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# The Taylor Formula

## 1.

Calculus++ Light## 2. The Taylor Formula

f (c)2

f ( x ) f (c) f (c)( x c)

( x c)

2!

(n)

( n 1)

f (c)

f

( )

n

n 1

( x c)

( x c) ,

n!

(n 1)!

where c ( x c), (0,1).

For instance,

0

0

e

e

2

3

e e e ( x 0) ( x 0) ( x 0)

2

!

3

!

0

e

e

n

n 1

( x 0)

( x 0) , where

n!

(n 1)!

x.

x

0

0

## 3.

Question 1. Ifx 3 x 1 a0 a1 ( x 2) a2 ( x 2) 2 a3 ( x 2)3 ,

for all real numbers x, then

a) a0 1, a1 12 , a2 0, a3 18

b) a0 1, a1 1, a2 0, a3 1

c) a0 7, a1 6, a2 10, a3 1

d ) a0 7, a1 11, a2 12, a3 6

e) a0 7, a1 11, a2 6, a3 1

Solution: Taylor’s formula tells us

0

f ( x) f (2) f (2)( x 2)

( 4)

f (2)

f (2)

f ( )

2

3

4

( x 2)

( x 2)

( x 2) .

2!

3!

4!

## 4.

x x 1 a0 a1 ( x 2) a2 ( x 2) a3 ( x 2) ,3

2

Therefore x x 1 7 (3x 1)

2

3

x 2

3

( x 2)

11

6

3

12 6 x x 2 ( x 2) ( x 2)

3!

6

2

a) a0 1, a1 12 , a2 0, a3 18

b) a0 1, a1 1, a2 0, a3 1

c) a0 7, a1 6, a2 10, a3 1

d ) a0 7, a1 11, a2 12, a3 6

e) a0 7, a1 11, a2 6, a3 1

## 5.

Answers to Questions from Light #7:Taylor’s Formula & Infinite Series

Question 2: e)

Question 3: L e

1

2

2

Question 4: S ln 2

## 6.

Calculus++Infinite Series

## 7.

Question 1. What is the greatest value of b forwhich any function f that satisfies the

properties (i), (ii), and (iii) must also satisfy

f (1) < 5?

(i) f (x) is infinitely differentiable for all x;

(ii) f (0) = 1, f (0) 1, and f (0) 2;

(iii) | f ( x) | b, for all x [0,1].

a) 1 b) 3 c) 6 d ) 12 e) 24

Solution: Taylor’s formula tells us

f (1) f (0) f (0)(1 0)

f (0)

f ( )

2

3

(1 0)

(1 0) .

2!

3!

## 8.

Using the properties (ii) and (iii) we obtainb

f ( )

3 .

f (1) 1 1 1

3!

3!

b

Solve the equation 3 5 to obtain b = 12.

3!

Therefore, the greatest value of b for which

f (1) < 5 is 12.

a) 1 b) 3 c) 6 d ) 12 e) 24

## 9.

e 1 x x x xx

1

2!

2

1

3!

3

1

n!

n

1

( n 1)!

n 1

e x .

In particular,

e 1 1

1

2!

1

3!

1

n!

1

( n 1)!

e ,

where (0,1).

That is

1 1 21! 31! n1! ( n 11)! e

1 1 21! 31! n1! ( n 11)! e

1 1 21! 31! n1! ( n 11)! 3.

## 10.

Question 2. Use the Taylor formula to showthat e is irrational.

k

Solution. Let us assume that e is rational, e .

m

Then the Taylor formula tells us that for any n:

k

1 1

m

1 1 21! 31! n1! ( n 11)! 3.

Take n max( m,3), and multiply both sides

1

2!

1

3!

1

n!

1

( n 1)!

of the double inequality by n! to obtain

k

2 n! 1

n!

m

2 n! n2!! n3!! 1 3 ( nn !1)! .

n!

2!

n!

3!

n!

( n 1)!

## 11.

Thereforek

n!

n!

3

n! 2 n! 2! 3! 1 n 1

m

k

0 n! 2 n! n2!! n3!! 1 1

m

1

n 1

an integer number

Contradiction!

Thus, our assumption that e is a rational

number leads to a contradiction.

Therefore, e is an irrational number.

## 12.

Question 3. Find lim n sin( 2 e n!).n

Solution. The Taylor formula tells us that for

1

1

1

1

1

any n: e 2 2! 3! n! ( n 1)! ( n 2)! e ,

where is a number between 0 and 1.

Multiply by 2 n! to obtain

2 e n! 2 (2n! 1)

n!

2!

n!

3!

2

n 1

2 n!

( n 2 )!

e.

= M, an integer number

Therefore

sin( 2 e n!) sin 2 M

sin

2

n 1

2

n 1

2

( n 1)( n 2 )

2 n!

( n 2 )!

e

e

## 13.

Hence, lim n sin( 2 e n!)n

lim n sin

n

2

n 1

2

( n 1)( n 2 )

e .

Since sin x is equivalent to x, when x is small,

we obtain

lim n sin( 2 e n!) lim

n

n

2 n

n 1

2 n

( n 1)( n 2 )

e 2 .

0

## 14.

Consider an infinite sequencea1 , a2 , a3 , a4 , , ak ,

If we add all the terms of this sequence we

obtain an infinite series

a1 a2 a3 ak

n

For example, consider the sequence an ( 1) :

1, 1, 1, 1, 1, 1, 1, 1,

The corresponding infinite series is

0

0

0

0

1 1 1 1 1 1 1 1 0 or ?! 1?

0

0

0

0

What is the value of this infinite series?

This infinite series does not have a value.

## 15.

The sum of the first n terms of an infinite series,Sn, is called the n-th partial sum of the series

Sn a1 a2 an .

An infinite series converges, if converges the

sequence of its partial sums:

lim Sn S .

n

The limit, S, of the sequence of partial sums is

the sum of the infinite series.

## 16.

Example. The geometric seriesab .

k

k 1

The n-th partial sum of the geometric series is

given by n

n 1

n

b b

.

S n ab a b a

1 b

k 1

k 1

k

k

If a = 0, the sequence Sn converges to 0.

If a 0 and |b| > 1, the sequence of partial sums

Sn diverges.

If |b| < 1, the sequence Sn converges.

n 1

ab

b b

.

lim a b lim a

n

n

1 b

1 b

k 1

n

k

## 17.

If b = 1, the sequence of partial sums Sn diverges(unless a = 0).

n

n

S n ab a 1 an sgn a .

k

k 1

k 1

If b = –1, the sequence of partial sums Sn also

diverges (again, unless a = 0).

a, if n is odd,

S n ab a ( 1) 0, if n is even.

k 1

k 1

n

n

k

k

## 18.

A necessary condition for convergence.If a series

a

k 1

k

converges, then lim ak 0.

k

Indeed, if the sequence of partial sums

n

S n ak converges, then (Cauchy criterion)

k 1

, N , n N and m 0 : | Sn m Sn | .

Set m = 1, then , N , n N : | an 1 | .

Therefore lim ak lim an 1 0.

k

n

## 19.

Question 5. Which of the following seriesconverge?

k

2

e ln k

I.

k!

k 1

III.

cos k

II. k

k

k 1

log

k 2

k

k

a. I only b. II only c. III only

d. I and II only e. II and III only

ln k

Solution: For series III: ak log k k

1.

ln k

lim ak 1 0. Hence, series III diverges.

k

## 20.

In series II: limk

k

k lim exp ln k

k

k

ln k

ln k

lim exp

exp lim

exp 0 1.

k

k

k k

The sequence cos(k) diverges as k .

cos k

Hence, the sequence ak k

k

does not converge to 0 as k .

Therefore, series II diverges.

a. I only b. II only c. III only

d. I and II only e. II and III only

## 21.

Important series.1 1

1

1

1 q q q q .

2 3

n

n 1 n

This series converges if q > 1, and it diverges if

q 1.

## 22.

Question 5. A certain ball has the propertythat each time it falls from a height h onto a

hard, level surface, it rebounds to a height

rh, where 0 < r < 1.

Suppose that the ball is dropped from an

initial height of H meters.

a) Assuming that the ball continues to bounce

indefinitely, find the total distance that it

travels.

Solution:

## 23.

A bouncing ball – total distance travelledd H 2rH 2r H 2r H

n

4

2r H H 2 H r

r

1 r n 1

H 2H

H

.

1 r

1 r

1 r

Answer : d H

1 r

2

h

H

rH

2

r H

3

r4 H

r H

3

«Ацкок»

1

2

## 24.

Question 7. A certain ball has the propertythat each time it falls from a height h onto a

hard, level surface, it rebounds to a height

r h, where 0 < r < 1.

Suppose that the ball is dropped from an

initial height of H meters.

b) Calculate the total time that the ball spends

bouncing.

Hint: A ball having zero velocity falls ½ gt2

meters in t seconds.

Solution: Actually, philosophers might find it

obvious that the ball never stop bouncing.

## 25.

A bouncing ball – total bouncing time2h

h gt t

h

g

2

3

2H

2rH

2r H

2r H

H T

2

2

2

g

g

g

g

rH

2

r H

3

r4 H

r H

1

2

2

«Ацкок»

1

2

## 26.

23

2H

2rH

2r H

2r H

T

2

2

2

g

g

g

g

2H

2H

2

g

g

r r r

2

3

2H

2H 1 r

2H

r

.

2

g

g 1 r

g 1 r

2H 1 r

Answer : T

g 1 r

## 27.

Question 7. c) Suppose that each time the ballstrikes the surface with velocity v, it

rebounds with velocity – kv, where 0 < k < 1.

How long will it take for the ball to come to

rest?

Solution:

The velocity of the ball when it hits the

ground for the first time is given by

2H

v gt , where t

, v 2Hg .

g

## 28.

Hence, the ball that rebounds with velocity v1will reach the height

2

v1

H1

.

2g

2 2

k v

2

In our case v1 kv. Hence H 1

k H.

2g

Therefore r k 2 ,

2H 1 k

2H 1 r

.

and T

g 1 k

g 1 r

2H 1 k

Answer : T

g 1 k