Let g(x) = log(f(x)) where f(x) is a twice di | vedclass

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(A) Given $g(x) = \log(f(x))$ and $f(x+1) = x f(x)$.Taking logarithm on both sides,$\log(f(x+1)) = \log(x) + \log(f(x))$.This implies $g(x+1) = g(x) +

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$-4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2N-1)^2}\right\}$

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(A) Given $g(x) = \log(f(x))$ and $f(x+1) = x f(x)$.Taking logarithm on both sides,$\log(f(x+1)) = \log(x) + \log(f(x))$.This implies $g(x+1) = g(x) + \log(x)$,or $g(x+1) - g(x) = \log(x)$.Differentiating twice with respect to $x$,we get $g^{\prime \prime}(x+1) - g^{\prime \prime}(x) = -\frac{1}{x^2}$.We want to find $g^{\prime \prime}\left(N+\frac{1}{2}\right) - g^{\prime \prime}\left(\frac{1}{2}\right)$.We can write this as a telescoping sum:$g^{\prime \prime}\left(N+\frac{1}{2}\right) - g^{\prime \prime}\left(\frac{1}{2}\right) = \sum_{k=1}^{N} \left[ g^{\prime \prime}\left(k+\frac{1}{2}\right) - g^{\prime \prime}\left(k-\frac{1}{2}\right) \right]$.Using the relation $g^{\prime \prime}(x+1) - g^{\prime \prime}(x) = -\frac{1}{x^2}$,we substitute $x = k - \frac{1}{2}$:$g^{\prime \prime}\left(k+\frac{1}{2}\right) - g^{\prime \prime}\left(k-\frac{1}{2}\right) = -\frac{1}{(k-\frac{1}{2})^2} = -\frac{1}{(\frac{2k-1}{2})^2} = -\frac{4}{(2k-1)^2}$.Summing from $k=1$ to $N$:$\sum_{k=1}^{N} -\frac{4}{(2k-1)^2} = -4 \left[ \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \ldots + \frac{1}{(2N-1)^2} \right]$.This simplifies to $-4 \left\{ 1 + \frac{1}{9} + \frac{1}{25} + \ldots + \frac{1}{(2N-1)^2} \right\}$.

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) = $

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