proving that $g(x)=0$ has one real root Given $g(x)=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\cdots+\dfrac{x^{2n+1}}{(2n+1)!}$,
Need to prove that $g(x)=0$ has one real root.
I thought to use the fact that $e^x<T_{2n}(x)$  for all $x<0$, but it didn't help me.
Thanks.
 A: $$f(x)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...+\frac{x^n}{n!}$$
with $n=0,$ we find that $1$ is always positive. With $n=1,$ we find $1+x$ has one real root, namely $-1.$
Induction, two cases, odd/even. For all smaller $n,$ even gives always strictlypositive, odd gives one real root.
If $n$ is odd, we know there is at least one real root. But $f'(x)$ is one of this same family of polynomials, and is always strictly positive. As a result, there is just one real root, the polynomial is negative on the left of that (smaller $x$), positive on the right of it. 
If $n$ is even, we know that $f'$ has one real root, negative on the left of that, positive on the right. This means that $f$ has one global minimum. Call the $x$ value $x_0.$ We know $f'(x_0) = 0.$ But we also know that
$$  f(x) = f'(x) +  \frac{x^n}{n!}. $$ So
$$  f(x_0) = f'(x_0) +  \frac{x_0^n}{n!}=  \frac{x_0^n}{n!}, $$ with $n$ even. Is it possible to have $x_0 = 0?$ No, as we always have  $f'(0) = 1.$ So, with $n$ even, we have  $x_0^n > 0,$ and $f(x_0) > 0.$
A: 
$$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots\\e^x=0,\text{ has no real root}$$but you are trying $e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots+\frac{x^{2n+1}}{(2n+1)!}$ it has 1 real root and $2n$ complex roots 
  to make a sense ,see picture above 
A: Let use induction method:
call

$${ P }_{ n }\left( x \right) =1+x+\frac { { x }^{ 2 } }{ 2! } +\frac { { x }^{ 3 } }{ 3! } +...+\frac { { x }^{ 2n+1 } }{ \left( 2n+1 \right) ! } $$

Hence polynomial with odd degree in $n=0$ ,${ P }_{ 0 }\left( x \right) =1>0\quad \\$ for every  $x\in R$.
Assume it is true for all x,less than $n\epsilon N $.Due to induction method ${ P }_{ n }^{ \prime  }\left( x \right) ={ P }_{ n-1 }\left( x \right) >0\quad \forall x\in R$ ,hence,${ P }_{ n }\left( x \right) $ is increasing and can't equal to zero no more one time.Because of ${ P }_{ n }\left( 0 \right) =1>0\quad $ and $\lim _{ x\rightarrow -\infty  }{ { P }_{ n }\left( x \right) =-\infty  } $ so continuous polynomial function ${ P }_{ n }\left( x \right) $ equel one time to zero. So polynomial function has one real root
