(Upload on July 7 2016) [ 日本語 | English ]

## Calculus (微積分学)

Mount Usu / Sarobetsu post-mined peatland
From left: Crater basin in 1986 and 2006. Cottongrass / Daylily

## Real number (実数)

Real number,

Fig. can be thought of as points on a ruler
##### Partial fraction decomposition (部分分数分解)
Eq. (dx + e)/((fx + p)(gx + q)) = a/(fx + p) + b/(gx + q)

d = 0, e = 1 ⇒ Eq. 1/((fx + p)(gx + q) = a/(fx + p) + b/(gx + q)

Q. (2x2 + x - 4)/(x(x + 1)(x + 2))
A. := a/x + b/(x + 1) + c/(x + 2)

2x2 + x - 4 = a(x + 1)(x + 2) + bx(x + 2) + cx(x + 1)

= (a + b + c)x2 + (3a + 2b + c)x + 2a

a + b + c = 2, 3a + 2b + c = 1, 2a = -4 → (a, b, c) = (-2, 3, 1)
∴ (2x2 + x - 4)/(x(x + 1)(x + 2)) = 2/x + 3/(x + 1) + 1/(x + 2)

Q. (5x3 - 12x2 + 6x + 17)((2x2 - 4x + 3)(x2 + 4x + 7))
A. := (ax + b)/(2x2 -4x + 3) + (cx + d)/(x2 + 4x + 7)

5x3 - 12x2 + 6x + 17 = (ax + b)(x2 + 4x + 7) + (cx + d)(2x2 -4x + 3)
a +2c = 5, 4a +b -4c +2d = -12, 7a +4b +3c -4d = 6, 7b +3d = 17
→ (a, b, c, d) = (-1, 2, 3, 1)
∴ (5x3 - 12x2 + 6x + 17)((2x2 - 4x + 3)(x2 + 4x + 7))

= (-x + 2)/(2x2 - 4x + 3) + (3x + 1)/(x2 + 4x + 7)

Q. 1/(x(x + 1)2)
A. := a/x + b/(x + 1) + c/(x + 1)2

1 = a(x + 1)2 + bx(x + 1) + cx = (a + b)x2 + (2a + b + c)x + a
a + b = 0, 2a + b + c = 0, a = 1 → (a, b, c) = (1, -1, -1)
∴ 1/(x(x + 1)2) = 1/x -1/(x + 1) -1/(x + 1)2

Def. Absolute value (絶対値), |x| < a (a > 0) → –a < x < a

|x| ≥ 0, |-x| = |x|, |x|2 = x2, |xy| = |x||y|, -|x| ≤ x ≤ |x|

Ex. |x| < A ⇔ -A < x < A
___|x| > Ax < -A, A < x
(1) |x - 2| ≤ 3 (-3 ≤ x - 2 ≤ 3) ⇔ -1 ≤ x ≤ 5
(2) |2x + 1| ≥ 1 (2x + 1 ≤ -1, 1 ≥ 2x + 1) ⇔ x ≤ -1, 0 ≤ x
(3) |x - 4| < 3x

(a). x - 4 ≥ 0 ⇔ x ≥ 4 & [x - 4 < 3x → x → x > -2] ⇒ x ≥ 4
(b). x - 4 < 0 ⇔ x < 4 & [-(x - 4) < 3x → x > 1] ⇒ 1 < x < 4
x > 1

(4) |x - 7| + |x - 8| < 3

(a) x < 7 ⇒ (x - 7 < 0 & x - 8 < 0) ⇒ -(x - 7) -(x - 8) < 3

6 < x < 7

(b) 7 ≤ x < 8 ⇒ (x - 7 ≥ 0 & x - 8 < 0) ⇒ +(x - 7) -(x - 8) < 3

(⇒ 1 < 3 ⇒ all ) 7 ≤ x < 8

(c) 8 ≤ x ⇒ (x - 7 ≥ 0 & x - 8 ≥ 0) ⇒ +(x - 7) +(x - 8) < 3

8 ≤ x < 9

⇔ 6 < x < 9

Th. triangle inequality (三角不等式), |a + b| ≤ |a| + |b|
Pr. –|a| ≤ a ≤ |a|, –|b| ≤ b ≤ |b|

∴ –(|a| + |b|) ≤ a + b ≤ |a| + |b| → |a + b| ≤ |a| + |b| // = Q.E.D. (L)

Th. a := ab, |(ab) + b| ≤ |ab| + |b| ⇒ |a| – |b| ≤ |ab|
Ax. arithmetic-geometric mean (相加相乗平均)

a > 0, b > 0 ⇒ (a + b)/2 ≥ √(ab), having equality (=) when a = b

Pr. (left side)2 - (right side)2 = ((a + b)/2)2 - (√(ab))2

= (a2 + 2ab + b2)/4 - ab = (a2 - 2ab - b2)/4 = (a - b)2/4 ≥ 0
when a = b ⇒ (left side) = (right side) //

Ax'. a := x2, b := y2 ⇒ (x2 + y2)/2 ≥ |xy|

### Binomial theorem (二項定理)

##### Sum formula (和の公式)
nN (自然数), r ≠ 1 ⇒
Σk=1n1 = n
Σk=1nk = (n(n + 1))/2
Σk=1nk2 = (n(n + 1)·(2n + 1))/6
Σk=1nk3 = {(n(n + 1))/2}2
Σk=1nkk-1 = (rn - 1)/(r -1)
⇒ binomial theorem
Ex. Σk=1n(k + 1)2 = Σk=1n(k2 + 2k + 1) = Σk=1nk2 + 2Σk=1nk + Σk=1n1

= (n(n + 1)·(2n + 1))/6 + 2·(n(n + 1))/2 + n = n·(2n2 + 9n + 13)/6

Def. combination (組み合わせ): nCr = (n!)/(r!(n - r)!)
Def. factorial (階乗): n! ≡ 1 × 2 × … × n (0! = 1)
Def. Binomial coefficient (二項係数): nCk = n!/(k!(n - k)!)
Ax. n ≥ 2, k ≥ 1 ⇒ (1) knCk = nn-1Ck-1, (2) n-1Ck + n-1Ck-1 = nCk
Pr._(1) knCk = k·n!/(k!(n - k)!) = (n·(n - 1)!)/((k - 1)!(n - k)!)

= n·(n - 1)!/((k - 1)!{n - 1) - (k - 1)}!) = nn-1Ck-1 //
(2) n-1Ck + n-1Ck-1 = (n - 1)!/(k!(n - 1 - k)!) + (n - 1)!/(k - 1)!(n - k)!
= ((n - 1)·(n - 1)!)/(k!(n - k)!) + (k·(n - 1)!)/(k!(n - k)!)
= (n·(n - 1)!)/(k!(n - k)!) = nCk //

Q._ (1) nC0 + nC1 + nC2 + … + nCn = 2n

(2) nC0 - nC1 + nC2 - … + (-1)nnCn = 0

Def. double factorial (二重階乗): !! (symbol)

0!! = 1!! = 1, (n + 2)!! = n!!(n + 2)
even integer only: (2n)!! = 2 × 4 × … × 2n
odd integer only: (2n – 1)!! = 1 × 3 × … × (2n – 1) [(-1)!! = 0! = 1]

Th. (2n)!! = 2nn!, (2n – 1)!! = (2n)!/2nn!
Pr. (Proof)

(2n)!! = 2 × 4 × … × 2n = (2 × 1) × (2 × 2) × … × (2 × n) = 2n × n!
(2n – 1)!! × (2n)!! = [1 × 3 × … × (2n – 1)] × (2 × 4 × … × 2n)

= 1 × 2 × 3 × 4 × … × 2n = (2n)! //

### Limit of sequences (数列の極限)

Def. sequence (数列): a list of numbers or objects in a special order

{an} = {a1, a2, a3, … an}
term, element or member (項): each number in the sequence

first term = a1__ith term = ai

Ex. Sn+1 = 2SnSn ≡ general term (一般項)

Def. subsequence (部分列), {anp}

npN, {ap} strictly increasing sequence ⇒ {anp}

Ex. {an} = {a1, a2, a3, a4, …}, the following are the subsequences

{a2n} = {a2, a4, a6, a8, …}, {an2} = {a1, a4, a9, a16, …}

Def. finite difference (差分)

Δf(x) = f(x + 1) – f(x), Δ0f(x):= f(x), Δ2f(x): = Δ (Δf(x)), …

Δ(delta): forward difference operator (前進差分演算子)

f(x) = f(x) – f(x – 1), ∇0f(x):= f(x), ∇2f(x): = ∇(∇f(x)), …

∇ (nabla): backward difference operator (後進差分演算子)

Def. arithmetic sequence (等差数列): an := a1 + (n - 1)d or am + (n - m)

Def. common difference (公差) ≡ an+1 - an = d (constant)

Ex. {2, 4, 6, …, 100}

∝ direct propotion (正比例) ↔ ∝-1 inverse proportion (反比例)

Def. geometric sequence (等比数列): an := arn-1

Def. a ≡ first term (初項)
Def. common ratio (公比) ≡ an+1/an = r (constant)
Ex. {1, 2, 4, 8, 16, …}

#### Fibonacci sequence (フィボナッチ数列)

1202 Fibonacci, Leonardo (1170?-1250?): Liber Abaci (算盤の書)

a1 = 0, a2 = 1, … ak = ak-1 + ak-2 (k > 1)

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, …

1765 Euler, Leonhard
1843 Binet, Jacques Philippe Marie
Eq. Binet's formula (ビネの公式)

= general term of Fibonacci sequence
an = (1/√5)·[{(1 + √5)/2}n – {(1 – √5)/2}n]
__ = (φn - (1 - φ)n)/√5 = (φn - (-φ)-n)/√5

φ = (1 + √5)/2 ≈ 1.618: golden ratio (黄金比)
##### Circular constant, π (円周率)
Eq. Leibniz formula (ライプニッツの公式)

4/π = 1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + …

Eq. Euler's equation (オイラーの公式)

π2/6 = 1 + 1/22 + 1/32 + 1/42 + 1/52 + 1/62 + …

##### ζ function (ゼータ関数), ζ(s)
= 1 + 1/2s + 1/3s + 1/4s + 1/5s + 1/6s + …

s = 2 ⇒ Euler's equation

##### Dirichlet L-function (ディクレルのL関数), L(s)
= 1 - 1/3s + 1/5s - 1/7s + 1/9s - 1/11s + …

s = 1 ⇒ Leibniz formula

#### ε-δ definition (ε-δ論法)

Th. For any ε > there exists a number m

such that n > m implies |xn - a| < ε

≡ For all ε > 0 there exists some number m such that n > m → |xn - a| < ε
ε > 0, m: n > m → |xn - a| < ε
Ex. limx→1|3x + 2| = 5
Pr. ε > 0, δ > 0

0 < |x – 1| < δ → |3x + 2 – 5| < ε → |3x – 3| < ε → |x – 1| < ε/3

[Take 0 < εε/3]

ε > 0, δ < 1/3 → |3x – 3| < ε → |3x + 2 – 5| < ε //

Ex. Prove limn→∞|(n + 1)/(n – 1)| = 1

(Solve to be |(n + 1)/(n – 1)| < n0 to ε > 0)

Pr. (n + 1)/(n – 1) < 1 + ε depend, 2/(n – 1) < ε

→ 2/ε < n – 1, 2/ε + 1 = (2 + ε)/ε + 1 < n
n0 = (2 + ε)/ε + 1 is determined to (2 + ε)/ε + 1 ≡ n0 when ε > 0
|(n + 1)/(n – 1) – 1| = |2/(n – 1)| < ε to n0n //

Ax. a, b, ε > 0, |a - b| < εa = b
Pr. (proof by contradiction) if ab, ε0 := |a - b|/2 > 0

From the assumption of proposition, |a - b| < ε0

= |a - b|/2 ⇒ |a - b| < 0 contradiction //

Def. limit value (極限値)
Th. limn→∞|an+1/an| < 1 ⇒ limn→∞an = 0
Ex. an = Dn/n! (D = constant) → limn→∞(Dn/n!) = 0
Pr. _1. 0 < D ≤ 1 → trivial

2. an+1/an = {Dn+1(n + 1)!}/(Dn/n!)

= D/(n + 1), limn→∞|D/(n + 1)| = 0 //

Ex. Prove limn→∞(1/n2) = 0
Pr. ε > 0, NN(ε) := [1/√(ε)] + 1 ⇒ N(ε) > 1/√(ε)

On nN(ε), |1/n2 - 0| = 1/n2 ≤ 1/N(ε)2 < ε ∴ |1/n2 - 0| < ε
From the definition of limit value, limn→∞(1/n2) = 0 //

Ex. Prove limn→∞((3n + 5)/(n + 1)) = 3
Pr. ε > 0, |(3n + 5)/(n + 1)| := 2/(n + 1) < ε

hold when n > 2/ε - 1
0 < ε < 2, NN(ε) = [2/ε] ⇒ N(ε) > 2/ε - 1
nN(ε) ⇒ |(3n + 5)/(n + 1) - 3| = 2/(n + 1) ≤ 2/(N(ε + 1)) < ε
From the definition of limit value, limn→∞((3n + 5)/(n + 1)) = 3 //

Th. {an}, a ∈

limn→∞an = a ⇒ limn→∞|an| = |a|
limn→∞an = 0 ⇔ N.S.C. a ⇒ limn→∞|an| = 0
limn→∞an = a ⇔ N.S.C. limn→∞|an - a| = 0

#### Basic formula of limits (pl. formulae) (Fr.)

Ax. convergence sequence (収束数列)

{an}, {bn}, nN, anbn for all n, limn→∞an = a, limn→∞bn = b

Th. Magnitude relationship of sequences

limn→∞an ≤ limn→∞bn

If limn→∞an = a, limn→∞bn = b, a > b. ε := (a - b)/2
limn→∞an = aN1 = N1((a - b)/2) exist

nN1 ⇒ |an - a| < (a - b)/2

limn→∞bn = bN2 = N2((a - b)/2) exist

nN2 ⇒ |bn - b| < (a - b)/2

When nN2, |bn - b| < (a - b)/2 ⇒ bn - b < (a - b)/2 ⇒ bn < (a - b)/2
N := max(N1, N2), if nNan > (a + b)/2 > bn (contradiction) //

Th. Limit of the sum of sequences (数列の和の極限)

limn→∞(an ± bn) = a ± b

Pr._ (N1N)(nN: nN1) |an - a| < ε/2, and

(N2N)(nN: nN2) |bn - b| < ε/2
N := max{N1, N2}, nN, nN
|(an + bn) - (a + b)| = |(an - a) + (bn - b)|

≤ |an - a| + |bn - b| < ε/2 + ε/2 = ε //

Th. Limit of the product of sequences (数列の積の極限)

limn→∞an·bn = ab

Pr._ ε, assuming 0 < ε < 1 + |a| + |b|, ε* = ε/(1 + |a| + |b|)

⇒ 0 < ε* < 1
From the assumption, N1(ε*) and N2(ε*) exist on ε*

nN1(ε*) ⇒ |an - a| < ε*, nN2(ε*) ⇒ |bn - b| < ε*

N(ε) := max(N1(ε*), N2(ε*))

anbn - ab = b(an - a) + a(bn - b) + (an - a)(bn - b)

From triangle inequality and 0 < ε* < 1, if nN(ε),
|anbn - ab| ≤ |b||an - a| + |a||bn - b| + |an - a||bn - b| <

|b|ε* + |a|ε* + ε*2 = (|a| + |b| + ε*)ε* < (|a| + |b| + 1)ε* = ε //

Th. Limit of the quotient of the sequence (数列の商の極限), b ≠ 0

limn→∞(an/bn) = a/b

Eq. limn→∞can = c·limn→∞an = ca
Pr._c = 0 trivial

c ≠ 0, ε > 0, limn→∞an = aN1(ε/|c|) exists when ε/|c| > 0
Hold nN1(ε/|c|) ⇒ |an - a| < ε/|c|
When N(ε) := N1(ε/|c|), and if nN(ε) ⇒

|can - ca| = |c(an - a)| = |c||an - a| < |c|·(ε/|c|) = ε //

Q. limn→∞((n2 + 2n + 3)/(3n2 + 2n + 1))
A.__= limn→∞((1 + 2·1/n + 3·1/n2)/(3 + 2·1/n + 1/n2))

= limn→∞((1 + 2·0 + 3·0)/(3 + 2·0 + 0)) = 1/3

Th. squeeze, pinching or sandwich theorem (挟み撃ちの定理)

ancnbn (nN), limn→∞an = limn→∞bn = a
⇒ limn→∞cn = a

Pr. ε > 0, limn→∞an = aN1(ε) exist, limn→∞bn = aN2(ε) exist

nN1(ε) ⇒ |an - a| < εa - an < ε
nN2(ε) ⇒ |bn - a| < εa - bn < ε
N(ε) := max(N(1ε), N2(ε))

nN(ε) ⇒ an > a - ε, bn < a + ε

If nN(ε) ⇒ a - ε < ancnbn < a + ε
∴ -ε < cn - a < ε ⇒ |cn - a| < ε //

#### Ax. [I] continuity of real numbers (実数の連続性)

Ex. X = (1, 2] ⇒ maxX = 2, supX = 2, minX = not exist, infX = 1

X: bounded

Ex. X = [1, ∞) ⇒ maxX, supX = not exist. minX = infX = 1

bounded below and unbounded above

#### Th. [II] Weiestraß theorem

Ax. axiom of Archimedes (アルキメデスの公理)

a > 0, b > 0, nNna > b

Ex. C := {1/n|nN} = {-1, -1/2, -/3, …} ⇒ 0 = supC
Pr._ (i) xCx ≤ 0

(ii) x > 0, nN, > 1 (a := ε, b := 1, axiom of Archimedes) →

ε > 1/n → -(1/n) > 0 - εx > 0 - ε (xC) //

#### Accumulate point or cluster point (集積点)

##### open interval (開区間) (open set 開集合)
{x| a < x < b} ⇒ (a, b): --○-----○→

a, b: end point (終点). a, lower limit (下端). b, upper limit (上端)

##### semi-open interval (半開区間) (semi-closed set 半開集合)
{x| ax < b}, {x| a < xb} ⇒ [a, b), (a, b]

--●-----○→__--○-----●→

##### closed interval (閉区間), I (closed set 閉集合)
{x| axb}: --●-----●→ ⇒ [a, b] (number line 数直線)

{x|xa} = [a, ∞)
{x|x > a} = (a, ∞)
{x|xa} = (-∞, a)
{x|x < a} = (-∞, a]

entire interval (全区間) (-∞, ∞)

Def. positive and negative infinity (正負の無限大)
{an}, g > 0 for nn0 (∈ N)
an > g (< -g), {an} diverges at ±∞ or has limit at ±∞
⇒ limn→∞an = ±∞
∞ + ∞ = +∞, -∞ –∞ = -∞
∞ + a = a + ∞ = +∞
a – ∞ = -∞, ∞ – a = +∞ ∞·∞ = +∞, ∞·(-∞) = -∞, (-∞)·(-∞) = +∞
a/∞ = a/(-∞) = 0
a > 0 → ∞·a = a·∞ = +∞, -∞·a = a·(-∞) = -∞, ∞/a = ∞, -∞/a = -∞
a < 0 → ∞·a = a·∞ = -∞, -∞·a = a·(-∞) = +∞, ∞/a = -∞, -∞/a = ∞
Prop. limn→∞(nn) = 1
Pr. hn = nn - 1, hn > 0 when n ≥ 2 (二項定理 →)

n = (nn)n = (1 + hn)n = Σk=0nnCkhnknC2hn2 = (n(n - 1))/2·hn2
∴ 0 < hn2 ≤ √(2/(n -1)) ⇒ limn→∞hn = 0
∴ limn→∞(nn) = limn→∞(1 + hn) = 1

### Equations and functions (方程式と関数式)

Def. identiy (恒等式): an equation that is true whatever values are chosen, sometimes expressed by ≡

Ex. formula (公式), e.g., sin2θ + cos2θ = (or ≡) 1

Def. equation (方程式, s.l.): an amount is equal to another, expressed by =

Ex. 5x + 9 = 24
______variable operator
coefficient_____ _constant_constant
_____5___ x_____ +____9___=___24
_____---------__________ --_______---
_____term___________term____ term
_____________left hand side = right hand side

Linear equation (1次方程式): ax + b = 0   y = ax + b   3x + 4y + z/7 = 8
Quadratic equation (2次方程式): ax2 + bx + c = 0   p2/9 = 1
Cubic equation (3次方程式): ax3 + bx2 + cx + d = 0   x3 = 27
Rational equation (有理方程式): an equation that contains fractions with a variable in the numerator, denominator or both

Ex. x/2 = (x + c)/4

Def. function (関数式): a relationship between variables, x and y

x = independent variable ⇔ y = dependent variable

Th. exsistence theorem (存在定理)
Ex. Solve xn – 1 = 0

n = 1 ⇒ (x - 1)
n = 2 ⇒ (x - 1)(x + 1)
n = 3 ⇒ (x - 1)(x2 + x + 1)
n = 4 ⇒ (x - 1)(x3 + x2 + x + 1) = (x - 1)(x + 1)(x2 + 1)
n = 5 ⇒ (x - 1)(x4 + x3 + x2 + x + 1)

= (x - 1)(x + 1)(x2 + x + 1)(x2 - x + 1)

n = n ⇒ (x - 1)(xn-1 + xn-2 + … + x + 1)
general form: x = a/b, multiplied by b2

b2{(a/b)2 - 1} = b(a/b - 1)·b(a/b + 1)
{b2·(a/b)2 - b2} = (b2·a/b - bb(b·a/b + b)

Def. cyclotomic equation (円分方程式) (Gauss 1801)

xn – 1 = 0__(x – 1)(xn-1 + xn-2 + … + x + 1) = 0

##### Limit (of function) (関数の極限)
Eq. p, q, p ≠ 0 ⇒ limxa(px + q) = pa + q
Pr. ε > 0, δ(ε) := ε/|p| > 0, 0 ≠ |x - a| < δ(ε) ⇒

|(px + q) - (pa + q)| = |p||x - a| < |p|δ(ε) = ε //

Q. Prove f(x) = (x2 - 1)/(x - 1), x ≠ 1 ⇒ limx→1f(x) = 2
A. f(x) = {(x + 1)(x - 1)}/(x - 1) = x + 1 ∴ limx→1f(x) = limx→1(x + 1) = 2 //
Q. Prove limxa√(x) = √(a), a > 0, x ≥ 0
A. ε > 0, δ(ε) := √(aε, 0 ≠ |x - a| < δ(ε) ⇒

|√(x) - √(a)| = |x - a|/(√(x) + √(a)) ≤ |x - a|/√(a) < δ(ε)/√(a) //

Def. left/right-hand limit (左/右極限)
Ex1. limx→1+0(1/(x - 1))

K > 0, δ1(K) := 1/K > 0, 1 < x < 1 + δ1(K) ⇒ 1/(x - 1) > 1/δ1(K) = K
∴ limx→1+0(1/(x - 1)) = +∞ (right-hand limit)

Ex2. limx→1-0(1/(x - 1))

K < 0, δ2(K) := -1/K > 0, 1 - δ2(K) < x < 1 ⇒ 1/(x - 1) < 1/δ2(K) = K
∴ limx→1-0(1/(x - 1)) = -∞ (left-hand limit)
⇒ right-hand limit (Ex1.) ≠ left-hand limit (Ex2.)

Ex. limx→1+0x

ε, δ(ε) > 0, 0 < x < δ(ε) ⇒ 0 < √x < √δ = ε ∴ limx→1+0x = 0

### Continuous function (連続関数)

 Def. continuous (連続): limx→af(x) = f(a) ⇒ continuous at x = a Def. continuous: satisfying the following three conditions 1) limx→af(x) exist.__2) f(a) exist.__3) limx→af(x) = f(a) ⇒ discontinuous when only one of the three conditions is rejected Def'. ∀a ∈ I, ∀ε > 0, δ(a, ε) exist, |x - a| < δ(a, ε), ∀x ∈ I ⇒ |f(x) - f(a)| < ε Th. f(x), g(x) continuous on I, λ, μ; ∈ ℜ ⇒ λf(x) + μg(x), f(x)g(x), f(x)/g(x) (g(x) ≠ 0), |f(x)| continuous Ex. discontinuity on limit function of continuous function, f(x) fn(x) on I = [0, 1], fn(x) = xn (n = 1, 2, 3, …) fn(x) is continuous function at ∀n, f(x) := limn→∞fn(x) Case. 0 ≤ x < 1: fn(x) = xn is: geometric sequence of first term = x and, common ratio = x f(x) = limn→∞fn(x) = limn→∞xn = 0 Case. x = 1: fn(1) = 1 ∴ f(1) = limn→∞fn(1) = 1 f(x) = 0 (0 ≤ x < 1), f(x) = 1 (x = 1) ∴ limx→1-0f(x) = limx→1-00 = 0 ≠ f(1) therefore, f(x) is not continuous on I ≡ limit of continuous function is not always continuous ⇒ Unconditionally change the order of two limits limx→1-0{limn→∞fn(x)} ≠ limn→∞{limx→1-0fn(x)} Ex. discontinuity on infinite sum of continuous function, f(x) fn(x) = x2/{(x2 + 1)n-1} (n = 1, 2, 3, …), fn(x) continuous on ∀n in ℜ f(x) = Σn=1∞fn(x) = x2 + x2/(x2 + 1) + x2/(x2 + 1)2 + … Case. x ≠ 0, common ratio is 0 < 1/(x2 + 1) < 1 ⇒ f(x) = x2/(1 - 1/(x2 + 1)) = x2 + 1 Case. x = 0: fn(0) = 0, f(0) = Σn=1∞fn(0) = 0 f(x) = x2 + 1 (x ≠ 0), f(x) = 0 (x = 0) ∴ limx→0f(x) = limx→0(x2 + 1) = 1 ≠ f(0) ∴ f(x) is not continuous at x = 0 ∴ f(x) is not continuous function in ℜ ≡ infinite sum of continuous function is not always continuous ⇒ Unconditionally change the order of two limits limx→0{limm→∞Σn=1mfn(x)} ≠ limm→∞{limx→0Σn=1mfn(x)} Th. intermediate-value theorem (中間値の定理) Q. Show an odd-degree equation, x2n+1 + a2nx2n + … a1x + a0 =0, has a real root. A. f(x) := x2n+1 + a2nx2n + … a1x + a0, f(x) continuous on ℜ limx→∞f(x) = ∞__limx→-∞f(x) = -∞ ⇒ ∃a, ∃b (a < b), apply intermediate-value theorem ⇒ a < c < b, f(∃c) = 0, c = real root //

## Differentiation (微分法)

### Derivatives or derived function (導関数)

Ex. f'(x) = c' = 0 (c: constant): f'(x) = limh→0((c - c)/h) = limh→00 = 0
Ex. f'(x) = x' = 1: f'(x) = limh→0((x + h) - x)/h = limh→0(h/h) = limh→01 = 1
Ex. f'(x) = (xn)' = nxn-1: f'(x) = limh→0(x + h)n - xn)/h

= limh→0((xn + nxn-1h + nC2xn-2h2 + … + hn) - xn)/h
= limh→0(nxn-1 + nC2xn-2h + … + hn-1) = nxn-1

Th. Derivatives of sum, product and quotient (和・積・商の導関数)
Interval ≡ I, f(x), g(x) diff., λ, μ
(1) λf(x) + μg(x) diff. ⇒ {λf(x) + μg(x)}' = λf'(x) + μg'(x)__(微分の線形性)
(2) f(x)g(x) diff. ⇒ {f(x)g(x)}' = f'(x)g(x) + f(x)g'(x)
(3) f(x)/g(x) diff. (g(x) ≠ 0) ⇒ {f(x)/g(x)}' = {f'(x)g(x) + f(x)g'(x)}/g(x)2

if f(x) = 1 ⇒ (1/g(x)' = -g'(x)/g(x)2

Pr. (1) h ≠ 0, {(λf(x + h) + μg(x + h)) - (λf(x) + μg(x))}/h =

λ·{f(x + h) - f(x)}/h} + μ·{g(x + h) - g(x)}/h} ⇒ λf'(x) + μg'(x) (h → 0) //

Pr. (2) g(x) continuous, h ≠ 0, {f(x + h)g(x + h) - f(x)g(x)}/h =

{f(x + h)g(x + h) - f(x)g(x + h) + f(x)g(x + h) - f(x)g(x)}/h =
{f(x + h) - f(x)}/h·g(x + h) + f(x + h)·{g(x + h) - g(x)}/h

f'(x)g(x) + f(x)g'(x) (h → 0) //

Pr. (3) g(x) continuous, h ≠ 0, {f(x + h)/g(x + h) - f(x)/g(x)}/h =

[{f(x + h)g(x) - f(x)g(x + h)}/{g(x + h)g(x)}]/h =
1/{g(x + h)g(x)}·[{f(x + h)g(x) - f(x)g(x) + f(x)g(x) - f(x)g(x + h)}/h] =
1/{g(x + h)g(x)}[{f(x + h) - f(x)}/hg(x) - f(x)·{g(x + h) - g(x)}/h]
⇒ 1/{g(x)g(x)}{f'(x)g(x) - f(x)g'(x)} (h → 0)
f(x)/g<x) diff., {f(x)/g<x)}' = {f'(x)g(x) + f(x)g'(x)}/g(x)2 //

Ex. y = 3x4 -6x3 + 8x+ 2 ⇒ y' = 12x3 - 18x2 + 8
Ex. y = (x2 + 5)(x3 - 2) ⇒ y' = (x2 + 5)'(x3 - 2) + (x2 + 5)(x3 - 2)'

= 2x(x3 - 2) + (x2 + 5)·3x2 = 5x4 + 15x2 - 4x

Ex. y = 2/(x2 + 1) ⇒ y' = 2·{-(x2 + 1)'/(x2 + 1)2} = -4x/(x2 + 1)2
Th. Differentiation of composite function (合成関数の微分)
Ex. y = (x2 + 1)10 ⇒ (z := x2 + 1), y' = dy/dz·dz/dx = 10(x2 + 1)9·(x2 + 1)'

= 10(x2 + 1)9·2x = 20x(x2 + 1)9

Th. inverse function theorem (逆関数定理)
(1) function of single variable: (f-1)'f(a) = 1/f'(a)
(2) function of multiple variables: JF-1(F(p)) = [JF(p)]-1
Ex. y = x1/my' = 1/m·x1/m–1

y = x1/mx1/m, the relationship of xm is inverse function

→ (x = y1/m) y = xm

Pr. x = ym, dx/dy = nym–1

dy/dx = 1/nym–1 = 1/m·1/(x1/m)m–1 = 1/m·x1/m–1

Ex. y = xn/my' = n/m·xn/m–1
Pr. y = (x1/m)n, t = x1/my = tn, dt/dx = 1/m·x1/m–1

dy/dx = dy/dt·dt/dx = ntn-1·(1/mx1/m–1

= n(x1/m)n–1·(1/mx1/m–1 = n/m·x(n–1)/m+1/m-1 = n/m·xn/m–1

Ex. a ≠ 1, a > 0, x = ayy = logax
Pr. (loga(x + h) – logax)/h = 1/h·(loga(x + h) – logax)

= 1/h·loga(1 + h/x) = 1/x … loga(1 + h/x)x/h

(∵ limx→±∞(1 + 1/x)x = e)

h → 0, y' → (1/x)·logae
dy/dx = logae //

Ex. y = sin-1xx = siny (-1 ≤ x ≤ 1, -π/2 ≤ xπ/2)
Pr. sin-1x/dx = 1/(dsiny/dy) = 1/cosy = 1/√(1 – x2) ∵ siny = x
Ex. a ≠ 1, a > 0, y = axx = logay
Pr. dy/dx = 1/(dx/dy) = 1/(dlogay/dy) = 1/(1/y·logae)

= 1/[(1/y)·(loge1/logea)] = y·logea = ax·logea

Ex. y = 3x (x > 0) ⇒ x = y3, dx/dy = 3y2 = 3(3x2)

y' = dy/dx = 1/(dx/dy) = 1/(3(3x2)

#### Logarithmic differentiation (対数微分法)

Th. f(x), g(x) [a, b] diff. ⇒

(f(x) ± g(x))' = f'(x) ± g'(x), (f(xg(x))' = f'(xg(x) + f(xg'(x)

Pr._ F(x) = f(x) ± g(x), G(x) = f(xg(x)

(F(x + h) – F(x))/h = {f(x + h) ± g(x + h) – (f(x) ± g(x))}/h

= (f(x + h) – f(x))/h ± (g(x + h) – g(x))/h

(G(x + h) – G(x))/h = (f(x + hg(x + h) – f(xg(x))/h

= (f(x + h) – f(x))/h·g(x) + f(x)·(g(x + h) – g(x))/h

#### Elementary function ( 初等関数 )

##### Differentiation of elementary functions (初等関数の微分)
Th. Diff. of log and exponential functions (対数・指数関数の微分)
Exponential function: (ax)' = axloga (a > 0)_____ a = e: (ex)' = ex
Logarithmic function: (loga|x|)' = 1/(xloga) (a > 0) ⇒ a = e: (log|x|)' = 1/x
Pr1. y := ax, logy = logax = xloga ∴ 1/y·dy/dx = loga

dy/dx = yloga = axloga //

Pr2. (loga|x|)' = (log|x|/a)' = 1/(xloga) //
Th. Differentiation of power function (冪関数の微分)
Th. Differentiation of a few elementary functions (諸初等関数の微分)
Differentiation of trigonometric functions (三角関数の微分)

(sinx)' = cosx
(cosx)' = -sinx
(tanx)' = (sinx/cosx)' = 1/cos2x = sec2x = 1 + tan2x
(cotx)' = (cosx/sinx)' = -cosec2x
(secx)' = (1/cosx)' = secxtanx
(cosecx)' = (1/sinx)' = -cosecxcotx

Differentiation of inverse trigonometric functions (逆三角関数の微分)

(sin-1x)' = 1/√(1 – x2)
(cos-1x)' = -1/√(1 – x2)
(tan-1x)' = 1/(1 + x2)
(cot-1x)' = -1/(1 + x2)
(sec-1x)' = 1/(|x|√(x2 – 1)
(cosec-1x)' = -1/(|x|√(x2 – 1)

Pr. for (sin-1x)' = 1/√(1 – x)2, y = sin-1x

x = siny (-1 < x < 1, -π/2 < y < π/2), x' = cosy > 0
∴ cosy = √(1 – sin2y) = √(1 – x2)
dy/dx = 1/(dx/dy) = 1/cosy = 1/√(1 – x)2

Ex. Differentiate 1/2·log(x2 + y2) = tan-1(y/x)

1/2·log(x2 + y2)·(x2 + y2)'
= 1/(2(x2+ y2)·(2x+ 2y·y') = (x + y·y')/(x2+ y2 = (y/x)'/(1 + (y/x)2
= [(x·y' - y)/x2]/(1 + y2/x2) = [(x·y' - y)/x2]/[(x2 + y2)/x2]
= (x·y' – y)/(x2 + y2)
∴ (x + y·y')/(x2 + y2) = (x·y' – y)/(x2 + y2) → y' = (x + y)/(xy) //

Ax. Differentiation of hyperbolic function (双曲線関数の微分)

(sinhx)' = coshx
(coshx)' = sinhx
(tanhx)' = 1/cosh2x

Pr. (sinhx)' = ((ex - e-x)/2)' = (ex + e-x)/2 = coshx. The same hereinafter //

### High-order derivatives (高次導関数)

 Ax. Equations of n-order derivatives (n次導関数公式) (1) (ex)(n) = ex (2) (sinx)(n) = sin(x + nπ/2) (3) (cosx)(n) = cos(x + nπ/2) (4) {(x + a)α}(n) = α(α - 1)(α - 2)…(α - n + 1)(x + a)α-n (5) {log(x + a)}(n) = (-1)n-1·(n - 1)!/(x + a)n (x + a > 0) Ax. a, b ∈ ℜ, f(x), g(x) n-order diff. ⇒ (1) {f(ax + b)}(n) = anf(n)(ax + b) (2) {af(x) + bg(x)}(n) = af(n)(x) + bg(n)(x) Pr. (1) {f(ax + b)}' = af'(ax + b), {f(ax + b)} = {af'(ax + b)}' = a2f''(ax + b), … ⇒ {f(ax + b)}(n) = anf(n)(ax + b) // (2) trivial because of the linearity of differentiation Law (generalized) Leibniz rule (ライプニッツの法則) f(x), g(x), n-order diff. ⇒ (f(x)·g(x))(n) = Σk=0nnCkf(n–k)(x)·g(k)(x) n = 1: {f(x)g(x)}' = f'(x)g(x) + f(x)g'(x) n = 2: {f(x)g(x)}'' = f''(x)g(x) + 2f'(x)g'(x) + f(x)g''(x) n = 3: {f(x)g(x)}''' = f'''(x)g(x) + 3f''(x)g'(x) + 3f'(x)g''(x) + f(x)g'''(x) Q. obtain y(n) of y = xe2x A. (x)(k) = 0 (k ≥ 2) → y(n) = Σk=0nnCk(x)(k)(e2x)(n-k) = x(e2x)(n) + n(x)'(e2x)(n-1) = x·2ne2x + n·1·2n-1e2x = 2n-1(2x + n)e2x

### Mean-value theroem (平均値の定理)

 Th. Taylor's theorem (Taylorの定理) Th. Maclaurin's theorem (Maclaurinの定理) [a = 0 on Taylor's theorem] Q. Express ex as f(x) = Σk=0n-1(f(k)(0)/k!·xk) + f(n)(θx)/n!·xn (0 < θ < 1) Then, express the sum of 5th-order polynomial and residual term. A. f(n)(x) = ex (n = 0, 1, 2, …), f(n)(θx) = eθx ∴ ex = Σk=0n-1(1/k!·xk) + eθx/n!·xn ∴ ex = 1 + x + x2/2 + x3/6 + x4/24 + x5/120 + eθx/720·x6

### Application of differentiation (微分の応用)

Def. Landau's symbol (ランダウの記号), o
Th. operation of Landau's symbol: x → 0

o(xn)o(xm) = o(xn+m)
xno(xm) = o(xn+m)
o(xn) ± o(xm) = o(xn) if mn
Co(xn) = o(xn), C = constant

Ax. Asymptotic expansion elementary function (初等関数の漸近展開)
When x → 0, equations shown below hold:

ex = Σk=0n(xk/k!) + o(xn)
sinx = Σk=0n((-1)kx2k+1)/(2k + 1)! + o(x2n+2)
cosx = Σk=0n((-1)kx2k/(2k)! + o(x2n+1)
log(1 + x) = Σk=1n((-1)k-1xk)/k + o(xn)
(1 + x)a = Σk=0n[a:k]xk + o(xn)

##### Extreme value (極値)
= local maximum (極大値) + local minimum (極小値)
Q. Obtain the extreme of f(x) = x2/ex using 2nd derivative
A. f'(x) = 2xe-x - x2e-x = x(2 - x)e-x

x = 0, 2 when f'(x) = 0
f''(x) = (2 - 2x)e-x - (2x - x2)e-x = (x2 - 4x + 2)e-x
f''(0) = 2 > 0, f''(2) = -2e-2 < 0
x = 0 → local minimum f(0) = 0, x = 2 → local maximum f(2) = 4/e2

{an} = n, n2, n3, √n, 2n, 3n or n!, n → ∞ ⇒ limn→∞an = ∞

increase rates are different among them

Q. Obtain limn→∞(2n/n!)
A. when n ≥ 4 ⇒ 0 < 2n/n! = 2/1·2/2·2/3·2/4 … 2/(n - 1)·2/n

< 2/1·2/n = 4/n → 0 (n → 0)__limn→∞(2n/n!) = 0 //

n! shows higher increase rate than 2n

Th. 0 ≤ a < 1, {an}, limn→∞|an+1/an| = alimn→∞an = 0
Pr. γ := (a + 1)/2, 0 ≤ a < γ < 1 (∵ a < 1)

ε = γ - a > 0, NN = N(γ - a) exist

nN ⇒ ||an+1/an| - a| < γ - a

here, if nN

|an+1/an| - a ≤ ||an+1/an| - a| ⇒ |an+1| < γ|an|

repeat using this inequality when n > N

0 ≤ |an| < γ|an-1| < γ2|an-2| < … < γn-N|aN|

from 0 < γ < 1, limn→∞γn-N|aN| = 0 //

Q. Show limn→∞(na/rn) = 0
A. bn := na/rn > 0

bn+1/bn = (n + 1)a/rn+1·rn/na = 1/r·(1 + 1/n)a ⇒ 1/r (n → ∞)
from 0 < 1/r < 1 and Th. shown above, limn→∞ = 0 //

rn shows higher increase rate than na

Q. Prove Solve f(x) = sin-1(x/√(1 + x2)) - tan-1x
A. y := x/√(1 + x2), f(x) = sin-1y - tan-1x

f'(x) = 1/√(1 - y2y' - 1/(1 + x2)

Here √(1 - y2) = √(1 - x2/(1 + x2)) = 1/√(1 + x2),

y' = (√(1 + x2) - x·(2x/2√(1 + x2)))/(√(1 + x2) = 1/((1 + x2)(√(1 + x2))
f'(x) = √(1 + x2)·1/((1 + x2)√(1 + x2)) - 1/(1 + x2) = 0
⇒ constant function, f(0) = sin-10 - tan-10 = 0 - 0 ∴ f(x) = 0 //

Th. Newton or Newton-Raphson method (ニュートン法)
Ex.a: f(x) = x2 - af'(x) = 2x, f''(x) = 2

xn+1 = xn - (xn2 - a)/2xn = (xn2 + a)/2xn

If a = 5 →

n = 1: (1 + 5)/2 = 3
n = 2: (32 + 5)/(2·3) ≈ 2.333
n = 3: (2.3332 + 5)/(2·2.333) ≈ 2.238
n = 4: (2.2382 + 5)/(2·2.238) ≈ 2.236

#### Limit of indeterminate forms (不定形の極限値)

Th. l'Hôpital's rule (ロピタルの定理)
Ex. limx→∞{(-3x2 + 7x + 4)/(2x + 5)} = limx→∞{(-3x + 7 + 4/x)/(2 + 5/x)} = -∞
Ex. limx→∞(x3 - 4x2 -5x -6) = limx→∞{(-3x + 7 + 4/x)/(2 + 5/x)} = -∞
Ex. limx→∞(logx/√x) = limx→∞((1/x)/(1/2√x)) = limx→∞(2/√x) = 0
Ex. limx→+0xlogx = limx→+0logx/(1/x) = limx→+0(1/x)/(-1/x2) = limx→+0(-x) = 0

These are not indeterminate forms

Def. indeterminate form (不定形): an expression involving two functions whose limit cannot be determined solely from the limits of the individual functions

seven indeterminate forms which are typically considered
limxaf(x)/g(x) = 0/0, ∞/∞, ∞ – ∞, ∞0, 00, or 1
i.e., limit value can not be obtained as it is

## Integration (積分法)

 Def. (Riemann) integration: a = ∫abf(x)dx ⇒ ∫aaf(x)dx = 0, ∫baf(x)dx = -∫abf(x)dx Ex. f(x) = c (c: constant) ⇒ I = [a, b] integrable, ∫abcdx = c(b - a) Ex. f(x) = x ⇒ I = [a, b] integrable, ∫abxdx = 1/2·(b2 - a2) Ex. f(x) = x2, 0 ≤ a < b ⇒ I = [a, b] integrable, ∫abx2dx = 1/3·(b3 - a3) Def. lower sum and upper sum (不足和・過剰和) I = [a, b], f(x) bounded, Δ = {xk}, Ik := [xk-1, xk] mk = infx∈Ikf(x), Mk = supx∈Ikf(x) (k = 1, 2, … n) ⇒ lower sum ≡ sΔ(f) = Σk=1nmk(xk - xk-1) upper sum ≡ SΔ(f) = Σk=1nMk(xk - xk-1) ⇒ sΔ(f) and SΔ(f) are determined only by Δ Σk=1nmk(xk - xk-1) ≤ Σk=1nf(ξk)(xk - xk-1) ≤ Σk=1nMk(xk - xk-1) ∴ sΔ(f) ≤ S(f;Δ, {ξk}) ≤ SΔ(f) a partition Δ' made by a new cutoff point c only one k, which supports xk-1 < c < xk, exists consider the upper sum: 0 ≤ SΔ(f) - SΔ'(f) ≤ (Mk - mk)(xk - xk-1) Note that SΔ'(f) ≤ SΔ(f) consider the lower sum: 0 ≤ sΔ'(f) - sΔ(f) ≤ (Mk - mk)(xk - xk-1) Note that sΔ(f) ≤ sΔ'(f) In general, sΔ(f) ≤ sΔ'(f) ≤ SΔ'(f) ≤ SΔ(f) (Δ ⊂ Δ') Δ = {xk}, Δ' = {yl} → Δ'' := {xk} ∪ {yl} ⇒ sΔ(f) ≤ sΔ''(f) ≤ SΔ''(f) ≤ SΔ'(f) ∴ sΔ(f) ≤ SΔ'(f) sΔ(f) bounded above on Δ and SΔ'(f) bounded below on Δ Def. upper integral and lower integral (上積分・下積分) I = [a, b], f(x) bounded ⇒ upper integral ≡ -∫abf(x)dx = infΔSΔ(f) on the greatest lower bound of Δ lower integral ≡ -∫abf(x)dx = supΔsΔ(f) on the least upper bound of Δ ⇒ Δ, Δ' → sΔ(f) ≤ SΔ'(f) ⇒ -∫abf(x)dx ≤ -∫abf(x)dx Th. amplitude (振幅), ω(f, I) Ex. I = [1, 3], f(x) = x2 ⇒ supx∈If(x) = 9, infx∈If(x) = 1 ∴ ω(f, I) = 9 - 1

### Indefinite integrals (不定積分)

Def. primitive function (原始関数) ≡ F(x)
G'(x) = f(x), F(x), G(x) diff., F'(x) = G'(x)

F(x) – G(x) = C (C: constant of integration 積分定数)
G(x) = F(x) + C = f(x)dx + C = f(x)dx

1) f(x), f'(x) ≡ 0, [a, b] diff. → f(x) = C (constant)
Pr. [a, b] ∋ af(x) = f(a) + (xa)f'(c) = f(a) = C //
2) f(x), g(x), [a, b] diff., f'(x) ≡ g'(x) → f(x) = g(x) + k
Pr. F(x) := f(x) – g(x) reduce (1)

F(x) = F(a) + (xa)F'(c) = {f(a) – g(a)} + (xa)(f'(c) – g'(c))
= (f(a) – (xa)f'(c)) + (g(a) – (xa)g'(c)) = f'(a) – g'(a) = k
F(x) = f(x) – g(x) = k //

##### Integration by substitution (置換積分)
Ex. Ex. I = (x/√(1 + x2))dx: t := 1 + x2, dt/dx = 2x

xdx = 1/2dtI = t·1/2dt = 1/2·2/3t3/2 = (1/3)·√(1 + x2)3

Ex. I = (ax + b)/√(x2 + 2px + q)dx (pq)

[I = (ax + b)/√{(x + p)2 + qp2}dx]
t = x + p
I = (at + bap)/√(t2 + qp2)dt

= at/√(t2 + qp2)dt(1) + (bap)/√(t2 + pq2)dt(2)

(1) I1: t2 + qp2 = u, 2tdt = du

I1 = a/(2√u)du = au = a√(t2 + q + p2)

(2) = I2 = (bap)log(t + √(t2 + qp2))
I1 + I2 = a√(t2 + q + p2) + (bap)log(t + √(t2 + qp2))

= a√(x2 + 2px + q) + (bap)log(x + p + √(x2 + 2px + q))

#### Th.integration by parts (部分積分)

f(x)g(x)dx = f(x)G(x) - f'(x)G(x)dx
Ex. xexdx = xex - xexdx = xex - ex + C

(C, omit – often done)

#### Basic formulae (基本公式)

Ax. 1) (f'(x)/f(x))dx = log|f(x)|
Pr. (log|f(x)|)' = f'(x)/f(x) //
Ax. 2) f(x)αf'(x)dx = f(x)α+1/(α + 1) (α ≠ -1)
0) {af(x) + bg(x)}dx = af(x)dx + bg(x)dx
1) xαdx = xα+1/(α + 1)
2) (1/x)dx = log|x|
3) 1/(a2 + x2)dx = 1/a·tan-1(x/a)
4) 1/(a2x2)dx = (1/2a)·log{(a + x)/(ax)} (|x| < a)
5) 1/(x2a2)dx = (1/2a)·log{(xa)(x + a)} (|x| > a)
6) 1/√(a2x2)dx = sin-1(x/a) (|x| < a)
7) 1/√(x2a2)dx = log(x + √(x2a2)) (|x| > a > 0)
8) 1/√(x2 + a2)dx = log(x + √(x2 + a2))
9) √(a2x2)dx = 1/2·{x√(a2x2) + a2sin-1(x/a)} (|x| < a)
10) √(x2a2)dx = 1/2·(x√(x2a2)) + a2log(x + √(x2 + a2))) (|x| > a > 0)
11) √(x2 + a2)dx = 1/2·(x√(x2 + a2)) + a2log(x + √(x2 + a2))
12) eaxdx = 1/a·eax
13) axdx = 1/loga·ax
14) sinaxdx = -1/a·cosax
15) cosaxdx = 1/a·sinax
16) 1/sin2axdx = -1/a·cotax
17) 1/cos2axdx = 1/a·tanax
18) tanaxdx = -1/a·log(cosax)
19) cotaxdx = 1/a·log(sinax)
##### Integration of rational functions (有理関数の積分)
Ax. f(u, v) rational functions on u and v, t := tan(x/2) ⇒

f(cosx, sinx)dx ⇒ expressed by the integral of rational function on t

Q. Solve 1/(2 + cosx)dx
A. 1/(2 + cosx)dx = 1/(2 + (1 - t2)/(1 + t2)·2/(1 + t2dt = 2/(3 + t2)dt

1/(2 + cosx)dx = 2/(3 + t2)dt

= 2/√3·tan-1(t/√3) = 2/√3·tan-1(1/√3·tan(x/2))

Q. Solve (cosx/(2 + sin2x))dx
A. t := sinxdt = cosxdx

(cosx/(2 + sin2x))dx

= (1/(t2 + 2)dt = 1/√2·tan-1(t/√2) = 1/√2·tan-1(sinx/√2)

##### Integration of irrational function (無理関数の積分)
Ex. (1/(√x + √(x + 1))dx = (√(x + 1) - √x)dx = 2/3·(x + 1)3/2 - 3/2·x3/2

### Definite integrals (定積分)

#### Basic equations of definite integrals

Th. f(x) continuous on I, F(x) := primitive function of f(x), α, βI

αβf(x)dx = F(β) - F(α) ≡ [F(x)]αβ

Pr. F(x) = axf(t)dt + C

F(β) - F(α) = (∫aβf(t)dt + C) - (∫aαf(t)dt + C) =

aβf(t)dt + αaf(t)dt = αβf(t)dt //

Th. f(x) diff. on I, f'(x) integrable on I, α and βIαβf'(x)dx = f(β) - f(α)
Th. f(x) continuous on I, a(x) and b(x) diff. on J, a(x) and b(x) ∈ I (xJ)

d/dxa(x)b(x)f(t)dt = f(b(x))b'(x) - f(a(x))a'(x)

Pr. F(x) := the primitive function of f(x) → a(x)b(x)f(t)dt = F(b(x)) - F(a(x))

differentiate both sides
d/dxa(x)b(x)f(t)dt = F'(b(x))b'(x) - F'(a(x))a'(x)

= f(b(x))b'(x) - f(a(x))a'(x) //

Ex. d/dxax2f(t)dt = f(x2)·(x2)' - f(aa' = 2xf(x2)

#### Solution of definite integral

Ex. 0πsinxdxπ/n := h
A.__SΔ = Σhsinnh,

2sin(h/2)·SΔ = Σhsin(nh)·2sin(h/2) = h{Σ2sin(nh)·sin(h/2)}

= -h[(cos3h/2 – cosh/2) + (cos5h/2 – cos3h/2)

+ (cos7h/2 – cos5h/2) +
… + {cos(2n + 1)h/2 – cos(2n – 1)h/2}]

∴ 2sin(h/2)·SΔ = -h{cos(2n + 1)h/2 – cos(h/2)},

SΔ = -h{cos(2n + 1)h/2 – cos(h/2)}/{2sin(h/2)}

n → ∞, h = π/n → 0
∴ -h/{2sin(h/2)} → 1, (2n + 1)h/2 = (2n + 1)π/2nπ
limn→∞SΔ = -(cosπ – cos0) = 2

2) Relationship with indefinite integral
Th. Characteristics of definite integrals
1) linearity (線形性)
2) abf(x)dx = acf(x)dx + cbf(x)dx

no matter how the relationships between a, b and c are large or small

Pr. c is a dividing point on Δ = {xk} →

Δ1 and Δ2 are divisions of [a, c] and [c, b], respectively
Set S[a,b]Δ(f) as interval considering upper sum →
S[a,b]Δ(f) = Σk=1nsupxIkf(x)(xk - xk-1)

= k=1l + Σk=l+1n)supxIkf(x)(xk - xk-1) = S[a,c]Δ1(f) + S[c,b]Δ2(f)

Here, |Δ1| ≤ |Δ|, |Δ2| ≤ |Δ|, select c as cutoff point, |Δ| < 0

⇒ |Δ1| < 0, |Δ2| < 0

-abf(x)dx = -acf(x)dx + -cbf(x)dx ⇒ true on upper integral

As well, true on lower integral

f(x) integrable on [a, b] ⇒ -abf(x)dx = -abf(x)dx

-acf(x)dx + -cbf(x)dx = -acf(x)dx + -cbf(x)dx

when -acf(x)dx = -acf(x)dx, -cbf(x)dx = -cbf(x)dx, the equation holds

f(x) integrable on [a, c] //

3) monotonicity (単調性)
4) absolute value inequalities (絶対値の不等式)

### Improper integral (特異積分)

Ex. 01(1/√x)dxt > 0 → limx→+0(1/√x) = ∞ ∴ t1(1/√x)dx = [2√x]t1 = 2 - 2√t

Here, limt→+0(2 - √t) = 2 (convergence) ∴ 01(1/√x)dx = 2

Ex. 01(1/x)dxt > 0 → t1(1/x)dx = [log|x|]t1 = -logt

limt→+0(-logt) = ∞ (divergence)

Ex. 01logxdxlimx→0logx = -∞, t > 0

t1logxdx = [xlogx - x]t1 = -1 - tlogt + t. From l'Hôpital's rule,

limt→+0tlogt = limt→+0t1logt/(1/t) = limt→+0((1/t)/(-1/t2)) = limt→+0(-t) = 0

01logxdx = limt→+0t1logxdx = limt→+0(-1 - tlogt + t) = -1

Ex. -11(1/x)dxlimx→+0(1/x) = ∞, limx→-0(1/x) = -∞

-11(1/x)dx = -10(1/x)dx + 01(1/x)dxdivergence → divergence

Integral test (収束判定法)
Comparison test (比較判定法)
Th'. Absolute converge of improper integral (広義積分の絶対収束)

f(x) integrable on [a, ∞], a|f(x)|dx convergence ⇒
af(x)dx convergence, as in any other intervals, (a, b], [a, b), (-∞ b]

Pr. Use Cauchy's convergence test (Cauchyの収束判定法)
Def. f(x) indefinite integrable on I, I = (a, b], [a, b), [a, ∞) or (-∞ b]
(1) I|f(x)|dx converge ⇒ If(x)dx converge ≡ absolute converge (絶対収束)
(2) I|f(x)|dx diverge, If(x)dx converge ⇒

If(x)dx ≡ conditional converge (条件収束)

### Application of integral (積分の応用)

##### Area (面積), S
Def. y = f(x) ≥ 0, [a, b] ⇒ S = abydx = abf(x)dx

≡ the area enclosed by y = f(x) and lines, x = a and x = b

Th. f(x), g(x) continuous, f(x) > g(x) (axb) ⇒ S = ab{f(x) - g(x)}dx

≡ the area enclosed by y = f(x), y = g(x) and lines, x = a and x = b

Q. Obtain the aera enclosed by f(x) = x/(x2 + 4) and g(x) = x/8
A. f(x), g(x) odd fucctions, S = double of area in x ≥ 0

x/(x2 + 4) - x/8 = (x(4 - x2)/(8(x2 + 4)) ≥ 0 ⇔ 0 ≤ x ≤ 2
S = 202{ x/(x2 + 4) - x/8}dx = [log(x2 + 4) - x2/8]02 = log2 - 1/2

Q. [polar coordinates] Obtain area (S) enclosed by r = 1 + cosθ,

r = sinθ and θ = 0 (0 ≤ θπ/2)

A. S = 1/20π/2(1 + cosθ)2 - 1/20π/2sin2θdθ = 1/20π/2

= 1/20π/2(1 + 2cosθ + cos2θ - sin2θ)
= 1/20π/2(1 + 2cosθ + cos2θ) = [θ + 2sinθ + sin2θ/2]0π/2 = 1 + π/4

##### Length (長さ), l
Th. Arc length of a curve (弧長), L:

Cx = φ(t), y = ψ(t) (αtβ), φ(t), ψ(t) C1-class ⇒

L = αβ√{(dx/dt)2 + (dy/dt)2}dt = αβ√{φ'(t)2 + ψ'(t)2}dt

Corollary. Arc length (弧長), L: y = f(x) C1-class (axb) ⇒

L = ab√(1 + f'(x)2)dx

Pr. x = t, y = f(x) ⇒ L = ab√((dx/dt)2 + (dy/dt)2)dt = ab√(1 + f'(x)2)dx //
Q. Obtain the arc length (L) of y = coshx (0 ≤ x ≤ log2)
A. y' = sinhx, coshx ≥ 0 → √(1 + y2) = √(1 + sinh2x) = √cos2x = coshx

L = 0log2coshxdx = [sinhx]0log2 = sinh(log2) = (2 - 1/2)/2 = 3/4

Co. Arc length on polar coords, L: C, r = f(θ) C1-class (αθβ) ⇒

L = αβ√(f(θ)2 + f'(θ)2)

Pr. r = f(θ) on xy coordinate → x = rcosθ = f(θ)cosθ, y = rsinθ = f(θ)sinθ

(dx/)2 + (dy/)2 = (f'(θ)cosθ - (f(θ)sinθ)2 + (f'(θ)sinθ - (f(θ)cosθ)2

= f'(θ)2(cos2θ + sin2θ) + f(θ)2(cos2θ + sin2θ) = f'(θ)2 + f(θ)2

L = αβ√((dx/)2 + (dy/)2) = αβ√(f(θ)2 + f'(θ)2) //

##### Volume and lateral (surface) area (体積・側面積), V and Sl
Q. Obtain V and Sl of sphere of which radius is a
A. the sphere ≡ solid revolution of which semicircle is y = √(a2 + x2)

V = π-aay2dx = π-aa(a2 - x2)dx = π/6·{a - (-a)}3 = 4/3·πa3

y' = -x/√(a2 - x2) → √(1 + y'2) = √(1 + x2/(a2 - x2)) = a/y

Sl = 2π-aay√(1 + y'2)dx = 2π-aaadx = 4πa2

[ sequence ]

## Series (級数)

Def. series (級数): S = Σan = Σn=1an = a1 + a2 + … + an + …

Def. nth item (第n項) = an
Def. sn: nth partial sum (第n部分和) = Sn

Def. finite series (有限級数) ⇔ infinite series (無限級数): n = ∞
Def. geometric series (等比級数/幾何級数), In:

the sum of geometric sequence between the first term to nth term when the first term = 1 and common ratio = p
(p = 1 → In = n)
In = 1 + p + p2 + … + pnpIn = p + p2 + p3 + … + pn+1
In - pIn = (1 – p)In = 1 – pn+1In = (1 – pn+1)/(1 – p)

Ex. p = 2 → I = (1 – 2n+1)/(1 – 2) = 2n+1 – 1

Cf. Mersenne number (メルセンヌ数)

Def. converge (収束) Σn=1ak = sdivergence (発散)
Ex._Σn=1xn, |x| > 1 → convergent to 1/(1 – x)

|x| ≤ 1 → divergence
x = -1 → oscillation (振動) ⊂ divergence

Prop. infinite geometric series, a ≠ 0

|r| < 1 ⇒ Σn=1arn-1 = a/(1 - r) convergence
|r| ≥ 1 ⇒ Σn=1arn-1 divergence

Th. Σn=1an = S, Σn=1bn = T convergence ⇒

Σn=1(aan ± bbn) = aΣn=1an ± bΣn=1bn

Pr. Sn, Tn nth partial sum ⇒ Σn=1(aan ± bbn) = aSn + bTn

aS + bT (n → ∞) //

Th'. Σn=1an = sΣn=1n1an + Σn=n1+1an = s
Q. obtain the sum of series Σn=1{1/(n(n + 1))}
A. conduct partial fraction decomposition: 1/(n(n + 1)) = 1/n - 1/(n + 1)

nth partial sum, Sn = Σk=1n{1/k - 1/(k + 1)}

= (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + … + (1/n - 1/(n + 1))
= 1 - 1/(n + 1) ∴ convergence

Σn=1{1/(n(n + 1))} = limn→∞Sn = limn→∞(1 - 1/(n + 1)) = 1

##### Sequence of functions (関数列)
Th. Weierstrass M-test (ワイエルシュトラウスのM判定法)
Ex. Σn=1(sinnx/n2): |sinnx/n2| ≤ 1/n2 for nN, x

Σn=1(1/n2) convergence (Weierstrass M-test)
Σn=1(sinnx/n2) convergence //

### Power series (整級数)

 Convergence radius or radius of convergence (収束半径) ≡ r A power series with radius of convergence r Ex. Obtain r of Σn=0∞((n + 1)/n!·xn) an := (n + 1)/n! → limn→∞|an/an+1| = limn→∞[(n + 1)/n!·{(n + 1)!/(n + 2)}] limn→∞{(n + 1)2/(n + 2)} = ∞ ∴ r = ∞ Th. Abel's continuity theorem (Abelの連続性定理) f(x) = Σn=0∞an convergence ⇒ limx→1–0Σn=0∞anxn = Σn=0∞an Pr. f(x) = Σn=0∞anxn (|x| < 1), sn = a0 + a1 + … + an, Σn=0∞an := s → Σn=0manxn = Σn=0m(sn – sn–1)xn = Σn=0msn(xn – xn+1) + smxm = (1 – x)Σn=0msnxn + smxm (s–1 := 0) |x| < 1, m → ∞ → f(x) = (1 – x)Σn=0∞snxn, 1 = (1 – x)Σsnxn 1 = (1 – x)Σxn, |x| < 1, f(x) – s = (1 – x)Σn=0∞(sn – s)xn = (1 – x)Σn=0∞rnxn, limn→∞rn = 0, ∃ε > 0, N < n → |rn| < ε ∴ 0 ≤ x < 1 → |f(x) – s| ≤ |(1 – x)Σn=0Nrnxn| + ε(1 – x)Σn=N+1∞xn |r0| + |r1| + … + |rN| = p → |f(x) – s| + ε(1 – x)·xN+1/(1 – x) δ = ε/p, 1 – δ < x < 1 → |f(x) – s| < ε + ε = 2ε // Abel transform (Abelの変換) Σn=0manxn = Σn=0m(sn – sn–1)xn = Σn=0msn(xn – xn+1)xn + smxm Th. Term-by-term integration theorem (項別積分定理) Th. Term-by-term differentiation theorem (項別微分定理) Q. Obtain the value of ∫01Σn=0∞{(n + 1)/3n·xndx A. an = (n + 1)/3n → r of Σn=0∞anxn = limn→∞|an/an+1| = limn→∞{(n + 1)/3n} = limn→∞{3(n + 1)/(n + 2} = 3, uniform convergence on [0, 1] ∴ integrable term by term ∫01Σn=0∞{(n + 1)/3n·xndx = Σn=0∞∫01{(n + 1)/3n·xndx = Σn=0∞[xn+1/3n]01 = Σn=0∞(1/3n) = 1/(1 - 1/3) = 3/2

### Fourier series (フーリエ級数)

 Trigonometric series (三角級数), f(x) = a0/2 + a1cosx + b1sinx + … + ancosnx + bnsinnx + … = a0/2 + Σn=1∞(ancosnx + bnsinnx) f(x) = f(x + 2π) i.e, a periodic function of period 2π → can be assumed as f(x) (–π ≤ x ≤ π) continuous → Fourier series an = 1/π∫-∞∞f(x)cosnxdx (n = 0, 1, 2, …), bn = 1/π∫-∞∞f(x)sinnxdx (n = 1, 2, …) Def. an, bn Fourier coefficient (フーリエ係数) f(x) ~ a0/2 + Σn=1∞(ancosnx + bnsinnx): ~ meaning of Fourier series Basic properties (基本性質) sn(x) = a0/2 + Σk=1n(akcoskx + bksinkx) = ∫-ππf(t)dt + Σk=1n(coskx·1/π∫-ππf(t)cosktdt + sinkx·1/π∫-ππf(t)sinktdt) = 1/π∫-ππf(t){1/2 + Σk=1ncos(t – x)}dt {1/2 + Σk=1ncos(t – x)}sin(t/2) = 1/2[sin(t/2) + Σk=1n{sin(k + 1/2)t – sin(k – 1/2)t}] = 1/2·sin(n + 1/2)t ∴ Sn(x) = 1/π∫-ππf(t)[{sin(n + 1/2)(t – x)}/2sin{(t - x)/2}]dt Dirichlet kernel (ディリクレ核), Dn(t) := {sin(n + 1/2)t}/2sin(t/2) → even function (偶関数) 1/π∫-ππDn(t)dt = 1 (f(t) ≡ 1) Dirichlet integral (ディリクレ積分), sn(x) = 1/π∫-ππf(t)Dn(t – x)dt = 1/π∫-ππf(t + x)Dn(t)dt Th. Riemann-Lebesgue: f(x) period = 2π continuous function, n → ∞ → Fourier coefficient = 0 (convergence), i.e., an → 0, bn → 0 Pr. (prove an → 0), πan = ∫-ππf(x)cosnxdx, x := t – π/n πan = ∫-π+π/nπ+π/nf(t – π/n)cosn(t – π/n)dt = ∫-ππf(t – π/n)cosntdt = –∫-ππf(x – π/n)cosnxdx 2πan = –2∫-ππf(x)cosnxdx – ∫-ππf(x – π/n)cosnxdx = ∫-ππ{f(x) – f(x – π/n)}cosnxdx f(x) [-π, π] uniformly cont., 0 < ε, 0 < N < ∀n, |f(x) – f(x – π/n)| < ε ∴ |2πan| ≤ ∫-ππ|f(x) – f(x – π/n)|dx < 2πε ∴ limn→∞an = 0, same applies to bn // Th. f(x) ~ a0/2 + Σn=1∞(ancosnx + bnsinnx), continuous ⇒ Σn=1∞an2, Σn=1∞bn2 converge together Pr. αn, βn J =Def ∫-ππ(f(x) – {α0 + Σn=1k(αncosnx + βnsinnx)})2dx = ∫-ππf(x)2dx – {αa02 + πΣn=1k(αn2 + βn2)} + π/2(a0 – α0)2 + πΣn=1k{(an – αn)2 + (bn – βn)2} → αn = an (n = 1, 2, …, k), βn = bn (n = 1, 2, …, k) in minimum, J ≥ 0 ∴ ∫-ππf(x)2dx ≥ πa02/2 + πΣn=1k(an2 + bn2) ≡ Bessel's inequality (ベッセルの不等式) //

## Partial differentiation (偏微分法)

 Th. lim(x, y)→(a, b)f(x, y) = α, lim(x, y)→(a, b)g(x, y) = β, λ, μ ∈ ℜ ⇒ (1) lim(x, y)→(a, b){λf(x, y) + μg(x, y)} = λα + μβ (2) lim(x, y)→(a, b)f(x, y)g(x, y) = αβ (3) lim(x, y)→(a, b){f(x, y)/g(x, y)} = α/β (β ≠ 0) (4) lim(x, y)→(a, b)|f(x, y)| = |α| Pr. λ = μ = 0 trivial. → (λ, μ) ≠ (0, 0), a = (a, b), ε-δ definition Th. (x, y) nbd of (a, b), (x, y) ≠ (a, b), f(x, y) ≤ g(a, b) lim(x, y)→(a, b)f(x, y) = α, lim(x, y)→(a, b)g(x, y) = β ⇒ α ≤ β Th. (squeeze theorem) (x, y) nbd of (a, b), (x, y) ≠ (a, b), |f(x, y) - α| ≤ g(x, y), lim(x, y)→(a, b)g(x, y) = 0 ⇒ lim(x, y)→(a, b)f(x, y) = α Ex. lim(x, y)→(0, 0)(xy/√(x2 + y2)) = 0 r := √(x2 + y2), |x| ≤ r, |y| ≤ r, (x, y) ≠ (0, 0) → |xy/√(x2 + y2) - 0| = |xy|/√(x2 + y2) = (|x|·|y)|/r ≤ (r·r)/r = r → 0 (r → +0)

### Partial derivative (偏導関数)

Def. Partial derivative (偏導関数)

fx(x, y) ≡ limh→0[{f(x + h, y) - f(x, y)}/h]
fy(x, y) ≡ limk→0[{f(x, y + k) - f(x, y)}/k]

Ex. z = f(x, y) = 4x2 + 5xy - 7y3zx = 8x + 5y, zy = 5x - 21y2
Ex. u = f(x, y, z) = sin(xyz)

ux = yzcos(xyz), uy = xzcos(xyz), uz = xycos(xyz)

Def. totally differential (全微分可能), df(x, y)
Ex. f(x, y) = xy (x > 0) → fx(x, y) = yxy-1, fy(x, y) = xylogx

df(x, y) = yxy-1dx + xylogxdy

#### High-order partial derivative (高次偏導関数)

Ex. obtain all 2nd-order partial derivative of z = f(x, y) = x3 + 3x2y + y2

z = f(x, y) C2-class → zx = 3x2 + 6xy, zy = 3x2 + 2y
zxx = 6x + 6y, zxy = zyx = 6x, zyy = 2

##### Partial differentiation of composite function (合成関数の偏微分)
Th. f(t) diff. on I, g(x, y) ∈ I partial diff. on D, (x, y) ∈ D

F(x, y) = f(g(x, y)) partial diff. on D,
Fx(x, y) = f'(g(x, y))gx(x, y), Fy(x, y) = f'(g(x, y))gy(x, y)

Pr. φ := g(x, y) → d/dx·f(φ(x)) = f'(φ(x))φ'(x)

φ'(x) = gz(x, y) → /∂x·F(x, y) = f'(g(x, y))gx(x, y). yas well //

Th. f(x, y) totally diff. on D, φ(t), ψ(t) diff. on I, (φ(t), ψ(t)) ∈ D (tI) ⇒

F'(t) = fx(φ(t), ψ(t))φ'(t) + fy(φ(t), ψ(t))ψ'(t)

z = f(x, y), x = φ(t), y = ψ(t) ⇒ dz/dt = ∂z/∂x·dx/dt + ∂z/∂y·dy/dt
Th. f(x, y) totally diff. on D, φ(s, t), ψ(s, t) partial diff. on E,

(φ(s, t), ψ(s, t)) ∈ D ((s, t) ∈ E) ⇒
F(s, t) = f(φ(s, t), ψ(s, t)) partial diff. on E

Fs(s, t) = fx(φ(s, t), ψ(s, t))φs(s, t) + fy(φ(s, t), ψ(s, t))ψs(s, t)
Ft(s, t) = fx(φ(s, t), ψ(s, t))φt(s, t) + fy(φ(s, t), ψ(s, t))ψt(s, t)

z = f(x, y), x = φ(s, t), y = ψ(s, t) ⇒

∂z/∂s = ∂z/∂x·∂x/∂s + ∂z/∂y/∂y/∂s, ∂z/∂t = ∂z/∂x·∂x/∂t + ∂z/∂y/∂y/∂t

Q. When f(x, y) C1-class on R2, x = u2 + v2, and y = uv, prove

g(u, v) = f(u2 + v2, uv) ⇒ vgu(u, v) + ugv(u, v) = 4yfx(x, y) + xfy(x, y)

A. gu(u, v) = /∂u·f(u2 + v2, uv)

= fx(u2 + v2, uv∂x/∂u + fy(u2 + v2, uv∂y/∂u = 2ufx(x, y) + vfy(x, y)

__gv(u, v) = /∂v·f(u2 + v2, uv)

= fx(u2 + v2, uv∂x/∂v + fy(u2 + v2, uv∂y/∂v = 2vfx(x, y) + ufy(x, y)
vgu(u, v) + ugv(u, v)

= v{2ufx(x, y) + vfy(x, y)} + u{2vfx(x, y) + ufy(x, y)}
= 4uvfx(x, y) + (u2 + v2)fy(x, y) = 4yfx(x, y) + xfy(x, y)

#### Application of partial derivative (偏導関数の応用)

Def. f(x, y) C2-class ⇒ ≡ Hessian matrix (ヘッセ行列)

C2-class → fxy = fyx ∴ Hesse matrix ⊂ symmetric matrix

Def. Hessian, H(x, y) ≡ determinant of Hessian matrix
Def. f(x, y) = 0, y = g(x), f(x, g(x)) = 0 ⇒

y = g(x) ≡ implicit function (陰関数)

Def'. f(x, y, z) = 0, z = g(x, y), f(x, y, g(x, y)) = 0 ⇒

z = g(x) ≡ implicit function (陰関数)

Q. Obtain y' and y'' of y = ex+y
A. diff. by x, y is the function of xy' = ex+y(1 + y') ∴ (1 - ex+y)y'

If 1 - ex+y ≠ 0, then y' = ex+y/(1 - ex+y)
y' = -(ex+y(1 + y'))/(1 - ex+y)2 = ex+y/(1 - ex+y)3

### Double integral and space (二重積分と空間)

##### Repeated or iterated integral (累次積分)
Q. Obtain the value of I = 01{∫x1ey2dy}dx
A. I = 01{∫y1ey2dx}dy = 01yey2dy = [1/2·ey2]01 = (e - 1)/2

## Differential equation (微分方程式)

##### Classification
Ordinary differential equation, ODE (常微分方程式)
Partial differential equation, PDE (偏微分方程式)
Ax. Laplace transform on elementary function
(1) L[eat](s) = 1/(s - a) (s > a)______(4) L[sinωt](s) = ω/(s2 + ω2) (s > 0)
(2) L[tn](s) = n!/sn+1 (s > 0)_______ (5) L[coshωt](s) = s/(s2 - ω2) (s > |ω|)
(3) L[cosωt](s) = s/(s2 + ω2) (s > 0) (6) L[sinhωt](s) = ω/(s2 - ω2) (s > |ω|)
Q. Solve y'' - 3y' + 2y = 4e3t, y(0) = 0, y'(0) = 1
A. L[y(t)] := Y(s) →

L[y] = Y
L[y'] = sY(s) - y(0) = sY
L[y''] = s2Y(s) - y(0)s - y'(0) = s2Y - 1
L[e3t] = 1/(s - 3)

∴ (s2Y - 1) - 3sY + 2Y = 4/(s - 3) after Laplace transform
Y = 1/{(s - 1)(s - 2)} + 4/{(s - 1)(s - 2)(s - 3)}
Y = 1/(s - 1) - 3/(s - 2) + 2/(s - 3)
→ inverse Laplace transformation: y = et - 3e2t + 2e3t

### Ordinary differential equation, ODE (常微分方程式)

 Q. Obtain the solution of: y'' - 2xy' - 2y = 0, y(0) = 1, y'(0) = 0 A. (apply term-by-term differentiation) → y = ex2