1. real roots of function

How do I show that the function

fn(x) = 1 + x + (x^2)/2! + ... + (x^n)/n!

has no real roots when n is an even number,
and has one real root when n is an odd number?

2. Originally Posted by weasley74
How do I show that the function

fn(x) = 1 + x + (x^2)/2! + ... + (x^n)/n!

has no real roots when n is an even number,
and has one real root when n is an odd number?
Suppose that $f_n(x)$ has the property that for any even $n$ we have no real roots. We can easy show that $f_n(x)$ has property that is has exactly one real root. Note that $f'_n(x) = f_{n-1}(x)$ where $n-1$ is even, and so the derivative is non-vanishing for $f_n(x)$, thus it is an increasing odd polynomial, thus, it has exactly one root.

There is an elementary way to prove that $f_n(x) > 0$ for $n$ even, which will complete what you want to show. Using this we can prove this. Let $F(x) = \frac{x^n}{n!}$ then $F(x)\geq 0$ if $n$ is even. This mean $f_n(x) = F(x)+F'(x)+...+F^{(n)}(x) \geq 0$.

I will try to come up with an approximating solution to this problem also, since I like putting bounds on functions.

3. Here is an approximation-type proof. First if $x>0$ then certainly $f_n(x) > 0$ and if $x=0$ then $f_n(x)\not = 0$. Thus, if is safe to assume that $x<0$. Let $f(x) = e^x$ by Taylor's theorem it means for any $x<0$ we have $e^x - f_n(x) = \frac{e^y}{(n+1)!} x^{n+1}$ where $x because $f_n(x)$ is a Taylor $n$ degree polynomial. Thus, $f_n(x) = e^x - \frac{e^y}{(n+1)!}x^{n+1}$. But $e^y>0 \mbox{ and }x^{n+1} < 0$ since $n$ is even and $x<0$. Thus, $e^x - \frac{e^y}{(n+1)!}x^{n+1} > e^x > 0$. This means, $f_n(x) > 0$ for all $x\in \mathbb{R}$ and $n$ even.

4. Isn't therse an easier way to prove this, cause frankly I don't really get this.. maybe i'm just dumb..?

5. Originally Posted by weasley74
Isn't therse an easier way to prove this, cause frankly I don't really get this.. maybe i'm just dumb..?
Are you familar with Taylor's theorem because that is all you need to know.