Tips For Solving The Problem Of Connecting The Dirichlet Kernel

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    Request If $L_n=displaystylefrac1piint_0^pi|D_n(t)|dt$, my spouse and I have $L_n=dfrac4pi^2log n+O(1) $.

    Proof Let’s start by dividing the interval $[0,pi]$ into subintervals, where $sinleft(n+dfrac 1 2right)$ ends with a sign. Since $sin t/2$ will always be positive in this region, our team can write }intlimits_0^pi $$frac1{pi D_nleft( t right) right means frac1 pi intlimits_0^frac2pi 2n + 1 fracsin t/2dt + frac1pi sumlimits_k is 1^n – 1 intlimits_frac2kpi 2n + 1^ frac2left ( k + 1 right)pi 2n + 1 frac sin left( n + frac12 right)t rightsin t/2dt + frac1pi intlimits_frac2n pi 2n + 1 ^ pi frac sin left( n + frac12 right)t rightsin t/2dt $$ The last integral helps you $0$ as $n$ goes to infinity for you. We can pay attention to the middle of the noise. The correct results $g(t)=dfrac1sin t/2 – dfrac1t/2$ are continuous with respect to $[0,pi]$, so the integrals mean $$c_n frac 1 pi int limits_0 ^frac2n pi 2n + 1 gleft( t $$ right)dt must exist and be bounded above by any constant for any $n$. Thenwe could assume that $$d_n=frac2pi sumlimits_k is 1^n – 1 intlimits_frac2kpi 2n + 1^frac2left( k + 1 right) pi 2n + 1 fracleftt dt$$ But $t^-1$ is positive and decreasing since we have limits $$frac2pi sumlimits_k equals 1^n – 1 frac2n + 12 left ( k + 1 right)pi intlimits_frac2kpi 2n + 1^frac2left( k + 1 right)pi 2n + left sin left( n + frac12 right ) t right dt leqslant d_n leqslant frac2pi sumlimits_k = 1^n – 1 frac2n + 12kpi intlimits_frac2kpi 2n + 1^frac2left( k + 1 right) pi 2n + 1 sin left( n + frac12 right)t right dt$$ But directly $$intlimits_frac2kpi 2n + 1^ dt= frac42n +1 $$ where $$ frac4pi ^2sumlimits_k = 1^n – 1 frac1k + 1 leqslant d_n leqslant frac4pi ^2sumlimits_k implies 1^n – 1 frac1k $$

    You probably know that $displaystylesumlimits_k = 1^n – 1 frac1k = log deborah + Oleft( 1 right)$ (just implement $=log +n+gamma + o ( 1)$), so we can say that $d_n = dfrac4pi ^2log s + Oleft( 1 right)$, i.e. $d_n-dfrac4pi^ 2 log n$ is bounded by a new constant of the form $ntoinfty$. Finally, we check the $$intlimits_0^frac2pi 2n + 1 left( fracsin n + frac12 right)tt dt$$ integral we got from our sum. The integrand is confident, stable and decreasing over the period under consideration, reaching a maximum $t=0$ at $n+dfrac definitely 2$, from which the following rough estimate can be given: $$int limit_0^ frac2pi 2n + 1 fracsin left( n + frac12 right)tt dt leqslant left( n + frac12 right)intlimits_0^frac2pi 2n + 1 dt implies pi $$

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    answered January 31, 2015 at 2:20 pm.

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