Is y=0 an Odd Function Despite Not Being Symmetric About the Origin?

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How is y=0 an odd function when it isn't symmetric to the origin?
 
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Sure it is. Consider f(x)=0 and recall the definition of an odd function.
 
Note here, that for any function f in general, it's not necessary that f is either odd or even; nor that it cannot be both. You've found an example of the latter :smile:
 
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CompuChip said:
Note here, that for any function f in general, it's not necessary that f is either closed or open; nor that it cannot be both. You've found an example of the latter :smile:
Did you mean "odd" and "even"? f(x)= 0 is clearly a closed function, certainly not open!
It is definitely both even and odd.
 
Heh, just read a topology thread. I was thinking "odd" and "even", but I wrote "open" and "closed". Still a bit sleepy probably :O
I'll change my post
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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