Introduction to calculus - HWfoss/F71SM1/CALCULUS/Elements_of... · 2009. 9. 4. · 1. Mathematical Induction In order to prove that a certain statement holds for all positive integers

Introduction to calculus

0. Notation we’ll use the following shortenings:

∀ means “for all”, “for any”, “for every”.

∃ means “there exists”, “there exist”.

iff means if and only if

colon “:” means “such that”.

For instance, here is an example of a “mathematical statement”:∀ x ∃ y : f(y) = g(x) iff ∃y : f(y) > 0 .

It should be read as follows:The following statements are equivalent:(i) for any x there exists y such that f(y) = g(x),and(ii) there exists y such that f(y) > 0.

1. Mathematical InductionIn order to prove that a certain statement holds for all positive integers

n ≥ n0, it is sufficient to prove that(i) it holds for n = n0, and(ii) if it holds for some n ≥ n0, it holds also for n + 1.

Example. Show that

1 + 2 + . . . + n =n(n + 1)

2for all n ≥ 1.

2

2. Absolute valueLet x be a real number. Its absolute value |x| is a non-negative number

which is defined as follows:

|x| = x for x ≥ 0 and |x| = −x for x < 0.

For all real numbers x and y, the following inequalities hold:

|x| − |y| ≤ |x + y| ≤ |x|+ |y|.

3

3. Maximum and supremum, minimum and infimumFor any finite set of numbers {x1, x2, . . . , xk}, the maximum max1≤i≤k xi

of these numbers is one of them, say, xi such that xi ≥ xj for all j 6= i. Themaximum over a finite set always exists.

Example. max{2, 3,−5, 1.5} = 3For any set of numbers X , its supremum y = supX is a number such that

y ≥ x for all x ∈ X , (1)

and y is the smallest number with property (1).If the supremum exist, a set is bounded from above. Otherwise we use a

convention that supX = ∞.If the supremum is an element of a set, it is also its maximum.Examples.

sup{0; 1/2; 2/3; . . . ; n/(n+1); . . .} = 1 and the maximum of this set does notexist.sup{1; 2; 3; . . . ; n; . . .} = ∞ and the maximum of this set does not exist.sup{1; 1/2; 1/3; . . . ; 1/n; . . .} = 1 = max{1; 1/2; 1/3; . . . ; 1/n; . . .}.

Minimum and infimum are defined by symmetry:For any finite set of numbers {x1, x2, . . . , xk}, the minimum min1≤i≤k xi

of these numbers is one of them, say, xi such that xi ≤ xj for all j 6= i. Theminimum over a finite set always exists.

For any set of numbers X , its infimum y = inf X is a number such that

y ≤ x for all x ∈ X , (2)

and y is the biggest number with property (2).If the infimum exist, a set is bounded from below. Otherwise we use a

convention that inf X = −∞.If the infimum is an element of a set, it is also its minimum.

4

4. Absolute and relative error.If a 6= 0 is an exact value of a measurement and r its approximate value,

then∆ = |x− a| is the absolute error andδ = ∆

|a| the relative error of a measurement.

5

Exercises.1. Prove by induction the following identities and inequalities.

(1.1) 12 + 22 + . . . + n2 =n(n + 1)(2n + 1)

6∀ n ≥ 1.

(1.2) 13 + 23 + . . . + n3 = (1 + 2 + . . . + n)2 ∀ n ≥ 1.

(1.3) 1 + 2 + 22 + . . . + 2n−1 = 2n − 1 ∀ n ≥ 1.

(1.4)1

2· 3

4· . . . · 2n− 1

2n<

1√2n + 1

∀ n ≥ 1.

(1.5) 2! · 4! · . . . · (2n)! > ((n + 1)!)n ∀ n ≥ 1

(here n! = 1 · 2 · . . . · n).

(1.6) 1 +1√2

+1√3

+ . . . +1√n

>√

n ∀ n ≥ 2.

(1.7) (2n)! < 22n(n!)2 ∀ n ≥ 1.

2. Absolute values.(2.1) Solve inequalities:

(i) |x + 1| < 0.01(ii) |x− 2| ≥ 10(iii) |x| > |x + 1|(iv) |2x− 1| < |x− 1|(v) |x + 2|+ |x− 2| ≤ 12(vi) |x + 2| − |x| > 1(vii) ||x + 1| − |x− 1|| < 1(viii) |x(1− x)| < 0.05.

(2.2) Prove inequalities:(i) |x− y| ≥ ||x| − |y||;(ii) |x + x1 + . . . + xn| ≥ |x| − (|x1|+ . . . + |xn|).

3. Supremum and Infimum.(3.1) Find a supremum and an infimum of a set of all rational numbers r

such that r2 < 2.(3.2) Let X + Y be a set of all sums x+y where x ∈ X and y ∈ Y . Prove

the following equalities:(i) inf (X + Y) = inf X + inf Y ;(ii) sup (X + Y) = supX + supY .

6

(3.3) Let XY be a set of all products xy where x ∈ X , y ∈ Y , and bothX and Y are subsets of the positive half-line [0,∞). Prove equalities:(i) inf (XY) = inf X · inf Y ;(ii) sup (XY) = supX · supY .

7

4. Sequences.A sequence x1, x2, . . . , xn (or, in short, a sequence {xn}) has a limit a, i.e.

limn→∞

xn = a or, equivalently, xn → a as n →∞

if, ∀ ε > 0 ∃ N = N(ε) :

|xn − a| < ε ∀n ≥ N.

A sequence is called convergent if it has a limit and divergent otherwise.Examples. A sequence xn = 1/n has a limit 0 (and is a convergent

sequence) while a sequence yn = (−1)n is divergent.

4.1. Criteria for existence of a limit.Criterion 1. (“rule of two policemen”). If

yn ≤ xn ≤ zn ∀ n

andlim

yn→∞= lim

zn→∞= c,

then limn→∞ xn also exists and equals c.Criterion 2. Any monotone (either increasing or decreasing) and bounded

sequence has a limit.Criterion 3. (Cauchy). A limit limn→∞ xn exists iff ∀ε > 0 ∃ an integer

N = N(ε) :

|xn − xn+m| ≤ ε ∀ n ≥ N and ∀ m ≥ 1.

Remark. In the sequel, we sometimes will write for short lim xn insteadof limn→∞ xn.

4.2. Main Theorems. If both limits a = lim xn and b = lim yn exist,then

1) if xn ≤ yn for all n, then a ≤ b;2) lim(xn + yn) = a + b and lim(xn − yn) = a− b;3) lim(xnyn) = ab;4) if b 6= 0, then lim (xn/yn) = a/b.

8

4.3. Number e. A sequence(1 +

1

n

)n

, n = 1, 2, . . .

has a finite limit

limn→∞

(1 +

1

n

)n

= e = 2.7182818284 . . .

4.4. Infinite limit. A symbolic writing

limn→∞

xn = ∞

(or, equivalently, xn →∞) means that, ∀ K > 0 ∃ N = N(K) :

xn ≥ K for all n ≥ N.

Similarly, we writelim

n→∞xn = −∞

(or, equivalently, xn → −∞) if ∀ K > 0 ∃ N = N(K) :

xn ≤ −K for all n ≥ N.

Limiting point of a sequence. A number c (or one of symbols ∞and −∞) is a limiting point of a sequence {xn} if there exists an increasingsequence of integers

1 ≤ m1 < m2 < . . .

such that a subsequence xmkhas a limit c.

Example. A sequence xn = (−1)n has two limiting points: 1 and −1.The minimal limiting point is called a lower limit, lim infn→∞ xn, and the

maximal limiting point is called an upper limit, lim supn→∞ xn.In the example above, −1 is a lower and 1 an upper limit.A limit exists iff the upper and the lower limits coincide.

9

Exercises.4.1. Find the following limits:

(4.1.1) limn→∞

10000n

n2 + 1

(4.1.2) limn→∞

(√n + 1−

√n)

(4.1.3) limn→∞

1 + a + . . . + an

1 + b + . . . + bnwhere |a| < 1, |b| < 1.

(4.1.4) limn→∞

(12

n3+

22

n3+ . . . +

(n− 1)2

n3

)(4.1.5) lim

n→∞

(1

1 · 2+

1

2 · 3+ . . . +

1

n(n + 1)

)4.2. Prove the following equalities:

(4.2.1) limn→∞

n

2n= 0

(4.2.2) limn→∞

2n

n!= 0

(4.2.3) limn→∞

nk

an= 0 where a > 1, k > 0

(4.2.4) limn→∞

an

n!= 0 ∀ a

(4.2.5) limn→∞

nqn = 0 if |q| < 1

(4.2.6) limnqn−1

(|q|+ c)n−1= 0 if |q| < 1 and 0 < c < 1− |q|

(4.2.7) limn→∞

a1/n = 1 if a > 0

(4.2.8) limn→∞

loga n

n= 0 if a > 1

(4.2.9) limn→∞

1

(n!)1/n= 0

4.3. Prove that

(4.3.1) limn→∞

(1

2· 3

4· . . . · 2n− 1

2n

)= 0

10

Hint: use a result from exercise (1.4).

(4.3.2) if limn→∞ xn = a, then limn→∞ |xn| = |a|.(4.3.3) if a sequence {xn} converges, then any its subsequence {xmk

} alsoconverges and has the same limit:

limn→∞

xn = limk→∞

xmk.

4.4. Find the supremum of a sequence {xn, n = 1, 2, . . .} if

(4.4.1) xn =n2

2n

(4.4.2) xn =

√n

100 + n

(4.4.3) xn =1000n

n!

4.5. Find the infimum of a sequence {xn, n = 1, 2, . . .} if

(4.5.1) xn = n2 − 9n− 100

(4.5.2) xn = n +100

n

4.6. For a sequence {xn, n = 1, 2, . . .}, find infn≥1 xn, supn≥1 xn, lim infn→∞ xn

and lim supn→∞ xn if

(4.6.1) xn = 1− 1/n

(4.6.2) xn = −n (2 + (−1)n)

11

5. Functionsbf Exercises.5.1. Assume that a variable x takes values in the interval 0 < x < 1.

Which values take a variable y if

(5.1.1) y = a + (b− a)x where a < b

(5.1.2) y =x

2x− 1

(5.1.3) y =1

1− x

(5.1.4) y√

x− x2

5.2. Find functions f(f(x)), f(g(x)), g(f(x)), and g(g(x)) if

(5.2.1) f(x) = x2 and g(x) = 1/x

(5.2.2) f(x) = x− 2 and g(x) = 2x

5.3. A function f(x) is odd if f(−x) = −f(x) for all values of x where fis defined, and even if f(−x) = f(x).

Determine which function is odd, which is even, and which is neither:

(5.3.1) f(x) = x2 − x4

(5.3.2) f(x) = x5 + x7 − x

(5.3.3) f(x) = 3x− x2

(5.3.4) f(x) = ax + a−x where a > 0

(5.3.5) f(x) = ln1− x

1 + xfor − 1 < x < 1

12

6. Limit of a functionA function f(x) is bounded in the interval a ≤ x ≤ b if both supx∈[a,b] f(x)

and infx∈[a,b] f(x) are finite.Let a and c be either numbers or simbol ∞ or symbol −∞. A function

f(x) has a limit c at point a,

limx→a

f(x) = c, or, equivalently f(x) → c as x → a

if, for any sequence xn → a, xn 6= a as n →∞,

f(xn) → c.

We say that a limit is finite if c is a finite number and infinite otherwise.A function f(x) is continuous at point a if limx→a f(x) exists and equals

f(a).In the same way as for sequences, one can introduce partial limits, lower

and upper limits.

13

Exercises6.1. Find the supremum and the infimum of a function f(x) if

(6.1.1) f(x) = x2 where − 2 ≤ x < 5

(6.1.2) f(x) =1

1 + x2where −∞ < x < ∞

(6.1.3) f(x) = x +1

xwhere 0 < x < ∞

(6.1.4) f(x) = 2x where − 1 < x < 2

6.2. Find the following limits:

(6.2.1) limx→0

x2 − 1

2x2 − x− 1

(6.2.2) limx→3

x2 − 5x + 6

x2 − 8x + 15

(6.2.3) limx→1

x3 − 3x + 2

x4 − 4x + 3

(6.2.4) limx→1

x + x2 + . . . + xn − n

x− 1

(6.2.5) limx→4

√1 + 2x− 3√

x− 2

(6.2.6) limx→∞

√1 + 2x− 3√

x− 2

(6.2.7) limx→∞

ln(1 + 3x

ln(1 + 2x)

(6.2.8) limx→−∞

ln(1 + 3x)

ln(1 + 2x)

6.3. Is a function f(x) continuous if

(6.3.1) f(x) = 2x for 0 ≤ x ≤ 1 and f(x) = 2− x for 1 < x ≤ 2

(6.3.2) f(x) = ex for x < 0 and f(x) = x + 1 for x > 0

14

7. DerivativesA derivative of a function y = f(x) at point x is a limit

y′=

dy

dx= f

′(x) =

df(x)

dx= lim

∆→0

f(x + ∆)− f(x)

∆

Main properties:1) c

′= 0

2) (cy)′= cy

′

3) (u + v)′= u

′+ v

′and (u− v)

′= u

′ − v′

4) (uv)′= u

′v + uv

′

5)(

uv

)′= u

′v−uv

′

v2

6) (un)′= nun−1u

′

7) (Chain Rule) if functions y = f(u) and u = g(x) have derivatives, then

dy

dx=

dy

du· du

dx

Main formulas:1) (xn)

′= nxn−1 where n is a constant

2) (sin x)′= cos x and (cos x)

′= − sin x

3) (tan x)′= 1

cos2 x

4) (arcsin x)′= 1√

1−x2

5) (arccos x)′= − 1√

1−x2

6) (arctan x)′= 1

1+x2

7) (ax)′= ax ln a and (ex)

′= ex

8) (loga x)′= 1

x ln afor a > 0, a 6= 1 and (ln x)

′= 1

x

15

Exercises7.1. Find a derivative y

′if

(7.1.1) y =2x

1− x2

(7.1.2) y =1 + x− x2

1− x + x2

(7.1.3) y =x√

1− x2

(7.1.4) y =(2− x2)(2− x3)

(1− x)2

(7.1.5) y =√

x + 1− ln(1 +√

x + 1)

(7.1.6) y = ex + eex

(7.1.7) y =1

4ln

x2 − 1

x2 + 1

7.2 Find a second derivative y′′

if

(7.2.1) y = x√

1 + x2

(7.2.2) y =x√

1− x2

(7.2.3) y = e−x

(7.2.4) y = e−x2

(7.2.5) y = x ln x

16

8. Increasing and decreasing functionsA function f(x) is increasing in an interval [a, b] if f(x1) ≤ f(x2) for all

a ≤ x1 ≤ x2 ≤ b.A function f(x) is decreasing in an interval [a, b] if f(x1) ≥ f(x2) for all

a ≤ x1 ≤ x2 ≤ b.

Let a function f be differentiable. Then it is increasing in [a, b] iff f′(x) ≥

0 for all x ∈ [a, b], and decreasing in [a, b] iff f′(x) ≤ 0 for all x ∈ [a, b].

Exercises8.1. Find intervals where a function y = f(x) is either increasing or de-

creasing:

(8.1.1) y = 2 + x− x2

(8.1.2) y = 3x− x3

(8.1.3) y =2x

1 + x2

(8.1.4) y =

√x

x + 100

8.2. Prove the following inequalities:

(8.2.1) ex > 1 + x ∀ x 6= 0

(8.2.2) x− x2

2< ln(1 + x) < x ∀ x > 0

17

9. Taylor formulaIf a function f(x) has n derivatives in an interval containing point x0,

then for any point x from this interval

f(x) =n∑

k=0

ak(x− x0)k + o(x− x0)

k

where o(x) is a function such that

o(x)/x → 0 as x →∞.

In particular, for any n = 1, 2, . . . and for any x,

(1) ex = 1 + x +x2

2!+

x3

3!+ . . . +

xn

n!+ o(xn),

(2) (1 + x)m = 1 + mx +m(m− 1)

2!x2 + . . . +

m(m− 1) · . . . · (m− n + 1)

n!xn + o(xn),

(3) ln(1 + x) = x− x2

2+ . . . + (−1)n−1xn

n+ o(xn).

Exercises9. Using expansions (1)-(3) above, find the following limits:

(9.1) limx→∞

x3/2(√

x + 1 +√

x− 1−√

x)

(9.2) limx→∞

(((x3 − x2 +

x

2

)e1/x −

√x6 + 1

)

18

10. Maximam and minimam values of a functionAssume that a function f is twice differentiable. It has a (strict local)

maximum at point x iff f′(x) = 0 and f

′′(x) < 0

and a (strict local) minimum iff f′(x) = 0 and f

′′(x) > 0.

Both maximal and minimal points are called extreme points.

Exercises10. Find and classify extreme points of the following functions:

(10.1) y = 2x − x2

(10.2) y = (x− 1)4

(10.3) y = xe−x

(10.4) y =√

2x− x2

(10.5) y =√

x ln x

(10.6) y = xae−x where a > 0

The following fact is of particular importance in statistical theory. Letg(x) be a strictly increasing function, i.e., for any x1 < x2, g(x1) < g(x2).Then, for any function f(x), the following are equivalent:(1) x is an extremal point of function f(x);(2) x is an extremal point of function g(f(x)).

In particular, the function g(x) = ln x is strictly increasing for x ∈ (0,∞).Exercises. Solve exercises (10.5) and (10.6) using the function g(x) =

ln x.

19

11. Indefinite integrals11.1. Definition. If functions f and F are such that F

′(x) = f(x) for

all x, then ∫f(x)dx = F (x) + C

where C is an arbitrary constant.

11.2. Basic properties.

d(∫

f(x)dx)

dx= f(x)∫

Af(x)dx = A

∫f(x)dx for any constant A∫

(f(x) + g(x))dx =

∫f(x)dx +

∫g(x)dx

11.3. Table of basic integrals.

(1)

∫xndx =

xn+1

n + 1+ C ∀n 6= −1

(2)

∫dx

x= ln |x|+ C if x 6= 0

(3)

∫dx

1 + x2= arctan x + C

(4)

∫dx

1− x2=

1

2ln

∣∣∣∣∣1 + x

1− x

∣∣∣∣∣+ C

(5)

∫dx√

1− x2= arcsin x + C = − arccos x + C

(6)

∫dx√

x2 + 1= ln |x +

√x2 + 1|+ C

(7)

∫axdx =

ax

ln a+ C where a > 0, a 6= 1, and

∫exdx = ex + C

11.4. Basic methods of integration.

20

11.4.1. Substitution (or change of variable). If∫f(x)dx = F (x) + C

and g(x) is a differentiable function, then∫f(g(x))g

′(x)dx = F (g(x)) + C

11.4.2. Decomposition.∫(f(x) + g(x))dx =

∫f(x)dx +

∫g(x)dx

11.4.3. Integration by parts.∫udv = uv −

∫vdu,

or, in other terms.∫f(x)g

′(x)dx = f(x)g(x)−

∫f

′(x)g(x)dx.

21

Exercises.11.1. Using the table of basic integrals, find the following integrals:

(11.1.1)

∫(3− x2)3dx

(11.1.2)

∫x2(5− x)4dx

(11.1.3)

∫x + 1√

xdx

11.2. Find the following integrals:

(11.2.1)

∫dx

x + a

(11.2.2)

∫(2x− 3)10dx

(11.2.3)

∫(1− 3x)1/3dx

(11.2.4)

∫dx√

2− 5x

(11.2.5)

∫dx

(1 + x)√

x

(11.2.6)

∫x2

x + 3dx

(11.2.7)

∫xe−x2

dx

(11.2.8)

∫xe−x2/2dx

22

11.3. Using integration by parts, find the following integrals:

(11.3.1)

∫ln xdx

(11.3.2)

∫x ln xdx

(11.3.3)

∫x2 ln xdx

(11.3.4)

∫ (ln x

x

)2

dx

(11.3.5)

∫xe−xdx

(11.3.6)

∫x2e−xdx

(11.3.7)

∫x3e−2xdx

(11.3.8)

∫xne−xdx for n = 3, 4, . . .

23

12. Definite integrals(A) Riemann integral. If a function f is defined on an interval [a, b]

and, for a fixed n = 1, 2, 3, . . .,

x0,n = a and xk,n = a + k∆n for k = 1, 2, . . . , n

where ∆ = b−an

, then ∫ b

a

= limn→∞

n−1∑k=0

f(xk,n) ·∆n. (3)

Graphic interpretation !

(B) Finding definite integrals using indefinite ones1. Here is the famous Newton-leibniz formula. If F

′(x) = f(x), then

∫ b

a

f(x)dx = F (b)− F (a) = F (x)

∣∣∣∣∣b

a

2. Integration by parts. If both functions f and g are differentiable in[a, b], then

∫ b

a

f(x)g′(x)dx = f(x)g(x)

∣∣∣∣∣b

a

−∫ b

a

f′(x)g(x)dx

3. Change of variable. If a function f is continuous and a function ϕis strictly monotone increasing (or monotone decreasing) and differentiable,and its derivative is continuous, then∫ b

a

f(x)dx =

∫ β

α

f(ϕ(t))ϕ′(t)df.

(C). Improper integralsIf a function f(x) is defined and integrable in any [a, b], then∫ ∞

a

f(x)dx = limb→∞

∫ b

a

f(x)dx

24

Similarly, if a function if defined and integrable in any [a, b− ε], then∫ b

a

f(x)dx = limε→0

∫ b−ε

a

f(x)dx

Similarly – with left limits.

(D). Convergence and divergence of integralsAssume that f is a non-negative function. If a limit (3) ix finite, the func-

tion f is integrable on [a, b] and integral∫ b

af(x)df is convergent. Otherwise

the function is non-integrable on [a, b] and the integral is divergent. Similardefinitions – for improper integrals.

Assume now that a function f is arbitrary. When an integral∫ b

af(x)dx

converges ? In general, this is a difficult uuestion not easy to answer. Hereis the following sufficient condition for convergence:

A function f is absolutely integrable, if∫ b

a|f(x)|dx converges. A function

is integrable if it is absolutely integrable.Comment...In the sequel, we will consider only nice cases.

25

Exercises.12.1. Using Newton-Leibniz formula, find the following definite integrals:

(12.1.1)

∫ 8

−1

√xdx

(12.1.2)

∫ 3

−3

exdx

(12.1.3)

∫ 2

0

|1− x|dx

12.2. Using integration by parts, find the following definite integrals:

(12.2.1)

∫ ln 2

0

xe−xdx

(12.2.2)

∫ e

1/e

ln xdx

(12.2.3)

∫ 3

0

x2e−xdx

12.3. Using an appropriate change of variable, find the following definiteintegrals:

(12.3.1)

∫ 1

−1

x√5− 4x

dx

(12.3.2)

∫ 2

1

ln xdx

(12.3.3)

∫ ln 2

0

√ex − 1dx

12.4. Find the following improper integrals:

(12.4.1)

∫ ∞

1

dx

x3

(12.4.2)

∫ 1

0

ln xdx

(12.4.3) In =

∫ ∞

0

xne−xdx for n = 1, 2, . . .

(12.4.4) In =

∫ ∞

−∞xne−x2/2dx for n = 1, 2, . . .

26

13. Series13.1. For a sequence {an, n = 1, 2, . . .}, a series

∞∑n=1

an =∞∑1

an

is a limit of partial sums∑N

1 an when N → ∞. A series converges if thelimit exists, and diverges otherwise.

Clearly, for any N ,

∞∑1

an =N∑1

an +∞∑

N+1

an,

and a series converges iff∑∞

N+1 → 0 as N → ∞. r k We can consider alsoseries

∑∞k an starting ftom any integer k.

Geometric Series: for |q| < 1,

∞∑0

qn = limN→∞

N∑0

qn = limN→∞

1− qN+1

1− q=

1

1− q

A series∑

an converges absolutely if∑|an| converges. Absolute Conver-

gence implies Convergence. We will deal mostly with absolutely convergentseries.

If an = fn(x) where all fn are continuously differentiable (what is this ?),∑fn(x) converges, and

∞∑1

|f ′

n(x)| converges,

then (∞∑1

fn(x)

)′

=∞∑1

f′

n(x).

13.2. Taylor series (expansions). If a function f(x) is analytic in aneighbourhood of a point x0 (“analytic” means “in infinitely differentiable ina neighbourhood plus a little more”), then, for all point x from this neigh-bourhood,

f(x) =∞∑

n=0

f (n)(x0)

n!9x− x0)

n.

27

In particular, you have to remember the following 5 Taylor expansions:

(1) ex =∞∑

n=0

xn

n!= 1 + x + . . . +

xn

n!+ . . . ∀ x

(2) sin x = x− x3

3!+ . . . + (−1)n−1 x2n1

(2n− 1)!+ . . . ∀ x

(3) cos x = 1− x2

2!+ . . . + (−1)n x2n

(2n)!+ . . . ∀ x

(4) (1 + x)m =∞∑

n=0

m(m− 1) · . . . · (m− n)

(n− 1)!xn−1 ∀ m and for − 1 < x < 1

(6) ln(1 + x)∞∑

n=1

(−1)n−1xn

n= x− x2

2+

x3

3− . . . for − 1 < x ≤ 1.

Here x0 = 1 and 0! = 1, by convention.

28

Example 1. For |q| < 1, find∑∞

1 nqn. There are various ways to do it(not only by differentiation !)

Informally, one can do the following:

∞∑1

nqn = q∞∑1

nqn−1

= q∞∑1

(qn)′

= q∞∑0

(qn)′

=(?) q

(∞∑0

qn

)′

= q

(1

1− q

)′

=q

(1− q)2.

We have to verify the equality with question mark.Here an = qn are continuously differentiable functions of q, and the series∑qn convergesSecond, the series

∑|nqn−1| converges too. Indeed, for for any c ∈

(0, 1−|q|), |nqn−1| < (|q|+ c)n−1 all n sufficiently large (see exercise (4.2.6)).Therefore,

∞∑n=N+1

|nqn−1| ≤∞∑

n=N+1

(|q|+ c)n−1 =(|q|+ c)N

1− |q| − c→ 0 as N →∞

Example 2. For any x > 0, find

∞∑1

nxn

n!

(Again, there are various ways to do it – not only by differentiation !)

29

∞∑1

nxn

n!= x

∞∑1

nxn−1

n!

= x

∞∑1

(xn

n!

)′

= x

∞∑0

(xn

n!

)′

=(?) x

(∞∑0

xn

n!

)′

= x (ex)′= xex

Juastify the question mark yourself !

Exercises(13.1) Find

(13.1.1)∞∑

n=1

nkqn if |q| < 1, k = 2, 3

(13.1.2)∞∑

n=1

nkxn

n!if k = 2, 3.

(13.2) Using the 5 main taylor expansions, give Taylor expansions for thefollowing functions:

(13.2.1) e−x2

(13.2.2)x10

1− x

(13.2.3)1

(1− x)2

(13.2.4)x√

1− 2x

30

14. Convergence of Series and of Definite IntegralsWe consider here only non-negative sequences and functions.For a sequence {an, n = 1, 2, }, introfuce a function f(x) on the half-line

[1,∞) by

f(x) = an if x ∈ [n, n + 1)

Then, for any N = 2, 3, . . .,

N∑1

an =

∫ N+1

1

f(x)dx.

Therefore,∞∑1

and

∫ ∞

1

f(x)dx

either converge or diverge simultaneuosly.Exercise. Show that

∞∑1

n−d

converges if d > 1 and diverges if d ≤ 1.

31

Exercises.(14.1) For which values of x the following series converge:

(14.1.1)∞∑1

n2

(3n + 4)c

(14.1.2)∞∑1

(3n− 5)c

(2n + 8)2c+1

(14.1.2)∞∑1

(4n2 + 5n− 7)3c−1

(5n3 + 2)c+8

(14.2) For which values of x and c the following series converge:

(14.2.1)∞∑1

xn

nc

(14.2.2)∞∑1

cnc

xn if 0 < c < 1

(14.2.3)∞∑1

xn

1n + cnif c > 0

(14.2.4)∞∑1

xn

c√

nif c > 0

32