How Can We Prove That the Exponential Function is Always Positive?

[Päällikkö]

[SOLVED] Showing exp is positive...

... well not quite. This isn't actually homework, but here's what I'd like to prove:
\forall n \in \mathbb{N} : \sum_{k=0}^{2n}\frac{x^k}{k!} \ge 0.

I tried going about it through induction, but got into trouble quite early on: I couldn't prove that
\sum_{k=0}^{2(n+1)}\frac{x^k}{k!} = \sum_{k=0}^{2n}\frac{x^k}{k!} + \frac{x^{2n+1}}{(2n+1)!} + \frac{x^{2n+2}}{(2n+2)!} \ge 0,
as using
\sum_{k=0}^{2n}\frac{x^k}{k!} \ge 0
here I'd get
\frac{x^{2n+1}}{(2n+1)!} + \frac{x^{2n+2}}{(2n+2)!} \ge 0,
which cannot be proven as it's false.

So obviously I'm in need of a different viewpoint. My brain is all tangled up, I do hope the problem isn't as easy as it looks at a first glance.

 

[rock.freak667]

Well try bringing the 2 fractions to the same base by saying that (2n+2)!=(2n+2)(2n+1)!

 

[Päällikkö]

Which equation are you talking about? I've tried that plenty before, to no avail.

I forgot to mention in the first post that I managed to show that
\frac{x^{2n+1}}{(2n+1)!} + \frac{x^{2n+2}}{(2n+2)!} \ge -\frac{(2n+1)^{2n+1}}{(2n+2)!},
but that lead to nowhere as the expression
\sum_{k=0}^{2n}\frac{x^k}{k!} \ge \frac{(2n+1)^{2n+1}}{(2n+2)!}
does not hold.

 

[Dick]

Try this. The Lagrange form of the Taylor series remainder is R(2n)(x)=exp(y)*x^(2n+1)/(2n+1)! for some y between 0 and x. So if I call your series P(2n)(x), then P(2n)(x)+R(2n)(x)=exp(x). Clearly you have nothing to worry about if x>0. If x<0 then R(2n)(x)<0, but exp(x)>0. Reach a conclusion about P(2n)(x).

 

[EnumaElish]

I'd start with S(1) = \sum_{k=0}^{2}{x^k}/{k!} = 1 + x + x^2/2. Assume x < 0, then show S(1) > 0 because the first and the third terms trump the second. Then generalize to S(n).

 

[Päällikkö]

Try this. The Lagrange form of the Taylor series remainder is R(2n)(x)=exp(y)*x^(2n+1)/(2n+1)! for some y between 0 and x. So if I call your series P(2n)(x), then P(2n)(x)+R(2n)(x)=exp(x). Clearly you have nothing to worry about if x>0. If x<0 then R(2n)(x)<0, but exp(x)>0. Reach a conclusion about P(2n)(x).

My thinking in the first post works for x > 0, but runs into trouble when x < 0 (More precisely, everything except -2n - 2 < x < 0 should work out directly (I haven't actually tried to prove that the assertion holds in the mentioned region separately, maybe that'll work)). You confronted this difficulty with a quite involved theorem about Lagrange remainders and the properties of exp. Although correct, I'm not completely satisfied: I'd rather have the proof not relying on the properties of exp.

I'd start with S(1) = \sum_{k=0}^{2}{x^k}/{k!} = 1 + x + x^2/2. Assume x < 0, then show S(1) > 0 because the first and the third terms trump the second. Then generalize to S(n).

This is indeed exactly what I've done in the first post. As I pointed out, the difficulty started when trying to prove the implication S(n) => S(n+1).

 

[Dick]

My thinking in the first post works for x > 0, but runs into trouble when x < 0 (More precisely, everything except -2n - 2 < x < 0 should work out directly (I haven't actually tried to prove that the assertion holds in the mentioned region separately, maybe that'll work)). You confronted this difficulty with a quite involved theorem about Lagrange remainders and the properties of exp. Although correct, I'm not completely satisfied: I'd rather have the proof not relying on the properties of exp.

I don't consider Taylor series remainder terms to be be rocket science, but maybe there is a more direct approach. I wish you luck in finding it. I couldn't come up with anything.

 

[EnumaElish]

This is indeed exactly what I've done in the first post. As I pointed out, the difficulty started when trying to prove the implication S(n) => S(n+1).

That must be because any "excess positivity" in S(n) actually gets to be used in S(n+1) to soak up (cancel out) the additional negativity introduced (together with some additional positivity) while one goes from n to n+1.

I.e. you have to assume not only that S(n) > 0 (which is what your attempt at induction assumes), but actually S(n) > M(n) > 0 so that S(n+1) > 0.

 

[Päällikkö]

That must be because any "excess positivity" in S(n) actually gets to be used in S(n+1) to soak up (cancel out) the additional negativity introduced (together with some additional positivity) while one goes from n to n+1.

I.e. you have to assume not only that S(n) > 0 (which is what your attempt at induction assumes), but actually S(n) > M(n) > 0 so that S(n+1) > 0.

If you'd read my posts, you would have noticed that that's what I've tried. To recap: there exists an x such that
\frac{x^{2n+1}}{(2n+1)!} + \frac{x^{2n+2}}{(2n+2)!} = -\frac{(2n+1)^{2n+1}}{(2n+2)!},
but
\sum_{k=0}^{2n}\frac{x^k}{k!} \ge \frac{(2n+1)^{2n+1}}{(2n+2)!}
does not hold for all x. That is, one cannot find an M(n) independent of x (of the kind that would help with the proof). This makes the proof trickier.

EDIT: Got it.

 

[Sourabh N]

Can you show the proof; I want to check whether my proof is same or not.

 

[Päällikkö]

It relies on the fact that:
\frac{d^2}{dx^2}\sum_{k=0}^{2(n+2)}\frac{x^k}{k!} = \sum_{k=0}^{2n}\frac{x^k}{k!}

The rest should easily follow.

 

[arildno]

The functional properties of the exponential function, that Exp(x+y)=Exp(x)*Exp(y) and Exp(0)=1 are sufficient to prove that Exp(x) are greater than 0 for ALL x, since Exp(-x)=Exp(0)/Exp(x) for arbitrary positive x.

 

[Sourabh N]

It relies on the fact that:
\frac{d^2}{dx^2}\sum_{k=0}^{2(n+2)}\frac{x^k}{k!} = \sum_{k=0}^{2n}\frac{x^k}{k!}

The rest should easily follow.

Mine is same.

 

[Dick]

Mine is same.

So to let the rest of us catch up, how did you do it using only that?

 

[Päällikkö]

\forall n \in \mathbb{N} : \sum_{k=0}^{2n}\frac{x^k}{k!} &gt; 0 \quad (1)
I changed the greater than or equal sign to a greater than (might as well go for the stronger version as the proof doesn't differ one bit).
So :
\sum_{k=0}^{2}\frac{x^k}{k!} = 1 + x + \frac{x^2}{2} &gt; 0.

Therefore we may assume that (1) holds for some n. Now
\frac{d^2}{dx^2}\sum_{k=0}^{2(n+1)}\frac{x^k}{k!} = \sum_{k=0}^{2n}\frac{x^k}{k!} &gt; 0
(Apparently I have a typo in my last post in the indices, that might be confusing)

It follows that the function has exactly one minimum, at the point x₀:
\frac{d}{dx}\sum_{k=0}^{2(n+1)}\frac{x_0^k}{k!} = 0

With the above equation, we can drop a whole lot of terms from the original:
\sum_{k=0}^{2(n+1)}\frac{x^k}{k!} &gt; \sum_{k=0}^{2(n+1)}\frac{x_0^k}{k!} = \frac{x_0^{2n+2}}{(2n+2)!} &gt; 0

This holds for all x.

 

[Dick]

Thanks.