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 $x,y\in\mathbf R$. Then $\sum_{n=1}^{100}\log_ef(n)$ is equal to :

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] Put $x=y=0$: $1=2f'(0)$, so $f'(0)=\frac12$. Put $y=0$: $f(x)=\frac12f(x)+f'(x)$, so $f'=\frac f2$ and $f(x)=e^{x/2}$. Then $\sum\log_ef(n)=\frac12\sum_{n=1}^{100}n=2525$. Answer (1).

Q. 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 $x,y\in\mathbf R$. Then $\sum_{n=1}^{100}\log_ef(n)$ is equal to :

Explanation

Reviewed by Dr. Abubacker

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