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

Question

Dr. Abubacker · Accepted Answer

(a)
Since, $f(x) = e^{g(x)}$
$\Rightarrow \ e^{g(x + 1)} = f(x + 1) = xf(x) = xe^{g(x)}$
and $g(x + 1) = \log{x + g(x)}$
$\Rightarrow$ $g(x + 1) - g(x) = \log x\$ …(i)
Replacing $x$ by $x - \frac{1}{2}$, we get
$g\left( x + \frac{1}{2} ight) - g\left( x - \frac{1}{2} ight) = \log{(x - \frac{1}{2})}$
$= \log{(2x - 1) - \log 2}$
$	herefore\ g^{''}\left( x + \frac{1}{2} ight) - g^{''}\left( x - \frac{1}{2} ight) = \  - \frac{4}{(2x - 1)^{2}}\ \ \$ …(ii)
On substituting, $x = 1,2,3,\ldots N$ in Eq. (ii) and adding, we get
$g^{''}\left( N + \frac{1}{2} ight) - g^{''}\left( \frac{1}{2} ight) = - 4\left\{ 1 + \frac{1}{9} + \frac{1}{25} + ... + \frac{1}{(2N - 1)^{2}} ight\}$

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

Explanation

Reviewed by Dr. Abubacker

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