elementary proof of growth of exponential function

elementary proof of growth of exponential function

Proposition 1

.

If

x

is a non-negative

real number

and

n

is a non-negative integer, then

(

1

+

x

)

n

≥

1

+

n

⁢

x

.

Proof.

When

n

=

0

, we have

(

1

+

x

)

0

=

1

≥

1

+

0

⋅

0

.
If, for some

natural number

n

, it is the case that

(

1

+

x

)

n

≥

1

+

n

⁢

x

then, multiplying both sides of
the

inequality

by

(

1

+

x

)

, we have

(

1

+

x

)

n

+

1

≥

(

1

+

x

)

⁢

(

1

+

n

⁢

x

)

=

1

+

(

n

+

1

)

⁢

x

+

n

⁢

x

2

≥

1

+

(

n

+

1

)

⁢

x

.

By

induction

,

(

1

+

x

)

n

≥

1

+

n

⁢

x

for every
natural number

n

.
∎

Proposition 2

.

If

b

is a real number such that

b

>

1

and

n

and

k

are non-negative integers, we have

b

n

>

(

b

-

1

b

⁢

k

)

k

⁢

n

k

.

Proof.

Let

x

=

b

-

1

. Write

n

=

m

⁢

k

-

r

where

m

and

r

are non-negative integers and

r

<

k

.

By the preceding

proposition

,

(

1

+

x

)

m

>

m

⁢

x

. Raising both
sides of this inequality to the

k

th

power, we
have

(

1

+

x

)

m

⁢

k

>

(

m

⁢

x

)

k

. Since

r

<

k

, we also
have

(

1

+

x

)

-

r

>

(

1

+

x

)

-

k

; multiplying both
sides by this inequality and collecting terms,

(

1

+

x

)

m

⁢

k

-

r

>

(

x

1

+

x

)

k

⁢

m

k

.

Multiplying the right-hand side by

k

k

/

k

k

and
rearranging,

(

x

1

+

x

)

k

⁢

m

k

=

(

x

(

1

+

x

)

⁢

k

)

k

⁢

(

m

⁢

k

)

k

.

Since

m

⁢

k

≥

m

⁢

k

-

r

, we also have

(

x

(

1

+

x

)

⁢

k

)

k

⁢

(

m

⁢

k

)

k

≥

(

x

(

1

+

x

)

⁢

k

)

k

⁢

(

m

⁢

k

-

r

)

k

.

Recalling that

m

⁢

k

-

r

=

n

and

1

+

x

=

b

, we conclude that

b

n

>

(

b

-

1

b

⁢

k

)

k

⁢

n

k

.

∎

Proposition 3

.

If

a

,

b

, and

x

are real numbers such that

a

≥

0

,

b

>

1

and

x

>

0

, then

b

x

>

(

(

b

-

1

)

a

b

a

+

1

⁢

(

a

+

1

)

a

)

⁢

x

a

.

Proof.

Let

k

and

n

be integers such that

a

≤

k

≤

<

a

+

1

and

x

≤

n

≤

x

+

1

. Since

x

+

1

>

n

, we have

b

x

+

1

>

b

n

. By the preceeding proposition, we have

b

n

>

(

b

-

1

b

⁢

k

)

k

⁢

n

k

.

Since

k

<

a

+

1

, we have

1

/

k

k

>

1

/

(

a

+

1

)

k

, so

(

b

-

1

b

⁢

k

)

k

>

(

b

-

1

b

⁢

(

a

+

1

)

)

k

.

Since

k

≥

a

≥

0

, we have

(

b

-

1

b

⁢

(

a

+

1

)

)

k

⁢

n

k

≥

(

b

-

1

b

⁢

(

a

+

1

)

)

a

⁢

n

a

.

Summarrizing our progress so far,

b

x

+

1

>

(

b

-

1

b

⁢

(

a

+

1

)

)

a

⁢

n

a

.

Dividing both sides by

b

and simplifying,

b

x

>

(

(

b

-

1

)

a

b

a

+

1

⁢

(

a

+

1

)

a

)

⁢

x

a

.

∎

Proposition 4

.

If

a

and

b

are real numbers and

b

>

1

, then

lim

x

→

∞

⁡

x

a

b

x

=

0

.

Proof.

Substituting

a

+

1

for

a

b

x

>

(

(

b

-

1

)

a

+

1

b

a

+

2

⁢

(

a

+

2

)

a

+

1

)

⁢

x

a

+

1

.

Dividing by

x

and rearranging,

0

<

x

a

b

x

<

(

b

a

+

2

⁢

(

a

+

2

)

a

+

1

(

b

-

1

)

a

+

1

)

⁢

1

x

Since

lim

x

→

∞

⁡

0

=

0

and

lim

x

→

∞

⁡

1

x

=

0

, we also have

lim

x

→

∞

⁡

x

a

b

x

=

0

by the

squeeze rule

.

∎

Title

elementary proof of growth of exponential function

Canonical

name

ElementaryProofOfGrowthOfExponentialFunction

Date of creation

2014-03-10 17:57:26

Last modified on

2014-03-10 17:57:26

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

28

Author

rspuzio (6075)

Entry type

Definition

Classification

msc 26A12

Classification

msc 26A06

Generated on Fri Feb 9 19:47:57 2018 by

LaTeXML