Assignment I

Name : Isnaini Nurul Noviana
NIM : 07305141025

CHAPTER I

INDUCTION OF MATHEMATICS AND BINOMIAL THEOREM

A. Induction of mathematics

Induction of mathematics is one of the proof method from many theorems in theory of number or in other mathematics. Whereas binomial theorem, besides as basic it used to derivation some binomial theorem and solution of mathematics problem. Because of that, the capable of this skill is necessary to who will learn mathematics, because there many treatises in mathematics which use that principal to derivate the theorem or to solve problem. Most of each next treatise use this both principal, either to prove the theorem or to solve the problems.

Induction of mathematics is one of the proof argumentation of a theorem or mathematics statement which whole object is set of integer number or especially set of original number. Pay attention to example mathematics statements bellow.

Example 1.1
1+2+3+…+n= ½ n(n+1) for each original number of n. Is this statement true? For replied this question, we can try to substitute n on this statement into any original number. If n = 1, then that statement become 1= ½ .1(1+1), or 1 = 1, that is got a true statement. If n = 2, then that statement become 1+2= ½ .2(2+1), , or 3 = 3, that is got a true statement. If n = 3, then that statement become 1+2+3= ½ .3(3+1), , or 6 = 6, that is got a true statement too.

The reader continueable to n = 4; 5; or other original number and usually will get true statement. Is with give some example with substitute some of original number into n from origin statement and got true statements can give proof about truth of this statement?

In mathematics, gift some example like that, is not proof of truth of statement which can used in the whole set. The statement in example above, the whole of set is set of all of original number. If we give an example for each original number of n in this statement and get true statement for each number, then that is as truth of proof from that statements.

But this is not efisien and imposible to do, because the element of set of original number unfinite. So, how to prove that statement? One of the way is view the first part from this statement as arithmetics series with first component a = 1, the difference b = 1, final component is Un= n and has n component. So the sum of this series is
Sn= ½ n(a+Un)
= ½ n(1+n)
= ½ n(n+1), that is second part from proved statement.

Representing the video of learning mathematics.

Representing the video of learning mathematics.
A. Solving Problem Graph Math
Problem :
13. The figure above shows the graph of . If the function h is defined by , what is the value of ?
Answer :
We are looking for

3
1
2




(substitute , in the function); for look at the graph.


So, the solution of
13. Let the function f be defined by , if . What is the value of ?
Answer :
Looking for , what is f when ?
The information is








Substitute to

So the answer of is 28.
17. In the xy-coordinate plane the graph of intersects line l at (0,p) and (5,t). What is the greatest possible value of the slope of l ?
Answer :
Looking for is greatest m


Line l




Line l :

x y
0 p
5 t

B. Factoring Polynomials
Algebraic long division
For example :
Is a factor of
make along division problem for mathematics school



- bringing down

-

-
0
no remainder
then
is a factor of

Is also a factor of
= ( )( ) can be foctor into
= ( ) , sharing the factor for the equation of
0 = ( )
Thus ever solving all equations , we get
= 0
Or

Or

The roots of are 3, -1, -2.

3 roots for this 3rd degree equation quadratic (2nd degree) equations always have at most 2 roots.
A 4th degree equation would have 4 or fewer roots and so on.
The degree of a polynomial equations always limits the number of roots.

Long division for a 3rd order polynomial =
1 Find a partial quotient of by dividing x into to get .
2 Multiply by the divisor and substract the product from the dividend.
3 Repeat the process until you either “clear it out” or reach a remainder.

Section D
Properties of polynomial graphs
Polynomial have even or odd degrees.

C. Pre-Calculus
Graph of a rational function
Can have discontinuities, because has a polynomial in denominator
For example :

When the function become
imposible
For this function choosing is a bad idea.
break in function graph
Graph
Insert x = 0

(0,-2)
Insert x = 1

that is imposible
x = 1

D.
Discontinuity







BREAK






Rational function don’t always this way
Not all rational functions will give zero in denominator.



never zero, because of the + 1


no break







Don’t forget
Rasional functions denominator can be zero!


For polynomial the graph is a smooth unbroken curve
Rational functions
x zero in thr denominator is apossible situation, there is no value for the functions so break in the graph

break 2 ways
missing point in the graph
for example
the graph like this

HOW TO EXPRESS MATHEMATICS

A. Empowering : give somebody the power or authority to do something
- Indonesia will be empowering education quality.

B. Pursuing : do something to try to achieve something over a period of time, continue to discuss or involved in something, follow or chase somebody/ something in order to catch them
- Education National Department was pursuing KTSP curriculum.

C. Pengayaan : enrichment, adding material
- Enrichment lesson is not always given.

D. Persamaan garis singgung : tangent equation
- Tangent equation is undefined.