Finding roots of 5 th degree taylor expansion of $e^x$ I need to find the roots [actually their count and sign] of :
$$
{\mathcal T}(e^x,5)=\frac{x^5}{5!}+\frac{x^4}{4!}+\frac{x^3}{3!}+\frac{x^2}{2!}+x+1$$


*

*It[Root] is clearly negative.

*It should have odd number of roots.

*Either One or Three or Five.

*It is easy to find the derivatives, but only the third derivative, a quadratic can be easily analyzed.


Any help?
Basics Only.
 A: Consider the derivative
$$\begin{aligned}
\frac{x^4}{4!} + \frac{x^3}{3!} + \frac{x^2}{2!} + x + 1 &= \frac{1}{4!}(x^4 + 4x^3 + 4x^2) + \frac{1}{12}x^2 + \left(\frac{x^2}{4} + x + 1\right)\\
&= \frac{x^2}{24}(x+2)^2 + \frac{x^2}{12} + \left(\frac{x}{2}+1\right)^2\\
&> 0.
\end{aligned}$$
So the fifth-order Taylor polynomial is strictly increasing, hence has exactly one real zero.
A: Hint
Fortunately (for me, at least !), you stopped at $x^5$. You can compute the derivative; this is a quartic polynomial we can solve with radicals. There is no real root to the derivative.
So the function increases and there is only one root.
A: Hint
My hint is not as well than Daniel Fisher but it always work.
$$T^{(5)}(x)=1>0$$ and thus $T^{(4)}$ is strict increasing. Therefore $T^{(3)}$ has a unique minimum (why ?). This minimum is strictly positive, therefore $T^{(2)}$ is strict increasing. You can show that $\lim_{x\to -\infty }T^{(2)}(x)=-\infty $ and $\lim_{x\to \infty }T^{(2)}(x)=+\infty $ and thus $T'$ has a unique minimum. This minimum is strictly positive and thus $T$ is strictly increasing. Show that $\lim_{x\to -\infty }T(x)=-\infty $ and $\lim_{x\to \infty }T(x)=+\infty $ and you'll got the unicity.
A: You can compute the third derivative -- it is a quadratic polynomial with no real roots.
Therefore the second derivative is a strictly increasing cubic polynomial with one real root; you can easily find a small rational interval that contains that root.
Now the first derivative is a quartic polynomial with one local (and thus global) minimum at the point where the second derivative had a root. If you have good enough bounds for the location of that root, you can prove (using the mean value theorem) that the global minimum is positive. Therefore the first derivative is everywhere positive.
Therefore the function itself has exactly one real root.
A: A similar way to @Daniel Fischer's answer (if not quite as clean) to show that the derivative is positive:
$$\begin{aligned}
\frac{x^4}{4!} + \frac{x^3}{3!} + \frac{x^2}{2!} + x + 1 &= \frac{1}{24}\big(x^4+4x^3+12x^2+24x+24\big)\\
&= \frac{1}{24}\big((x^4+4x^3+6x^2+4x+1) + (6x^2+20x+23)\big)\\
&> \frac{1}{24}\big((x^4+4x^3+6x^2+4x+1) + (6x^2+20x+\frac{50}{3})\big)\\
&= \frac{1}{24}\big((x+1)^4 + 6(x+\frac53)^2\big)\\
&> 0.
\end{aligned}$$
A: This one is pretty standard. A generalization is given by Hardy in Pure Mathematics. I will quote his argument verbatim.
Let $$f_{n}(x) = 1 + x + \frac{x^{2}}{2!} + \cdots + \frac{x^{n}}{n!}\tag{1}$$ then $f_{n}(x) = 0$ has one real root if $n$ is odd and none if $n$ is even.
"Assume this proved for $n = 1, 2,\ldots, 2k$. Then $f_{2k + 1}(x) = 0$ has at least one real root, since its degree is odd, and it cannot have more since, if it had, $f'_{2k + 1}(x)$ or $f_{2k}(x)$ would have to vanish at least once. Hence $f_{2k + 1}(x) = 0$ has just one root, and so $f_{2k + 2}(x) = 0$ cannot have more than two. If it has two, say $\alpha$ and $\beta$, then $f'_{2k + 2}(x)$ or $f_{2k + 1}(x)$ must vanish once at least between $\alpha$ and $\beta$, say at $\gamma$; and $$f_{2k + 2}(\gamma) = f_{2k + 1}(\gamma) + \frac{\gamma^{2k + 2}}{(2k + 2)!} > 0$$ but $f_{2k + 2}(x)$ is also positive when $x$ is large (positively or negatively), and a glance at a figure will show that these results are contradictory. Hence $f_{2k + 2}(x) = 0$ has no real roots."
The above paragraph completes Hardy's concise and fully rigorous argument. Now let's see the analytic equivalent of the italicized statement in above paragraph (i.e. obtain the contradiction without graph). Since $f_{2k + 2}(x)$ has only two roots $\alpha,\beta$ and say $\alpha < \beta$ and $\lim_{x \to \pm\infty}f_{2k + 2}(x) = +\infty$ and $f_{2k + 2}(\gamma) > 0$, it follows that the function $f_{2k + 2}(x)$ is positive in $(-\infty, \alpha), (\alpha, \beta), (\beta, \infty)$. It follows that $\alpha, \beta$ are points of minima of $f_{2k + 2}(x)$ and there is at least one maximum in the interval $(\alpha, \beta)$. It follows that there are three distinct points of local extrema for $f_{2k + 2}(x)$ and hence its derivative $f'_{2k + 2}(x) = f_{2k + 1}(x)$ vanishes thrice and this is contrary to what we have proved above. The proof is now complete by induction over $n$.
Your current question deals with $f_{5}(x) = 0$ and clearly it has only one root.
