bai bao cao khoa hoc tai bac kinh 2009 thescientificreportatcm 2009

8/3/2019 Bai Bao Cao Khoa Hoc Tai Bac Kinh 2009 Thescientificreportatcm 2009

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The 14th ASIAN TECHNOLOGYCONFERENCE IN MATHEMATICS,

BEIJING, CHINA 2009

BEIJING NORMAL UNIVERSITY

December 17-21, 2009

THE MINH TRAN

HCMC UNIVERSITY OF TECHNOLOGY

VIETNAM


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TITLE : USING SCIENTIFIC

CALCULATORS TO SOLVEMATHEMATICAL PROBLEMS

Authors : THE MINH TRAN

DEPARTMENT OF APPLIED MATHEMATICS

HCMC UNIVERSITY OF TECHNOLOGY (HCMUT)

VIETNAM


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Introduction

Using a calculator is an important skill forstudents. As they grow and progress throughtheir education.

They will learn advanced math skills that

certainly require them to use a calculator . A few quick steps can help anyone use the

device and solve mathematical problems .

I have organized exams : Excellent students use

scientific calculators to solve mathematicalproblems

Therefore, I have built math-calculator subjectand technology in mathematics


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WE CAN USE SCIENTIFICCALCULATORS :

1.VIETNAM CALCULATORVN- 570RS 3.CASIO FX-570MS

4.CASIO FX-570ES2.TEXAS INSTRUMENTS83 Plus


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A. ARITHMETIC

We calculate : 3281978 8791803

We input 3281978 8791803 on the screen

Press the = key. Answer : 2.885450403

We know that answer have 14 digits but the digit 3 canhave rounded

We delete the digit 3 at the first number and the digit 8 atthe second number then we multiply two numbers together


We continue to delete the digit 2 at the first number andthe digit 7 at the second number then we multiply 81978and 91803 to get 7525826334

Result: 3281978 8791803 = 28854504026334

Note : when you multiply two numbers, you need tothoroughly check digits

1310

1.Multiply two numbers

1110


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2. Find x, know that :

We input 381978 382007 on the screen


We continue to press : 3 - 8 then we press 9 timesthe = key. The final result have saved to the Ans key

Now,we have then we press : Ans - 1 =

Result : x = - 1.11963298

3

83

83

8 3

83

83

83

83

83

81

1 x

381978

382007

1

x

xAns

1

1 1x


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3. Find a natural number n , know that thenumber contains the first four digits and the

end four digits are digit 1

Step 1 :Find the number contains the end four digits are digit 1.

Solution :The number contains the end digit is digit 1

The number contains the end two digits are digit 1The number contains the end three digits are digit 1The number contains the end four digits are digit 1

Answer : the number contains the end four digits aredigit 1

n1 = 8471

3n

31

371

3471

38471

38471


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Step 2 : Find the number contains the first fourdigits are digit 1.

Solution :

We have ( we get the number 223 )

We try ( answer contains two digit

are digit 1 ) . we do not choose the number 223

The number contains the first two digit are digit 1We have ( we get the number 480 )We try ( answer contains four digits

are digit 1 )

We choose the number 480 . Answer : n2 = 480If we do not find n2, we can calculate

to find n2

Result: n = 4808471and

19310...1216.12238471

7.4801111011113 203

10...11178.14808471

11111000111;1111001111 33

14.223111111113

322308471

34808471 1111. . .1111


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4. Find natural number x,y, know that :( and y include a digit)

Solution : we guess : x is digit 2 or 3

When x = 2 : (we do not choose x = 2 )

We choose the digit 3 , we have :

2

2 25 541 1 ( )4481448( )

x

y y y x x y

2x

4 4(5 ) 59 12117361y

9 99

5410109448144809 5 541 1 94481448 9 5419199448144899y y y y

55, 99303467 5 56, 00348007 6y y

Result: x = 3 , y = 6 and 5416169448144896

Furthermore , we can find y by the SOLVE key.We input

9 13 12 11 10 9 2

50 541 10 10 1 10 10 9 10 4481448 10 90y y y y on the screen. We press SHIFT SOLVE key and press 1 key

6y

956


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5. Find the fourteenth digit after of thedecimal point of

We have : the digit 5 can have rounded

We try the digit 5 can not haverounded

We set :

The second power of two sides of equation :

We have :

(1)

We use function of calculator to solve equation :

We have roots :We only get root

12

12 3.464101615.. .Solution :10

12 3.464101615 1.4 10 0

12 3, 464101615 ( 0)x x

2 22 3.464101615 3.464101615 12 0 (1)x x

2 103.464101615 12 9,54391775 10

.

2 102 3.464101615 9,54391775 10 0x x

10

11, 37754 10x

2 6.92820323x

1x

12 3, 464101615137754...

Result : 5


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Exercise : Find the eighteenth digit after ofthe decimal point of

We use calculators :

VietnamCalculator Vn-570RS,Vn-500RS,CASIO fx 570MS,

fx 570ES , fx 991ES,fx-82ES,

HP- 10S,HP-300S

337

INTRODUCTION


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6.Calculate :

We press :

Input 1 for A , we press 1 SHIFT STO A

Input 7 for B ,we press 7 SHIFT STO BInput 7 for C ,we press 7 SHIFT STO C

We input letters on the screen : A = A +1:B = 10B + 7 : C =C + B

Press the = key until we have seen A = 17 on thescreen, then we press the = key two times

2

17 7

7 77 777 ... 77......77 293972367

digits are digit

P

B. PROGRAMMING ONVIETNAMCALCULATOR VN-570RS ;

CASIO FX- 570MS & ES , FX-991ES

Solution :


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168, 641975309 10C

We continue to press 2ALPHA C - 293972367

Answer : P = 526800000

Answer :

we continue to find the end five digits of P and we guess thatthe digit 8 can have rounded, so we only calculate

7 77 777 7777 77777 13 272367

=> The end five numerals of P are :

P = 1019739 - 82689 = 37050

Result : P = 526837050(we are sure that the digit 8 is true because theafter digit of thedigit 8 is the digit 3)

Answer : 1019739We calculate : Answer : 5236982689


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7. solve linear equations with four unknowns :

53605030702

4887465,0

100

tzyx

tzyx

tzyx

know that :tzyx ,,,

100,,,0 tzyx

We have

53605030702

4887465,0100

tzyx

tzyxtzyx

12901271787613711

tzytzy

4146 yt1000 t 8669 y

11 7 13 876y z t

7

1311876 tyz

Using the X ,Y key for x,y variable and the A, B key for z, tvariablePress 69 SHIFT STO YWe input letters on the screen :Y = Y + 1 : B = 6Y 414 :

A = ( 876

11Y

13B ) 7 : X=100

Y

B

A

Solution :

We have known :


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Press the = key until we have seen Y = 70 on the screen

We have seen :

B = 6 after we press the = keyA = 4 after we press the = keyX = 20 after we press the = keyResult : x = 20 ; y = 70 ; z = 4 ; t = 6


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C. INTRODUCE PROGRAMMING ON TI83,84,86,89 CALCULATORS

SOLVE UNLINEAR EQUATIONS WITH FOUR UNKNOWS

We use Jacobi theory

The expression in right side have solved with

The linear equation just is feature case

It can use for n roots

This program have run on TI calculators from 83 to 89 and

It can solve all equations

Cl H


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: ClrHome

: Lbl 2

: Input F(X, Y, Z, T) = 0 =, Str1

: Input G(X, Y, Z, T) = 0 =, Str2

: Input H(X, Y, Z, T) = 0 =, Str3

: Input K(X, Y, Z, T) = 0 =, Str4

: Lbl 1

: Input Xo = , A

: Input Yo = , B

: Input Zo = , C

: Input To = , D

: 1 N


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: String Equ(Str1, Y1): String Equ(Str2, Y2): String Equ(Str3, Y3): String Equ(Str4, Y4): While abs(X-A) > 1E-12 or abs(Y-B) > 1E

12 or abs(Z-C) > 1E 12 or abs(T-D) >1E-12

: Radian

: N+1 N: if (N


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: Disp CONTINUE

: Goto 1

: Else: A X: B Y: C Z: D T: nDeriv(Y1, X, X) E: nDeriv(Y1, Y, Y) F: nDeriv(Y1, Z, Z) G: nDeriv(Y1, T, T) H: nDeriv(Y2, X, X) I

: nDeriv(Y2, Y, Y)

K


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: nDeriv(Y2, Z, Z) L: nDeriv(Y2, T, T) M: nDeriv(Y3, X, X) O: nDeriv(Y3, Y, Y) P: nDeriv(Y3, Z, Z) Q: nDeriv(Y3, T, T) R: nDeriv(Y4, X, X) S: nDeriv(Y4, Y, Y) U: nDeriv(Y4, Z, Z) V: nDeriv(Y4, T, T) W: det([[E, F, G, H][I, K, L, M][O, P, Q, R][S,

U, V, W]]) J: if(J=0)


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: Then: Disp NO ROOT, MULTIPLICITY ROOT.

OR.XO, YO

: Goto 1: Else: X det ([[Y1, F, G, H][Y2, K, L, M][Y3, P,

Q, R][Y4, U, V, W]])/J A: Y det ([[E, Y1, G, H][I, Y2, L, M][O, Y3,

Q, R][S, Y4, V, W]])/J B: Z det ([[E, F, Y1, H][I, K, Y2, M][O, P,

Y3, R][S, U, Y4, W]])/J C: T det ([[E, F, G, Y1][I, K, L, Y2][O, P, Q,

Y3][S, U, V, Y4]])/J A


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: End

: End

: End: Disp X= ,X

: Disp Y =, Y

: Disp Z =,Z: Disp PRESS QUIT TO EXIT (1)

: Goto 1


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Ex : Solve the following unlinear equations :

Solution : The TI calculator will have result in 2 minutes

x=1.073805344; y=1.896586209; z=13.99719752;t=3.999983857

x=0.5624970866; y=1.514473157; z=17.34656344;t=3.995849184

x=0.2226617235 ; y=1.081330382; z=23.60485634 ;t=3.992903209

x=0.3465996175 ; y=1.267952118; z=20.3744967 ;t=3.994052946

x=2.130906063 ; y=0.6207813991 ; z=11.11177199 ;

bai bao cao khoa hoc tai bac kinh 2009 thescientificreportatcm 2009

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