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 1

1α PP K

K⋅⋅⋅⋅ αα2

2 then

( )1 . n 的正整數個數為(+1)( 2α +1)…( kα +1)

ex: n=P 1

1α , P0

1 | n , P1

1 | n, …P1

1

α

| n ⇒ ( 1α +1)個

( )2 .n 的正因數和

=(1+P1 +…+P 1

1α )(1+P2 +…+P 2

2α )…(1+PK +…+P K

Kα )

=∏= −

− +k

i i

i

PP i

1

)1

1(

1α

其 k

k

ii aaaa ⋅⋅⋅••=∏

=21

1

( )3 .n 的正整數乘積:

= ( )( ) ( )n k 11121

21 +⋅⋅⋅++ ααα = ( )nk

ii∏

=

+1

121 α

6.輾轉相除法:

設 a、b∈N, 若存在 q、r∈Z,使得 a=bq+r, br p≤0 ,

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 aaZZ 2121 =⇒= , bb = 21

( )2 .If ZZ 21+ = ( )aa 21+ + ( )bb 21+ i,

( )3 .If ZZ 21• = ( ) ( )bababbaa 12212121 ++− i,

( )4 .If baZ

Z2

2

2

22

1 1+

= 〔 ( ) ( )bababbaa 21122121 −++ i〕

( )5 . baZ −= i

8.㆒元㆓次方程式的解:

02 =++ cbxax , (a≠0 , a、b、c ∋ R)

a cabx +

+

2

2-ab4

2

2

2

+

abx = 24

1a

( )acb 42 −

x = a

acbb2

42 −±−

( )1 . acb 42 − > 0 ⇒ 相異兩實根.

( )2 . acb 42 − < 0 ⇒ 相等兩實根.

( )3 . acb 42 − = 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) 220 bar +=≤ <∞ , θ≤0 < 2π

(a,b) ≤0 sinθ≡rb≤ 1

x ≤0 cosθ≡ra≤ 1

∞− <tanθ≡ab <∞

rθ

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 ρ≤0 <∞ x=ρcosθ

θ≤0 < 2π y =ρsinθ

θ y ∞− < z <∞ z = z

x

( )3 .球座標(r、θ、φ)

x= rsinθcosφ

y= rsinθsinφ

y z= rcosθ

x

r≤0 <∞

θ≤0 <π

ϕ≤0 < 2π

ρθ

r

θ

φ

11.分點公式:

設 A( x1 , y1), B( x2 , y2

),則

( )1 . AB 之點座標

++

2,

22121 yyxx

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

++

++

nmmyny

nmmxnx 2121 ,

12.質心:

( )1 .若 A，B 兩點的質量均為 M,則質心為

++

2,

22121 yyxx

( )2 .若 A，B 兩點的質量比為 m : n,則質心為

++

++

nmnymy

nmnxmx 2121 ,

( )3 .若 ,, 21 PP …PN 的質量均為 M,則質心為

( ) ∑=

=++=N

iiNcm xxxxx NN 1

21

1...1

∑=

=N

iicm yy N 1

1

( )4 .若Pi 的質量為 im∆ , for i =1,2,3…N

∑∑ =

=

∆∆

=N

iiiN

ii

cmm

mxx

1

1

1

∑∑ =

=

∆∆

=N

iiiN

ii

cmm

myy

1

1

1

( )5 . N ∞→ , im∆ dm→ , MdmmN

ii =→∆ ∫∑

=1

∫Ω

= xdmMxcm1 Ω為其分布空間

∫Ω

= ydmMycm1

13.微分與積分

( )1 .微分:

∆ s = 12 ss −

∆ t = 12 tt −

11,ts 平均速度: v=ts

∆∆

順時速度: dtds

tsv

t=

∆∆

=→∆

lim0

( )2 .積分:

F = F(x)

作功 W= ( )dxxFx

x∫'

= ( ) ii

N

iNxxF ∆∑

=∞→ 1

lim

x 'x ix∆ =Nxx −'

, ( ) ( ) ( )( )121

++= iii xFxFxF

14.微分運算:

f(x)

( ) ( ) ( )x

xfxxfdxxdf

x ∆−∆+

=→∆ 0

lim

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

dxdx =

0lim→∆x x

xxx∆

−∆+ =1

( )2 . f(x) = 2x , ( )dxxdf =

dxdx2

=0

lim→∆x

( )x

xxx∆

−∆+ 22

=0

lim→∆x

( )xxxx

∆∆+∆ 22 =

0lim→∆x

2x+ x∆ =2x

( )3 . f(x)= nx , ( )dxxdf = n 1−nx

22 ,ts

15.積分運算:

F(x)=x , ( )dxxFx

x∫2

1 = dxx

x

x∫2

1 = 2

21 x | 2

1

xx = ( )xx 2

1

2

221

−

F( 2X ) ( )dxxFx

x∫2

1

F( 1X ) = ( ) ( ) ( )( )121221 xFxFxx +− = ( )xx 2

1

2

221

−

1X 2X

F(x)=e ax2−

( )2

∫

∞

∞−dxxF = dxdyee ayax

∫ ∫∞

∞−

∞

∞−

−− 22

= ( )dxdye yxa∫ ∫∞

∞−

∞

∞−

+− 22 22 yx + = 2r ,dx dy = r dr dθ

= θπ

rdrde ar∫ ∫∞ −

0

2

0

2

=π 2

0

2

dre ar∫∞ − t= a r 2

=aπ dte t∫∞ −

0

=aπ (-e t− )| ∞0 =

aπ

22

dxe ax∫∞

∞−

− = aπ

16.極座標積分

柱座標積分

球座標積分

θ dθ

R rdθ

dr

dA rd drθ=

2

0 0

2

02

12 22

R

R

A dA

rd dr

rdr R

R

πθ

π π

π

=

=

= =

=

∫

∫ ∫

∫

L

0

2

0 0 02

L R

dV rd drdz

V dV

rd drdz

L R

π

θ

θ

π

=

=

=

=

∫

∫ ∫ ∫

θ

rsinθ dφ rdθ

dr

2

2 20 0 0

20 0

20

3

( sin )( )( )

sin

sin

2 sin

4

43

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

=

=

=

=

=

=

=

∫

∫ ∫ ∫

∫ ∫

∫01

1

sin

cos 2d

πθ

θ−

= − =

∫

∫

Q

17.向量分析

z

zyx AkAjAi

zkjyxiA^^^

1

^^

11

^

++=

++=→

y 1z11 zAyAxA yx === ，，

θ

φθφθ

cos

sinsincossin

)()()(

AA

AAAA

BAkBAjBAiBA

z

y

x

zzyyzz

=

==

±+±+±=±→→→→→

scalar product

θcosABBABABABA zzyyxx =++=•→→

θcos2

2

)()(

222 ABBAC

BABBAA

BABACC

BAC

++=

•+•+•=

+•+=•

+=

→→→→→→

→→→→→→

→→→

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→ → →

→ → →

→ → →

→ →

= × ≡

= + +

= − + − + −

= − ×

sinr F

rFτ

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

，

ijkε =

^

^ ^ ^3 1 21 2 2 3 3 1

^ ^ ^2 3 11 3 2 1 3 2

^ ^ ^3 1 21 2 2 1 2 3 3 2 3 1 1 3

, , 1, 2,3

) ( ) ( )

kijk i jA 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 mlkm l ijk i j lkm ijk l i j

mli mj lj mi l i j

i jj 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 kx y z

i j kx y zφ φ φφ

∂ ∂ ∂∇ = + +

∂ ∂ ∂

∂ ∂ ∂∇ = + +

∂ ∂ ∂

ex

( , , )f f r θ φ= chain rule

所以

^ ^ ^

^

1 ( )( ) ( )

( )

d f rf r i x j y k zr d r

d f rrdr

∇ = + +

=

2 2 2

^ ^ ^

( )( ) ( ) ( )( )

( ) ( ) *

( ) 1 2( )2

( )*

r x y zf f r

f r f r f rf r i j kx y z

f r df r rx dr xdf r xdr rx df rr dr

= + +

=

∂ ∂ ∂∇ = + +

∂ ∂ ∂∂ ∂

=∂ ∂

=

=

f f r f fx r x x xdf rdr x

θ φθ φ

∂ ∂ ∂ ∂ ∂ ∂ ∂= + +

∂ ∂ ∂ ∂ ∂ ∂ ∂∂

=∂

|| 0

|| 0

20.Circular motion

Rvaa

2

0lim ==

−

→θ向心加速度

hw1：prove 1sinlim0

=→ θ

θθ

21.Simple harmonic oscillation

0 0

cos cos( )sin sin( )x

x R R tv v v t

t

θ ωθ ω

θ ω

= == − = −=

ω：角頻率(速度)dtdθω =

0 ( )

cos( ) cos( )x

ds d dv R R R Rdt dt dt

dx d dv R t R tdt dt dt

θω θ ω

ω ω

= = = = =

= = =

θ θ R

θ

θ

1 2

2 1

2 1

2 1

2 1

2

2 sin2

2 sin sin2

v

v vat t

v v vRt tv

v va R R

θθ

θ θθ θ

−

−

−=

−− =

− =

= =

平均加速度

R

θ

θ

])sin()cos(1)cos()[sin(lim

)sin()sin(cos)cos(sinlim

)sin()sin(limsindxd

0

0

0

xxx

xxx

xxxxxx

xxxxx

x

x

x

∆∆

+∆

−∆=

∆−∆+∆

=

∆−∆+

=

→∆

→∆

→∆

| | | |

xcos= 0 1

Hw2：prove 01)cos(lim0

=∆

−∆→∆ x

xx

Hw3：prove yxyxyxyxyxyx

sinsincoscos)cos(sincoscossin)sin(

−=++=+

22.Euler’s formula

θθθ sincos iei +=

pf：

θθ

θθ

θθ

θ

θ

cosysin

sinycos1

))((*122

22

==

==+=

+=−+==

−=

+=

−

−

，

，

xorx

yx

yxiyxiyxee

iyxe

iyxe

ii

i

i

Pick’s thm

0sin0cos10 iei +==

So θθθ sincos iei +=

1

θ

cosθ

sinθ

23.

)sin(coscossin

sincos

θθθθθ

θθθ

θ

iiidde

iei

i

+=+−=

+=

θθ

θθθ

ii

eiidde

=+= sincos)(

let xi =θ

x

xx

xx

exd

dedxde

edxde

ααα

αα

α

==

=

)(1

24.

)()()()(

)()]()([lim)()]()([lim

)()()()()()()()(lim

)()()()(lim

)()()()()]()([

00

0

0

xgxfxgxfx

xfxgxxgx

xxgxfxxfx

xgxfxxgxfxxgxfxxgxxfx

xgxfxgxxf

xgxfxgxfxgxfdxd

xx

x

x

′+′=∆−∆+

+∆

∆+−∆+=

∆−∆++∆+−∆+∆+

=

∆−∆+

=

′+′=

→∆→∆

→∆

→∆

其 dxxdfxf )()( ≡′

25.Chain rule

let )(xfy =

Ex：

1.

xx

dxxdh

xxh

αα

α

2*)cos()()sin()(

2

2

=

=

2.

)ln1(

)1(ln

)ln()()(

ln

ln

lnln

xx

xe

xxdxde

dxxdh

eexxh

x

xx

xx

xxxx x

+=

+=

=

===

)())(()()(

*)()(lim

)()(lim

)()(lim)(

)())(()())(()(

0

0

0

xfxfgxfyg

xy

yygyyg

xygyygx

xhxxhdxxdh

xfxfgdxxdh

xfgxh

x

x

x

′′=

′′=∆∆

∆−∆+

=

∆−∆+

=

∆−∆+

=

′′=

=

→∆

→∆

→∆

)()()(lim

*)()()()(

)()(

)()()(

))(()(

0yg

yygyyg

xx

xfxxfxfxxfy

yxfxxfyy

ygxhyyg

xxfgxxh

x′=

∆−∆+

∆∆

−∆+=

−∆+=∆∆+=

∆+=∆+=

∆+=∆+=∆+

→∆

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

∆ABC 之面積為 31231213322121 yxyxyxyxyxyx −−−++

Hw5： )31(21 i+−=ω 求 ?20015251 =+++ ωωω L

Hw6： 21 << x

451

2222

2 ?x

log2loglog+

=則

長為㆒直角㆔角形之㆔邊，，若 xxx

Hw7：

1. αββαβαβα 2222 coscossinsin))sin(sin( −=−=−+

2. αββαβαβα 2222 sincossincos)cos()cos( −=−=−+

Hw8：prove that

1. ㆓倍角公式 θθθ

θθ cossin2tan1tan22sin 2 =

+=

θθθθθ 2

222

tan1tan1sincos2cos

+−

=−=

2. ㆔倍角 θθθ 3sin4sin33sin −= ； θθθ cos3cos43cos 3 −=

半角 2cos1

2cos

2cos1

2sin 22 θθθθ +

=−

=

Hw9：求 ?)(?de2x

=+= xxdxd

dx

α

Hw10： prove that 2)]([)()(

)()(]

)()([

xgxfxg

xgxf

xgxf

dxd ′

−′

=