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Results tagged with zeroes
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user 21724
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Accepted
The $n$-th derivative has $n$ zeros. Can such a function be unbounded?
We then assume that $f^{(n)}$ has exactly $n$ distinct simple zeroes $x_n < x_{n-1} < \dots < x_1$ for some $n \geq 1$. … Together with the zeroes of $f^{(n+1)}$ between the $x_i$'s, we get $\geq n+1$ distinct zeroes (and therefore exactly $n+1$ zeroes). …
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