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单词 ElementaryProofOfGrowthOfExponentialFunction
释义

elementary proof of growth of exponential function


Proposition 1.

If x is a non-negative real number and nis a non-negative integer, then (1+x)n1+nx.

Proof.

When n=0, we have (1+x)0=11+00.If, for some natural numberMathworldPlanetmath n, it is the case that(1+x)n1+nx then, multiplying both sides ofthe inequalityMathworldPlanetmath by (1+x), we have

(1+x)n+1(1+x)(1+nx)=1+(n+1)x+nx21+(n+1)x.

By inductionMathworldPlanetmath, (1+x)n1+nx for everynatural number n.∎

Proposition 2.

If b is a real number such that b>1 and n andk are non-negative integers, we havebn>(b-1bk)knk.

Proof.

Let x=b-1. Write n=mk-r where m and rare non-negative integers and r<k.

By the preceding propositionPlanetmathPlanetmathPlanetmath, (1+x)m>mx. Raising bothsides of this inequality to the kth power, wehave (1+x)mk>(mx)k. Since r<k, we alsohave (1+x)-r>(1+x)-k; multiplying bothsides by this inequality and collecting terms,

(1+x)mk-r>(x1+x)kmk.

Multiplying the right-hand side by kk/kk andrearranging,

(x1+x)kmk=(x(1+x)k)k(mk)k.

Since mkmk-r, we also have

(x(1+x)k)k(mk)k(x(1+x)k)k(mk-r)k.

Recalling that mk-r=n and 1+x=b, we conclude that

bn>(b-1bk)knk.

Proposition 3.

If a, b, and x are real numbers such that a0,b>1 and x>0, then

bx>((b-1)aba+1(a+1)a)xa.
Proof.

Let k and n be integers such that ak<a+1and xnx+1. Since x+1>n, we have bx+1>bn. By the preceeding proposition, we have

bn>(b-1bk)knk.

Since k<a+1, we have 1/kk>1/(a+1)k, so

(b-1bk)k>(b-1b(a+1))k.

Since ka0, we have

(b-1b(a+1))knk(b-1b(a+1))ana.

Summarrizing our progress so far,

bx+1>(b-1b(a+1))ana.

Dividing both sides by b and simplifying,

bx>((b-1)aba+1(a+1)a)xa.

Proposition 4.

If a and b are real numbers and b>1, then

limxxabx=0.
Proof.

Substituting a+1 for a

bx>((b-1)a+1ba+2(a+2)a+1)xa+1.

Dividing by x and rearranging,

0<xabx<(ba+2(a+2)a+1(b-1)a+1)1x

Since limx0=0 and limx1x=0, we also have limxxabx=0 by the squeeze rule.

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