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# Mathematical Induction

## 1.

Calculus++ Light## 2. Mathematical Induction

Let Sn, n = 1,2,3,… be statements involvingpositive integer numbers n.

Suppose that

1. S1 is true.

2. If Sk is true, then Sk +1 is also true.

Then Sn is true for all positive integer numbers n.

## 3.

Question 1. Let x1 = 1 and xn 1 7 6 xnfor all positive integers n. If it is assumed

that the sequence {xn} converges, then a ) 1

lim xn

b) 0

Solution. First, we find fixed points of the

c ) 13

relationship xn 1 7 6 xn .

That is, we find solutions of the equation d ) e

e) 7

x 7 6x.

n

x 7 6 x x1 1, x2 7.

2

Next, we calculate a few terms of the

sequence. x2 7 6 1 13 3 x2 7

x3 7 6 x2 7 6 3 x3 7 6 7

5 x3 7.

## 4.

x4 7 6 x3 7 6 5 x4 7 6 76 x4 7

a) 1

Our first statement S1: 6 x4 7.

b) 0

The general statement Sn: 6 xn 3 7.

The first statement S1 is correct.

Let us assume that a statement Sk :

6 xk 3 7 is also correct. Then

c ) 13

d) e

e) 7

xk 4 7 6 xk 3 7 6 6 xk 4 7 6 7

6 xk 4 7

Therefore, the principle of mathematical

induction tells us that 6 xn 3 7 for any

positive integer n.

## 5.

Since we know that the sequence xn converges6 lim xn 3 7

n

But, what if this is a free-response

question?

Then we can pass to the limit n

in both parts of the relationship

xn 1 7 6 xn .

L lim xn 1 lim 7 6 xn .

n

a) 1

b) 0

c ) 13

d) e

e) 7

n

Since f ( x) 7 6 x is a continuous function

for x 76 , we obtain

L 7 6 lim xn 7 6 L . L 7.

n

## 6.

Answers to Questions from Light #2:Sequences and Limits

1 ( 1)

Question 1: An

, Bn n

2

Question 2: d ) b1 36

n

Question 3a: N min 17

Question 3b: N min Integer part of

1

5

Question 4: 2

Question 5: 4

k

You have to show that k 2

k 1 2

5 8

## 7.

Calculus++Also known as

Hysterical Calculus

## 8.

Question 1b. Find the limit of the sequencen

n a n b

.

yn

2

n

n a n b

Solution. We have

2

n

n a n b

n

n

a

b

n ln

exp ln

exp

2

2

Using the equivalence n a ~ 1 1n ln a we obtain

n

n a n b

1 ln a ln b

~ exp n ln 1

n

2

2

## 9.

Using now the equivalence ln 1 1n c ~ 1n c,we obtain

n

a b

1 ln a ln b

~ exp n

exp ln ab

n

2

2

ab

.

n

n a n b

ab .

Therefore, lim

n

2

n

n

## 10.

Question 2b. Find the following limit17

20

L lim

.

20

17

x 1 1 x

1 x

Solution. Actually, the equivalence

x (1 ( x 1)) ~ 1 m( x 1),

m

m

gives an incorrect result for this limit.

20

17

L lim

x 1 1 1 20( x 1)

1 1 17( x 1)

1

1

lim

lim 0 0. Incorrect!

x 1

x 1 x 1 x 1

To obtain a correct result we have to use

x ~ 1 m( x 1) m(m 1)( x 1) .

m

1

2

2

## 11.

The last equivalence yields20

L lim

2

x 1 1 1 20( x 1) 1 20 19( x 1)

2

17

2

1

1 1 17( x 1) 2 17 16( x 1)

1

1

lim

2

1

x 1 ( x 1) 1 19( x 1) 2

( x 1) 2 16( x 1)

2

2

2

1

1

( x 1) 2 16( x 1) ( x 1) 2 19( x 1)

lim

x 1 ( x 1) 1 16( x 1) 2 ( x 1) 1 19( x 1) 2

2

2

2

1

3

2 (19 16)( x 1)

lim

2

2

1

x 1 ( x 1) 1 16( x 1)

(

x

1

)

2

2

2 19( x 1)

## 12.

Question 3. Find the following limitsin tan x tan sin x

lim

.

x 0 arcsin arctan x arctan arcsin x

Solution. We have to find precise equivalent

functions for each term at x = 0

sin tan x 0 x 0 x x 0 x 4 ( 3) 51! x 5

2

1

6

3

0 x ( 275) x

6

1

7!

7

tan sin x 0 x 0 x x 0 x ( 3) x

2

1

6

0 x ( 107) x

6

4

3

1

7!

7

Therefore, sin tan x tan sin x 168 71! x 7 .

1

5!

5

## 13.

A similar (but even more difficult) calculationyields

arcsin(arc tan x) x x 13 x 341 x

3

5

arctan(arc sin x) x x 13 x 173 x

1

3!

1

3!

3

1

5!

1

5!

5

1 7

7!

1 7

7!

Therefore

arcsin arctan x arctan arcsin x 168 x .

1

7!

7

sin tan x tan sin x

and lim

1.

x 0 arcsin arctan x arctan arcsin x

## 14. Picture of the Week

## 15.

Question 8. The initial location of a snail isthe point S1 = (0,0).

The first 3 turning points, keeping the snail

confined inside a unit square, are given by

S2 = (½, 1), S3 = (1, ½), and S4 = (½, 0).

After that the snail always heads towards the

midpoint of the next path segment that it

sees, without crossing its own path.

That is, the coordinates (xn, yn) of the turning

points Sn are given by

xn 3 xn 4

yn 3 yn 4

and yn

xn

,

2

2

for n = 5,6,7,…

## 16.

a) Show that xn+3 + xn+2 + xn+1 + ½ xn = 2,and find a similar relationship for the ycoordinates.

b) Find the x and y coordinates of the limiting

point S of the snail path.

c) Repeat parts a) and b) for a snail confined

inside an isosceles triangle.

The initial location of the snail is S1 = (0,0).

The first 2 turning points are S2 = (¾, ½) and

S3 = (½, 0).

## 17.

Snail path in the xy – plane.y

1

x

1

## 18.

Solution. Write down the characteristicequation for the formula

xn 3 xn 4

xn

.

2

xn 1 xn

4

xn 4

0 12 12 0.

2

Unfortunately, this equation has only two real

solutions, but it definitely has a solution 1 1.

Divide the characteristic polynomial by 1.

( 1)( 12 ).

Denote z n xn 3 xn 2 xn 1 12 xn ,

then z n 1 z n xn 4 xn 3 xn 2 12 xn 1

xn 3 xn 2 xn 1 12 xn xn 4 12 ( xn 1 xn ) 0.

4

1

2

1

2

3

2

1

## 19.

Therefore z n 1 z n z n 1 z n 2 z1.z n xn 3 xn 2 xn 1 12 xn

x4 x3 x2 12 x1

12 1 12 12 0 2.

If the sequence xn converges to x , then

lim xn 3 lim xn 2 lim xn 1 lim xn x .

Thus

n

n

n

n

Passing to the limit n in the relationship

xn 3 xn 2 xn 1 12 xn 2

we obtain x x x 12 x 2.

4

Hence x .

7

## 20.

Snail path in the xy – plane.y

1

x

1

## 21.

Answers to Questions from Seminar 2.1

3

Questions 1b and 1c: lim yn , lim zn .

n

2

2 n

Question 8: n = 20182 – 1.