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The exponential function
Arun Ram Department of Mathematics University of Wisconsin, Madison Madison, WI 53706 USA [email protected] Version: September 6, 2006
Define the exponential function as
ex^ = 1 + x + x
2 2!
3 3!
4 4!
5 5!
6 6!
7 7!
where k-factorial is
k! = k(k − 1)(k − 2) · · · 3 · 2 · 1 , for k = 1, 2 , 3 ,....
Why would anyone be so crazy as to write down such a horrible mess??
Example: Is there a function
f (x) = c 0 + c 1 x + c 2 x^2 + c 3 x^3 + c 4 x^4 + c 5 x^5 + · · ·
that changes addition into multiplication??,
f (x)f (y) = f (x + y).
If so
f (x + y) = c 0 + c 1 (x + y) + c 2 (x + y)^2 + c 3 (x + y)^3 + c 4 (x + y)^4 + c 5 (x + y)^5 + · · · = c 0
- c 1 x + c 1 y+
- c 2 x^2 + 2c 2 xy + c 2 y^2 +
- c 3 x^3 + 3c 3 x^2 y + 3c 3 xy^2 + c 3 y^3 +
- c 4 x^4 + 4c 4 x^3 y + 6c 4 x^2 y^2 + 4c 4 xy^3 + c 4 y^4 +
- · · ·
must be equal to
f (x)f (y) = (c 0 + c 1 x + c 2 x^2 + c 3 x^3 + c 4 x^4 + · · ·)(c 0 + c 1 y + c 2 y^2 + c 3 y^3 + c 4 y^4 + · · ·) = c^20 + c 0 c 1 x + c 0 c 2 x^2 + c 0 c 3 x^3 + c 0 c 4 x^4 + · · ·
- c 0 c 1 y + c^21 xy + c 1 c 2 x^3 y + c 1 c 4 x^4 y + · · ·
- · · ·.
2 a. ram
Comparing terms in these two expressions gives
c^20 = c 0 , c 0 c 1 = c 1 , c 0 c 2 = c 2 , c 0 c 3 = c 3 , c 0 c 4 = c 4 ,... , c 0 c 1 = c 1 , c^21 = 2c 2 , c 1 c 2 = 3c 3 , c 1 c 3 = 4c 4 , c 1 c 4 = 5c 5 ,... ,
So
c 0 = 1, c 2 = c
(^21) 2
, c 1 c
(^21) 2
= 3c 3 , c 1 c
(^31) 3 · 2
= 4c 4 , c 1 c
(^41) 4 · 3 · 2
= 5c 5 ,....
So
c 0 = 1, c 2 = c
(^21) 2
, c 3 = c
(^31) 3 · 2 · 1
, c 4 = c
(^41) 4 · 3 · 2 · 1
, c 5 = c
(^51) 5 · 4 · 3 · 2 · 1
So
f (x) = 1 + c 1 x +
c^21 2 x
(^2) + c^31 3! x
(^3) + c^41 4! x
(^4) + c^51 5! x
= 1 + c 1 x + (c^1 x)
2 2
3 3!
4 4!
5 5!
= ec^1 x.
So, if f (x + y) = f (x)f (y) then f (x) = ec^1 x.
Example: Is there a function
f (x) = c 0 + c 1 x + c 2 x^2 + c 3 x^3 + c 4 x^4 + c 5 x^5 + · · ·
whose derivative is itself, df dx
= f ???
If so, f (x) = c 0 + c 1 x + c 2 x^2 + c 3 x^3 + c 4 x^4 + c 5 x^5 + · · ·
must be equal to
df dx
= c 1 + 2c 2 x + 3c 3 x^2 + 4c 4 x^3 + 5c 5 x^4 + 6c 6 x^5 + · · ·.
Comparing terms in these two expressions gives
c 1 = c 0 , 2 c 2 = c 1 , 3 c 3 = c 2 , 4 c 4 = c 3 , 5 c 5 = c 4 , 6 c 6 = c 5 ,...
So c 2 = c^0 2
, 3 c 3 = c^0 2
, 4 c 4 = c^0 3 · 2
, 5 c 5 = c^0 4 · 3 · 2
So
f (x) = c 0 + c 0 x + c^0 2
x^2 + c
(^20) 3!
x^3 + c^0 4!
x^4 + c^0 5!
x^5 + · · ·
= c 0 (1 + x + x
2 2
3 3!
4 4!
5 5!
= c 0 ex.
4 a. ram
The exponential function is the function ex^ such that
dex dx =^ e
x (^) and e (^0) = 1.
Figure out what ex^ is: Suppose ex^ = a 0 + a 1 x + a 2 x^2 + a 3 x^3 + · · · Then e^0 = a 0 + 0 + 0 + · · · = 1. So a 0 = 1. dex dx =^ a^1 + 2a^2 x^ + 3a^3 x
(^2) + 4a 4 x (^3) + · · ·
= ex^ = a 0 + a 1 x + a 2 x^2 + a 3 x^3 + · · ·
So a 1 = 0, 2a 2 = a 1 , 4a 3 = a 2 , 4a 4 = a 3 ,.. .. So
a 0 = 1, a 1 = 1, a 2 =
2 ,^ a^3 =^
2 · 3 ,^ a^4 =^
2 · 3 · 4 ,^ a^5 =^
So ex^ = 1 + x +^1 2
x^2 + 1 2 · 3
x^3 + 1 2 · 3 · 4
x^4 + 1 2 · 3 · 4 · 5
x^5 + · · ·.
Factorials
7! = 7 · 6 · 5 · 4 · 3 · 2 · 1 = 5040 5! = 5 · 4 · 3 · 2 · 1 = 120 3! = 3 · 2 · 1.
So
ex^ = 1 + x +
x^2 2! +^
x^3 3! +^
x^4 4! +^
x^5 5! +^ · · · So e^1 = 1 + 1 +
24 +^
120 +^ · · ·^ = 2.^781828...
e−^3 = 1 + (−3) + (−3)
2 2
3 6
4 24
= 1 − 3 +^3
2 2
3 6
+^3
4 24
input x −→^ e
x (^) −→ output ex Note: By the chain rule
d dx e
2+x (^) = e2+x (^) · d(2 +^ x) dx =^ e
2+x (^) · 1 = e2+x
and e2+0^ = e^2.
the exponential function 5
So
e2+x^ = e^2 x + e
(^2) x 2 2!
(^2) x 3 3!
since, in this case a 0 = e^2 , a 1 = a 0 , 2 a 2 = a 1 , 3 a 3 = a 2 ,...
if e2+x^ = a 0 + a 1 x + a 2 x^2 + · · ·.
So e2+x^ = e^2 ex.
Similarly, e10+x^ = e^10 ex^ and e642+x^ = e^542 ex
and ey+x^ = ey^ ex^.
Since e−xex^ = e−x+x^ = e^0 = 1 e−x^ =
ex^.
Since e^10 x^ = ex+x+x+x+x+x+x+x+x+x = exex+x+x+x+x+x+x+x+x = exexex+x+x+x+x+x+x+x = exexexex+x+x+x+x+x = exexexexex+x+x+x+x = exexexexexex+x+x+x = exexexexexexexexexex^ = (ex)^10
Summary: ex^ is the function such that
dex dx =^ e
x (^) and e (^0) = 1.
Then ex+y^ = exey e−x^ = 1 ex enx^ = (ex)n.
