TITLE: Proving Modified Taylor expansion of $e^x$ (negative sign for odd powers) is always positive when n is EVEN.
QUESTION [4 upvotes]: Can we prove that the following sequence is always positive for EVEN $n$ and for positive real $x$. The sequence is very similar to Taylor expansion of $e^x$, but negative signs for odd exponents:
$$\sum_{i=0}^n \frac{x^i(-1)^i}{i!}=1-\frac{x^1}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!} ... $$
for small $n$, it is easy to show, but I need proof for every EVEN $n$. I tried Induction, but I cannot finish. If somebody knows how to prove by Induction or by any other method, please explain it to me.
Thanks for Everyone!
This is my first post, sorry for any inconvenience

REPLY [0 votes]: First note that your polynomial is the Taylor approximant of $e^{-x}$. Let us define:
$$
f_n(x) = \sum_{k=0}^n \frac{x^k}{k!}.
$$
Your polynomial is simply $f_n(-x)$. Now by Taylor's remainder formula in the form of Lagrange, there exist a $\xi\in (0,x)$
$$
f(x):=e^{x} = f_n(x) +  \frac{ e^\xi x^{(n+1)}}{(n+1)!}.
$$
Now send $x\to -x$ to obtain
$$
e^{-x} = f_n(-x) + \frac{ e^\xi (-x)^{(n+1)}}{(n+1)!}.
$$
for $\xi \in (-x,0)$. From this, if $n$ is even, we derive
$$
f_n(-x) = e^{-x} + \frac{ e^\xi x^{(n+1)}}{(n+1)!},
$$
which is a sum of two positive quantities when $x$ is positive.
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