Teachers Teaching with Technology™ Research & Exploration in the classroom Koen Stulens Hasselt University

Teachers Teaching with Technology™

Research & Explorationin the classroom

Koen StulensHasselt University


European Competitive Knowledge Economy

Gross Domestic Product

Lissabon Declaration (2000)Increase Investment in Research

20001,9 %

20103 %

+ 700 K Researchers

+ 15 % graduates in Science and Engineering


-10

-5

0

5

10

15

20

Comp ScienceMathScienceEngineer

Flanders Netherlands France Germany

Annual mean decrease/increase of graduates


99 - 00 1.223

00 - 01 1.214

01 - 02 1.034

02 - 03 863

03 - 04 814

04 - 05 782

Generation Students in Flanders

Mathematics, Physics, Biology, Chemistry, Computer Science

Researchers

GenerationStudents

= 2,7


Technology


Technology


Technology


TechnologyLess Math?More Math!

Different Approaches

Problem Solving SkillsReasoning Skills

Improve Research Skills

More Active Student Engagement

Gain a better insight in Math

Improve Student Achievement



Successive Integers

Armando M. Martínez Cruz


Successive Integers

Choose four successive integers a, b, c and d.

Make the sum a + b2 + c3.

Divide the sum by d.

Conclusion?

a=1, b=2, c=3, d=4

a + b2 + c3 = 1 + 4 + 27 = 32

(a + b2 + c3) / d = 32 / 4 = 8

Divisible!

Numerical Exploration


Successive Integers

Choose four succussive integers a, b, c and d.

Make the sum a + b2 + c3.

Divide the sum by d.

Conclusion?

a=1, b=2, c=3, d=4

a + b2 + c3 = 1 + 4 + 27 = 32

(a + b2 + c3) / d = 32 / 4 = 8

Divisible!

Symbolical GeneralizationX

1X 3X2X

2 3( 1) ( 2)

3

X X X

X

X 1X 3X2X

2 3( 1) ( 2)

3

X X X

X

Symbolical Generalization


Successive Integers

2 3( 1) ( 2) X X X

? ?2( 1)( 3) X X

X 1X 3X2X

2 3( 1) ( 2)

3

X X X

X

Symbolical Generalization

2 31 ( 1) ( 2) Y X X XGraphical Exploration


Successive Integers

Graphical Exploration

Conjecture And See


Conjecture And SeePlot a cubic function with three zeros a, b and c.

Plot the tangent line in the midpoint between a and b. Conclusion?



Optimization


Optimization

Classical Box Problem

Geometrical exploration Numerical representationGraphical representation


OptimizationClassical Box Problem

GRAPHICAL EXPLORATION & MODELLING


Optimization

x

V

x

RESUD

Classical Box ProblemSTATISTICAL EXPLORATION


Optimization

3 2( ) (8 2 )(4 2 ) 4 24 32 V x x x x x x x

2'( ) 4(3 12 8) 0

2 3 2 32 0.8453 2 3.1457

3 3

V x x x

x x

2'( ) 4(3 12 8) 0

2 32 0.8453

3

V x x x

x

''( ) 24( 2)

2 3''(2 ) 16 3

3

V x x

V

2 3 64 3(2 ) 12,3168

3 9 V

Classical Box ProblemSYMBOLICAL EXPLORATION


OptimizationAngle of View


Optimization

Angle of View


Optimization

A numerical solution

( ) sin( )( 2 ) O x x x

'( ) cos( )( 2 ) 2sin( ) O x x x x



Code Theory


Caesar code

100-44 BC

w i s k u n d e

z

l

v

n x q g h

a b c d e f g h i j k l m n o p q s t u v w x y zr

d

e

f

a b c

g h i j k l m n o p q s t u v w x y zr

+ 3

w i s k u n d e

+ ?+ 23

w i s k u n d e

22

8 18

10 20 13 3 4

l

v

n x q g h

z

25 11 21 13 23 16 6 7

0 1 2 25

- 3

x y z

23 24 25

BUT

26 27 28

+ 3

0 1 2


9 + 5 = 29 + 5 = 149 + 5 = 2

+ 2+ 1+ 4+ 3+ 5

9 + 1 = 109 + 2 = 119 + 3 = 129 + 4 = 19 + 5 = = 2

14 2mod12mod(14,12) 2

Modulo


0a

iParta

x a nn

0a

iPart 1a

x a nn

mod( , )a n


inString(Str1,Str2)-1-KA26NprgmMOD (A modulo N A)sub(Str1,A+1,1)Str2

Decoding

Caesar code & TI-84 Plus

Input Character & Key

Definition Characters

Coding

Output Code

"ABCDEFGHIJKLMNOPQRSTUVWXYZ"Str1

Disp "MESSAGE"Input Str2Disp "KEY"Input K

inString(Str1,Str2)-1+KA26NprgmMOD (A modulo N A)sub(Str1,A+1,1)Str2

Disp Str2

MStr2

20K

32A

GStr26A

G

-14

12M


Disp Str0

inString(Str1,Str3)-1+KA26NprgmMOD (A modulo N A)Str0+sub(Str1,A+1,1)Str0

Caesar code & TI-84 Plus

Input Message & Key

Definition Characters

Coding

Output Code

"ABCDEFGHIJKLMNOPQRSTUVWXYZ"Str1

Disp "MESSAGE"Input Str2Disp "KEY"Input K

inString(Str1,Str2)-1+KA26NprgmMOD (A modulo N A)sub(Str1,A+1,1)Str2

Disp Str2

WISStr2

20K

42A

Str0 + Q16A

Q

" "Str0

Endsub(Str0,2,length(Str0)-

1)Str0

For(θ,1,length(Str2)) sub(Str2,θ,1)Str3

θ = 1WStr3θ = 2IStr328A

2AStr0 + C

θ = 3SStr338A

12AStr0 + M

QCQCM


Str0


Public Key Systems

PublicFuncX C

SecretFuncX

M

M

Y X

PublicFunc SecretFunc

Condition

SecretFuncX(PublicFuncX(M))=M


PublicFuncX C

SecretFuncX

M

M

Y X

PublicFunc SecretFunc

RSA code

Secret e = 23 en n = 55

Public d = 7 en n = 55

M d mod n

C e mod n

? ?

1978 - Rivest, Shamir & Adleman


RSA code

Two Primes

p = 5 & q = 11 n = pq = 55 en φ(n) = 40

Secret KeyDetermine e : 1 < e < φ(n) = 40 and gcd(e,φ(n)) = 1 e = 23 & n = 55

Public keyDetermine d : 1 < d < φ(n) = 40 and e.d = 1 mod φ(n)

e φ(φ(n)) ≡ 1 mod φ(n) e.eφ(φ(n)) - 1 ≡ 1 mod φ(n)

e φ(φ(n)) - 1 = 2315 and 2315 ≡ 7 mod φ(n) d = 7 en n = 55

Euler’s theorem

If gcd(a,n) = 1 than aφ(n) ≡ 1 mod n


RSA code M

Secret e = 23 en n = 55

Public d = 7 en n = 55 PublicFunc(x) = x d mod n

SecretFunc(x) = x e mod n

M d

(M d ) e

(M d ) e ≡ M d e ≡ M 1 + kφ(n) ≡ M.(M φ(n)) k

e.d = 1 mod φ(n)

≡ M mod n


RSA code



Boyle’s Law


Boyle’s Law

Volume versus Pressure



THE END

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Teachers Teaching with Technology™ Research & Exploration in the classroom Koen Stulens Hasselt University