)1 .2 的倍數⇔個位數為偶數

㆒. 數與座標系:

1.實數:實數座標系㆖的點係由有理數與無理數組成,其㆗有理數具有稠 密性,亦即在任意兩個不相等的有理數㆗間至

少存在㆒個有理數存在.

若 X、Y、Z 均為實數,則

( )1 .㆔㆒律:X<Y、X=Y、X>Y 恰有㆒個成立.

( )2 .遞移律:若 X<Y,Y<Z,則 X<Z.

( )3 .X<Y^⇔X+Z < Y+Z.

( )4 .若 Z>0,則 X<Y^⇔X˙Z < Y˙Z ( )5 .若 Z<0,則 X<Y^⇔X˙Z > Y˙Z

2.整數,為有理數的㆒個子集,任意兩的不相等整數差的絕對值都大 於或等於㆒,這個性質稱為整數的離散性.

3.常見的因數、倍數判斷法:

( )1 .2 的倍數^⇔個位數為偶數.

Ex : abc = a˙100+b˙10+c ⇔c 為偶數 ( )2 .3 的倍數^⇔數字和為 3 的倍數.

Ex: abc = a˙100+b˙10+c = (a˙99+b˙9)+(a+b+c) ( )3 .4 的倍數^⇔末兩位為4 的倍數

Ex: abc = a˙100+(b˙10+c) ( )4 .5 的倍數⇔末位為 5 的倍數.

Ex: abc = a˙100+b˙10+c ( )5 .8 的倍數^⇔末㆔位為 8 的倍數.

Ex: abcd = a˙1000+b˙100+c˙10+d ( )6 .9 的倍數^⇔末㆔位為9 的倍數.

Ex: abc = a˙100+b˙10+c = (a˙99+b˙9)+(a+b+c) ( )7 .11 的倍數^⇔(奇位數字和)-(偶位數字和)=11 的倍數.

Ex: abcd = a˙1000+b˙100+c˙10+d

=(99+1)˙(11-1)˙a+(99+1)˙b+(11-1)˙c+d =(b+d)-(a+c)+(99˙11+11-99)˙a+99˙b+11˙c 4.因數倍數的性質:

if a、b、c ∈Z

then ( )1 . a|b ,b|c ^⇒a|c

( )2 . a|b ,a|c ⇒ a | bm+cn

(m,n) are arbitrary integers

ex:密立根油滴實驗㆗,藉著求得各油滴的帶電量,並求其最大公因數, 即為基本電荷.

5.標準分解釋的應用:

if n =

P

^1α1•

P

^α22

⋅⋅ ⋅⋅ P

^αKKthen

( )1 . n 的正整數個數為(+1)(α2+1)…(αk+1)

ex: n=

P

^1α1,

P

^{10 | n ,}

P

^{11| n, …}

P

^{1α1| n ⇒ (α1+1)個}

( )2 .n 的正因數和

=(1+

P

^1+…+

P

^1α1)(1+

P

^2+…+

P

^2α2)…(1+

P

^K+…+

P

^αKK)

=

∏

= −

− ⁺

k

i i

i

P P

ⁱ

1

1 ) (1

α 1

其㆗ k

k

i

i a a a

a = • •⋅ ⋅⋅

∏

= 1 2 1

( )3 .n 的正整數乘積:

=

n

^{12(α1+1)(α2+1) (⋅⋅⋅αk+1) =}

n

^{( )}

k

i i

∏=

+

1

2 1

1 α

6.輾轉相除法:

設 a、b^∈N, 若存在 q、r^∈Z,使得 a=bq+r, 0≤r p^b, 則最大公因數 (a , b) = (bq+r , b) = (b , r)

7.複數:

Z = a + bi , a、b 為實數 ( )1 .If

Z

¹⁼

Z

^2⇒

a

¹⁼

a

^{2 ,}

b

¹

= b

²

( )2 .If

Z

¹⁺

Z

²⁼

( _a

₁⁺

_a

₂

)

+

( _b

₁⁺

_b

₂

)

i,

( )3 .If

Z

^1•

Z

²⁼

( _a

₁

_a

₂⁻

_b

₁

_b

₂

) (

⁺

_a

₁

_b

₂

_{+ a}

₂

_b

₁

)

i, ( )⁴ .If

b Z Z a

2 2 2 2 2

1 1

= + 〔

( _a

₁

_a

₂⁺

_b

₁

_b

₂

) (

⁺

_a

₂

_b

₁

_{− a}

₁

_b

₂

)

i〕

( )5 .

Z = a − b

_i

8.㆒元㆓次方程式的解:

ax²+bx+c=0 , (a≠0 , a、b、c ∋ R)

a ^c

a x b  +



 

 + ²

2 -

a b 4

2

²

2 



 

 + a

x b = ₂

4 1

a

(

^{b2 −4ac}

)

x =

a ac b

b 2

2−4

±

−

( )1 . ^b2−4ac > ⁰ ⇒ 相異兩實根.

( )2 . ^b2−4ac < ⁰ ⇒ 相等兩實根.

( )3 . ^{b2 −4ac} = ⁰ ⇒ 兩共軛虛根.

9.直角座標(x,y)

y

(a,b) Z = a + bi 可以在㆓維㆗表示.

x

極座標(r,^θ)

y (r,^θ) 0≤r= a²+b² <∞ , 0≤θ < 2π

(a,b) ^0≤sin^θ≡

b ≤1r

x ^0≤cosθ≡

a ≤1r

^−∞<^tanθ≡

a b<∞ r

θ

a b

10.㆔維座標系:

( )1 .直角座標系(x、y、z) z

y x

( )2 .柱座標系(^ρ、θ、z)

z ^0≤ρ<^∞ x=^{ρcosθ 0≤θ} < 2π y =^ρsinθ θ y ^−∞< ^{z <∞} z = z x

( )3 .球座標(^r、θ、φ)

x= ^rsinθcosφ y= ^rsinθsinφ y z= ^rcosθ x

^0≤r<^∞ 0≤θ <^{π 0≤ϕ}< ^2π

ρ θ

r θ

φ

11.分點公式:

設 A(

x

^1,

y

₁^{), B(}

x

^2,

y

₂^),則

( )1 . AB 之㆗點座標 _^









 + +

, 2 2

2 1 2

1

x y y

x

( )2 . 若 P 介於 A，B 之間且 AP :PB= m : n,則 P 為





 





+ + +

+

n m

my ny n m

mx

nx_{1 2 1 2}

,

12.質心:

( )1 .若 A，B 兩點的質量均為 M,則質心為 _^









 + +

, 2 2

2 1 2

1

x y y

x

( )2 .若 A，B 兩點的質量比為 m : n,則質心為 



 





+ + +

+

n m

ny my n m

nx

mx_{1 2 1 2}

,

( )3 .若

P

₁^,

P

₂^,…

P

^{N的質量均為M,則質心為}

( ) _∑

=

= +

+

= ^N

i i

N

cm

x x x x

x

_{N N}

2 1 1

... 1 1

∑

=

= ^N

i i

cm

y

y

_N

1

1

( )4 .若

P

^{i的質量為∆mi} , for i =1,2,3…N

∑

∑

⁼

=

∆

∆

= ^N

i i i N

i i

cm m

m

x

x

1

1

1

∑

∑

⁼

=

∆

∆

= ^N

i i i N

i i

cm m

m

y

y

1

1

1

( )5 . N^→∞, ∆mi →dm , ^{N m dm M}

i

i → =

∆

∫

∑

=1

∫

Ω

= xdm

x

^cm M^{1 Ω為其分布空間}

∫

Ω

= ydm

y

_cm M¹

13.微分與積分 ( )1 .微分:

^∆s = s2 −s1

^∆t = t2 −t1

1 1,t

s 平均速度: ^v=

t s

∆

∆

順時速度:

dt ds t v s

t

∆ =

= ∆

→

lim

∆ 0

( )2 .積分:

F = F(x)

作功 W= ^xF( )xdx

∫

x^'

= ^N ( )_{i i}

N

∑

i F x ∆x

∞ =

→ 1

lim

x ^{x' ∆x}_i=

N x

x^'− , ( ) ( ( ) ( )₁ )

2 1

+ +

= _{i i}

i F x F x

x F

14.微分運算:

f(x)

( ) ( ) ( )

x x f x x f dx

x df

x ∆

−

∆

= +

→

∆lim0

( )1 . f(x) = x , ( )

dx x df =

dx dx=

lim0

→

∆x x

x x x

∆

−

∆

+ =1

( )2 . f(x) = x²_{, ( )}

dx x df =

dx dx² =

lim0

→

∆x

( )

x x x x

∆

−

∆

+ ^{2 2}

=

lim0

→

∆x

( )

x x x x

∆

∆ +

∆ ²

2 =

lim0

→

∆x 2x+∆x=2x

( )3 . f(x)= xⁿ_, ^{( )}

dx x

df = ⁿxⁿ⁻¹

2 2,t s

15.積分運算:

F(x)=x , ^x F

( )

x dx

∫

x₁^{2 =}

∫

x^x₁^{2 xdx = 1}₂^{x2|xx12 =}

( _x

²₂

_x

₁²

)

2

1 −

F(^X₂) ^x F

( )

xdx

∫

x₁²

F(X1) = ( _{2 1}) ( ) ( )( _{2 1} )

2

1 x −x F x +F x =

( _x

²₂

_x

₁²

)

2

1 −

X1 ^X₂ F(x)=

e

^−ax2



∫

−^∞∞F( )x dx²=

∫ ∫

−^∞∞

e

^ax

e

^aydxdy

∞

∞

−

−

− ^{2 2}

=

∫ ∫

−^∞∞

e

^{a(x y )}dxdy

∞

∞

−

+

− ^{2 2} x²+y²=r² ,dx dy = r dr dθ

=

∫ ∫

0^{∞ π}

e

^−ar rdrdθ

2 0

2

=^{π 2}

0

2dr

e

^ar

∫

^{∞ −} t= a r²

=

a

π

∫

0^∞

e

^−tdt

=

a

π (-

e

^{−t)|∞0 = π}_a

2 2

e

^axdx

∫

−^∞∞

− =

a π

16.極座標積分

柱座標積分

球座標積分

θ

d

θ

R

rd

θ

dr

dA rd dr=

θ

2 0 0

2 0

2

2 2 1

2

R

R

A dA

rd dr

rdr R

R

π

θ

π π

π

=

=

= =

=

∫

∫ ∫

∫

L

0

2 0 0 0

2 L R

dV rd drdz

V dV

rd drdz L R

π

θ

θ π

=

=

=

=

∫

∫ ∫ ∫

θ

rsin

θ

^d

φ

^rd

θ

dr

2

2 2

0 0 0 2 0 0

2 0

3

( sin )( )( )

sin

sin

2 sin

4 4 3

R

R

R

dV r d rd dr

r drd d

V dV

r drd d

r drd

r dr R

π π π

θ φ θ

θ θ φ

θ θ φ

π θ θ

π π

=

=

=

=

=

=

=

∫

∫ ∫ ∫

∫ ∫

∫

0 1 1

sin

cos 2

d

π

θ

−

θ

= − =

∫

∫

Q

17.向量分析

z

z y

x j A k A

A i

z k jy x

i A

^

^

^

1

^

^ 1 1

^

+ +

=

+ +

→ =

y A_x = x，₁ A_y = y，₁ A_z = z₁

θ φ θ

φ θ

cos sin sin

cos sin

) (

) (

) (

A A

A A

A A

B A k B

A j B

A i B A

z y x

z z y

y z

z

=

=

=

± +

± +

±

=

±^{→ → → →}

→

scalar product

θ

cos AB B

A B

A B

A B

A•^→ = _{x x} + _{y y} + _{z z} =

→

θ cos 2

2

) (

) (

2 2

2 A B AB

C

B A B

B A A

B A B

A C

C

B A C

+ +

=

• +

• +

•

=

+

• +

=

• +

=

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

vector product

θ

C A (x1，y1，z1)

x

A

θ

A

B

B

sin

( ) ( ) ( )

x y z

y z y z z x z x x y x y

C A B AB

i C j C k C

i A B B A j A B B A k A B B A B A

→ → → θ

→ → →

→ → →

→ →

= × ≡

= + +

= − + − + −

= − ×

sin

r F rF

τ θ

→ → →

= ×

= 力矩

θ

r F

18.Triple scalar product

^ ^ ^ ^ ^ ^

( )

( ) [ ( ) ( ) ( )]

( ) ( ) ( )

( ) ( )

( ) ( ) ( )

x y z y z y z z x z x x y x y

x y z y z y z x z x z x y x y

A B C

A i A j A k i B C C B j B C C B k B C C B A B C C B A B C C B A B C C B

B C A C A B

A C B B A C C B A

→ → →

→ → → → → →

→ → → → → → → → →

• ×

= + + • − + − + −

= − + − + −

= • × = • ×

= − • × = − • × = − • ×

triple vector product

( )

( )

( ) ( )

( ) ( ) 0

( ) ( )

A B C D

D B C

D B C

A D A B A C

A B A C

A C A B

D B A C C A B

α β

α β

α β

α β

→ → → →

→ → →

→ → →

→ → → → → →

→ → → →

→ → → →

→ → → → → → →

× × ≡

⊥ ×

= +

• = • + •

• + • =

= • = − •

= • − •

所 以

，

ε

_ijk=

^

^ ^ ^

3 1 2

1 2 2 3 3 1

^ ^ ^

2 3 1

1 3 2 1 3 2

^ ^ ^

3 1 2

1 2 2 1 2 3 3 2 3 1 1 3

, , 1, 2, 3

) ( ) ( )

ijk i j k

A B A B e i j k A B e A B e A B e A B e A B e A B e

A B A B e A B A B e A B A B e

→ → ε

× = =

= + +

− − −

− + − + −

，

=(

1 123，231，312

-1 132，213，321

B C

A B×C

A×(B×C)

^ ^

^

^ ^

( )

( )

( ) ( )

m m

lkm l ijk i j lkm ijk l i j li mj lj mi l i j m

i j

j i j i i j

A B C

A B C e A B C e

A B C e A B C e A B C e

A C B A B C

ε ε ε ε

δ δ δ δ

→ → →

→ → → → → →

× ×

= =

= − −

= −

= • − •

19.Gradient，∇

^ ^ ^

^ ^ ^

i j k

x y z

i j k

x y z

φ φ φ

φ

∂ ∂ ∂

∇ = + +

∂ ∂ ∂

∂ ∂ ∂

∇ = + +

∂ ∂ ∂

ex

f = f r( , , )

θ φ

chain rule

所以

^ ^ ^

^

1 ( )

( ) ( )

( )

d f r

f r i x j y k z

r d r d f r

r d r

∇ = + +

=

2 2 2

^ ^ ^

( )

( ) ( ) ( )

( )

( ) ( )

* ( ) 1 2

( )2

* ( )

r x y z

f f r

f r f r f r

f r i j k

x y z

f r df r r

x dr x

df r x

dr r

x df r r dr

= + +

=

∂ ∂ ∂

∇ = + +

∂ ∂ ∂

∂ = ∂

∂ ∂

=

=

f f r f f

x r x x x

df r dr x

θ φ

θ φ

∂ = ∂ ∂ + ∂ ∂ + ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂ ∂

= ∂

∂

||

0

||

0

20.Circular motion

R a v a

2

lim0 =

= ⁻

θ→

向心加速度

hw1：prove

sin 1

lim

0

=

→

θ

θ

θ

21.Simple harmonic oscillation

0 0

cos cos( )

sin sin( )

x

x R R t

v v v t

t

θ ω

θ ω

θ ω

= =

= − = −

=

ω

：角頻率(速度)

dt d

θ ω

=

0 ( )

cos( ) cos( )

x

ds d d

v R R R R

dt dt dt

dx d d

v R t R t

dt dt dt

ω θ θ ω

ω ω

= = = = =

= = =

θ θ

R

θ

θ

1 2

2 1 2 1 2 1

2 1

2

2 sin 2

2 sin sin

2 v

v v a t t

v v v

t t R

v

v v

a R R

θ θ

θ θ

θ θ

−

−

= −

−

− =

− =

= =

平均加速度

R

θ

θ

)] )sin(

1 cos(

) ) cos(

[sin(

lim

) sin(

) sin(

cos )

cos(

lim sin

) sin(

) lim sin(

dx sin d

0 0

0

x x x

x x x

x

x x

x x

x

x

x x

x x

x x

x

∆ + ∆

∆

−

= ∆

∆

−

∆ +

= ∆

∆

−

∆

= +

→

∆

→

∆

→

∆

| | | |

x

= cos 0 1

Hw2：prove cos( ) 1 0

lim0 =

∆

−

∆

→

∆ x

x

x

Hw3：prove

y x y

x y

x

y x y

x y

x

sin sin cos

cos )

cos(

sin cos cos

sin )

sin(

−

= +

+

= +

22.Euler’s formula

θ

θ cos

θ

isin eⁱ = + pf：

θ θ

θ θ

θ θ

θ θ

cos y

sin

sin y

cos 1

) )(

(

* 1

2 2

2 2

=

=

=

= +

=

+

=

− +

=

=

−

= +

=

−

−

，

，

x or x

y x

y x iy x iy x e

e

iy x e

iy x e

i i

i i

Pick’s thm

0 sin 0

cos

0 1 i

eⁱ = = + So e^iθ = cos

θ

+isin

θ

1

θ

cos

θ

sin

θ

23.

) sin (cos

cos sin

sin cos

θ θ

θ θ θ

θ θ

θ θ

i i

d i de

i e

i i

+

= +

−

=

+

=

^θ

θ θ

^θ

θ

ⁱ

i

e i i

d

de = cos + sin = )

(

let i

θ

= x

x x x

x x

e x

d de dx

de dx e de

α α

α

α

α α

=

=

=

) 1 (

24.

) ( ) ( ) ( ) (

) ( )]

( ) (

lim [ )

( )]

( ) (

lim [

) ( ) ( ) (

) ( ) (

) ( ) (

) lim (

) ( ) ( ) ( ) lim (

) ( ) ( ) ( ) ( )]

( ) ( [

0 0

0 0

x g x f x g x f

x

x f x g x x g x

x x g x f x x f

x

x g x f x x g x f x x g x f x x g x x f

x

x g x f x g x x f

x g x f x g x f x

g x dx f

d

x x

x x

+ ′

= ′

∆

−

∆ + +

∆

∆ +

−

∆

= +

∆

−

∆ + +

∆ +

−

∆ +

∆

= +

∆

−

∆

= +

+ ′

= ′

→

∆

→

∆

→

∆

→

∆

其㆗ dx

x x df

f ( )

)

( ≡

′

25.Chain rule

let y = f (x)

Ex：

1.

x dx x

x dh

x x

h

α α

α

2

* ) ) cos(

(

) sin(

) (

2 2

=

=

2.

) ln 1 (

) 1 (ln

) ln ) (

( ) (

ln

ln

ln ln

x x

x e

x dx x

e d dx

x dh

e e

x x h

x x x

x x

x x x

x ^x

+

=

+

=

=

=

=

=

) ( )) ( (

) ( ) (

)* ( ) lim (

) ( ) lim (

) ( ) lim (

) (

) ( )) ( ) (

(

)) ( ( ) (

0 0

0

x f x f g

x f y g

x y y

y g y y g

x

y g y y g

x

x h x x h dx

x dh

x f x f dx g

x dh

x f g x h

x x

x

′

= ′

′

= ′

∆

∆

∆

−

∆

= +

∆

−

∆

= +

∆

−

∆

= +

′

= ′

=

→

∆

→

∆

→

∆

) ) (

( ) lim (

)* ( ) (

) ( ) (

) (

) (

) ( ) (

) (

)) (

( ) (

0 g y

y

y g y y g

x x

x f x x f

x f x x f y

y x

f

x x f y y

y g x h

y y g

x x f g x x h

x = ′

∆

−

∆ +

∆ ∆

−

∆

= +

−

∆ +

=

∆

∆ +

=

∆ +

=

∆ +

=

∆ +

=

∆ +

=

∆ +

→

∆

Hw4： A=(x₁，y₁)，B=(x₂，y₂)，C=(x₃，y₃)， prove that

∆ ABC 之面積為 _{1 2 2 3 3 1 2 1 3 2 1 3}

2

1 x y + x y + x y − x y − x y − x y

Hw5： ( 1 3 )

2

1 − + i

ω

= 求

ω

⁵¹ +

ω

⁵² +_L+

ω

²⁰⁰¹ = ? Hw6：1< x < 2

4 5 1

2 2 2

2

2

? x

log 2

log log

+

= 則

長 為㆒直角㆔角形之㆔邊

，

，

若 x x x

Hw7：

1. sin(α + β)sin(α − β) = sin ²α − sin² β = cos ² β −cos ² α 2. cos(α + β)cos(α − β) = cos ² α −sin ² β = cos ² β − sin ²α Hw8：prove that

1. ㆓倍角公式 θ θ

θ

θ θ 2sin cos

tan 1

tan 2 2

sin ₂ =

= +

θ θ θ

θ

θ ^{2 2} ₂²

tan 1

tan sin 1

cos 2

cos +

= −

−

=

2. ㆔倍角 sin 3θ = 3sinθ − 4sin³θ ；cos3

θ

= 4cos³

θ

−3cos

θ

半角

2 cos 1

cos 2 2

cos 1

sin

²

θ 2 θ

²

θ + θ

− =

=

Hw9：求 de ^x2 =? ( x+ x) =? dx

d dx

α

Hw10： prove that ₂

)]

( [

) ( ) ( )

( ) ] (

) (

) [ (

x g

x f x g x

g x f x

g x f dx

d ′

′ −

=