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Let $\displaystyle t'=t-2c$(a)$\displaystyle f(t-2c) = f(t)$
so $\displaystyle f(t)=f(t'+2c)=f((t'+c)+c)=f(t'+c)=f(t')=f(t-2c)$
b) is false, it makes nonsense...(b)$\displaystyle f(t+ \frac {1}{2}c) = f(\frac {1}{2}t)$
(c)$\displaystyle f(\frac {1}{2}(t+c)) = f(\frac {1}{2}t)$
c) is false, because it would mean it has a period $\displaystyle \frac c2$. But c is the smallest number such that $\displaystyle f(t+c)=f(t)$
Ok...part (a)...I just don't seek how you got that. Where did you get $\displaystyle f(t'+2c)$ and how does $\displaystyle f((t'+c)+c)=f(t'+c)$?Let $\displaystyle t'=t-2c$
so $\displaystyle f(t)=f(t'+2c)=f((t'+c)+c)=f(t'+c)=f(t')=f(t-2c)$
Why is it nonsense? What makes it false?(b) is false, it makes nonsense...
Because I let $\displaystyle t=t'+2c$
It's easier if you make a substitution, as you'll report to the formula you know
Because c is the period !and how does $\displaystyle f((t'+c)+c)=f(t'+c)$?
Again, if you can't see it, let T=t'+c, you'd have f(T+c)=f(T), which is f((t'+c)+c)=f(t'+c)
Whatever you have, if you add c, it will give the same !
Hmmm I've been looking for an explanation... I think I found itWhy is it nonsense? What makes it false?
$\displaystyle f\left(\tfrac t2\right)=f \left(\tfrac t2+c\right)$
$\displaystyle =f \left(\tfrac 12 (t+2c)\right)=f \left(\tfrac 12 (t+c)+\tfrac c2\right)$
Can it possibly equal $\displaystyle f \left(\tfrac 12(t+c)\right)$ ?
Only if :
- c=0 (which is not possible otherwise it would not make sense for the periodicity)
- f is c/2 periodic. But as I explained for c), it's not possible.
Thus b) is false