Let f:S S, where S=(0,), be a twice differentiable function such that f(x+1)=xf(x).

Question

Let $f:S	o S$, where $S=(0,\infty)$, be a twice differentiable function such that $f(x+1)=xf(x)$. If $g:S	o\mathbb{R}$ is defined as $g(x)=\log_e f(x)$, then the value of $|g''(5)-g''(1)|$ is equal to

Accepted Answer

$f(x + 1) = xf(x)$

$$\ln (f(x + 1)) = \ln x + \ln f(x)$$

$$g(x + 1) = \ln x + g(x)$$

$$g(x + 1) - g(x) = \ln x$$ ..... (i)

$$g'(x + 1) - g'(x) = {1 \over x}$$

$$g''(x + 1) - g''(x) = {{ - 1} \over {{x^2}}}$$

$$g''(2) - g'(1) = {{ - 1} \over 1}$$ .... (ii)

$$g''(3) - g''(2) = {{ - 1} \over 4}$$ .... (iii)

$$g''(4) - g''(3) = {{ - 1} \over 9}$$ ..... (iv)

$$g''(5) - g''(4) = {{ - 1} \over {16}}$$ ....(v)

Adding (ii), (iii), (iv) \& (v)

$$g''(5) - g''(1) =  - \left( {{1 \over 1} + {1 \over 4} + {1 \over 9} + {1 \over {16}}} \right) = {{ - 205} \over {144}}$$

$$|g''(5) - g''(1)|\, = {{205} \over {144}}$$

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