it.
odd-function / odd-function = even-function
even-function / even-function = even-function

No. General statements such as these ones will be settled only when you
*prove* that they are true in general (which is extraordinarily easy to
do in these cases).

well
when using f(x) = x for an odd-function and f(x) = x^2 for an
even-function, testing the hypothesis gives:
odd-function / odd-function = 1
even-function / even-function = 1
which is f(x)=1.
now lets test this.
f(-x) = 1 = f(x) even-function
does this settles it, otherwise I am not able to look outside the
box. and thus I will need help because I give up.

(Groan) Here's a proof of your first observation:
odd-function / even-function = odd-function

Proof. Let f be an odd function and g be an even function. f being odd
means
f(-x) = -f(x), for all x (in the domain of f), and g being even means
g(-x) = g(x)
for all x (in the domain of g).

To show that f/g is odd, we need to show that (f/g)(-x) = (f/g)(x),
for any x
(in the domain of f/g), where
(f/g)(y) = f(y)/g(y).

Choose a x (in the domain of f/g). Then we calculate:

(f/g)(-x) = f(-x) / g(-x) = -f(x) / g(x),

since f is even and g is odd,

[equation continued] = - [f(x) / g(x)] = - (f/g)(x).

Since x was arbitrary, f/g is odd. QED.

Now you do it.

--- Christopher Heckman

.

Relevant Pages

  • Re: even/odd functions
    ... even-function so that I can apply the and test on them? ... does that settles it. ... even part odd part ... even/odd functions in class, ...
    (sci.math)
  • Re: even/odd functions
    ... even-function so that I can apply the and test on them? ... does that settles it. ... even part odd part ... even/odd functions in class, ...
    (sci.math)
  • Re: even/odd functions
    ... even-function so that I can apply the and test on them? ... does that settles it. ... Let f be an odd function and g be an even function. ... Choose a x (in the domain of f/g). ...
    (sci.math)
  • Re: even/odd functions
    ... even-function so that I can apply the and test on them? ... even part odd part ... even/odd functions in class, ... algebraic relationship. ...
    (sci.math)
  • Re: even/odd functions
    ... even-function so that I can apply the and test on them? ... even part odd part ... we did notice the stability of parity for n iterates ... do you know functions m without any parity ...
    (sci.math)
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