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# Mathematical Induction

## 1.

Calculus++ Light## 2.

Question 0. A continuous function f is definedon the interval [−1,1], and f 2(x) = x 2 for

each x from the interval [−1,1].

How many functions like that exist?

Solution. The function f (x) = 0 only if x = 0.

At all other points either f (x) = x or f (x) = – x.

Since the function f is continuous, it obtains

values of the same sign on each of the a ) 1

intervals [−1,0) and (0,1].

b) 2

Therefore, there are exactly four possible c ) 3

cases: f1 ( x) x, f 2 ( x) x,

f 3 ( x) | x |, f 4 ( x) | x | .

d) 4

e) 20

## 3.

Question 0+. A function f is defined on theinterval [−1,1], and f 2(x) = x 2 for each x

from the interval [−1,1].

How many functions like that exist if it is

known that x = ½ is the only point where f is

not continuous?

## 4. 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.

## 5.

Question 1. Using the Principle ofMathematical Induction show that

n(n 1)( 2n 1)

for

any

n

=

1,2,3,….

m

6

m 1

n

2

Solution. Step 1. The formula is correct in the

1(1 1)( 2 1)

case n = 1, because 1

.

6

2

Step 2. Let as assume that the formula is

k (k 1)( 2k 1)

correct for n = k: m

.

6

m 1

k

2

Our aim is to show, that in this case the

formula is also correct for n = k + 1.

## 6.

k 1We have

k

m m

2

m 1

2

(k 1)

2

m 1

k (k 1)( 2k 1)

2

(k 1)

6

k (2k 1)

(k 1)

k 1

6

2

2k 7 k 6

(k 2)( 2k 3)

(k 1)

(k 1)

6

6

(k 1)(( k 1) 1)( 2(k 1) 1)

.

6

Now, the principle of mathematical induction

tells us that our formula is correct for any n.

## 7.

Answers to Questions from Light #3:Functions and Limits

Question 2: e) the open first quadrant

Question 3: lim x

x 0

ln1x

1

e .

Question 4: e) I and III

Question 5: b) a nonconstant function

Question 6: f (x) is positive for all x > 0

## 8.

Calculus++Also known as

Hysterical Calculus

## 9.

Question 1b. Using the Principle ofMathematical Induction show that

sin( 2n x)

cos x cos(3x) cos(( 2n 1) x)

.

2

sin

x

for any n = 1,2,3,….

Solution. Step 1. The formula is correct in the

case n = 1, because sin( 2 x) 2 sin x cos x.

sin( 2 x)

And hence cos x

.

2 sin x

Step 2. Let as assume that the formula is

correct for n = k:

sin( 2k x)

cos x cos(3x) cos(( 2k 1) x)

.

2 sin x

## 10.

Our aim is to show, if our formula is correctfor n = k, then it is also correct for n = k + 1.

cos x cos(3x) cos(( 2k 1) x)

sin( 2k x)

cos(( 2k 1) x)

cos(( 2k 1) x)

2 sin x

sin( 2k x) 2 sin x cos(( 2k 1) x)

2 sin x

sin( 2k x) sin(( 2k 1) x x) sin(( 2k 1) x x)

2 sin x

sin( 2(k 1) x)

2 sin x

Now, the principle of mathematical induction

tells us that our formula is correct for any n.

## 11.

Question 3a. Calculate the following sumn

mq

m

.

m 0

Solution. We have

n 1

d 1 q

d

m

mq q q q

dq 1 q

dq m 0

m 0

n 1

n 1

(1 q ) (1 q) (1 q )(1 q)

q

2

(

1

q

)

n

n 1

(n 1)q (1 q) (1 q )

q

2

(1 q)

n 1

n

1 nq (n 1)q

q

2

(1 q)

n

m

n

## 12.

Question 5. Using the formula for thederivative of inverse function derive explicit

formulae for the derivatives of arcsin x,

arccos x, arctan x, and arccot x.

1

Solution. Using the formula g ( x)

f ( y ) y g ( x )

we obtain the following formula for the

derivative of arcsin x:

d

1

1

arcsin x

.

dx

(sin y ) y arcsin x cos y y arcsin x

The range of arcsin x is the interval [ 12 , 12 ].

Therefore cos(arcsin x) is always non-negative.

## 13.

Hence1

d

1

.

arcsin x

2

2

dx

1 x

1 sin (arcsin x)

Similar calculations yield the following

formula for the derivative of arccot x.

d

1

2

arccot x

sin y

y arccot x

dx

(cot y ) y arccot x

1

1

2

2

1 cot y y arccot x

1 cot (arccot x)

1

.

2

1 x

## 14.

Question 6. Use the Cauchy criterion to shown

1

that the sequence xn 2 , n 1,2, ,

k 1 k

converges.

Solution: It is sufficient to show that the

sequence xn is fundamental:

0, N , n N , m 0 : xn m xn .

n m

n m

1

1

We have xn m xn 2

k n 1 k

k n 1 ( k 1) k

n m

1 1

1

1

1

1

k n n 1 n 1 n 2

k n 1 k 1

1

1

1

1

n m 1 n m

n 2 n 3

## 15.

m1

1

1

.

n n m n ( n m) n

1

Thus xn m xn .

n

1

Therefore 0, N , we set N

,

n N , m 0 :

1 1

1

xn m xn 1 .

n N

Thus, the sequence xn is fundamental, and

therefore it converges to some limit L.

In fact, L

2

6

.

## 16. Picture of the Week

All ICEF students are of thesame height

## 17.

Question 4. Let f (x) be a differentiablefunction such that the derivative f (x) is a

continuous function and f (f (x)) = x for any

x. Furthermore, let f (0) = 1, and f (1) = 0.

a) Is it possible that there exists a number a

such that f (a) 0 ?

Solution: Differentiate the equation f (f (x)) = x

to obtain f ( f ( x)) f ( x) 1.

Therefore, there are no points a such that

f (a) 0,

because at a point like that

f ( f (a)) f (a) 0 1.

## 18.

b) Is it possible that there exists a number asuch that f (a) 0 ?

Solution: The Mean Value Theorem tells us

that

f (1) f (0) 0 1

f (c)

1 0

1 0

1.

for some point c: 0 < c < 1.

Therefore, there are no points a such that

f (a) 0,

otherwise there would be a point b

somewhere between c and a such that

f (b) 0.

## 19.

c) Let x1 be a solution of the equation f (x) = x.Find f ( x1 ).

Solution: If f (x1) = x1, then

1 ( f ( f ( x))) x x f ( f ( x1 )) f ( x1 )

1

f ( x1 ) f ( x1 ) f ( x1 ) 1.

## 20.

Answers to Questions from Seminar 3.1

a

Questions 1a: lim xn b .

n

Questions 2a: 2

1 ln 2

2

Questions 7: c) I and II only

Question 8a: yn 3 yn 2 yn 1 12 yn 32 .

3

Question 8b: lim yn .

n

7 1

Question 8c: xn 2 xn 1 2 xn 54 , lim xn 12 .

n

yn 2 yn 1 12 yn 12 , lim yn 15 .

n