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
Top Related