# How do I show that this function is always $> 0$

Show that $$f(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} > 0 ~~~ \forall_x \in \mathbb{R}$$

I can show that the first 3 terms are $> 0$ for all $x$:

$(x+1)^2 + 1 > 0$

But, I'm having trouble with the last two terms. I tried to show that the following was true:

$\frac{x^3}{3!} \leq \frac{x^4}{4!}$

$4x^3 \leq x^4$

$4 \leq x$

which is not true for all $x$.

I tried taking the derivative and all that I could ascertain was that the the function became more and more increasing as $x \rightarrow \infty$ and became more and more decreasing as $x \rightarrow -\infty$, but I couldn't seem to prove that there were no roots to go with this property.

• Use the result on the first two terms you have and see that $f^{'''}(x)>0$. Further when $f^{'''}(x)>0$, then $f^{''}(x)$ is always increasing, so is $f^{''}(x)$ and last but not least $f(x)$. Jun 9, 2012 at 16:16
• Or use the solutions of the Quartic polynomial and check whether it has real roots. Jun 9, 2012 at 16:29
• The argument of Andrew Salmon works if we truncate the series for $e^x$ at any even exponent, call that $f_{2n}(x).$ The trick is that $f_{k}' = f_{k-1},$ so $f_k(x) = f_{k-1}(x) + \frac{x^k}{k!}.$ Back to $k$ even, we get $$f_{2n}(x) = f_{2n}'(x) + \frac{x^{2n}}{(2n)!}.$$ At any local minimum, the derivative is $0.$ As the minimum does not occur at $x=0,$ the minimum is positive. Some of this involves induction on $n.$ Jun 9, 2012 at 18:51

Hint: $$f(x) = \frac{1}{4} + \frac{(x + 3/2)^2}{3} +\frac{x^2(x+2)^2}{24}$$

• Oh! Very nice and short. I'd +1 twice if I could. Jun 9, 2012 at 16:56
• Completing the square solves everything that ever existed! Jun 9, 2012 at 17:13
• @stariz77, yes, it is true for single-variable polynomials but not for several variables: en.wikipedia.org/wiki/Hilbert's_seventeenth_problem
– lhf
Jun 9, 2012 at 19:03

$f$ is a polynomial, and therefore, is differentiable at all points. Furthermore, as $x\to\infty$ or $x\to-\infty$, $f(x)\to+\infty$. Thus, if $f(x)\le0$ for some $x$, then $f(x)\le0$ for some relative minimum.

$$f'(x)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}$$

$f'(x)=0$ for all relative minima. However, if $f'(x)=0$, then $$f(x)=f'(x)+\frac{x^4}{4!}$$

Thus, $f(x)=\frac{x^4}{4!}>0$ for all relative minima $x\not=0$. $x=0$ is not a relative minimum, because $f'(0)\not=0$, so this equation holds for all relative minima of $f$. This contradicts our assumption, so $f(x)>0$ for all $x\in \mathbb R$.

• +1 Nice, but I'd edit the answer, pointing out simply: since $$f(x)=f'(x)+\frac{x^4}{4!}\,\,\forall\,x\in\mathbb{R}$$, if we have a minimum at $\,x_0\,$ then $$f(x_0)=f'(x_0)+\frac{x_0^2}{4!}=\frac{x_0^2}{4!}>0$$as clearly $\,x_0\neq 0\,$ , which contradicts our initial assumption that $\,f\leq 0\,$ at some minimum... Jun 9, 2012 at 17:01
• @DonAntonio We can polish it up even more, the contradiction is not required, the entire answer could be: $x=0$ is not a minimum, so if $x_0$ is a minimum then $f(x_0) = f'(x_0) + x^4_0/4! > 0.$ So $f> 0.$ Jun 9, 2012 at 17:45
• This argument shows that, if we truncate the series for the exponential ending with any even exponent, the resulting polynomial is always positive. And, by induction, positive second derivative and a single global minimum. Jun 9, 2012 at 18:20

You have had some good ideas so far. You tried to see when this was true: $$\frac{x^3}{3!} \leq \frac{x^4}{4!}.$$

You rearranged this to $4x^3\leq x^4$ but you made an incorrect conclusion when you divided by $x^3$ (if $x<0$ then the inequality sign should flip). Instead, lets divide by $x^2$ to get $4x \leq x^2$ or $x(x-4)\geq 0.$ This is true when $x\leq 0$ or $x\geq 4$ so the desired inequality is true in that range.

For $0< x < 4$ we don't have $\frac{x^3}{3!} \leq \frac{x^4}{4!}$ but lets see if the other terms can save us. To do this, we need to see exactly how large $g(x) = x^3/3! - x^4/4!$ can be in $(0,4).$ We calculate that $g'(x) = -(x-3)x^2/6$ so $g$ increases when $0\leq x\leq 3$, the maximum occurs at $g(3)=9/8$, and then it decreases after that.

This is good, because the $1+x+x^2/2$ terms obviously give at least $1$ from $x=0$, and will give us more as $x$ gets bigger. So we solve $1+x+x^2/2=9/8$ and we take the positive solution which is $\frac{\sqrt{5}-2}{2} \approx 0.118.$ So the inequality is definitely true for $x\geq 0.12$ because $g$ is at most $9/8$ and $1+x+x^2/2$ accounts for that amount in that range.

Remember that $g$ was increasing between $x=0$ to $x=3$, so the largest $g$ can be in the remaining range is $g(0.12) = 873/315000 <1$, which is less than the amount $1+x+x^2/2$ gives us. So the inequality is also true for $0\leq x\leq 0.12$, so overall, for all $x.$

So all in all, the only trouble was for $x$ in $(0,4)$ and the contribution from the other terms was always enough to account for $x^3/3!$ when $x^4/4!$ wasn't enough.

Observe that

$$e^x = f(x) + \frac{x^5}{5!} + \cdots$$

Then show that for $x<0$ $$f(x) > \frac{x^5}{5!} + \cdots$$

But $e^x > 0, \forall x\in R$

So, $$f(x) > e^x - f(x) \Rightarrow 2f(x)>e^x>0$$

• Well, showing $\,\displaystyle{f(x)>\frac{x^5}{5!}}\,$ may prove to be tricky... Jun 9, 2012 at 16:45
• Well, not so much (unless I'm mistaken) but I don't want to give away the answer. Jun 9, 2012 at 16:49
• Well, perhaps I'm wrong...I'll expect your posting on this after some time has ellapsed. Jun 9, 2012 at 16:52
• $f(x)\gt\frac{x^5}{5!}\dots$ is incorrect. Think of a sufficiently large $x$. This would imply that a polynomial function can be strictly greater than an exponential one.
– A.S
Jun 9, 2012 at 16:52
• @DonAntonio Sure :) Jun 9, 2012 at 16:53