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Prove that $a^n = e^{n \ln(a)} \gt \dfrac{(n \ln(a))^m}{m!}$ for all $m\in \mathbb{N}$ and $a>1$ without using taylor series/calculus.

I am trying methods that use binomial theorem but I can't get far. I tried induction on $m$. For $m=1$, we have $e^{n\ln(a)}>n\ln(a)$, which is true since $e^x\geq 1+x>x.$ Suppose the proposition is true for $m=t,$ then we have $e^{n\ln(a)}>\dfrac{{(n\ln(a))^t}}{t!}.$ Then in order to prove $m=t+1$ is true we must could show that $e^{n\ln(a)}>\dfrac{e^{n\ln(a)}n\ln(a)}{t+1}.$ But after this step I am stuck. Any hints/ideas will be much appreciated.

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  • $\begingroup$ May I wrong, but it's trivial! $\endgroup$ – Nosrati Oct 15 '17 at 15:36
  • $\begingroup$ @MyGlasses what do you mean? $\endgroup$ – model_checker Oct 15 '17 at 15:38
  • $\begingroup$ No Taylor Series! $\endgroup$ – model_checker Oct 15 '17 at 15:40
  • $\begingroup$ Could you please post your proof? $\endgroup$ – model_checker Oct 15 '17 at 15:41
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    $\begingroup$ It's in general not easy to avoid calculus, since the definition of the various functions involved is always done with calculus. How do you define $e$, $\ln$, etc. without calculus? $\endgroup$ – Thomas Andrews Oct 15 '17 at 16:30
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Prove with induction. $P(n): e^x>\dfrac{x^n}{n!}$ then $P(0): e^x>1$ is trivial. With assumption $P(m): e^x>\dfrac{x^m}{m!}$ be true, then $$\int_0^x e^x>\int_0^x \dfrac{x^m}{m!}=\dfrac{x^{m+1}}{(m+1)!}$$ shows $P(m+1)$ is true.

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  • $\begingroup$ This is an incredibly slick answer. $\endgroup$ – Stella Biderman Oct 15 '17 at 16:08
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    $\begingroup$ But it uses calculus! $\endgroup$ – gammatester Oct 15 '17 at 16:09
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    $\begingroup$ Nice, but you shouldn't use $x$ in two ways like that. It should be $\int_0^x e^t\,dt$. $\endgroup$ – Thomas Andrews Oct 15 '17 at 16:28
  • $\begingroup$ Thanks. yes. Can make confusion for some people! $\endgroup$ – Nosrati Oct 15 '17 at 16:31
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    $\begingroup$ Sorry, but if writing things correctly creates confusion, then they need to be confused so that they learn to understand things written correctly. $\endgroup$ – Thomas Andrews Oct 15 '17 at 16:33
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We assume two results:

  1. $e^x>1+x$ for $x>0$.
  2. $(e^y)^n= e^{yn}$ for an real $y$ and positive integer $n$.

and prove:

Theorem: If $x>0$ and $m$ a non-negative integer, then $e^x>\frac{x^m}{m!}.$

Assumption (1) implies this result for $m=0,1.$

Given any $x>0$, and any integer $k>m$, (1) means $e^{x/k}\geq 1+\frac{x}{k}$ and thus, by $(2)$:

$$e^x = (e^{x/k})^k>\left(1+\frac{x}{k}\right)^k\geq 1+\binom{k}{m}\frac{x^m}{k^m}$$

By the binomial theorem.

Now $$\frac{1}{k^m}\binom{k}{m}=\frac{(1-1/k)(1-2/k)\cdots(1-(m-1)/k)}{m!}>\frac{(1-m/k)^m}{m!}$$

Now you need to show that if you pick $\epsilon>0$ you can find $k$ so that:

$$(1-m/k)^m>1-\epsilon$$

(We will assume $\epsilon<1$.)

Then we want $1-\frac{m}{k}>(1-\epsilon)^{1/m}$ or

$$k>m\left(1-(1-\epsilon)^{1/m}\right)^{-1}$$

So this means:

$$e^x>1+\frac{1-\epsilon}{m!}x^m=\frac{x^m}{m!}+\left(1-\frac{x^m}{m!}\epsilon\right)$$ for any $\epsilon>0$.

But then pick $\epsilon <\min\left(1,\frac{m!}{x^m}\right)$ so we get:

$$e^x>\frac{x^m}{m!}$$


None of these steps used calculus, but (1) and (2) depend on our definition of $e$ (or of $e^x$).

If you define $e^x=\lim_{n\to\infty} \left(1+\frac xn\right)^n$ then you get (1) and (2) pretty easily, but you'd have to prove that this limit exists.

If you define $e^x$ in terms of the power series expansion, then you get the theorem much more directly.

Other ways to define $e^x$ or $e$ are with integrals (definition of natural logarithm) or derivatives, which are calculus. I guess you could "hide" the calculus by defining natural log as a limit of the Reimann sums, and brute force from there.

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    $\begingroup$ This is really avoiding calculus (derivatives, integrals so to speak). It uses two fundamental ideas : $e^{x} \geq 1+x,e^{x+y}=e^{x}e^{y}$. It is interesting to note that these two properties characterize the exponential function. +1 $\endgroup$ – Paramanand Singh Oct 15 '17 at 17:37
  • $\begingroup$ See math.stackexchange.com/a/2383082/72031 $\endgroup$ – Paramanand Singh Oct 15 '17 at 17:38
  • $\begingroup$ @ParamanandSingh Yeah. Technically, I don't use the case $x<0$, so $c^x$ satisfies this inequality for any $c\geq e$. While $e^x> 1+x$ for all $x\neq 0$, the general inequality is not true. For $x=-2, m=2$ you get $e^{-2}<\frac{(-2)^2}{2!}=2$, for example. $\endgroup$ – Thomas Andrews Nov 1 '17 at 15:58
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Let prove by induction that $$e^x >\frac{x^m}{m!} ~~~, \forall ~~m~~ x>0$$

  1. for $n=0$, we have $e^x > 1 =\frac{x^)0}{0!}$
  2. Asumme that $e^x >\frac{x^m}{m!} ~~~, x>0$

consider $$(0,\infty)\ni x\mapsto f(x) = e^x -\frac{x^{m+1}}{(m+1)!}$$

Then by Assumption we have $$ f'(x) = e^x -\frac{x^{m}}{m!}>0$$ that is $f$ is strictly increasing on $(0,\infty)$ therefore,

$$ 0=f(0)<f(x) =e^x -\frac{x^{m+1}}{(m+1)!}$$

hence for all $m$ we have $$e^x -\frac{x^{m}}{m!}>0$$

In particular $a>1$ then, taking, $x= n\ln a>0$ one gets $$a^n = e^{n \ln(a)} \gt \dfrac{(n \ln(a))^m}{m!}$$

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