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[itex]e^{x} > x[/itex]

  1. Dec 25, 2011 #1
    When you see the graphs of the functions [itex]f(x)=e^{x}[/itex] and [itex]f(x)=x[/itex] it is obvious that the former is always greater than the latter (when x is a real number) for the portion of the graph you are observing. In elementary algebra, this was the "proof" that the following inequality is true for all real numbers x: [itex]e^{x} > x[/itex].

    By using calculus, you can prove that that the inequality [itex]e^{x} > x[/itex] is true for all real number x. One method is to employ the properties of exponents to show that [itex]e^{x} > x[/itex] for real numbers less than or equal to 0 and then show that [itex]e^{x} \leq x[/itex] results in a contradiction when you let x increase without bound. Another method, which is much simpler than the first, is to show that the power series of [itex]e^{x}[/itex] necessarily implies [itex]e^{x} > x[/itex] for all real numbers x.

    Is it possible to prove [itex]e^{x} > x[/itex] for all real numbers x without resorting to calculus? Can you use elementary algebra to prove it? Is the previous question not possible because the function [itex]e^{x}[/itex] is transcendental?
  2. jcsd
  3. Dec 25, 2011 #2


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    As you noted the exponential function is not algebraic, so algebra can't be used by itself. The simplest was is using the power series for x > 0 and note that for x < 0, the exponential is still positive.
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