Firstly, I'm aware of the conditions that need to be satisfied for odd and even functions; $f(-x)=-f(x)$ and $f(-x)=f(x)$ respectively.

I'm also aware that the product of an odd function with another odd function is an even function. In the proof for this, nothing is said about whether the two functions have to be the same or different.

So I will assume that the functions can both be the same (and hence say $x^3 \times x^3= x^6$ which is even, as required).

But now consider this function: $$g(x)= \frac{4x}{\pi}-\frac{4x^2}{\pi^2}=x\left(\frac4\pi-\frac{4x}{\pi^2}\right)=f_1(x)\times f_2(x)$$

here $f_1(x)=x\qquad$ (which is an odd function),

and $f_2(x)=\dfrac4\pi-\dfrac{4x}{\pi^2}\qquad$ (which is an odd function).

So does this mean that $g(x)$ is even? Well actually the function is neither odd or even according to $f(-x)=-f(x)$ and $f(-x)=f(x)$.

Could someone please explain why the example I gave does not satisfy the odd times odd = even function relation?

  • 4
    $\begingroup$ Wait! $f_2$ is not odd! To be odd, a function needs to pass through the origin. No "exception" can be found for a proof :) $\endgroup$ – Tommaso Seneci Apr 3 '18 at 13:32
  • 3
    $\begingroup$ $f_2$ is strange but not odd. ;-) $\endgroup$ – Przemysław Scherwentke Apr 3 '18 at 13:33
  • 2
    $\begingroup$ But $f_2$ isn't odd. There's that constant term. $f_2(-x) = 4/\pi + 4x/\pi^2 \neq -4/\pi + 4x/\pi^2 = -f_2(x).$ $\endgroup$ – Shirish Kulhari Apr 3 '18 at 13:33
  • 2
    $\begingroup$ So it seems like my mistake is assuming that adding or subtracting a constant to an odd function is that the overall function is still odd? $\endgroup$ – user395550 Apr 3 '18 at 13:36
  • 3
    $\begingroup$ @user395550: Yes $\endgroup$ – Shirish Kulhari Apr 3 '18 at 13:37

There are no exceptions for this rule.

Let 2 odd functions be $ f(x) $ and $ g(x) $

So, $ f(-x)=-f(x), g(-x)=-g(x) $

Multiplying $ f(-x) $ and $ g(-x) $ we get $ f(-x)g(-x) = (-f(x))(-(g(x))) = f(x)*g(x) $

Now you can assume $ f(x)g(x) $to be another function $ h(x) $ which is even because of the above result.

Since this result is obtained by taking any 2 general functions it must be true without any exceptions.

In your example, f2 is not an odd function. Check it again


It is not possible since

$$f(-x)=-f(x)\quad g(-x)=-g(x)\implies fg(-x)=fg(x)$$

Note also that in your example $f_2(x)$ is not odd.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.