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Digital download I have achieved a first-class in Math by taking these notes. First-year uni math notes in SMA for those who struggle through the course or for uni applications interview. Hope can help you pass the exam Cheers












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Uploaded on
October 15, 2022
Number of pages
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Written in
2021/2022
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Lecture notes
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, Topic I
1. 1 Real numbers ,
Algebraic manipulations
Number field :




Real numbers ,
IR : All numbers I e.
go.li
2 it , } ,
- I - . .

I




Integers , 2 : positive ,
negative integers
and zero f- . .
-3 ,
-2 ,
-1
,
a , I ,
2,3 - . .


}

Non -



negative integers , IN :
positive integers
and zero
f o
,
I , 2,3 . - .




}
Operations :





Add att defined for
- subtract a- to all red a and I

at
Multiply





-

Division £ defined for all real a and b- ,
except
tota


Comparisons :




acts a > hi , ast A7T att
, , ,




shorthand :


"


a = a. a -
a - . .
a for all real a
,
HE IN and N 70
-
n times


. [ =/ for a f- 0

X x-

y
As when a -_ a
I = a

a× f- is undefined .




Ix I
?
=



× '
i. do is undefined yet

-




i. = a =
I


Summations
I -2
, proof by induction

Summations :



¥2 nlhtll 12h41 )
'

j =


Mtn ) 1+2+3-1 .tn ? ,
i. = . .

6

M In ) = 1+2+3-1 . . . th n
hlhtl )
I
Mint htlh 1) 1- In -21 -1 + I j =

2
j= ,
= -
.
. .




a. Min) = Intl ) -14+111-4+1 ) + . . . -1 ( htt )

n times

hlhtl )
i. Min ) =

2



z
Summation Notation :


II
.




flk )

fill + f- 121 + f- 13 ) + . . .
+ flkl is written as ,



.
More
generally we define , for a ,t integer acts

II. aflkl =
flat + flat 't + . .
.
+ fitt

properties
• :



of label
Independent
:
1




eflkl =
IIe till =
¥Ea f- I

' '

k l called
, , is a
dummy
hmm
index


② off
peel terms :




II.afoot =
flkl + fitt

=
flat -1 . .
.
+ fib -
It

3 We can shift the label :




II. f- 1kt =
II ,
fit - thi ) =
f Im -
a) = final + flattest . . .




+ f- latte -
a)
= flat -14111 . - .
=
f- 155-55) + fltb-551
+ f- 114 f- . -

tf 165-55)

, 22
j⇐
' '


't
'


Example pin ) I + + t n
j
=
• : = o . . .




works for
only
Proof by induction :
integers

Suppose we want to
prove that same result is true for all +
integers ,
h .
If it can be shown that the



result holds for n=I and also that ,
for all N71 , if the result is true for n = N , then it must also


be true for n = Ntl , then follows that the result is true for all
positive n



Inland )
:=€=a for
'
Qin )
k equal Intl )
negative integer
=

Prove that pin ) is to :
any non n
Question
-
.

:




i. = stands for a
definition
0
' z


when n =
0 pie) = [ok = a = a



Q a) =
to 1410+1110-11 ) = 0 i. 1710 ) =
Q lol i . True for h= 0


i. Pll) = QU )


True for h =/

Assume it is true for n=N

i. PIN ) = QIN )

when n= Ntl when u = Ntl

PINHI = ET bi
① 1µg =
LNHI 121*1+11 IN -11+1 )
k= I
'
6
= FINI t Will =
to IN -11 ) 12N -13 ) IN -14
'

=
QINI + IN -111
2

=
IN IZNHIINH ) 1- IN -111 i. PINT ) = QI Ntl)

= '
Ntl ) ( IN lsNH ) + IN-111 ) since true for n=o ,
then it is t ru e for all IN

=
INH ) I NL2N-1II -161N -111 )
=
to INH ) I ZN2-1TN -16 )
= to IN -11112 # 3) IN -12 )


General template
:



all E IN
Claim f- In )= gin ) for n



proof : .
show f- lot =
gio )
-

show that for n
any
i




if f- In ) =
gun) ,
then final =
glntl )



1.3 Binomial theorem

Binaural coefficients :




i.
Definition Factorial function : n ! = 1×2×3 × . . . X In 2)
-
X In -
1) X h ,
h E IN
ht
0 ! =
I In -11 ! =
n i . It -11 ! = = I = 0 !


Factorial function is
only defined for -heintegemm






Factorial function is never zero




Binomial coefficient
2.
Definition :



For all h .
k E IN , h 7k ,

-
h

(1) with)
I


Ck
'

n choose k =

!
=




> .
Pascal 's triangle :




.to/--lnn/-- I

n!
proof :
(2) =
d. in -4 ! =
I
h !
(1) =

h ! In -
n ) !
=
I

i. 141=111






properties of Pascal 's
triangle
:





symmetry :
121=11 )

, h !
proof :
II ) =

k :( n -
KI !


(Ik )
n !
=

i. in -
ntkl !
=

4-kl ! k !
= II )

② I 1) = III ) + 1hL ) ,
For all in > k -
the number is the sum of two numbers above

k > o


In-1 ) ! In-11 !

Proof :
RHS =

µ -11 ! In -1 htt ) !
-
X k ! In -
I -
kl !


In -11 ! In -

1) !
As In k ) ! (n k-11 ! In k )
=
+
-
-
=
1k -11 ! 1h kl ! k ! In tkl !
-


- -




In -11 ! In -11 !
= +

1k -11 ! In -
tell ! Intel klk-11 ! In k -11 ! -




kin -11 ! + In-11 ! In k ) -


=



klk-11 ! In k -11 !
-
In - k)

= h In -11 !

talk-11 ! In -
k -
1) ! In k -
)

!
%)
n
=

In )
=
I box hand
k ! - k
II
-
The little on the
right
'
of proof
'
side means end



Binaural Theorem :




i. Binaural Theorem :
For any non -


negative integer IN
, nza


-1121
"
171 't -111)a" th
"

't
" " " ' "

(att ) =
a + a a + . . .
t . .
. + to

=
If (1) a
"-
kfk
-
Binaural coefficient

Question :
Express (text 't IHX 15 as a
polynomial in ×

'
HX15 5) 1+5×+10×2 -110×7+5×4 -1×5 )
'

4- X ) + I =
(I -5×+10×2 - lax -15×4 -
+ + I


= 21 It 10×2+5×4 )

induction
proof by
:
:
a.




• when n = 0
,




II. "kt_
'


4th ) =
I
,
IL) a 111111111=1

True for h = 0



Just to be sure
,




when u =/


(att )
'
= att ,
LEO 1k ) atktk =
4) a
'

b-
°

+ 11 )a°t
'




=
att

i. True for n= I


Assume it is true for h=n


¥
k
1k ) and
"
i. lattt =

.
b-


sub n = h t I
n
""

(att ) = late ) ( att )
=
4th ) ¥411)a"ktk expand the bracket
get
¥4111 ⇐ 111 antebkti
"'
th
"
-




+

+In-n+IfI
= a



k
an "tk
k
>
LEO (1) 12,1 "zk
"
-
-
=
+
a
-
shift the variables

the same

€ 1L) and € 11,1 and"zk+
"" " k htt
↳ > =
a + b- + b-
,
2 ,
"'

EI,[ (1) II ) )
""
*
= a
"'
+ + ,
an thet

property ②
111=1111 -11nF )

) and"zk
'

£2
htt
= and +
,
I + b-


II 1h11 anti kzk
-


=
£6.39
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