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!}+\dots+\frac{x^n}{n!}?
For friends who have already studied calculus, this problem is very simple; one might even say that f(x) > h_n (x) is "obviously true" (because all subsequent infinite terms in the expansion of e^x are positive). However, when this problem appeared in the 2012 Guangzhou First Mock Exam for Science Mathematics, it became less straightforward, as it required a proof using elementary methods. Recently, I have developed a habit: when I receive a math paper, I don’t start with the multiple-choice questions, but rather with the final problem. Thus, while taking the Guangzhou First Mock Exam, I spent half an hour solving the final problem (this one) first. Below are the three solutions I came up with.
I. Mathematical Induction
This is the method adopted in the official solution and also the one I wrote on my answer sheet (I didn’t expect us to coincide).
First, use functional methods (differentiation) to prove e^x > 1+x as the base case for the induction (this was 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 discover a very interesting fact: the derivative of g_{k+1} (x) is g_{k+1}' (x)=g_k (x). Thus, we have g_{k+1}' (x)=g_k (x) > 0, which means that under the inductive hypothesis, g_{k+1} (x) is an increasing function on (0, +\infty). Consequently, g_{k+1} (x) > g_{k+1} (0) = 0, proving that it also holds for n=k+1.
This is indeed a good method. Mathematical induction rarely has the chance to appear in these "exam-oriented education problems," so being able to apply it to the grand finale problem is quite exciting! I hope to see more problems involving mathematical induction, proof by contradiction, and the like (and I hope they appear in the National College Entrance Exam, haha).
II. Quotient Differentiation
However, the method above was not the first one I thought of. I initially considered taking the quotient of the two and then differentiating. Unfortunately, I didn’t stick with it at the time, only to realize later that this is likely the simplest method. First, define the function: g_n (x)=\frac{h_n (x)}{f(x)}=e^{-x}+xe^{-x}+\frac{1}{2!}x^2 e^{-x}+\dots+\frac{1}{n!}x^n e^{-x}
Can you guess what happens after differentiating? \left(\frac{1}{n!}x^n e^{-x}\right)'=\frac{1}{(n-1)!}x^{n-1} e^{-x}-\frac{1}{n!}x^n e^{-x}
This is exactly what is known as "telescoping cancellation"! You will find that after summing all the derivative terms, you get: g_n '(x)=-\frac{1}{n!}x^n e^{-x}
This is clearly a negative number! Therefore, g_n (x) is monotonically decreasing on (0, \infty), so g_n (x) < g_n (0)=1. This is equivalent to f(x) > h_n (x).
This is the most direct method (to my knowledge). While mathematical induction is clever, its procedure is quite lengthy and inconvenient to write on an answer sheet. However, from the perspective of pure mathematical appreciation, both are equally stunning! Because they are both beautiful!
III. Repeated Integration
This method might not be very easy for the average high school student to understand, but for me, it is almost obvious. Of course, this method is probably not suitable for use on an exam paper, but the articles in Scientific Space are not meant for exams anyway; we are here to appreciate the beauty of mathematics!
First, we must know that if a function f(x) is always greater than 0, then \int_0^a f(x) dx > 0. From the definition of the definite integral, this is very easy to understand.
Applying this to the problem, we start with e^x > 1. Consider the integral \int_0^x (e^t -1)dt > 0, which yields e^x - x - 1 > 0. Continue by considering the integral \int_0^x (e^t - 1 - t)dt > 0, which yields e^x - 1 - x - \frac{1}{2}x^2 > 0; and so on. This is actually just a sneaky use of a derivation of the Taylor series. This also belongs to the category of inductive (recursive) methods.
The appearance of this type of problem in the Guangzhou First Mock Exam reflects two points: first, many final math problems are the simplification of higher mathematics into elementary forms; second, ideas such as mathematical induction are receiving increasing attention. If this is truly the case, I would be very happy. In fact, I dislike those high-intensity calculation problems; they are meaningless for promoting our thinking. Only those problems that are challenging and novel in thought, yet easy to understand in their statement, are truly good problems!
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