Let $\displaystyle f(x)= 1+ x + \frac{x^2}{2!} + \frac{x^3}{3!} +...+ \frac{x^{2n}}{(2n)!}. $
How many multiple roots does the equation $\displaystyle f(x)=0$ have?
Can someone pls explain how to solve this? Your help is very much appreciated.
Let $\displaystyle f(x)= 1+ x + \frac{x^2}{2!} + \frac{x^3}{3!} +...+ \frac{x^{2n}}{(2n)!}. $
How many multiple roots does the equation $\displaystyle f(x)=0$ have?
Can someone pls explain how to solve this? Your help is very much appreciated.
Suppose $\displaystyle a$ is a multiple root of $\displaystyle f(x)=0$, then $\displaystyle f(a)=0$ and $\displaystyle f'(a)=0$. This implies $\displaystyle f(a)-f'(a)=a^{2n}/(2n)!=0$ or equivalently $\displaystyle a=0$. But $\displaystyle f(0)\neq 0$, that is, $\displaystyle f(x)=0$ has no multiple roots.