Lecture Notes: Friday, October 23, 1998

Sixth Homework Assignment - for discussion at recitations Tuesday October 20, and Thursday October 22. The sixth homework also includes your first MAPLE ASSIGNMENT.

Seventh Homework Assignment - for discussion at recitations Tuesday October 27, and Thursday October 29.

Chapter 5

• Section 1 (Exponential Functions)
• Section 2 (Logarithmic Functions)

Logarithmic Functions

For c > 0, a = logb(c) if and only if b^a = c (here, as above, b > 0, b not equal to 1)

In the text, log with no other indication of base means log to the base 10 and "ln" means log to the base e. However, in MAPLE both ln and log mean base e. To get base 10 in MAPLE you use log[10], so log[10](1000) = 3.

Properties:

logb(x*y) = logb(x) + logb(y)

logb(x/t) = logb(x) - logb(y)

logb(x^y) = y*logb(x)

logb(1) = 0

logb(b) = 1

Inverse functions. g and f are inverse functions if g(f(x)) = x and f(g(x)) = x. In this case, (a,b) is on the graph of f if and only if (b,a) is on the graph of g, so the graphs are reflections of each other about the line y = x.

logb(x) and b^x are inverse functions.

e^(ln(x)) = x for x > 0. ln(e^x) = x for all x. So exp(x) and ln(x) are inverse functions. We looked at the graphs in MAPLE.

Problems:

5.1 # 24

5.2 # 14, 24, 26, 40, 44

Finally, the first MAPLE homework problem suggests the use of induction. It is not required that you give a proof by induction. However, I would like you to know about induction. If you can do the proof by induction, that would be great.

MATHEMATICAL INDUCTION says this: Suppose you have a collection of statements P(n), one for each positive integer n. Suppose

• P(1) is true and
• P(k+1) is true whenever P(k) is true.

Then P(n) is true for all n.

Typical Example: I claim 1+ ... + n = n(n+1)/2. This is our statement P(n).

Proof by induction:

• P(1) just says 1 = 1*2/2 which is clearly true.
• Assume P(k) is true. Then
1+...+k + (k+1) = k(k+1)/2 + (k+1) (by the induction assumption)
=(k+1)(k+2)/2 (by algebra)

But this is the statement P(k+1). Thus P(k) true implies P(k+1) true.

Then induction says that P(n) is true for all n.

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