MHB Half wave symmetry and integrals

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A periodic wave x(t) with half-wave symmetry satisfies the condition x(t + T0/2) = -x(t). This property leads to the question of whether the integral of the wave, X(t), also exhibits half-wave symmetry, specifically if X(t + T0/2) = -X(t). The discussion confirms that integrating both sides of the symmetry condition supports this conclusion. Overall, the relationship holds true for various examples considered in the discussion.
fernlund
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If I have a periodic wave x(t) with half-wave symmetry, it means that:

x(t + T0/2) = -x(t)

where T0 is the period of the wave. Would this automatically lead to the conclusion that

X(t + T0/2) = -X(t)

where X'(t) = x(t), i.e X(t) is the integral of x(t).

?
 
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fernlund said:
If I have a periodic wave x(t) with half-wave symmetry, it means that:

x(t + T0/2) = -x(t)

where T0 is the period of the wave. Would this automatically lead to the conclusion that

X(t + T0/2) = -X(t)

where X'(t) = x(t), i.e X(t) is the integral of x(t).

?

I think so. Can you integrate $x(t+T_0/2)=-x(t)$ on both sides w.r.t. $t$, and obtain your result?
 
Ackbach said:
I think so. Can you integrate $x(t+T_0/2)=-x(t)$ on both sides w.r.t. $t$, and obtain your result?

Yeah, it's correct for all of the examples I can think of. Thanks!
 

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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