Let $g(x)=\log f(x)$, where $f(x)$ is twice differentiable positive function on $(0, \infty)$ such that $f(x+1)=x f(x)$. Then, for $N=1,2,3, ....g^{\prime \prime}\left(N+\frac{1}{2}\right)-g^{\prime \prime}\left(\frac{1}{2}\right)=$
Answer & explanation
Correct answer: option 1
We have,
$g(x)=\log f(x)$ and $f(x+1)=x f(x)$
$\Rightarrow g(x+1)=\log f(x+1)$
$\Rightarrow g(x+1)=\log (x f(x))$ $[∵ f(x+1)=x f(x)]$
$\Rightarrow g(x+1)=\log x+\log f(x)$
$\Rightarrow g(x+1)-g(x)=\log x$
$\Rightarrow g^{\prime \prime}(x+1)-g^{\prime \prime}(x)=-\frac{1}{x^2}$
Putting $x=\frac{1}{2}, \frac{3}{2}, \frac{5}{2}, \frac{7}{2}, ..., \frac{2 N-1}{2}$, we get
$g^{\prime \prime}\left(1+\frac{1}{2}\right)-g^{\prime \prime}\left(\frac{1}{2}\right)=-4$
$g^{\prime \prime}\left(2+\frac{1}{2}\right)-g^{\prime \prime}\left(1+\frac{1}{2}\right)=-\frac{4}{9}$
$g^{\prime \prime}\left(3+\frac{1}{2}\right)-g^{\prime \prime}\left(2+\frac{1}{2}\right)=-\frac{4}{25}$
. . .
. . .
. . .
$g^{\prime \prime}\left(N+\frac{1}{2}\right)-g^{\prime \prime}\left(N-\frac{1}{2}\right)=-\frac{4}{(2 N-1)^2}$
Adding all these, we get
$g^{\prime \prime}\left(N+\frac{1}{2}\right)-g^{\prime \prime}\left(\frac{1}{2}\right)=-4\left(1+\frac{1}{9}+\frac{1}{25}+\frac{1}{(2 N-1)^2}\right)$