HOW DIFFICULT TO COMMUNICATE MATHEMATICS

HOW DIFFICULT TO COMMUNICATE MATHEMATICS
ABOUT POLYNOMIALS IN ENGLISH

Date : Sunday, January 5th 2009
Place : Ardhita Kirana Rukmi’s Home

It’s difficult to communicate mathematics language to the other. That is not as easy as read or understand any topic in mathematics. We have to be smart to find the right words, not so hard word, so that others can understand what we said. The difficulties to communicate this mathematics language become the challenge for university students of Mathematics in Yogyakarta State University to complete an English assignment lectured by Dr. Marsigit, M. A.. We have to communicate any topic about mathematics, absolutely in English, in pairs. I am with Ardhita Kirana Rukmi.
I take The Polynomials from mathematics book of 11th grade Senior High School, second semester, published by Erlangga.
Firstly, I give her the definitions of the polynomials. Then, I show her an example of the polynomials write. She seems not in trouble in understanding what I said. So, I move to give her several questions about the definitions of the polynomials as follows:
Mention the variable, the degree, and the coefficients for each following polynomials:
a. (4+3t-2t^2+t^3+10t^4-2t^5)
b. (2x^3+5x^2-10x+7)
c. (b^8-6b^7+5b^6-16)
She can do it well. I told her the next subsection about the value of the polynomials after the examples. There are two methods to find the value of the polynomial in this subsection. First method is the substitution and the second is the scheme method. From this two methods, Ardhita like the first method better. She said that she always used the substitution method since the first time she got it to find the value of the polynomials. Here’s the example:
Determine the value of the polynomial f(x)=x^3+3x^2-x+5 if x is replaced by x=m (m element of R).
Answer: For x=m we obtain f(m)=m^3+3m^2-m+5.
I get a little bit difficulties for the scheme method because the explanation from the book uses an alphabet so it seems so abstract. I told Ardhita step by step on scheme method. Unfortunately, she’s not so understands about what I give. It maybe caused by my language that difficult to understand. But, she said that she had understands after I show the picture of the scheme.
I only take the subject matter until the second subsection. I give the exercise after finishing explaining her to see how far she knows about the polynomials. I give her three questions:
1. With the substitution method, find f(1,y) if f(x,y)=x^3y^4+xy^3+y+2x^2+3
2. Find f(2,-1) if f(x,y)=x^2y+y^2-2x+10 with the substitution method
3. Find the value of the polynomials from number 1 and two with the scheme method
She did the two first questions well. For the last question using the scheme method she needs a little longer time than the first.
From this activity I can conclude that Ardhita understanding about the polynomials better than before. The first reason is that she had learned the lesson about the polynomials since junior high school. The second reason is the definition of the polynomial reminding her to the very beginning we learned it. Or on the words it will remind all of the things that maybe passed by her.
Scheme method rarely used by Ardhita to solve the problems. She keen on the substitution method better because this method always she use. It becomes one problem to understand the scheme method quickly. She said that she uses the scheme method in factoring polynomials, not to find the value of the polynomials.
The main challenge for me is in communicate this polynomials topic in English. A little number of mathematics vocabularies is the problem that makes our communication heard not so smooth. Mathematics term maybe rarely heard. But it must be an ordinary thing for mathematics student like me.
Other problem that I face is translating this matter. I take it from the book in Indonesian. There are several term that not existing in the dictionary. I ask to my friend for sometimes or browse from the internet to handle it. The problems can be cleared then.
This activity is so good and gives so many advantages for student to know how far our communication capability in English about mathematics. I motivated to be more diligent to learn and learn, English especially, because nowadays we have to being a part of an international society from the globalization.

THE VIDEOS

1. Pre Calculus Graph
Let’s begin by discussing the graph of a rational function which can have discontinuities. A rational function has a polynomial in the denominator which means you are dividing by something that is valuable quantity.
It’s possible that some value of x will meet to division be zero. Example:
If f(x)=(x+2)/(x-1), when x=1, the function value become f(x)=(1+2)/(1-1)=3/0 with 0 in the denominator. For this function, choosing x=1 is a bad idea.
When there is a bad choice of x when it makes the bottom of the rational function, it shows up a break in a function graph. For example, suppose you to finding the graph f(x)=(x+2)/(x-1). Start with inserting 0 for x. So now we have f(0)=(0+2)/(0-1)=2/(-1)=-1. So, you put off point down on the graph at (0,-2).

Next you try x=1. This time you get f(1)=(1+2)/(1-1)=3/0. That is you know is impossible and it means that you can not compute y-value when x=1.
It also mean that the graph of this function will not have any point for x=1. The graph is separated into two disconnected pieces at x=1.

Take the graph of f(x)=1/(x^2+1). No matter what numbers you choose for x, the denominator will never zero. The graph is smooth and unbroken.

Don’t forget that the general rule on the rational function, you must expect the possibility that the denominator may turn out to be zero.
Review: For polynomials, the graph is a smooth unbroken curve. For a rational function, sometimes the value of x may be zero in the denominator. That is an impossible situation because there is no y for that x. At that point, there is no value for the function and there is a break in the graph.
A break can show up in two ways. The simpler type of a break is just a missing point. This function is an example of this type of break: y=(x^2-x-6)/(x-3).

The gate that appear in the graph is at points where x=3. If you try to substitute 3 for x in the equation, the result is y=(3^2-3-6)/(3-3)=0/0. That is not possible, not feasible, and not allowed. So, there is no y for x=3. That is the typical example of the missing point syndrome and conveniently it always goes with the result like 0/0.
When you see the result of 0/0, it also tells you that it should be possible to factor the top and the bottom of the rational function and simplify. For an example: y=(x^2-x-6)/(x-3)=(x-3)(x-2)/(x-3)=(x+2).
For this kind of the break, the missing point is a loop hole. For the original function without simplify, x=3 is a bad point because it leads to the division be zero, y=0/0. But if you simplify first, then there’s no problem with x=3, y=x+2. It can be one idea or key in calculus.
Removable singularity appears simplest missing points on the graph when x leads to 0/0. For this kind of a break, if you factor and simplify the rational function, the division by zero can be avoided.

2. Kata Kerja

Kata kerja menunjukan sebuah kegiatan atau untuk menjelaskan sebuah kejadian. Tentang apa yang sedang dilakukan oleh suatu kalimat.
Dalam sebuah kalimat sederhana:
Dave berlari
Kata kerja
Karena berlari menunjukan apa yang sedang dilakukan Dave.
Dalam tata Bahasa Inggris, perubahan bentuk kata kerja menunjukan siapa yang sedang melakukan kegiatan tersebut.
I do, you do, he does, we do, they do.
Contoh : Dave runs, tetapi kita mengatakan I run
Kata kerja berubah karena orang lain yang nelakukan kegiatan tersebut
I, You, We, They He, She, it

run runs
kata kerja - to be
I am
You are singular sbyek/kata ganti tunggal, diikuti kata kerja tunggal
She is
It is
We are
They are kata ganti jamak, diikuti kata kerja jamak
Contoh :
Nyonya Midori adalah kata ganti tunggal, jadi menggunakan :
Mrs Midori Yodels bentuk yang digunakan untuk kata ganti tunggal.
Saudara-saudara Midori seperti Else, Gretel, Heidi. Mereka adalah kata ganti jamak, maka menggunakan kata kerja jamak :
Midori’s sister yodel.


3. Basic Trigonometry
Trigonometry (from g reek, trigon and metron). Trigonometry is relly study of rectangle and the relationship between the side and the angle of rectangle.

Function
Sin Ǿ = ?
cos Ǿ = ?
tan Ǿ = ?
to solve them, use triy function soh, cah, toa.
Soh : sine is opposite over hypotenuse
Cah : cosine is adjacent over hypotenuse
Toa : tangent is opposite over adjacent

- Sin Ǿ = 4/5 the other trigfunctions : 1. Cosecant Ǿ

- Cos Ǿ = 3/5 2. Secant Ǿ

- Tan Ǿ = 4/3 3. Catangent Ǿ

If the angle is x, so tan x = ¾ (the invers of tan Ǿ)

4. Kalimat Majemuk

Contoh :
Klausa 1 : It’s the end of the world as we know it, and
Klausa 2 : I feel fine
Ada 2 klausa yang dihubungkan dengan 1 kata penghubung, yaitu “and” ketika satu kalimat digunakan sebagai bagian dalam kelompok yang lebih besar, kalimat yang lebih kecil disebut klausa.
Ketika sebuah klausa dpaat berdiri sendiri dalam sebuah kalimat, maka disebut “independent clause”. Jika memiliki 2 klausa, maka disebut dengan kalimat majemuk. Untuk menggabungkan 2 independent klausa menggunakan tanda titik dua (:), klausa yang kedua menjelaskan klausa yang pertama.
- I love my two sisters. They bake me pie.
Untuk menggabungkan kedua kalimat tersebut, menggunakan tanda titik dua
- I Love my two sister : they bake me pie
• Tanda titik koma (;)
Contoh kalimat :
It’s the end of the world as we know it and I feel fine
Dapat disingkat menjadi :
It’s the end of the world as we know it, I Feel fine.
• Garis penghubung (-)
Terdiri dari beberapa elemen yang mengejutkan. Kita menggunakan garis penghubung karena klausa kedua dihubungkan dengan klausa pertama
Untuk menghubungkan kalimat terdapat 4 cara :
- Kata penguhubung
- Titik dua
- Titik koma
- Garis penghubung
• Kalimat Fragmen
Jika kita mendapati suatu kalimat tidak dapat berdiri sendiri sebagai kalimat lengkap. Contoh kalimat lengkap:
My pet Komodo dragon is a lamb
Contoh kalimat fragmen :
Because he has no teeth
Klausa dependent : - tidak bisa berdiri sendiri
- klausa dependent terdapat pada klausa independent
- bukan kalimat lengkap
Contoh :
Although Tom sleeps regulary, he is constantly tired
Klausa dependent klausa independent
Kalimat diatas disebut kalimat kompleks.

5. Limit By Inspection

There are two conditions :
1. X goes to positive or negative infinity
2. Limit involves a polynomial divide by a polynomial
For example :
Lim x3 + 4 = ~
x2 + x+1
X→ ~
(Because highest power of x in number is 3 greater than highest power in denominator) since all the number are positive and x is positive infinity
The key to defermining limits by inspection is in looking at powers of x in the numerator ana the denominator. To apply these rules : - must be divining by polynomial
- x has to be approaching infinity

1. First shortcut rule
If the highest power of x in numerator is 3 greather than highest power in denominator) since all the number are positive and x is going to positive infinity if you can’t tell it the answer is positive or negative:
- substitute a large number for x
- see if you end up with a positive or negative number
- whatever sign you get is the sign of infinity for the limit.

2. Second shortcut rule:
If the highest power of x is in the denominator, then limit is zero
Lim x2 + 3
x → ~ x3 + I , x2 + 3 = highest power of x in numerator is 2
x3 + 1 = highest power of x in denominator is 3

3. Last shortcut rule
Used when : highest power of x in numerator is same as highest power of x in denominator.
Lim = the quotient on the efficient on the two highest powers
x → ~
-Remember : coefisient → the number that goes with a variable
Ex : 2 is the coefficient of 2x2
75 is the coefficient of 75 x 4
Sho that is no way if x = 3

When you see the result and also tell you direction be possible factor top and bottom of rasional function and simplify

6. Trig Function

Trig function : really study of right triangle

To remember you can use : s o h c a h t o a
i p y o d y a p d
n p p s j p n p j
e o o i a g
s t n c e
i e e e n
t n n t
u t
s
e
Trig Function
- ratios of different sides of triangles
- With respect to an angle
- Only need to know values of sides to find measure of an angles figure of all part of triangle
Trig function : 1. Sine 4. Cosecant
2. cosine 5. Secant
3. tangent 6. cotangent


Six basis trig function defined by :
1. Sides of a triangle
2. angle being measured
opp : side opposite to theta
adj : side adjacent to theta
hyp : hypotenuse
Six trig function : 1. Sin Ө 3. Tan Ө 5. secӨ
2. cos Ө 4. Csc Ө 6. Cot Ө
Trid shortcut

According to this rule :
Lim → coefficients of x³ is over cach other
Lim
x 4

7. Kalimat Sederhana

Kebanyakan tipe kalimat adalah kalimat sederhana, sederhana karena semua elemen didalam kalimat adalah bagian dari subjek dan predikat. Subjek menunjukan kegiatan dari kata kerja utama. Subjek sederhana adalah kata benda khusus yang menunjukan sebuah kegiatan.
The happy little child kicked the gnome over the fence
Subyek
Predikat dari sebuah kalimat terdiri dari : main verb kata kerja utama dan apapun yang mengikutinya. Gabungan keduanya disebut predikat yang lengkap. Contoh :
The happy little child kicked the gnme over the fence predikat yang lengkap.
Karena “kecked the gnome over the fence” menunjukan tentang apa yang ditendang dan bagaimana tendangan itu secara lebih jelas lag.
- Kalimat sederhana dapat diperoleh tanpa subjek dan predikat.
- Kalimat perintah adalah kalimat yang ditunjukan langsung kepada orang kedua, yaitu “kamu”, berfungsi untuk memerintahkan seseorang agar melakukan sesuatu.
Kick that gnome over the fence
- Tidak ada subjek
- Siapa yang sebenarnya melakukan kegiatan tersebut?”kamu”
Bisa ditulis : “Hey you, kick that gnome over the fence.”
Tidak perlu ditulis karena sudah tersirat

POLYNOMIALS

POLYNOMIALS

Subject:
5-1 Definitions of polynomials, the value of the polynomials, and polynomials operations
5-2 The division of polynomials
5-3 Remaining Theorem
5-4 Factoring theorem
5-5 Polynomials Equation solution

Algebraic form had been learned at junior High School which is discussing about the definition of a term, factor, coefficient, and constant. Monomial, binomial, and polynomial in same or different variable, solving the operations of the polynomial, solving the division of the same term, and factoring algebraic terms had been studied too. That matter will be discussed again deeper, then will be developed to the division, the polynomials, remaining theorem, and the roots of the polynomials.
The basic competent of the “Polynomial” is using the division algorithm, remaining theorem, and factoring theorem on problem solving Realization of the competent will be show through study result : using polynomials division algorithm to decide the division result and division remain, using the theorem and the factor to solve the problems, and proof the remaining and the factoring theorem.
To support the success of the competent in this chapter, the indicators of the success are, the student can:
- explain the polynomials division algorithm
- decide the degree of division result polynomials remain by linear or quadratic form
- determine the division result and division remain by linear or quadratic form
- determine the polynomials division remain by linear or quadratic form using the remaining theorem
- determine the linear factor of the polynomials using factoring theorem
- proof the remaining theorem and the factoring theorem

5-1 Definitions of polynomials, the value of the polynomials, and polynomials operations

5-1-1 Definition
Look at there algebraic form:
(i) x2-3x+4
(ii) 4x3 + x2-16x+2
(iii) X4+3x3-12x2+10x+5
(iv) 2x5-10x4+2x3+3x2+15x-6
Those algebraic forms are called polynomials in x.
The degree/ exponent of the polynomial in x decided by the highest exponent of x.
For example:
(i) x2-3x+4 is a second degree polynomial because the highest exponent of x is 2
(ii) 4x3 + x2-16x+2 is a third degree polynomial because the highest exponent of x is 3
(iii) X4+3x3-12x2+10x+5 is a fourth degree polynomial because the highest exponent of x is 4
(iv) 2x5-10x4+2x3+3x2+15x-6 is a fifth degree polynomial because the highest exponent of x is 5
So, the polynomial’s degree in x, generally wrote as:

anxn + an-1xn-1 + an-2xn-2+…+a2x2 + a1x + a0

Where:
- an, an-1, an-2, …, a2, a1, a0, are real number with an  0. an is the coefficient of xn, an-1 is the confident of xn-1, an-2 is the coefficient of xn-2, …, and soon, a0 called as a constant.
- n is a counting number which represent the degree of the polynomials
the terms of the polynomials above are beginning by the term that the variable has he highest power ansn, then followed by the terms with decreasing power of x, an-1xn-1, an-2xn-2, …, a2x2, a1x1, and ended by the constant 90. The polynomials written on that way called arranged in decreasing power rule on variable of x. Remember that the variable must not be x, but can be a,b,c…, y and z.
For illustration, look at these polynomials:
a) 6x2-3x2+10x+4 is a third degree polynomial in x. The coefficient of x3 is 6, for x2 15-3, for x is 10, and the constant is 4.
b) 9y4-y3+5y2-2y-13 is a fourth degree polynomial in x. The coefficient of y4 is 9, for y3 is -1, for x is 5, for y is-2, and the constant is -13.
c) (x-1)(x+1)2=x3+x2-x-1 is a third degree polynomial in x. The coefficient of x3 is 1, for x is -1, and the constant is -1.
d) (t+1)2(t-2)(t+3)=t4+3t3-3t2-11t-6 is a fourth degree polynomial in t. The coefficient of t4 is 1, for t3 is 3, for t3 is -3, for t is-11, and the constant is -6.
Polynomials above are polynomials in one variable called a univariable polynomials. The polynomials in more than one variable called a multivariable polynomials. For illustration, look at following polynomials.
a) x3+x2y4-4x+3y2-10 is a polynomial in two variable x and y. this is a third degree polynomial in x or fourth degree polynomial in y
b) a3+b3+c3+3ab+3ac-bc+8 is a polynomial in three variables a, b, and c. this is a third degree polynomial in a, b, or c.

THE DEFINITION OF THE POLYNOMIALS

EXERCISE 1
1. Mention the variable, the degree, and the coefficient for each following polynomials:
a) 2x3+5x2-10x+7
b) x3-4x+2
c) 10-2y
d) 4+3y-5y2
e) 4a7-3a2+3a-10
f) 3-2a5
g) p4+2p-5
h) 4+3t-2t2+t3+10t4-2t5
2. Repeat question number 1 for following polynomials:
a) 3x4+10x3-5x2+ ¼ x+4
b) 6x7-4x6+2x5-10x4+x3-4x2-5x+8
c) 4-y+3y2-5y3+10y4-y5
d) 3+y-10y2+3y3
e) b8-6b7+5b6-16
f) 4+2b2+3b3-5b4+10b5+6b6-9b+2b10
g) q12+4q8-q6+q4-5q2+q+6
h) 10-q+2q3+4q5- 7q7+10 q9+7 q11+ q13
i) 6-4s+s2-10s3+6s7+s16+3s15
j) z20-4z18+5z16-13z14+4z12+10z7-5z2-z+3
3. Determine the number of variables for each polynomial, then determine the degree (according to the variable)
a) x5y+xy3+4x-5y+12
b) a5b5-a4b3+a3b3+a+b2-6
c) (2x-y)4-(2y+3z)3+(2z-x)6
d) P4+q4+r4+3pq-5pr+6qr+4
4. Repeat question number 3 for following polynomials:
a) x5y5-2x4y3+3x2y2-10xy2-4x+3y-10
b) a7b6-a6b5-6a5b3-8a4+3a2b+5ab2+5a2-6b2+10
c) (2x-y)3-4(y+3z)4+(2z-x)2
d) 3a5+2b5-5c5+3a4-b4+3c4+2abc+4ab-8c+9
5. Multiply these polynomial product (state the result in decreasing power rule then, determine the degree and coefficients
a) (x-4)(x+2)
b) (x2-1)(x+4)
c) (y2-3)(y2+3)
d) (y2-4)(y2+2y+1)
e) (a2+2)(a-4)2
f) (a+4)2(2a-1)
g) (b-1)2(b-2)2
h) (2b-1)2(2-b)2
6. Find the coefficient of the following statements:
a) x2 in (x-1)2(x+2)(x+1)
b) y3 in (2y+y2)(4y2-2y+1)
c) z in z(z-1)(z-2)(2+3)
d) t4 in (t2+2t-1)3


5-1-2 The value of the polynomials
The polynomials can be written in the form of function of the variable, based on the fact that polynomial is an algebraic form in a variable. Polynomials in x can be written as a function of x. for example, general form of the polynomials can be state in a function form as:

f(x)=anxn+ an-1xn-1+ an-2xn-2+…+a2x2+a1x+a0

Notes:
Polynomial function above is state in f (x), sometimes state as:
- S(x) that shows polynomial function in x, or
- P(x) that shows polynomial function in x

Denote the polynomials as a function in x, then the polynomials value can be determined. Generally, polynomials value of f(x) for x = k is f(k) where k is a real number. Then, the value of f(k) can be found with two methods:

Substitution method, or

Scheme method

A. Substitution Method
To explain the way to find the polynomials value with substitution, look at this third degree polynomial in x:
f(x) = x3+3x2-5x+2
- The value of f(x) for x =-1 is:
f(-1) = (-1)3+3 (-1)2-5(-1)+2=-1+3+5+2=9
- The value of f(x) for x =0 is:
f(0) = (0)3+3 (0)2-5(0)+2=0+0-0+2=2
- The value of f(x) for x =1 is:
f(1) = (1)3+3 (1)2-5(1)+2=1+3-5+2=1
- The value of f(x) for x =2 is:
f(2)= (2)3+3 (2)2-5(2)+2=8+12-10+2=12

Based on the description above, the value of the polynomials for a certain variable can be found by the rule of substation method as follows:
The polynomials value of f(x)=anxn+ an-1xn-1+ an-2xn-2+…+a2x2+a1x+a0 for x = k
(k  real numbers) determined by:
f(x)=an(k)n+ an-1(k)n-1+ an-2(k)n-2+…+a2 (k)2+a1 (k)+a0

The value of f(x) for x = k obtained by substitute the value of k to the variable of x on the polynomials of f(x). Hence, counting the polynomials value as above called as a substitution method. Look at these example to understanding the way to counting polynomials value with substitution method.

Example 1:
Determine the value of the polynomial f(x) = x3+3x2-x+5 if x is replaced by:
a) x = 0
b) x = 1
c) x =-1
d) x =2
e) x =-2
f) x = m(m element of R)
g) x = m-2(m element of R)
h) x = m+1(m element of R)
Answer:
f(x) = x3+3x2-x+5, hence
a) for x = 0, we obtain:
f(0) = (0)3+3(0)2-(0)+5=0+0-0+5=5
so, the value of f(x) for x = 0 is f (0) = 5
b) for x = 1, we obtain
f(1) = (1)3+3 (1)2-(1)+5=1+3-1+5=8
so, the value of f(x) for x = 1 is f (1) = 8
c) for x =-1, we obtain
f(-1) = (-1)3+3 (-1)2-(-1)+5=-1+3+1+5=8
so, the value of f(x) for x = -1 is f (-1) = 8
d) for x = 2, we obtain
f(2) = (2)3+3 (2)2-(2)+5=8+12-2+5=23
so, the value of f(x) for x = 2 is f (2) = 8
e) for x = -2, we obtain
f(-2) = (-2)3+3 (-2)2-(-2)+5=-8+12+2+5=11
so, the value of f(x) for x = -2is f (-2) = 8
f) for x = m(mR), we obtain
f(m) = (m)3+3 (m)2-(m)+5=m3+3m2-m+5
so, the value of f(x) for x = m is f (m) = m3+3m2-m+5
g) for x = m-2(mR), we obtain
f(m-2) = (m-2)3+3 (m-2)2-(m-2)+5=m3-3m2-m+11
so, the value of f(x) for x = m-2 is f (m-2) = m3-3m2-m+11
h) for x = m+1(mR), we obtain
f(m+1) = (m+1)3+3 (m+1)2-(m+1)+5=m3+6m2+8m-8
so, the value of f(m+1) for x = m+1 (mR)is f (m+1) = m3+6m2+8m+2

Example 2:
Given a polynomial in 2 variables x and y
f(x,y)=x2y+xy2+3x-4y+2
find
a) f(4,y)
b) f(-1,y)
c) f(x,3)
d) f(x,-2)
e) f(4,2)
f) f(-2,3)
Answer:
f(4,y)=x2y+xy2+3x-4y+2
a) f(4,y) means variable of x = 4 and variable of y constant
f(4,y)= (4)2y+(4)y2+3(4)-4y+2
=16y+4y2+12-4y+2
=4y212y+14
So, f(4,y)=x2y+xy2+3x-4y+2 is polynomial in y
b) f(-1,y) means variable of x = 4 and variable of y constant
f(4,y)= (-1)2y+(-1)y2+3(-1)-4y+2
=y-y2-3-4y+2
=-y2-3y-1
So, f(-1,y)= -y2-3y-1 is polynomial in y
c) f(x,3) means variable of x is constant and variable of y is 3
f(x,3)= x2(3)+x(3)2+3x-4(x)+2
=3x+9x+3x-12+2
=3x2+12x-10
So, f(x,3)= 3x2+12x-10 is a polynomial in x
d) f(x,-2) = x2(-2)+x(-2)2+3x-4(-2)+2
=-2x+4x+3x+8+2
=-2x2+7x+10
So, f(x, -2)= -2x2+7x+10 is a polynomial in x
e) f(4,2) means variable of x = 4 and y = 2
f(4,2)= (4)2(2) +(4) (2) 2+3(4)-4(2) +2
=32+16+12-8+2
= 54
So, f(4,2)= 54 is a real number
f) f(-2,3) means variable of x = -2 and y = 3
f(-2,3)= (-2)2(3)+(-2)(3)2+3(-2)-4(3)+2
= 12-18-6-12+2
= -22
So, f(-2,3)= -22 is a real number
What conclusions can you get based on the results above?


B. Scheme Method
To describe the way to find the polynomials value by scheme method, look at this fourth degree polynomial.
f(x) = a4x4+a3x3+ a2x2+a1x+a0
With substitution method, the polynomial value of f(x) for x = k determine by:
f(x) = a4k4+a3k3+ a2k2+a1k+a0
That f(x) can be arranged with the operation of multiplication addition as follows:
f(x) = a4k4+a3k3+ a2k2+a1k+a0
f(x) = (a4k4+a3k3+ a2k2+a1)k+a0
f(x) = ((a4k4+a3k3+ a2)k+a1)k+a0
f(x) = [{(a4k4+a3)k3+ a2}k2+a1]k+a0

Based on the last equation, we can see that the value of f(k) can be found step by step with an algorithms as follow:
- First step:
Multiply a4 with k, then sum the result with a3
a4k + a3
- Second step
Multiply the result of the first step (a4k + a3) with k, then sum the result with a2
(a4k + a3)k + a2 = a4k2 + a3k + a2
- Third step
Multiply the result of the first step (a4k+a3k+a2 ) with k, then sum the result with a1
(a4k+a3k+a2 )k+a1= a4k2 + a3k + a2k+a1
- Fourth step
Multiply the result of the first step (a4k+a3k+a2k+a1) with k, then sum the result with a0
(a4k+a3k+a2k+a1)k+a0= a4k+a3k+a2k+a1k+a0
The result of this fourth step is the value of f(x)= a4k+a3k+a2k+a1k+a0 for x=k
The processes of the algorithms above can be displayed on a scheme x = k is wrote in the first row of the scheme, then followed by polynomials coefficients. These coefficients arranged from the highest exponent coefficients to the smallest exponent. Look at this scheme:

x=k a4 a3 a2 a1 a0
+ + + +
a4k a4k2+a3k a3k3+a3k2+a2k a4k4+a3k3+a2k2+a1k
a4k a4k2+a3k a4k2+a3k3+a2k a3k3+a3k2+a2k+a1k a4k2+a3k3+a2k+a1k+a0
= f(k)

Finding the value of the polynomials as above called as a scheme method. This name is given because of the scheme that used.

Notes:
(1) Substitution method is suitable to find the value of the polynomials with a simple form for a small value and integer of x
(2) Scheme method can be used o find the value of all forms of the polynomials and for arbitrary x  E

Here are the applications examples to find the value of polynomials with scheme method:
Example 3:
Find the value of these polynomials with scheme method
a) f(x) = x4-3x3+4x2-x+10 for x = 5
b) f(x) = x5-x2+4x-10 for x = 2
Answer:
a) The coefficients of f(x) = x4-3x3+4x2-x+10 are a5=1, a4=a3=0,a2=-1,a1=4, and a0 = -10
For x =5, so k = 5.The scheme is shown on picture 5-2a. Based on the scheme, the value of f(x) = x4-3x3+4x2-x+10 for x=5 is f(5) = 355
b) The coefficients of f(x) = x5-x2+4x-10 are a5=1, a4=a3=0,a2= -1,a1=4, and a0 = -10
For x =2, so k = 5.The scheme is shown on picture 5-2a. Based on the scheme, the value of f(x) = x5-x2+4x-10 for x=2 is f(2) = 26

5 a4 a3 a2 a1 a0
+ + + +
5 10 70 345
1 2 14 69 355


2 a5 a4 a2 a1 a0
+ + + +
2 4 8 36
1 2 4 ¬18 26

Example 4:
Find the value of this polynomial with scheme method:
f(x,y)=x2y+x3 y2+x2+3y+2 for x = 2
Answer:
To find the value of f(x,y) for x=2, f(x,y) is considered as a polynomial in x with the form of decreasing exponents:
f(x,y)=y2x3 + (y+1)x2+(3y+2)
The coefficients of f(x,y) are a3=y2, a2 = y+1, a1=0, and a0 =3y+2. x=2 means k=2. Based on the scheme, the value of f(x,y)=x2y+x3 y2+x2+3y+2 for x = 2 is f(2,y)=8y+7y+6
The value of f(x) = x5-x2+4x-10 for x=2 is f(2) = 26

2 a3 a2 a1 a0
y2 y+1 0 3y+2
+ + +
y2 2y2 4y2+2y +2 8y2+7y +6
y2 2y2y+1 4y2+2y +2 8y2+7y +6



THE VALUE OF THE POLYNOMIALS

EXERCISE 2
1. With the substitution method, find:
a) f(1), if f(x) = x2-3x2+4x-2
b) f(-1), if f(x) = 2x2-4x2+5
c) f(2), if f(x) = x4-2x3+5x2+6
d) f(-2), if f(x) = 2x+2x3-5x2+6x-8
e) f(m), if f(x) = x3-4x2+3
f) f(m-3), if f(x)=x3-2
g) f(m+2), if(x) = x3-2x2+4x+3
h) f(1,y), if f(x,y)=x3y4+xy3+y+2x2+3
i) f(-2,y), if f(x,y)=x3y4+xy3+y+2x2+3
j) f(x,3), if f(x,y)=x3y4+xy3+y+2x2+3
k) f(x,-1), if f(x,y)=x3y4+xy3+y+2x2+3
l) f(2,-1), if f(x,y)=x3y4+xy3+y+2x2+3
m) f(-4,-3), if f(x,y)=x3y4+xy3+y+2x2+3
2. Find the value of these polynomials for given variable value with the substitution method:
a) X3-25 for x = 2
b) x4-4x3-x2+5x-1 for x = 4
c) a3+2a2-a+8 for a=1
d) a6-a for a=-1
e) y3+2y2-5y+10 for y = 4
f) y6-70 for y = 2
g) 4x3y3+5xy2+6x2+6x2-8y2+4 for x = 1
h) 4x3y3+5xy2+6x2+6x2-8y2+4 for x = 2 and y = 3
i) x3+y3z3-3xyz for x = y
j) x3+y3z3-3xyz for x = z
k) x3+y3z3-3xyz for x = -(y+z)
3. Find the value of the polynomials from number 1 and 2 above with the scheme method
4. Use the scheme method to find these values:
a) f(1), if (x)=x2-3x+10
b) f(2), if (f(x) = x3+10x2-5x+4
c) f(10), if(x)=x4-10x3+x-8
d) f(2), if f(x)=x5-4x2+x+4
e) f(1,y), if f(x,y) = 4x3y2-5x2y2+6x2-y2+2
f) f(x,3), if (x,y) = 4x3y2-5x2y2+6x2-y2+2
5. a) Use the substitution and the scheme method to find the value of these polynomial
(i) 4x4+10x2-2x2+5 for x = -0,75
(ii) 8x4+4x3+2,55x+5x+2 for x =-0,75
b) Based on the results of a) above, what conclusions you can get about those two methods?

Bibliography:
Wirodikromo,Santoso. Matematika Jilid 4 untuk SMA kelas XI Semester 2. Jakarta: Erlangga, 2001

YOU HAVE TO BELIEVE IN YOURSELF

YOU HAVE TO BELIEVE IN YOURSELF

The boy are try to motivate us to build our confidence. Anyone can do anything, can be anything, and become anything. No matter where we come from, we are just the same each other. We got to trust to ourselves that we can do what we want to do.
When there is no belief in our heart, we will be nothing. The key to be confident, to be survive in our life is be believe in ourselves. We can be success in our life if we do trust that we can do anything and be anything.
No one have no belief. The boy is a good motivator. He can persuade everybody around him to be come in his speech.

Revision:

DO YOU BELIEVE?

There is a boy named Dalton Sherman. He came from Charles Rice Learning Center. He tried to promote the Charles Rice Learning Center to the audience in Dallas ISD.
Firstly, he asked to the audience, do they believe that he can stand up on the stage, fearless, and talk to them. He said that everyone must believe in him because there is some deals that he can do anything, be anything, and become anything.
Then, he asked to the Dallas ISD, do they believe in his classmate and do they believe that everyone of them can graduate for workplace or college.
He said, they better do believe, because next week they will show up the Dallas ISD that they can reach their highest potential, no matter where they come from, they better not give up on them.
The audiences are the one who feed them and who wipe their tears. They are the one who love them when no one does.
Next, he asked do they believe in their colleagues. They came to the audience’s school because they want to be developed also. No matter what they are, he needs them.
He tried to persuade then to trust to their colleagues, so they will do to. What are they doing is not only for his generation, but to the next generation also.
The boy and his classmate need the audience more than ever. So, the audience needs to believe in themselves.
Then he asks do they believe that every child in Dallas needs to be ready for college? They must believe that they can do. They have to believe in their colleagues, in themselves, and in their goals.
Finally, he say thank you to them who does so many things for him and others. They have to believe in him because he believes in himself. The audience helped him get to where he is today.

THE DEFINITION, EXPLANATION, AND THE EXAMPLE OF THE TERMS

The definition, explanation, and the examples of the terms questioning by…

1. Titik Pangkal: Initial Point
Sentence: Graph y = x^2 +1 from the initial point x = 1.

2. Pasangan Berurutan: Consecutive Pair
Sentence: A consecutive pairs from C = (x, y, z) and D = (1, 2, 3) are {(x,1),(y,2),(z,3)}

3. Bilangan Cacah: Counting Number
Sentence: In math, numbers are grouping in several number sets, such as integers, counting number, etc.

4. Bilangan Berpangkat: Exponential Number
Sentence: 2^3 is called as an exponential number with the power or the exponent of 3.

5. Deret Tak Hingga: Infinitive Row
Sentence: 1+2+3+… is an infinitive row.

6. Menggapai Impian: To Reach the Dream
Sentence: We have to work hard to reach our dream.

7. Mendalami Jati Diri: To Identify the Identity
Sentence: No one knows about us better than ourselves, so we have to identify our identity well.

8. Garis Keturunan: Origin Line/ Genetic Line
Sentence: Adam is the very first man in the origin line of human being.

9. Naik Sepeda Berboncengan: To Drive Together In A Bicycle
Sentence: Be careful when you drive together with your friend in a bicycle.

10. Serius: Serious
Sentence: He does the test seriously.

11. Saling Bersalam-Salaman: To Exchange Greetings/ To Shake Hands
Sentence: They shake their hands when they meet at school.

12. Jangan Melirak-Lirik: Do Not Look Aside
Sentence: Do not look aside when you are driving, for your safety.

13. Pohon Itu Melambai-Lambai: The Tree Is Waving
Sentence: I see the trees are waving along the beach.

REPRESENTING THE VIDEO OF LEARNING MATHEMATICS

1. Solving Problem Graph

The problem is question number 13 on page 411. The question is:

The figure shows the graph of y=g(x), in the function h is defined by h(x)=g(2x)+2, what is the value of h(1)?

Answer: We are looking for h(1). The first information is the graph. The next piece of the information is that h(x)=g(2x)+2. What is h equal when x is equal to 1? So let’s substitute 1 into h(x)=g(2x)+2 à h(1)=g(2)+2. Now, how we figure out g(2) or what is g when x is equal to 2? We have the graph above. When x is equal to 2, y is equal to 1. Then g is equal to 1 when x is equal to 2. So, h(1)=g(2)+2 become h(1)=1+2 and we got h(1)=3.

The next problem is another question 13 on page 534:

Let the function f be defined by f(x)=x+1, if 2f(p)=20, what is the value of f(3p)?

Answer: We are looking for f(3p) or in another word what is f when x is equal to 3p? The first piece of information is equation of f(x)=x+1 and the next information is 2f(p)=20. To figure out what is f when x=3p, we need to figure out what p is. Let’s start with the equation of 2f(p)=20. Divide both side by 2, we get f(p)=10. Then, f(p) is just what is function f(x) when x is equal to p. Write that equation. Plug in p in the equation f(x)=x+1à f(p)=p+1=10àp=9. That is not the answer because we are looking for f when x=3p. So if we take p=9 to x=3p, we get x=27. We have the equation for function f which is x+1. We know that x=27, so f(27)=27+1=28.

Let’ move to the next question. Question 17 on page 412:

In the xy-coordinate plane, the graph of x=y^2-4 intersect line l at (0,p) and (5,t). What is the greatest possible value of the slope of l?

Answer: We are looking for the greatest slope (m). In the xy-coordinate plane, the graph of x=y^2-4 is:

x=y^2-4 intersect line l at two points (0,p) and (5,t). It does not define exactly where those points are. We have the general idea of the line l on this graph.

Rewrite those two points and put it in a little table to help us.

What is the greatest slope? How do we know about the slope. We know that the slope at the line is going to be m=(y2-y1)/(x2-x1). We can put the values of x and y from the coordinate table that apply to line l. So the slope is going to be m=(t-p)/(5-0)=(t-p)/5. If we want to maximize the slope, we need to maximize the numerator (t-p). How we figure out (t-p)? We have the equation x=y^2-4 and the coordinates (0,p) and (5,t) applied to the equation. The points are the intersection of the graph and the line. So we can plug in the coordinates into the equation of x=y^2-4.

2. Factoring Polynomials

One way to define factor of the polynomials is the rule of algebraic long division that looks like the long division you learn, only harder. For an example, let’s try to see:

Is x-3 a factor of x^3-7x-6?

Answer: When dividing x-3 into x^3-7x-6, first setup the problem like a long division problem from elementary school. That is, you dividing x-3 into x^3-7x-6. Now the zero is in there because there is no second degree term. Now you must ask yourself, what times x gives you x^3? Of course it is x^2. So you write x^2 as the part of the quotient and then multiply x-3 by x^2 which gives you x^3-7x, which you subtract from x^3-0x^2 to get 3x^2. Bringing down the next term -7x, you have 3x^2-7x.

Now we begin again dividing x-3 into 3x^2-7x. Just looking at the first term x goes into 3x^2, 3x. The next part of the answer is 3x. Multiply x-3 by 3x for a product of 3x^2-9x. Subtracting with 3x^2-7x you are left with 2x-6. Now we see that x-3 divides evenly into 2x-6 which equal 2, with no remainder. So the solution to the long division problem x^3-7x-6 divided by x-3 is x^2+3x+2.

Since x-3 divided into x^3-7x-6 evenly with no remainder then x-3 is a factor of x^3-7x-6. The quotient which is x^2+3x+2 is also a factor of x^3-7x-6. We know that x^3-7x-6=(x-3)(x^2+3x+2). The quadratic expression (x^2+3x+2) can be factored into (x+1)(x+2). So x^3-7x-6=(x-3)(x+1)(x+2). Setting the factored form of the equation x^3-7x-6 to zero, we get 0=(x-3)(x+1)(x+2). Those either x-3=0, or x+1=0, or x+2=0. Solving all of this equation for x, we get x=3, x= -1, and x= -2. The roots to x^3-7x-6 are 3, -1, and -2.

Now there are three roots for the third degree equation of x^3-7x-6. On the quadratic or second degree equation we look that always have at most two roots. A fourth degree equation would have four or fewer roots and so on. The degree of a polynomial equation always limits the number of roots.

Let’s summarize the long division process for third order polynomial:

1. Find a partial quotient of x^2, by dividing x into the first term x^3 to get x^2.

2. Multiply x^2 by the divisor and subtract the product from the dividend.

3. Repeat the process until you either “clear it out” or reach a remainder.

3. Pre-Calculus Graph

Let’s begin by discussing the graph of a rational function which can have discontinuities. A rational function has a polynomial in the denominator which means you are dividing by something that is valuable quantity.

It’s possible that some value of x will meet to division be zero. Example:

If f(x)=(x+2)/(x-1), when x=1, the function value become f(x)=(1+2)/(1-1)=3/0 with 0 in the denominator. For this function, choosing x=1 is a bad idea.

When there is a bad choice of x when it makes the bottom of the rational function, it shows up a break in a function graph. For example, suppose you to finding the graph f(x)=(x+2)/(x-1). Start with inserting 0 for x. So now we have f(0)=(0+2)/(0-1)=2/(-1)=-1. So, you put off point down on the graph at (0,-2).

Next you try x=1. This time you get f(1)=(1+2)/(1-1)=3/0. That is you know is impossible and it means that you can not compute y-value when x=1.

It also mean that the graph of this function will not have any point for x=1. The graph is separated into two disconnected pieces at x=1.

Take the graph of f(x)=1/(x^2+1). No matter what numbers you choose for x, the denominator will never zero. The graph is smooth and unbroken.

Don’t forget that the general rule on the rational function, you must expect the possibility that the denominator may turn out to be zero.

Review: For polynomials, the graph is a smooth unbroken curve. For a rational function, sometimes the value of x may be zero in the denominator. That is an impossible situation because there is no y for that x. At that point, there is no value for the function and there is a break in the graph.

A break can show up in two ways. The simpler type of a break is just a missing point. This function is an example of this type of break: y=(x^2-x-6)/(x-3).

The gate that appear in the graph is at points where x=3. If you try to substitute 3 for x in the equation, the result is y=(3^2-3-6)/(3-3)=0/0. That is not possible, not feasible, and not allowed. So, there is no y for x=3. That is the typical example of the missing point syndrome and conveniently it always goes with the result like 0/0.

When you see the result of 0/0, it also tells you that it should be possible to factor the top and the bottom of the rational function and simplify. For an example: y=(x^2-x-6)/(x-3)=(x-3)(x-2)/(x-3)=(x+2).

For this kind of the break, the missing point is a loop hole. For the original function without simplify, x=3 is a bad point because it leads to the division be zero, y=0/0. But if you simplify first, then there’s no problem with x=3, y=x+2. It can be one idea or key in calculus.

Removable singularity appears simplest missing points on the graph when x leads to 0/0. For this kind of a break, if you factor and simplify the rational function, the division by zero can be avoided.

4. Inverse Function

We will talk about the inverse function. To talk about it, we need to review the definition of the function. If we have the relation of f(x,y)=o, then function y=f(x). That is 1.1 function if we can represent it in x=g(y). The function y=f(x) is “VLT” (Vertical Line Test). Every vertical line intersect the graph in at most one point. The 1.1 function satisfy “HLT” (Horizontal Line Test).

If we look at function y=x^2, this is not 1.1 function because if we graph any horizontal line on the graph, we get two intersection points.

To make it as a 1.1 function, we need a domain of 0<=x, so the graph will be:

We can figure out the other squared root of the left side so if we have function of 1.1 then the function is “invertible”.

Let’s start out with the function y=2x-1. And let’s look at the graph of that function.

This is a straight line with y intersect (-1) and x intersect (1/2). Look at the line y=x. That line intersect the graph of y=2x-1. So, we get x=2x-1à 1+x=2x à 1=x. So, the intersection of the line y=x and y=2x-1 with x=1 is (1,1).

See this relation for y=x. Write 2x-1=y à 2x=y+1 à x=(y+1)/2 à x=(y/2)+1/2. Changing x to y in the last equation, we get y=(x/2)+1/2. Looking back on the graph, then we get another line. That line contain the point (1,1). The x-intercept is (0,-1) and the y-intercept is (0,1/2). So, the invers line to the given line y=2x-1 passes through the same point.

We have f(x)=2x-1 and on the other hand, we have g(x)=(x/2)+1/2. We want to compute f(g(x)). f(g(x))=2(…)-1=2((x/2)+1/2)-1à x+1-1=x. g(f(x))=1/2(…)+1/2à g(f(x))=x-1/2+1/2=x.

So, the important problem is g=f-1 à f(g(x))=f(f^(-1)(x))=x and g(f(x))=f^(-1)(f(x))=x.

For another example, take y=(x-1)/(x+2). This is the line with a vertical asymptot at the line x=2 and x=1. The x-intercept is equal to 1, so the point is (1,0). The y-intercept is equal to (0,-1/2). And otherwise the function graph is a hyperbola in a second and fourth quadrant and respect to its asymptoth.

Take a look at the equation of y=(x-1)/(x+2). Multiply both side with (x+2), we get y(x+2)=(x-1) à yx+2y=x-1 à yx-x=-1-2y à (y-1)x=-1-2y à x=(-1-2y)/(y-1). Now, interchange the x to y and y to x à y=(-1-2x)/(x-1). When x=0, we get y=-1 and when y=0 we get -1-2x=0 à -2x=-1 à x= -1/2.

With the vertical asymptot at x=1 and horizontal asymptot at y=-2, again we get the hyperbola. If we look carefully, we will see two functions that reflect each other.

The invers function of y=2^x is: .

TRANSLATE A MATHEMATICS ARTICEL

1. Translate an English mathematics article into Indonesian

ARITHMETIC

Arithmetic or arithmetics (from the Greek word αριθμός = number) is the oldest and most elementary branch of mathematics, used by almost everyone, for tasks ranging from simple day-to-day counting to advanced science and business calculations. In common usage, the word refers to a branch of (or the forerunner of) mathematics which records elementary properties of certain operations on numbers. Professional mathematicians sometimes use the term (higher) arithmetic when referring to number theory, but this should not be confused with elementary arithmetic.

a. History

The prehistory of arithmetic is limited to a very small number of small artifacts indicating a clear conception of addition and subtraction, the best-known being the Ishango bone from central Africa, dating from somewhere between 18,000 and 20,000 BC.

It is clear that the Babylonians had solid knowledge of almost all aspects of elementary arithmetic by 1800 BC, although historians can only guess at the methods utilized to generate the arithmetical results - as shown, for instance, in the clay tablet Plimpton 322, which appears to be a list of Pythagorean triples, but with no workings to show how the list was originally produced. Likewise, the Egyptian Rhind Mathematical Papyrus (dating from c. 1650 BC, though evidently a copy of an older text from c. 1850 BC) shows evidence of addition, subtraction, multiplication, and division being used within a unit fraction system.

Nicomachus (c. AD60 - c. AD120) summarised the philosophical Pythagorean approach to numbers, and their relationships to each other, in his Introduction to Arithmetic. At this time, basic arithmetical operations were highly complicated affairs; it was the method known as the "Method of the Indians" (Latin "Modus Indorum") that became the arithmetic that we know today. Indian arithmetic was much simpler than Greek arithmetic due to the simplicity of the Indian number system, which had a zero and place-value notation. The 7th century Syriac bishop Severus Sebhokt mentioned this method with admiration, stating however that the Method of the Indians was beyond description. The Arabs learned this new method and called it "Hesab" or "Hindu Science". Fibonacci (also known as Leonardo of Pisa) introduced the "Method of the Indians" to Europe in 1202. In his book "Liber Abaci", Fibonacci says that, compared with this new method, all other methods had been mistakes. In the Middle Ages, arithmetic was one of the seven liberal arts taught in universities.

Modern algorithms for arithmetic (both for hand and electronic computation) were made possible by the introduction of Hindu-Arabic numerals and decimal place notation for numbers. Hindu-Arabic numeral based arithmetic was developed by the great Indian mathematicians Aryabhatta, Brahmagupta and Bhāskara I. Aryabhatta tried different place value notations and Brahmagupta added zero to the Indian number system. Brahmagupta developed modern multiplication, division, addition and subtraction based on Hindu-Arabic numerals. Although it is now considered elementary, its simplicity is the culmination of thousands of years of mathematical development. By contrast, the ancient mathematician Archimedes devoted an entire work, The Sand Reckoner, to devising a notation for a certain large integer. The flourishing of algebra in the medieval Islamic world and in Renaissance Europe was an outgrowth of the enormous simplification of computation through decimal notation.

b. Decimal arithmetic

Decimal notation constructs all real numbers from the basic digits, the first ten non-negative integers 0,1,2,...,9. A decimal numeral consists of a sequence of these basic digits, with the "denomination" of each digit depending on its position with respect to the decimal point: for example, 507.36 denotes 5 hundreds (10²), plus 0 tens (10¹), plus 7 units (10⁰), plus 3 tenths (10^-1) plus 6 hundredths (10^-2). An essential part of this notation (and a major stumbling block in achieving it) was conceiving of zero as a number comparable to the other basic digits.

Algorism comprises all of the rules of performing arithmetic computations using a decimal system for representing numbers in which numbers written using ten symbols having the values 0 through 9 are combined using a place-value system (positional notation), where each symbol has ten times the weight of the one to its right. This notation allows the addition of arbitrary numbers by adding the digits in each place, which is accomplished with a 10 x 10 addition table. (A sum of digits which exceeds 9 must have its 10-digit carried to the next place leftward.) One can make a similar algorithm for multiplying arbitrary numbers because the set of denominations {...,10²,10,1,10^-1,...} is closed under multiplication. Subtraction and division are achieved by similar, though more complicated algorithms.

c. Arithmetic operations

The traditional arithmetic operations are addition, subtraction, multiplication and division, although more advanced operations (such as manipulations of percentages, square root, exponentiation, and logarithmic functions) are also sometimes included in this subject. Arithmetic is performed according to an order of operations. Any set of objects upon which all four operations of arithmetic can be performed (except division by zero), and wherein these four operations obey the usual laws, is called a field.

(Taken from http://en.wikipedia.org/wiki/Arithmetic)

In Indonesian:

ARITMETIKA

Aritmetika berasal dari bahasa Yunani αριθμός yang berarti “bilangan”. Aritmetika merupakan cabang matematika paling dasar yang tertua. Cabang ini digunakan oleh hampir setiap orang untuk melakukan perhitungan sederhana hingga bisnis dalam kemajuan ilmu pengetahuan. Aritmetika merupakan cikal bakal matematika yang menghasilkan operasi dasar pada bilangan. Ahli matematika profesional kadang menggunakan istilah aritmetika untuk mengartikan kata teori bilangan yang seharusnya tidak rancu dengan aritmetika dasar.

1. Sejarah

Prasejarah aritmetika ditemukan pada sedikit artifak yang mengindikasikan konsep yang jelas tentang penjumlahan. Hasil terbaik yang diketahui ditemukan pada tulang Ishago di Afrika Tengah sekitar tahun 18.000-20.000 SM.

Orang-orang Babilonia mempunyai pegetahuan yang kuat pada hampir setiap aspek dasar aritmetika sekitar tahun 1800 SM. Akan tetapi, ahli sejarah hanya dapat menduga kegunaan metode yang digunakan untuk membangkitkan hasil dari aritmetika, seperti pada tablet tanah Plimpton 322 yang kemudian menjadi daftar tripel Pythagoras, tanpa cara kerja untuk menunjukkan bagaimana daftar tersebut dapat dihasilkan pada awalnya. Papirus Rhind Mesir (tahun 1650 SM, merupakan salinan dari teks yang lebih tua pada 1850 SM) menunjukkan bukti dari penjumlahan, pengurangan, perkalian, dan pembagian yang digunakan pada sistem pecahan satuan.

Nicomachus meringkas filosofi Pythagoras dengan bilangan dan hubungan antara keduanya pada bukunya yang berjudul “Introduction to Arithmetic”. Pada masa ini, operasi dasar aritmetika memiliki tingkat kesulitan yang tinggi. Metode orang India “Modus Indorum”-lah yang kemudian menjadi aritmetika yang kita kenal saat ini. Aritmetika India jauh lebih sederhana dari aritmetika Yunani, seperti yang terlihat pada sistem bilangannya (India) yang mempunyai angka nol dan notasi nilai tempat. Pada abad ke-7, uskup Siria bernama Severus Sebhokt memuji metode sistem bilangan ini dengan menyatakan bahwa metode India adalah deskripsi yang luar biasa. Orang-orang Arab mempelajari metode baru ini dan menamakannya “Hesab” atau “Ilmu Pengetahuan Hindu”. Fibonacci (dikenal sebagai Leonardo dari Pisa) mengenalkan metode India ini ke Eropa pada tahun 1202. Pada bukunya “Liber Abaci”, Fibonacci menyatakan bahwa semua metode memiliki kesalahan jika dibandingkan dengan metode India. Pada zaman pertengahan, aritmetika merupakan satu dari tujuh seni bebas di universitas-universitas.

Algoritma modern pada aritmetika (baik perhitungan manual atau elektronik) dapat disusun berkat pengenalan notasi tempat desimal dan angka Hindu-Arab pada bilangan. Aritmetika angka dasar Hindu-Arab dikembangkan oleh para matematikawan besar India, yaitu Aryabhatta, Brahmagupta, dan Bhaskara. Aryabhatta berusaha untuk membedakan notasi nilai tempat dan Brahmagupta menambahkan nol pada sistem bilangan India. Brahmagupta mengembangkan perkalian, pembagian, penjumlahan, dan pengurangan modern yang berdasar pada angka Hindu-Arab. Walau kini dianggap sebagai dasar, penyederhanaan ini merupakan puncak dari ribuan tahun perkembangan matematika. Archimedes, seorang matematikawan, telah mencurahkan seluruh hasil kerjanya, “The Sand Reckoner”, untuk menemukan notasi pada angka yang lebih besar. Perkembangan aljabar di dunia Islam dan Renaissance Eropa merupakan perkembangan penyederhanaan besar dari perhitungan notasi desimal.

b. Aritmetika desimal

Notasi desimal menggagasi semua bilangan real dari digit yang paling dasar, yaitu 10 bilangan positif pertama: 0, 1, 2,…,9. Bilangan desimal berisi rangkaian dari digit dasar ini dengan “denominasi” (penyebutan) dari tiap digitnya yang tergantung pada posisinya. Misalnya bilangan 507,36 merupakan 5 ratusan (10²), ditambah 0 puluhan (10¹), ditambah 7 satuan (10⁰), ditambah 3 seperpuluhan (10^-1), ditambah 6 seperratusan (10^-2). Bagian penting dari notasi ini (dan hambatan utama untuk mencapainya) adalah penyusunan nol sebagai bilangan yang dapat diperbandingkan dengan digit dasar lainnya.

Algoritma terdiri dari semua aturan dalam menyajikan perhitungan aritmetika dengan menggunakan sistem desimal untuk menampilkan bilangan, dimana bilangan yang ditulis dengan 10 simbol bernilai 0-9 dikombinasikan memakai sistem nilai tempat (notasi posisi). Setiap simbol tersebut mempunyai bobot 10 kali lipat dari satu simbol ke simbol lainnya ke arah samping kanannya. Notasi ini memperbolehkan penjumlahan bilangan sebarang dengan menambah digit pada tiap tempatnya yang disempurnakan dengan tabel penjumlahan 10x10 (jumlah digit yang melebihi 9, 10 digitnya dibawa ke posisi kirinya). Angka 1 dapat membentuk algoritma yang mirip untuk mengalikan bilangan sebarang karena himpunan dari denotasi {…,10².10,1,10^-1,…} mengakhiri sebuah perkalian. Pengurangan dan pembagian didapat dengan cara serupa dengan algoritma yang lebih sukar.

c. Operasi Aritmetika

Yang termasuk operasi aritmetika tradisional adalah penjumlahan, pengurangan, perkalian, dan pembagian. Operasi yang lebih modern (seperti manipulasi persentase, akar kuadrat, eksponensial, dan fungsi logaritma) terkadang termasuk dalam operasi aritmetika tradisional. Aritmetika ditampilkan tergantung pada perintah operasi. Himpunan dimana keempat operasi aritmetika ini dapat ditampilkan (kecuali pembagian dengan nol) dan mengikuti aturan yang berlaku dinamakan himpunan semesta.

2. Translate an Indonesian mathematics article into English

BILANGAN KOMPLEKS

Dalam matematika, bilangan kompleks adalah bilangan yang berbentuk:

a+bi

dimana a dan b adalah bilangan riil, dan i adalah bilangan imajiner tertentu yang mempunyai sifat i ² = −1. Bilangan riil a disebut juga bagian riil dari bilangan kompleks, dan bilangan real b disebut bagian imajiner. Jika pada suatu bilangan kompleks, nilai b adalah 0, maka bilangan kompleks tersebut menjadi sama dengan bilangan real a.

Sebagai contoh, 3 + 2i adalah bilangan kompleks dengan bagian riil 3 dan bagian imajiner 2.

Bilangan kompleks dapat ditambah, dikurang, dikali, dan dibagi seperti bilangan riil; namun bilangan kompleks juga mempunyai sifat-sifat tambahan yang menarik. Misalnya, setiap persamaan aljabar polinomial mempunyai solusi bilangan kompleks, tidak seperti bilangan riil yang hanya memiliki sebagian.

Dalam bidang-bidang tertentu (seperti teknik elektro, dimana i digunakan sebagai simbol untuk arus listrik), bilangan kompleks ditulis a + bj.

a. Notasi dan operasi

Himpunan bilangan kompleks umumnya dinotasikan dengan C, atau . Bilangan real, R, dapat dinyatakan sebagai bagian dari himpunan C dengan menyatakan setiap bilangan real sebagai bilangan kompleks: a = a + 0i.

Bilangan kompleks ditambah, dikurang, dan dikali dengan menggunakan sifat-sifat aljabar seperti asosiatif, komutatif, dan distributif, dan dengan persamaan i ² = −1:

(a + bi) + (c + di) = (a+c) + (b+d)i

(a + bi) − (c + di) = (a−c) + (b−d)i

(a + bi)(c + di) = ac + bci + adi + bd i ² = (ac−bd) + (bc+ad)i

Pembagian bilangan kompleks juga dapat didefinisikan (lihat dibawah). Jadi, himpunan bilangan kompleks membentuk bidang matematika yang, berbeda dengan bilangan real, berupa aljabar tertutup.

Dalam matematika, adjektif "kompleks" berarti bilangan kompleks digunakan sebagai dasar teori angka yang digunakan. Sebagai contoh, analisis kompleks, matriks kompleks, polinomial kompleks, dan aljabar Lie kompleks.

b. Definisi

Definisi formal bilangan kompleks adalah sepasang bilangan real (a, b) dengan operasi sebagai berikut:

(a, b) + (c, d) = (a + c, b +d)

(a, b) . (c, d) = (ac - bd, bc + ad)

Dengan definisi diatas, bilangan-bilangan kompleks yang ada membentuk suatu himpunan bilangan kompleks yang dinotasikan dengan C.

Karena bilangan kompleks a + bi merupakan spesifikasi unik yang berdasarkan sepasang bilangan riil (a, b), bilangan kompleks mempunyai hubungan korespondensi satu-satu dengan titik-titik pada satu bidang yang dinamakan bidang kompleks.

Bilangan riil a dapat disebut juga dengan bilangan kompleks (a, 0), dan dengan cara ini, himpunan bilangan riil R menjadi bagian dari himpunan bilangan kompleks C.

Dalam C, berlaku sebagai berikut:

1) Identitas penjumlahan ("nol"): (0, 0)

2) Identitas perkalian ("satu"): (1, 0)

3) Invers penjumlahan (a,b): (−a, −b)

4) Invers perkalian (reciprocal) bukan nol (a, b): [a/(a^2+b^2),-b/(a^2+b^2)]

(Taken from http://id.wikipedia.org/Bilangan_Kompleks)

In English:

COMPLEX NUMBER

In mathematics, the complex number are the number with the form:

a+bi

where a and b are real numbers, and i is the imaginary number which has a characteristic i² = -1. The real number a is called the real part of the complex number and b is the imaginary part. The complex number is equal to the real number a if the value of b is 0.

For example, 3 + 2i is a complex number with the real part 3 and the imaginary part 2.

Complex number can be added, subtracted, multiplied, and divided as the real number, but it has some interesting additional characteristics. Such as, polynomial algebraic equation has a solution of the complex number, unlike the real number which has a half only.

In some diciplines (such as electrical engineering, where i is a symbol for current), the imaginary unit are written as a + bj.

a. Notation and Operation

The set of the complex number is denoted by C, or . The real number, R, can be called as a subset of C by considering every real number as a complex number :

a = a + 0i.

Complex number are added, suntracted, multiplied, and divided with the laws of algebra such as associative, commutative, and distributive, with the equation i² = -1 :

(a + bi) + (c + di) = (a+c) + (b+d)i

(a + bi) − (c + di) = (a−c) + (b−d)i

(a + bi)(c + di) = ac + bci + adi + bd i ² = (ac−bd) + (bc+ad)i

The division of the complex number can be defined. So that, the set of the complex number can build a mathematics plane that different with the real number as an enclosed algebra.

In mathematics, the adjective of the “complex” has the meaning that the complex number is used as the based of used number theory. For example, complex analysis, complex matrix, complex polynomial, and complex Lie algebra.

b. Definition

The formal definotion of the complex number is a pair of the real number (a, b) with the operation of:

(a, b) + (c, d) = (a + c, b +d)

(a, b) . (c, d) = (ac - bd, bc + ad)

The complex numbers forms the set of complex number denoted by C from above definition.

Since complex number a + bi is an uniquely specified by the real number pair (a, b), the complex number has a relation of of on-on-one corespondence with the points on a plane called the complex plane.

The real number a can be called by the complex number (a, 0). Through this way, the set of the real number R be the subset of the complex number set C.

In C, there is:

1) An additive identity (‘zero”) : (0,0)

2) A multiplicative identity (“one”) : (1,0)

3) An additive inverse (a, b) : (-a, -b)

4) A multiplicative inverse (reciprocal) nonzero (a, b) : [a/(a^2+b^2),-b/(a^2+b^2)]