Let $g(x)=\log f(x)$, where $f(x)$ is a twice differentiable positive function on $(0, \infty)$ such that $f(x+1)=x f(x)$. Then, for $N=1,2,3, \ldots . g^{\prime \prime}\left(N+\frac{1}{2}\right)-g^{\prime \prime}\left(\frac{1}{2}\right)$ is equal to

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Aicharya · Accepted Answer

Since, $\quad f(x)=e^{g(x)}$ $ \begin{aligned} \Rightarrow \quad e^{g(x+1)} & =f(x+1) \ & =x f(x) \ & =x e^{g(x)} \end{aligned} $ and $ g(x+1)=\log x+g(x) $ $ \Rightarrow \quad g(x+1)-g(x)=\log x $ Replacing $x$ by $x-\frac{1}{2}$, we get $ \begin{aligned} g\left(x+\frac{1}{2}ight)-g\left(x-\frac{1}{2}ight) & =\log \left(x-\frac{1}{2}ight)=\log (2 x-1)-\log 2 \ 	herefore g^{\prime \prime}\left(x+\frac{1}{2}ight)-g^{\prime \prime}\left(x-\frac{1}{2}ight) & =-\frac{4}{(2 x-1)^2} \end{aligned} $ Substituting, $x=1,2,3, \ldots, N$ in Eq. (ii) and adding, we get $ g^{\prime \prime}\left(N+\frac{1}{2}ight)-g^{\prime \prime}\left(\frac{1}{2}ight)=-4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N-1)^2}ight\} . $

Let $g(x)=\log f(x)$, where $f(x)$ is a twice differentiable positive function on $(0, \infty)$ such that…

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