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For $p=0,1,2,\dots$ and $n=0,1,2,\dots,$, let $f_{n,p}(z)=\sum_{k=0}^n k^p z^k$ be a sequence of polynomials. Restricted to the unit circle, the functions $g_{n,p}(t):=f_{n,p}(e^{it})$ are trigonometric polynomials.

I would like to estimate the decay of $\left|g_{n,p}(t)\right|$ "away from" $t=0$ for $n\gg 1$. For example, if $p=0$ then $$\left|g_{n,0}(t)\right|^2=\frac{\sin^2\frac{(n+1)t}{2}}{\sin^2 \frac{t}{2}}.$$ While $g^2_{n,0}(0)=(n+1)^2$, for $t>2\arcsin{1\over\sqrt{n}}$ we have $\left|g_{n,0}(t)\right|<n$. Numerical evidence seems to suggest that similar estimate holds for $p>0$. That is, $$\left| g_{n,p}(0)\right|^2=\biggl(\sum_{k=0}^n k^p\biggr)^2=\mathcal{O}(n^{2p+2}), $$ and is it true that for $t>2\arcsin\frac{1}{\sqrt{n}}$ we have already $$ \left| g_{n,p}(t) \right|^2 < n^{2p+1}? $$

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  • $\begingroup$ The functions are periodic in $t$, in what exact ranges are you interested? $\endgroup$ – Asaf May 4 '14 at 17:21
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    $\begingroup$ I don't know about the exact inequalities, but if constants are not bothering you, then summation by parts should give $g(t)=O(n^pt^{-1})$. $\endgroup$ – Christian Remling May 4 '14 at 17:48

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