task4 present
TRANSCRIPT
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TASK 4:Sequences,
Mathematical Induction and Recursion
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Definition :
ordered set of mathematical objects.
SEQUENCES
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Find the sum of all the integers from 1 to 1000.
The sequence of integers starting from 1 to 1000 is given by
1 , 2 , 3 , 4 , ... , 1000
The above sequence has 1000 terms. The first term is 1 and the last term is 1000 and the common difference is equal to 1. We have the formula that gives the sum of the first n terms of an arithmetic sequence knowing the first and last term of the sequence and the number of terms (see formula above).
s1000 = 1000 (1 + 1000) / 2 = 500500
EXAMPLE PROBLEM
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Sequeance
Arithmetic Sequences
Geometric Sequences
Special Integers
Sequence
TYPE OF SEQUENCE
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In an Arithmetic Sequence the difference between one term and the next is a constant.
In other words, you just add the same value each time ... infinitely.
WHAT IS ARITHMETIC SEQUENCE
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In General you could write an arithmetic sequence like this:
{a, a+d, a+2d, a+3d, ... }
where:
a is the first term, and
d is the difference between the terms (called the "common difference")
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In a Geometric Sequence each term is found by multiplying the previous term by a constant.
In General you could write a Geometric Sequence like this:
{a, ar, ar2, ar3, ... }
where:
a is the first term, and
r is the factor between the terms (called the "common ratio")
Geometric Sequence
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• This Triangular Number Sequence is generated from a pattern of dots which form a triangle.
Triangular Numbers
• Square numbers, better known as perfect squares, are an integer which is the product of that integer with itself.
Square Numbers
• The Fibonacci Sequence is found by adding the two numbers before it together.
Fibonacci Numbers
• A cube number sequence is a mathematical sequence consisting of a sequence in which the next term originates by multiplying the number 3 times with itself, or in other words, raising it to the power of three
Cube Numbers
SPECIAL INTEGER SEQUENCE
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An example of this type of number sequence could be the following:
1, 3, 6, 10, 15, 21, 28, 36, 45, …
This sequence is generated from a pattern of dots which form a triangle. By adding another row of dots and counting all the dots we can find the next number of the sequence.
Triangular Numbers
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Continue….
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1, 4, 9, 16, 25, 36, 49, 64, 81, …
The next number is made by squaring where it is in the pattern.
The second number is 2 squared (22 or 2×2)
The seventh number is 7 squared (72 or 7×7) etc
Square Numbers
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0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …
The Fibonacci Sequence is found by adding the two numbers before it together.
The 2 is found by adding the two numbers before it (1+1)
The 21 is found by adding the two numbers before it (8+13)
The next number in the sequence above would be 55 (21+34)
Fibonacci Numbers
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1, 8, 27, 64, 125, 216, 343, 512, 729, …
The next number is made by cubing where it is in the pattern.
The second number is 2 cubed (23 or 2×2×2)
The seventh number is 7 cubed (73 or 7×7×7) etc
Cube Numbers
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Give 1 example sequences use in computer programming
Sequence in computer programming is the default control structure, instructions are executed one after another. They might, for example, carry out a series of arithmetic operations, assigning results to variables
Example:-
Make a list number of 100 students that attend “seminar keusahawanan”.
Pseudocode
BEGIN
Declare a list as an array with 100 elements
for (int student = 0, student<100; student++){
list[student] = student + 1 (arithmetic operation)
print list[student]
}
END
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Answer:
No.Student = 0
Print list[student] = 1
No.Student = 1
Print list[student] = 2
.
.
.
.
Print list[student] = 99
No.Student = 100
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Source Code
public class sequences {
public static void main(String[] args) {
int list [] = new int[100];
for(int student = 0; student<100; student++) {
list[student] = student + 1;
System.out.println(list[student]);
}
}
}
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PRINCIPLE OF MATHEMATICAL INDUCTION
To prove that P(n) is true for all positive integers n, where P(n) is a propositional funtion
METHOD OF PROOF
1) BASIS STEP: We verify that P(1) is true.
2) INDUCTIVE STEP: We show that the conditional statement
P(k) → P(k + 1) is true for all positive
integers k.
PRINCIPLE OF MATHEMATICAL INDUCTION &
METHOD OF PROOF
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Problem1:
Show that if n is a positive integer, then
1 + 2 + ... + n = n(n + 1)/2.
BASIC STEP: P(1),
1 = 1(1 + 1)/2
1 =1 (RHS=LHS)
INDUCTIVE STEP: we assume that 1 + 2 + ... + k = k(k + 1)/2
it must be shown that P(k + 1) is true,
1 + 2 + ... + k + ( k +1) = (k + 1)[( k+ 1) (k +1) + 1] /2 = (k + 1)(k + 2)/ 2
Add k +1 to both sides of the equation P(k)
1 + 2 + ... + k + (k +1) = k(k +1) /2 + (k +1)
= k(k +1 ) + 2 (k +1 )/2
= (k + 1) (k + 2)/2
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Problem 2:
For all n≥0
= n(n+1)(2n+1)/6
Basis:
=0
Induction:
= [n+1][(n+1)+1][2(n+1)+1]/6
= (2n3 +3n +n)/6
= (2n3 +9n +13n+6)/6
() + (n+1) = (2n3 +9n +13n+6)/6
(2n3 +3n +n)/6 + n +2n+1 = (2n3 +9n +13n+6)/6