On the behaviour of conjugate derived Fourier sequence by Taylor-Cesàro product summability method (Q2775724)

Let \(\sum_{n=0}^\infty u_n\) be an infinite series and the \(n\)th partial sum of this series be denoted by \(S_n\). Then, the \(n\)th Taylor mean of order \(r\) of the sequence \(\{S_n\}\) is given by \(\sigma^r_n=\sum_{k=0}^\infty a_{nk}S_k\) where, \(\frac{(1-r)^{n+1}\theta^n}{(1-r\theta)^{n+1}}= \sum^\infty_{k=0} a_{nk}\theta^k\), \(| r\theta|<1\). This method is regular if and only if \(0\leq r<1\). Since \(r=0\) corresponds to ordinary convergence, it will be assumed here that \(0<r<1\).NEWLINENEWLINE In this paper, summability by Taylor-Cesàro product means or summability by \((T)(C,1)\) has been investigated. Let for a \(2\pi\)-periodic function \(f(x)\in L(-\pi,\pi)\) its Fourier series be given by NEWLINE\[NEWLINEf(x)\sim\tfrac12\, a_0+\sum_{n=1}^\infty(a_n\cos nx+b_n\sin nx)=\sum_{n=1}^\infty A_n(x).NEWLINE\]NEWLINE Then the series \(\sum_{n=1}^\infty(a_n\sin nx-b_n\cos nx)=\sum_{n=1}^\infty B_n(x)\) is called conjugate series. A new theorem on Taylor-Cesàro product summability of a conjugate derived Fourier sequence has been established.NEWLINENEWLINE Theorem: If \(\Phi(t)=\int^t_0|\phi(u)|\,du= o\left(\frac{t}{\log\frac1t}\right)\) as \(t\to +0\) then the conjugate derived Fourier sequence \(\{nA_n(x)\}\) is Taylor-Cesàro product summable \((T)(C,1)\) to zero.