Using (1.1), we obtain that a1= a0, a2 = a1/2, a3= a2/3 and so forth: (1.2) an= 1 nan−1, n ≥ 1

1. Exponential Function

In this note, we are going to solve the following differential equation

(1.1) d

dxy = y, y(0) = 1.

Let us introduce an infinite series

y = a₀+ a₁x + a₂x²+ · · · =

∞

X

n=0

a_nxⁿ. Assume that we can differentiate y with respect to x:

d

dxy = a₁+ 2a₂x + 3a₃x²+ · · · =

∞

X

n=0

(n + 1)a_n+1xⁿ. Using (1.1), we obtain that a₁= a₀, a₂ = a₁/2, a₃= a₂/3 and so forth:

(1.2) a_n= 1

na_n−1, n ≥ 1.

Using y(0) = 1, we have a₀ = 1.

Lemma 1.1. If (an) is a sequence of numbers obeying (1.2) with a0= 1, then

(1.3) a_n= 1

n! n ≥ 0.

Proof. We know that a₀ = 1, and hence (1.3) holds for n = 0. Assume that (1.3) holds for n = k, i.e. ak= 1/k!. Then

a_k+1= ak

k + 1 = 1 k + 1 · 1

k! = 1 (k + 1)!.

Hence (1.3) holds for n = k + 1. By Mathematical induction, (1.3) holds for all n ≥ 0.  Hence given any x, we can define (formally) an infinite series:

(1.4)

∞

X

n=0

xⁿ n!.

Lemma 1.2. For any x ∈ R, the infinite series (1.4) is absolutely convergent.

Proof. Let b_n= xⁿ/n!. Then bn+1

b_n = xⁿ⁺¹ (n + 1)!

n!

xⁿ = x n + 1. Then

n→∞lim    

b_n+1 bn

= 0 < 1.

By ratio test, the infinite series (1.4) is absolutely convergent (and hence convergent).  Hence the assignment

x 7→

∞

X

n=0

xⁿ n!

is a real-valued function. We denote this function by exp(x) or by e^x.

1

2

Proposition 1.1. For any x, y ∈ R, we have

(1.5) exp(x + y) = exp(x) exp(y).

Proof. We know

exp(x + y) =

∞

X

n=0

(x + y)ⁿ n! . Using Binomial theorem,

(x + y)ⁿ=

n

X

k=0

n k



x^ky^n−k. By definition,

n k



= n!

k!(n − k)!. Hence we obtain

∞

X

n=0

(x + y)ⁿ

n! =

∞

X

n=0 n

X

k=0

1 n!

n k

 x^ky^n−k

=

∞

X

n=0 n

X

k=0

1 n!

n!

k!(n − k)!x^ky^n−k

=

∞

X

n=0 n

X

k=0

1

k!(n − k)!x^ky^n−k

=

∞

X

n=0 n

X

k=0

x^k k!

y^n−k (n − k)!. Let j = n − k, we obtain

∞

X

n=0 n

X

k=0

x^k k!

y^n−k (n − k)! =

∞

X

n=0

X

k+j=n

x^k k!

y^j

j! = exp(x) exp(y).

We conclude that (1.5) holds for any x, y ∈ R. 

Now, we need to check that y = exp(x) is indeed the solution to (1.1) and at first, we need to show that the function is differentiable at all x ∈ R. To do this, we compute the difference quotient: using Proposition 1.1, we have

exp(x + h) − exp(x)

h = exp(x) exp(h) − exp(x)

h = exp(x)exp(h) − 1

h .

To show lim

h→∞

exp(x + h) − exp(x)

h exists, we only need to show that lim

h→∞

exp(h) − 1

h exists.

By definition,

exp(h) = 1 + h 1!+h²

2! + · · · , we see that

exp(h) − 1

h = 1 + h

2! +h²

3! + · · · = 1 +

∞

X

n=1

hⁿ (n + 1)!. Hence intuitively, we find

h→0lim

exp(h) − 1

h = 1.

3

Hence we find

d

dxexp(x) = lim

h→0

exp(x + h) − exp(x) h

= lim

h→0exp(x)exp(h) − 1 h

= exp(x) lim

h→0

exp(h) − 1 h

= exp(x).

In other words, y = exp(x) is differentiable at all x ∈ R and satisfies (1.1).