<s>[INST] Let $f(x)=(x+1)(x+2)..(x+100)$ and $g(x)=f(x)f’’(x)-(f’(x))^2$. Find number of roots $n$ of $g(x)=0$ After relevant simplification ie. Log on both sides and then differentiating, the expression received is
$$\frac{f’(x)}{f(x)} =\sum_{k=1}^{100} \frac{1}{x+k}$$
Differentiating again will give
$$\frac{g(x)}{(f(x))^2}=-\sum_{k=1}^{100} \frac{1}{(x+k)^2}$$
So in turn, the roots of $g(x)=0$ are simply the roots of $f(x)=0$, so ans should be 100, but that’s not correct.
Where am I going wrong?
 [/INST] $$\frac{f’(x)}{f(x)} =\sum_{k=1}^{100} \frac{1}{x+k}$$
D.w.r.t. $x$ both sides
$$\frac{f(x)f''(x)-f'^2(x)}{f^2(x)}=-\sum_{k=1}^{100} \frac{1}{(x+k)^2}<0 \ne 0.$$
So $$g(x)=f(x)f''(x)-f'^2(x)=\sum_{k=1}^{n}\frac{f^2(x)}{(x+k)^2}$$
Then $$-g(x)=(x+2)^2(x+3)^2(x+4)^2....(x+n)^2+ (x+1)^2(x+3)^2(x+4)^2....(x+n)^2+......>0 \ne 0,$$ has no real root.
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