proof of equivalence of formulas for exp

proof of equivalence of formulas for exp

We present an elementary proof that:

∑

k

=

0

∞

z

k

k

!

=

lim

n

→

∞

⁡

(

1

+

z

n

)

n

.

There are of course other proofs, but this one has the advantage that it carries verbatim for the

matrix exponential

and the operator

exponential

.

At the outset, we observe that

∑

k

=

0

∞

z

k

/

k

!

converges

by the

ratio test

.
For definiteness, the notation

e

z

below will refer to exactly this series.

Proof.

We expand the right-hand in the straightforward manner:

(

1

+

z

n

)

n

=

∑

k

=

0

n

(

n

k

)

⁢

(

z

n

)

k

=

∑

k

=

0

n

n

⋅

(

n

-

1

)

⁢

⋯

⁢

(

n

-

k

+

1

)

n

k

⁢

z

k

k

!

=

∑

k

=

0

n

π

⁢

(

k

,

n

)

⁢

z

k

k

!

,

where

π

⁢

(

k

,

n

)

denotes the

coefficient

1

⁢

(

1

-

1

n

)

⋅

(

1

-

2

n

)

⁢

⋯

⁢

(

1

-

k

-

1

n

)

.

Let

|

z

|

≤

M

.
Given

ϵ

>

0

, there is a

N

∈

ℕ

such that whenever

n

≥

N

, then

∑

k

=

n

+

1

∞

M

k

/

k

!

<

ϵ

/

2

,
since the sum is the tail of the

convergent series

e

M

.

Since

lim

n

→

∞

⁡

π

⁢

(

k

,

n

)

=

1

for

k

, there is also a

N

′

∈

ℕ

, with

N

′

≥

N

, so that whenever

n

≥

N

′

and

0

≤

k

≤

N

, then

|

π

⁢

(

k

,

n

)

-

1

|

<

ϵ

/

(

2

⁢

e

M

)

.
(Note that

k

is chosen only from a

finite

set.)

Now, when

n

≥

N

′

, we have

|

∑

k

=

0

n

π

⁢

(

k

,

n

)

⁢

z

k

k

!

-

∑

k

=

0

∞

z

k

k

!

|

=

|

∑

k

=

0

n

(

π

⁢

(

k

,

n

)

-

1

)

⁢

z

k

k

!

-

∑

k

=

n

+

1

∞

z

k

k

!

|

≤

∑

k

=

0

n

|

π

⁢

(

k

,

n

)

-

1

|

⁢

M

k

k

!

+

∑

k

=

n

+

1

∞

M

k

k

!

=

∑

k

=

0

N

|

π

⁢

(

k

,

n

)

-

1

|

⁢

M

k

k

!

+

∑

k

=

N

+

1

n

|

π

⁢

(

k

,

n

)

-

1

|

⁢

M

k

k

!

+

∑

k

=

n

+

1

∞

M

k

k

!

<

ϵ

2

⁢

e

M

⁢

∑

k

=

0

N

M

k

k

!

+

∑

k

=

N

+

1

n

M

k

k

!

+

∑

k

=

n

+

1

∞

M

k

k

!

(In the middle sum, we use the bound

|

π

⁢

(

k

,

n

)

-

1

|

=

1

-

π

⁢

(

k

,

n

)

≤

1

for all

k

and

n

.)

<

ϵ

2

⁢

e

M

⋅

e

M

+

ϵ

2

=

ϵ

.

∎

In fact, we have proved

uniform convergence

of

lim

n

→

∞

⁡

(

1

+

z

n

)

n

over

|

z

|

≤

M

.
Exploiting this fact we can also show:

(

1

+

z

n

+

o

⁢

(

1

n

)

)

n

=

(

1

+

z

+

o

⁢

(

1

)

n

)

n

→

∑

k

=

0

∞

z

k

k

!

(pointwise, as

n

→

∞

)

Proof.

|

z

|

<

M

.
Given

ϵ

>

0

, for large enough

n

, we have

|

(

1

+

w

n

)

n

-

e

w

|

<

ϵ

/

2

uniformly for all

|

w

|

≤

M

.

Since

o

⁢

(

1

)

→

0

, for large enough

n

we can set

w

=

z

+

o

⁢

(

1

)

above.
Since the exponential is

continuous

1

1

follows from uniform convergence on

bounded

subsets of either expression for

e

z

, for large enough

n

we also have

|

e

z

+

o

⁢

(

1

)

-

e

z

|

<

ϵ

/

2

. Thus

|

(

1

+

z

+

o

⁢

(

1

)

n

)

n

-

e

z

|

≤

|

(

1

+

z

+

o

⁢

(

1

)

n

)

n

-

e

z

+

o

⁢

(

1

)

|

+

|

e

z

+

o

⁢

(

1

)

-

e

z

|

<

ϵ

.

∎

Title

proof of equivalence of formulas for exp

Canonical name

ProofOfEquivalenceOfFormulasForExp

Date of creation

2013-03-22 15:22:52

Last modified on

2013-03-22 15:22:52

Owner

stevecheng (10074)

Last modified by

stevecheng (10074)

Numerical id

11

Author

stevecheng (10074)

Entry type

Proof

Classification

msc 30A99

Related topic

ComplexExponentialFunction

Related topic

ExponentialFunction

Related topic

MatrixExponential

Generated on Fri Feb 9 20:14:09 2018 by

LaTeXML