Mathematics - Functions Question with Solution | TestHub

Since, $f(x)=e^{g(x)}$ $$\begin{array}{rl}\Rightarrow e^{g(x+1)}& =f(x+1)\\ & =xf(x)\\ & =x{e}^{g(x)}\end{array}$$ and $$g(x+1)=\log x+g(x)$$ $$\Rightarrow g(x+1)-g(x)=\log x$$ Replacing $x$ by $x-\frac{1}{2}$, we get $$\begin{array}{rl}g\left(x+\frac{1}{2}\right)-g\left(x-\frac{1}{2}\right)& =\log \left(x-\frac{1}{2}\right)=\log (2x-1)-\log 2\\ \therefore g^{\prime \prime }\left(x+\frac{1}{2}\right)-g^{\prime \prime }\left(x-\frac{1}{2}\right)& =-\frac{4}{(2x-1{)}^2}\end{array}$$ Substituting, $x=1,2,3,\dots ,N$ in Eq. (ii) and adding, we get $$g^{\prime \prime }\left(N+\frac{1}{2}\right)-g^{\prime \prime }\left(\frac{1}{2}\right)=-4\left\{1+\frac{1}{9}+\frac{1}{25}+\dots +\frac{1}{(2N-1{)}^2}\right\}.$$