Let f(x) be a real differentiable function such that f(0)=1 and f(x+y)=f(x)f^'(y)+f^'(x)f(…

Question

Let $f(x)$ be a real differentiable function such that $f(0)=1$ and $f(x+y)=f(x){f}^{'}(y)+{f}^{'}(x)f(y)$ for all $x,y\in R$. Then $\sum _{n=1}^{100}{\log }_{e}f(n)$ is equal to :

Accepted Answer

$\begin{aligned} & f(x+y)=f(x) f^{\prime}(y)+f^{\prime}(x) f(x) \ & \text { Put }=x=y=0 \ & f(0)=f(0) f^{\prime}(0)+f^{\prime}(0) f(0) \ & f^{\prime}(0)=\frac{1}{2} \ & \text { Put } y=0 \ & f(x)=f(x) f^{\prime}(0)+f^{\prime}(x) f(0) \ & f(x)=\frac{1}{2} f(x)+f^{\prime}(x) \ & f^{\prime}(x)=\frac{f(x)}{2}\end{aligned}$

$$ \frac{d y}{d x}=\frac{y}{2} \Rightarrow \int \frac{d y}{y}=\int \frac{d x}{2} $$

$$ \Rightarrow \ln y=\frac{x}{2}+c $$

$\begin{aligned} & \because \mathrm{f}(0)=1 \Rightarrow \mathrm{C}=0 \ & \ell \mathrm{ny}=\frac{\pi}{2} \Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{e}^{\mathrm{x} / 2} \ & \ln \mathrm{f}(\mathrm{n})=\frac{\mathrm{n}}{2} \ & \sum_{\mathrm{n}=1}^{100} \ell \mathrm{f}(\mathrm{n})=\frac{1}{2} \sum_{\mathrm{n}=1}^{100} \mathrm{n}=\frac{5050}{2} \ & =2525\end{aligned}$

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