Which Is Larger: the Exponential Function or Its Expansion?

指数函数及其展开式孰大孰小？ 苏剑林 · 2012-03-18 · Mathematics exponent function

For x>0, which is larger, the exponential function $f(x)=e^x$ or the power function $h_n (x)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...+\frac{x^n}{n!}$?

For those who have already studied calculus, this problem is quite simple—one might even say $f(x) > h_n (x)$ is "obviously true" (because all the remaining terms in the expansion of $e^x$ are positive). But this problem appeared on the 2012 Guangzhou Mock Exam No. 1 (science track), where it isn't quite so simple, since it has to be proved using elementary methods. I've recently gotten into the habit that, when I pick up a math exam paper, I don't start with the multiple-choice questions but go straight to the last problem. So when I sat the Guangzhou Mock Exam, I spent half an hour solving the last question (i.e., this one) first. Below are three solutions I came up with.

Method One: Mathematical Induction

This is the method used in the official answer, and also the method BoJone wrote on the answer sheet (didn't expect we'd end up on the same page).

First, use calculus (differentiation) to prove $e^x > 1+x$ as the base case of the induction (this is actually the first sub-question). Then assume it holds for n=k, i.e., $g_k (x)=f(x) - h_k (x) > 0$. The reader will notice something interesting: the derivative of $g_{k+1} (x)$ is $g_{k+1}' (x)=g_k (x)$, so we get $g_{k+1}' (x)=g_k (x) > 0$, meaning that, under this assumption, $g_{k+1} (x)$ is increasing on $(0,+\infty)$, so $g_{k+1} (x) > g_{k+1} (0) =0$, which proves the case n=k+1 as well.

This is indeed a nice method. Mathematical induction rarely gets a chance to appear in this kind of "exam-oriented" question, so being able to apply it to the final, most difficult problem is genuinely exciting! I hope to see more problems involving mathematical induction, proof by contradiction, and the like (and I'd love to see them on the Gaokao too, ha).

Method Two: Take the Ratio and Differentiate

However, the above method wasn't actually the first one I thought of. My first instinct was to take the ratio of the two sides and differentiate. Unfortunately, I didn't push through with it at the time, only realizing afterward that it's actually the simplest approach. First, define the function:

$$g_n (x)=\frac{h_n (x)}{f(x)}=e^{-x}+xe^{-x}+\frac{1}{2!}x^2 e^{-x}+...\frac{1}{n!}x^n e^{-x}$$

Can you guess what happens when we differentiate?

$(\frac{1}{n!}x^n e^{-x})'=\frac{1}{(n-1)!}x^{n-1} e^{-x}-\frac{1}{n!}x^n e^{-x}$

This is exactly a "telescoping" pattern! If you sum up all the derivative terms, you get:

$$g_n '(x)=-\frac{1}{n!}x^n e^{-x}$$

which is clearly negative! So $g_n (x)$ is monotonically decreasing on $(0,\infty)$, giving $g_n (x) < g_n (0)=1$. This is equivalent to $f(x) > h_n (x)$.

This is the most direct method (as far as I know). Mathematical induction is certainly clever, but it involves a lot of steps, which makes it inconvenient to write out on an answer sheet. Still, from a purely mathematical appreciation standpoint, both methods are equally admirable! Because they're both beautiful!

Method Three: Repeated Integration

This method might not be so easy for the average high school student to understand, but to me it feels almost obvious. Of course, this approach probably isn't suitable for use on an actual exam paper, but articles on Scientific Spaces were never meant for exam prep in the first place—we're here to appreciate the beauty of mathematics!

First, we need to know that if a function f(x) is always greater than 0, then $\int_0^a f(x) dx > 0$ must hold—this is easy to understand from the definition of the definite integral.

Applying this to our problem: we start with $e^x > 1$, then consider the integral $\int_0^x (e^t -1)dt >0$, which gives $e^x -x -1 >0$. Next, consider the integral $\int_0^x (e^t -1-t)dt >0$, which gives $e^x -1 -x-\frac{1}{2}x^2>0$; continuing in this fashion completes the proof. This is essentially borrowing a step from the derivation of the Taylor series. It, too, is a kind of inductive (recursive) method.

**This kind of problem appearing on the Guangzhou Mock Exam No. 1 reflects two things: first, that many of these capstone math problems are essentially "elementarized" higher mathematics; and second, that ideas like mathematical induction are increasingly being emphasized. If that's really the trend, BoJone would be very pleased—frankly, I dislike those high-computation-load problems, which do nothing to sharpen our thinking. Only problems that are genuinely challenging, conceptually fresh, yet still easy to understand on their face are truly good problems!