Calculus: Analytic Geometry and Calculus, with Vectors

9.2 Derivatives and integrals of exponentials and logarithms 503

F and each of the derivatives F(l), n = 1, 2,3, ,is continuous over the
whole infinite interval - ao < x < oo and, moreover,

(2) Fn) (0)= 0 (n = 1, 2, 3, ...).

Remark: This is one of the famous functions of mathematical analysis. Note
that we have no formulas into which we can put x = 0 to obtain the numbers
F(n)(0). After we have learned that F(k)(0) = 0 when k = n, we must use the
definition
F(n+1) (0)= limFcn) (x)-Fcn) (0)
X-0 x

to learn that F(k)(0) = 0 when k = n + 1.
22 Supposing that a,, a2, a3, are positive numbers, that n is an integer
for which n > 1, that x > 0, and that

ai+a2+ +an-i+x

aja2. an_ix

prove that fn(x) is a minimum when and only when

(2) x =

a1 + a2 + + a,,_1

n

Hint: Use the fact that fn(x) is a minimum when log fn(x) is a minimum. Show

that

(3) Id log fn(x)

_ n-1 a,+a2+ ... +an-,

nx(ai + a2 + ... + an-1 + x) Lx - n - 1 1

Remark: Our result shows that

(4) fn

a, + a2 + ... + an-1)

n < fn an)

and that equality holds only when an is the arithmetic mean (or ordinary average)

of the numbers a1, a2, ,a,,_l. Interest in this matter can start to develop

when we observe that

a1+a2+ +an

n

n n

a1a2 an

and hence that fn(an) is the ratio of the arithmetic mean of the numbers a1, a2,

, an to the geometric mean of the same numbers. By using (1) to obtain

an expression for the left member of the formula

a, + a2 + + an-1 cn-i>tn

(6)

a1+a2+ + an_1n-1

fn n (ala2 ..

[fn-1(an-1)lcn-1)In

and applying some quite elementary algebraic operations to simplify the result,

we can derive the formula. From (6) and (4) we obtain the formula

(7) [fn-1(an-1)]n-1 C ln(an)l.