Series

So far, we have been learning about sequences. Today, we will take this further to learn about series. But before I do that, let me show the answers to the questions I posted yesterday.

1) 131072/5   2) 6   3) 12

Now to series. What exactly is a series? No, it is not a series of books, but rather it is the sum of all the terms in a sequences. The series of an infinite arithmetic sequence, such as 1, 2, 3…, is impossible to calculate, but it is always possible to find the series of a finite sequence. One simple but time-consuming way to solve for the series is to simply add up the terms one by one. But this doesn’t make math beautiful; in fact, it makes it seem clumsy. Rather, however, in certain types of sequences there are beautiful formulas that can be used to calculate a series.

Carl Gauss

Mathematician Carl Gauss was the first one to find a formula for an arithmetic series. Legend says that in his second grade class, he and his class was given the problem 1+2+3…+98+99+100=?. This is essentially just finding the series of the arithmetic sequence 1,2,3….98,99,100. While all his classmates were trying to solve this problem the stupid way, genius Gauss solved it in three seconds. How? Well, let’s think about it. If we took a1 which is 1 and summed up it with the last term, 100, we get 101. If we took the second term 2 and added it up to the second to last number 98 we also get 101. We can keep on adding the terms like this, until we approach to the very center. By the time we do that, we will have 50 101s. So, to get the total sum do 50 (101)=5050. Wala, this is the series!

So, from here, we can assume a basic formula for an arithmetic series, which is given a is a term and n is the number of terms and d is the constant difference

This formula happens to not only work for even number of terms but also for odd number of terms.

Well, that was the formula for an arithmetic series. How about a geometric series? Well, from our work with sequences, we know that a general geometric series would be:

a+a*r+a*r^2+…+a*r^(k-2)+a*r^(k-1) for k terms. Let this equal S or series. The next terms would be a*r^k, so if we add it to both sides:

a + a *r + a* r^2 +…+ a* r^(k−2) + a* r^(k−1) + a* r^k = S + a* r^k  (Now move a to the other side)

a *r + a* r^2 +…+ a* r^(k−2) + a* r^(k−1) + a* r^k = S + a* r^k – a (Now factor an r at the left side)

r  (a + a* r +…+ a* r^(k−3) + a* r^(k−2) + a* r^(k−1))= S + a* r^k − a  [Now notice that the sum within the parentheses is equal to S, so..]

r*S= S + a* r^k − a   [Now simply solve for S using algebra to get…]

where in this case n=k

Magic! This is the formula for the geometric finite series. How about geometric infinite series? Well, it turns out the only type of infinite series you can solve are infinite geometric whose terms’ absolute value keep on going lower and lower. One example would be the series of the sequence 1/2. 1/4. 1/8…. Here, the sequence approaches 0 but never meets it. However, to use the previous formula to solve this kind of series, you should substitue (1-r^n) with 0, thus the formula being a/(1-r).

Well, that’s it for series. If you want to know more, I suggest more research on y0ur own. For my next post, I will perhaps talk and discuss about a short story that I will introduce tomorrow.

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Sequences Part 2

Let us continue with introducing the two types of sequences. The first one is the arithmetic sequence, defined as a sequence with a constant difference between consecutive terms. The constant difference is represented by the variable d and the first term by the variable a. For instance, the sequence of even numbers is an arithmetic sequence, because all the terms are consecutive with a difference of 2, which is d. Variable a would be 0, the first even number. As you can expect, there are formulas for an arithmetic sequence, recursive and direct. Xi = Xi-1 + is the recursive formula, as you can see that Xi-1 is the previous term and must be added a certain amount d in order to get to the next term. For direct, the formula is Xi = X1 + (i – 1)d.

Geometric sequence is a sequence with a constant ratio between consecutive terms. In other words, in order to get from one term to the next, you have to multiply or divide by a constant r. One example is 2,4,8,16,32, etc. Here, r is 2 because in order for one term to go on to the next, you have to multiply by 2. This all translates into a geometric sequence’s recursive formula: Xi = r(Xi-1). The direct formula in this case is Xi = a(ri-1), where r is the constant ratio and a is the first term.

Now it’s time to put your sequences knowledge from yesterday and today to the test. Here are a few questions for you to try out:

1) What is the 20th term in the geometric sequence 1/20, 1/10, 1/5, 2/5…?

2) If the 6th term of an arithmetic sequence is 21 and the constant difference is 3, then what is the first term?

3) Finally, there is a geometric sequence an with n= 1,2,3…etc. Assume a1=2 and a3=4. Find the value of a7. 

Tomorrow, I will post up just the answers. If you want to know the process, contact me tomorrow after I post up the answers. I will also for tomorrow introduce series and present the proof of the formula of a geometric series. That’s all for today.

Sequences Part I

In one of my past math posts (12/21), I talked about permutations and combinations. Today, we will be stepping off of that and getting into a little bit of series/sequences.

First of all, we need to know what sequences are. A sequence is a list of objects presented in a particular order. The objects in a sequence are called the terms of the sequence. For example, 2,4,6… is a sequence of positive even numbers. The first term here is 2, the second term is 4, and so on. Pretty simple concept. Now, another simple thing to know is that the index of a term is the position of a term. In our previous example, 4’s index is 2 because it is the second term. Mathematicians notate this as x2=4, with the index being the subscript.

Sometimes, there is a relationship between the index and the term. For instance, notice that in even numbers, the term is two times its index. For instance 4 is two times its index-2, and 6 is two times its index-3. We can notate this as { xi=2i}. Let’s try out this question:

What sequence is generated by {xi=i^2}?

Well, lets start solving for the first term, which would equal x1. Meaning i=1. So we plug in 1 into i^2, which would be 1^2=1. So 1 is our first term. For our second term, we also simply go through the same process, plugging 2 into i^2, which would equal 4. For the third, we plug in 3, equaling 9. For 4, we plug it in equaling 16. So the sequence goes like this: 1,4,9,16….

Try out this question for yourself. What formula will generate the sequence 3,7,11, 15…? Do the formula in terms of  xi. If you can’t answer and want to know how, contact me.

So far, all the sequences we have considered are infinite sequences, meaning they go on forever. However, not all sequences are like this. Some do end, or are finite. For instance, let’s say we shorten the infinite sequence of positive even numbers to start at 2 and stop at 10. We would notate this the same way as we did with the infinite sequence of even numbers, but with a few exceptions. After the end bracket, we would add a subscript (starting with i=) to denote the starting value as the index. In this case, the subscript would be i=1, because 2 is the first term. We would also add a superscript, denoting the ending value of the index, which in this case is 5 because 10 is the 5th term.

In this type of subject, there are two kind of formulas: recursive and direct formulas. Recursive formulas are formulas for a sequence that declare the starting value for that sequence and how a subsequent term is made from the previous term. For instance, take the sequence 3,7,11,15,19… We first declare the starting value like this: x1=3. Then we add a semicolon. We know that in order to get from one term to the next ,we add four. So let xbe the term we want to find. The previous term would be xi-1 and for the term to move onto xi, we add four so the recursive formula is xi= xi-1+4. Note that in order to use this, we have to know the previous term xi-1.

So what are direct formulas? It is any formula where xis expressed not in terms of xi-1, for instance like {xi=i^2}. It is a formula that you state directly and can be used without knowing the previous term.

For tomorrow, I will introduce the two types of sequences.

Permutations/Combinations

For today, I will answer the questions from yesterday’s post. But before I do that, I will need to present some mathematical concepts.

The first concept is a simple one. It is the Multiplication Principle. Let’s say there are five apples on a table. How many ways is it possible for me to choose three apples? Well, for my first apple, I have five choices. For my second, I now have four, since I took one away for my first apple. For my third, I am left with three choices. Thus, I multiply 5x4x3 to get the total possibilities of choosing three apples. In textbook terms,  when listing out all the possibilities for k items, the total number of entries in this list is given by n1 ⋅n2⋅n3 ⋅ ⋅ ⋅ nk , where nk is the number of possibilities for the kth item. For example, n3 is the number of possibilities for the third item.

Another important concept is the factorial, symbolized by “!” A factorial is the product of all whole numbers starting from the number indicated all the way down to 1. For example, 5! or five factorial is 5x4x3x2x1, as 3! is 3x2x1.

So what is a permutation? A permutation is an arrangement of objects from a group where no object can be used more than once and the order of selection matters. For example, let’s say I have to make a 5-letter code without replacement (meaning I can’t use a letter more than once). Using the Multiplication Principle, I have a total of 26x25x24x23x22 choices. This is a permutation because no letter can be used more than once, and the order of selection matters because placing a letter Z as the first letter is different from placing the letter Z as the last letter. You will see this type of situation happening a lot and there is a formula for it. It states that the total number of permutations P of k objects from a group containing n objects is given by the formula

I can derive this formula, but since this is a blog post, I will not do so. Perhaps for a later time.

We of course need to know what combinations are. Combinations are arrangements of objects from a group where no object can be used more than once and the order of selection does not matter. One example would be choosing four candidates out of ten for a committee. Say the four candidates were Joe, Moe, Steve, and Larry. First, one candidate cannot be selected for a position more than once. Second, it wouldn’t matter if the order was Joe, Moe, Steve, and Larry or Moe, Steve, Joe and Larry or any other order. It is still the same committee, and thus the order of selection does not matter. By definition, this is a combination. Just like permutation, there is also a combinations formula. The total combinations C of k objects from a group containing n objects is given by the formula

Figure 2. Combinations formula

 Notice the middle part of the formula above, which has a parentheses  This is a notation for combinations and is read as “n choose k.”

I gave you the concepts; now let’s used them to solve our questions. The first question being:

At the beginning of math class every day, Mr. Smith selects students to write up homework problems on the board. These problems can be discussed as a class. There are 26 students in Mr. Smith’s math class, and he randomly selects with replacement a student to write up each of the first five problems. How many different ways can students be assigned to the problems?

Note that permutations and combinations occur without replacement. In this case it is not, so none of these situations occur. So what do we use? The Multiplication Principle, that is. To select the first student, Mr. Smith has 26 choices since he has 26 students. For the 2nd, he also has 26 choices because of the words “with replacement.” He also has 26 choices for the 3rd, 4th, and 5th students. Therefore, the equation is 26x26x26x26x26 or 26^5, which equals 11881376 different ways.

First question solved. Just remember that when permutations or combinations can’t be applied, always start off with the Multiplication Principle. Now onto the second question:

A regular polygon is a polygon with equal angle measures and equal side lengths. A diagonal of a polygon connects two non-adjacent vertices. How many diagonals are there in a regular heptadecagon (17-sided polygon)?

Instead of taking this in terms of geometry, let us take it in terms of what we are learning. Since there are 17 angles, let each angle be A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, and Q. A diagonal is connecting two non-adjacent angles, so we are trying to find all the possible ways two angles can be selected. Notice that this is a combination, because no pair can be made more than once and that order of selection also doesn’t matter, since AC and CA are the same diagonal. Now, in order to use the combination formula, we first have to know what n and k is. N is the total no. of objects, which in this case is 17 angles. K is how much we select from these 17 angles, which in this case is 2, because two angles make up a diagonal. So let’s plug it in.

17!/(17-2)!2!=(17*16)/2=136

 Wala! The answer is 136. Just kidding. It is in fact not. Why? Because we forgot to take into account that this only works when each angle has 16 diagonals, or in other words, when any two angles can form a “diagonal”. However, this is restricted by the fact that diagonals are only formed by non-adjacent angles. Meaning only each angle has 14 diagonals, in fact. So what do we do in this case? Well, we don’t want those “diagonals” formed between adjacent angles, so why not subtract them. Note that these “diagonals” are in fact the sides of the polygon. Using this fact, we do the equation 136-17 which equals 119 diagonals in a heptadecagon. (If you don’t get this, contact me on what you don’t understand.)

Now, the last question: George is traveling to New York City and has created a list of ten different possible sightseeing activities in which he is interested. He will be in New York City for three days, but will only have time for two different activities each day. How many different sightseeing plans can George create? (Assume each day is treated separately, and clearly George will not want to complete each activity more than once.)

The first thing to notice is that this is a permutation, because one activity cannot be done more than once, and whether Activity Y is the first on the agenda or second can result in different plans, and thus order of selection matters. Now, we must know what n and k is. N in this case is the total no. of events, which is 10. However, George can choose only 6 events, with 2 events/day for 3 days. So, let’s plug n and k into the permutations formula:

10!/(10-6)!= 151200

 Therefore, there are 151,200 different possible sightseeing plans.

Well, this is just a basic overview over permutation/combinations. To see more in-depth, I advise you to continue researching on this amazingly complex subject. Somewhere soon into the future, we will step off of what we learned today and begin to learn about series.

A Few Math Problems To Start With

Hello guys. I initially wanted to post a few problems from the IMO, but I decided to start off easy. So for today, try out these relatively simple problems.

1) At the beginning of math class every day, Mr. Smith selects students to write up homework problems on the board. These problems can be discussed as a class. There are 26 students in Mr. Smith’s math class, and he randomly selects with replacement a student to write up each of the first five problems. How many different ways can students be assigned to the problems?

2) A regular polygon is a polygon with equal angle measures and equal side lengths. A diagonal of a polygon connects two non-adjacent vertices. How many diagonals are there in a regular heptadecagon (17-sided polygon)?

3) George is traveling to New York City and has created a list of ten different possible sightseeing activities in which he is interested. He will be in New York City for three days, but will only have time for two different activities each day. How many different sightseeing plans can George create? (Assume each day is treated separately, and clearly George will not want to complete each activity more than once.)

Just to note, all these problems have to do with permutations/combinations. For tomorrow, I will introduce some concepts and go over these 3 questions.