Math 3AH, Fall 2011
Gonzalez
Section 10045
Assignment #3: Mathematical Induction
Directions: This assignment is due no later than Monday, November 28, 2011, at the
beginning of class. Late assignments will not be graded. You will be graded on
exactly what is asked for in the instructions below. You should submit your work on
a separate sheet of paper in the order the questions are asked. You will not only be
graded on your mathematics, but also on your organization, proper use of English,
spelling, punctuation, and logic.
Purpose: In this assignment you will study a form of mathematical proof called
mathematical induction. In general, induction is used to prove statements that
involve an integer n in their statement. Examples of such statements are:

The sum of the first n integers:

The sum of all odd integers up to a certain integer:

Integrating the sum of n functions:
∫∑
∫
∑∫
Introduction: The simplest and most common form of mathematical induction
proves that a statement involving a natural number n holds for all values of n. The
proof consists of two steps:
The Base Case: Here we show that the statement holds true when n is equal to the
lowest possible value that n can be given in the question. Usually that’s n  0 or n  1 .
The Inductive Step: Here we show that if the statement holds for some arbitrary
value of n  k , then the statement also holds when n  k  1 is substituted for n.
The assumption in the inductive step that the statement holds for some arbitrary
n  k is called the induction hypothesis (or inductive hypothesis). To perform the
inductive step, one assumes the induction hypothesis and then uses this assumption
to prove the statement for n  k  1 .
This method works by first proving the statement is true for a starting value, and
then proving that the process used to go from one value to the next is valid. If these
are both proven, then any value can be obtained by performing the process
Math 3AH, Fall 2011
Gonzalez
Section 10045
repeatedly. It may be helpful to think of the domino effect; if one is presented with a
long row of dominoes standing on end, one can be sure that:
The first domino will fall
Whenever a domino falls, its next neighbor will also fall,
so it is concluded that all of the dominoes will fall, and that this fact is inevitable.
Example 1: One application of mathematical induction is to prove a variety of sum
formulas. For example, the sum of the whole numbers starting with 1 up to some
arbitrary integer n obeys the following pattern.
1 2
1
2
23
1 2  3 
2
3 4
1 2  3  6 
2
45
1  2  3  4  10 
2
n   n  1
2
This pattern motivates the following proof by induction.
1 2  3  4 
n
[Proof] I will prove that the sum of the first n positive integers can be given by the
formula
n   n  1
1 2  3  4   n 
.
2
1 2 n   n  1

Base Case: Lets show it’s true for n  1 . Above we saw that 1 
, and
2
2
thus our formula holds for the first possible value of n.
Inductive Step: Let’s suppose our statement is true for an arbitrary value n  k and
let’s show it’s true for n  k  1 . That is, we want to show that
1 2  3  4 
 k   k  1 
 k  1    k  1  1
2
.
Math 3AH, Fall 2011
Gonzalez
Section 10045
With a little algebraic manipulation we see that
1 2  3  4 
 k   k  1  1  2  3  4 
 k    k  1
k  k  1
  k  1
2
k  k  1 2  k  1


2
2
 k  1 k  2 

2
 k  1   k  1  1

2
Therefore our statement is true for n  k  1 . Since the basis and inductive step have
been proved, our statement is true by mathematical induction.

Example 2: Another example of a sum formula dependent on an integer n is that of
the sum of the first n odds. The pattern is as follows:
1  1  12
1  3  4  22
1  3  5  9  32
1 3  5 
  2n  1  n 2
This pattern motivates the following proof by induction.
[Proof] I will prove that the sum of the first n odd integers can be given by the
formula
1  3  5    2n  1  n2
Base Case: Let’s show it’s true for n  1 .
1  12  n2
Inductive Step: Let’s suppose our statement is true for an arbitrary value n  k and
let’s show it’s true for n  k  1 . That is, we want to show that
Math 3AH, Fall 2011
Gonzalez
1 3  5 
Section 10045
  2k  1   2  k  1  1  1  3  5 

  2k  1    2  k  1  1
  2  k  1  1 
k2
 k 2  2k  1
  k  1
2
Therefore our statement is true for n  k  1 . Since the basis and inductive step have
been proved, our statement is true by mathematical induction.
Now it’s time for you to try some!
1. Prove the following formulas by mathematical induction.
n  n  1 2n  1
6
a) 12  22  32 
 n2 
b) 1  2  3 
 n  n  1 
n  

 2 
3
3
3
2
3
Example 3: Another application of mathematical induction is to extend properties
about two objects to properties about n objects. For example, if f and g are both
functions differentiable at x , then the function  f  g  is differentiable at x and
d
d
d
 f  x   g  x     f  x    g  x  .
dx
dx
dx
However, it seems reasonable that differentiation should distribute over addition no
matter how many functions are being added. That is,
d n
d
f j  x    f1  x  

dx j 1
dx
f n  x  
d
d
 f1  x   
 f n  x  
dx
dx 
n
d
   f j  x  
j 1 dx

Here’s a proof of this fact using induction.
Math 3AH, Fall 2011
Gonzalez
Section 10045
[Proof]
Base Case: The fact that we need at least two functions to add implies we should
start with n  2 . However this is trivial since this is the property about
differentiation that motivated us in the first place:
d
d
d
 f1  x   f 2  x     f1  x    f 2  x 
dx
dx
dx
Inductive Step: Let’s suppose our statement is true for an arbitrary value n  k and
let’s show it’s true for n  k  1 . That is let’s assume that
k
d k
d
 f j  x  
f
x





j
dx j 1
j 1 dx
()
This isn’t so bad! Let’s just apply our Base Case knowledge and our assumption.

d k 1
d  k
f
x



  f j  x   f k 1  x   ,

j
dx j 1
dx  j 1

 d
d  k
   f j  x     f k 1  x   ,
dx  j 1
 dx
k
d
d
   f j  x     f k 1  x   ,
dx
j 1 dx
by our Base Case
by ()
k 1
d
 f j  x   .
j 1 dx

Therefore our statement is true for n  k  1 . Since the basis and inductive step have
been proved, our statement is true by mathematical induction.
Now it’s time for you to try some!
2. Prove the following statements about functions using mathematical induction.
a) Suppose f1 , f 2 , , f n are all continuous functions and c1 , c2 ,..., cn are constants.
Then the function
n
f  x    ck f k  x 
k 1
is also continuous. Here you should use facts proven in assignment #2.
Math 3AH, Fall 2011
Gonzalez
b) Suppose f1 , f 2 ,
integrable and
Section 10045
, f n are all integrable functions. Then their sum is also
∫∑
∑∫
c) Generalized Product Rule: Suppose f1 , f 2 , , f n are all differentiable
functions at x . Then their product is also differentiable at x and
d
 f1  x   f 2  x  
dx 
d
 f1  x    f 2  x  
dx 
d
 f1  x    f 2  x   
dx
 f n  x   

 f1  x   f 2  x  
 fn  x 
 fn  x 

d
 f n  x  
dx 