When you first start
Sometimes when you are counting books at a library for parking, you select them off the shelves in fives or fours and you know you have now 105 books as you knew you had 100 before, for instance. When you have counted 40 times, you know you now have 80. So you need to store the counts only when you are parking the stuff, instead of putting the whole park in your head.
Now in the field of biology, cell that live by themselves called bacteria multiply asexually using mitosis. This means that they simple divide themselves into two when they become larger than life. Now say there were just one bacteria in a petri dish or culture, and it feeds and grows. Finally it splits in two, and not stopping there these two grow and splits after 20 secs. How many will you have after a hundred seconds? You'll have have 1, 2, 4, 8, 16, 32 cells. These may now be more formidable than helpful...
The two processes above are called progressions; with the first one called an Arithmetic Progression and the second one called a Geometric Progression.
The progression is called a 'sequence' and could be a 'finite series' or an 'infinite series'. Both processes above are finte because realistically, you can't continue counting books forever or bacteria can't keep growing in any specific location due to food exhaustion, predators and breaks in time caused by funny factors like temperature and static. But in theory there are 'infinite series' which we could model using numbers only if you wanted.
Using Jumps or Cuts
Arithmetic progression is a process resulting in increases such that a specific, whether a positive or a negative number, can be added anytime you needed to get the next number in the sequence. Look below at the examples:
2, 4, 6, 8, 10, 12, 14........
100, 50, 0, -50, -100. -150, -200.......
10, 7, 4, 1.....
You can see that the first example has a positive difference and the other two has negative differences. The 'common difference' as it is called is added to get the next 'term' in the sequence. The first term of the first sequence is the number 2, and the fifth term of the third sequence is is the number ... -2, because 1 +(-3) is -2. So in the third sequence, -3 is added first to the fir term which is 10, then added until the person writing it is OK with he or she got or as outlined by a problem.
One each one
There is a formula to get each term of an Arithmetic Progression. The formula is written below:
Tn = T1 + (n-1)(common difference)
Tn - for the nth term
T1 - the first term
n - the term position to be found in the sequence
The common difference could be represented by d
consider the following problem:
Find the fourth, eleventh and twentieth term of the sequence:
10, 7, 4.......
You solve this problem by listing out the terms or parts of the problem:
Tn = ?
T1 = 10
n = 4
The common difference = 7-10 = -3
Then solving it by placing the values in the formula
T4 = 10 + (4-1)(-3)
T4 = 10 + (-9)
T4 = 1
The other problems I will leave you to solve by your selves.
Aggregation
What about when you want to find the sum of the series, that is, you want to know how much totals you have packed up? There is a formula for that too.
Sn = (n/2)(2(T1) + (n-1)(common difference))
Sn - sum for the n terms
T1 - the first term
n -> the term position to be summed up to in the sequence
The common difference could be represented by d
What this means is that when we have 2, 4, 6, 8, 10, 12, 14........, what is the sum up to 12 for instance, or 14. this will be equivalent to doing 2 + 4 + 6 + 8 + 10 + 12. The answer here is 42. If we use the formula we get
S6 = (6/2)(2(2) + (6-1)(2))
= (3)(4+(5)(2))
= (3)(4 + 10)
= (3)(14) = 42
For the problem we used for each term, we could find the sum of the number until the 14th term.
Sn = ?
n =14
T1 = 10
The common difference = 7-10 = -3
Using a consistent Scaling
When you have a list of numbers like 4, 16, 64, ... at least from these three you can see a relationship growing. There is a four problem: the list starts with 4 and is multiplied on and on by 4, at least for those three. What about 20, 5, 5/4, 5/16, 5/64...? In this case you still have a consistent multiplicative but it is a fraction, and you could rightly say that you have a consistent divisor. But we are talking about cases where the multiplicative number is never changed. This kind of sequence is called a Geometric Progression.
There is a formula for this progression too:
Tn = arn
Tn - the nth term
a - the first term
r - the multiplicative constant or 'common ratio'
The following example can illustrate how to use this formula: If we have these numbers 24, ....., 2/12, x, 1/96...
can you find the multiplicative constant here? Solved below:
a = 24
r = x/(2/12)=(1/96)/x
x2 = 2/((12)(96))
x = 1/24
therefore r = 1/4
Summing the Parts
When you need to find the sum of a geometric progression, starting from the beginning there is a formula for that too:
Sn = a(1-rn)/(1-r)
Sn - the sum until n terms
a - the first term
r - the muiplicative constant or 'common ratio'
The ratio (1-rn)/(1-r) could be (rn-1)/(r-1), when r is greater than 1, not that it will make any difference in the ratio's value.
We could solve for the sum of the progression above until the 10th term.
a = 24
r = 1/4
S10 = 24(1-(1/4)10)/(1-(1/4))
S10 = 24(1-(1/1048576))/(1-(1/4))
S10 = 24(1048575/1048576)/(3/4)
S10 = 24(3145725/262144)
Try to work that out...
Proof for the formulea above will be added...
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