Odd Functions and Their Derivatives: A Theorem

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The discussion confirms that the derivative of an odd function is indeed an even function, and vice versa. It demonstrates this relationship using the properties of odd functions, where f(-x) = -f(x), and applies the chain rule to derive the conclusion. The derivation shows that f'(-x) = f'(x), establishing that the derivative of an odd function is even. This insight highlights a fundamental theorem in calculus regarding the behavior of odd and even functions. The conversation concludes with a sense of satisfaction in understanding the theorem.
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It seems the derivate of an odd function (f(-x)=-f(x)) is an even function (f(-x)=f(x)), and vice versa. Is there a theroem about this?
 
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Suppose f is odd. We have that (f(-x))' = (-f(x))' = -f'(x). But by the chain rule, (f(-x))' = -f'(-x). Thus -f'(-x) = -f'(x) <=> f'(-x) = f'(x) <=> f' is even.
 
Ah, that was easy. Thank you. :)
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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