Series and the Sigma Notation (Part 1)
Sometimes, when we're working on problems, we run into situations where we have to add together a list of things. For example, a polynomial like x^{4} + x^{3} + x^{2} + x + 1 is just the sum of five monomials. Writing out long sums can get tedious so mathematicians have a short hand way to do it that's based on the Greek letter sigma (Σ). Using this notation has four parts:
The a_{i} is the same as it was when we were talking about sequences - a formula that tells you what the individual terms of the sum look like. To write out the sum, you would evaluate the formula for every integer starting at m and counting up to n then add the results together.
If we start with a sequence, {a_{i}}, and add its terms together, the result is called a series.
Example 1 - Writing Out a Series
Write out the series $\sum_{i=1}^4 (i + 1)$.
The i = 1 part tells us that i is going to start at 1 and the 4 at the top tells us that it's going to go up to 4. If I evaluate (i + 1) for every integer between 1 and 4 ,I get
i | i + 1 |
1 | 1 + 1 = 2 |
2 | 2 + 1 = 3 |
3 | 3 + 1 = 4 |
4 | 4 + 1 = 5 |
To get the value of the sum, all we have to do is add up the values in the right hand column:
$$\sum_{i=1}^4=(i + 1)2 + 3 + 4 + 5 = 14$$Example 2 - Writing Out a Series
Write out the series $\sum_{i=1}^5 x^i$.
i | x^{i} |
1 | x^{1} |
2 | x^{2} |
3 | x^{3} |
4 | x^{4} |
4 | x^{5} |
If we add those five terms together, the sum becomes:
$$\sum_{i=1}^5 x^i= x+x^2+x^3+x^4+x^5$$Examples 1 and 2 illustrate an important fact about series that you should be aware of: Sometimes they simplify to a number and sometimes the expanded version will still have a variable in it.
Example 3 - Writing Out a Series
Write out the series $\sum_{i=2}^4i^2$
Instead of writing out the table, I'm going to take the more common approach and write the series out on a single line. To help you see where the term comes from, I'm going to write the value of i underneath it.
= | 2^{2} | + | 3^{2} | + | 4^{2} | |
i = 2 | i = 3 | i = 4 |
Now we can get the final result by adding those values together.
$$\sum_{i=2}^4i^2 = 2^2 + 3^2 + 4^2 = 4 + 9 + 16 = 29$$Example 4 - Writing Out a Series
Write out the series $\sum_{i=0}^3 2i x^{i+1}$.
This time, I'm going to write the result the way we would usually expand a series by just writing out the terms. Each of the terms comes from a value of i starting with 0 and ending with 3.
$$\sum_{i=0}^3 2i x^{i+1} = (2 \cdot 0)x^{0 + 1} + (2 \cdot 1)x^{1 + 1} + (2 \cdot 2)x^{2 + 1} + (2 \cdot 3)x^{3 + 1}$$ $$\sum_{i=0}^3 2i x^{i+1} = 0x^{1} + 2x^{2} + 4x^{3} + 6x^{4}$$ $$\sum_{i=0}^3 2i x^{i+1} = 2x^{2} + 4x^{3} + 6x^{4}$$Sometimes, you'll find yourself in situations where you start with a sum, e.g. 1 + 2 + 3 + 4 + 5 + 6, and you want to simplify it by writing it using the Σ notation. Doing this can significantly simplify a series but it often takes some creativity.
Example 5 - Writing a Series
First, notice the ... in the expression. That's a shorthand way of saying, "I'm not going to write out all the terms. Just assume that they follow the same pattern." In this kind of question, identifying that pattern is the key to simplifying the expression.
The first thing I see is that the exponents on the variable start at 1 (on the x) term and increase by 1 in each term up to 12. That makes me think that part of the formula is going to be x^{i}. The next thing I see is that the numbers in front of the x start at 1 and increase by 1 in each term up to 12. (Think of the x as being 1x.) This makes me think that I can get the finished formula by putting an i in front of it: $ix^i$.
The sum starts at 1 and ends at 12 which makes my final result
$$\sum_{x=1}^12 x^i$$Example 6 - Writing a Series
To answer this kind of question, you have to bring some creativity to the table. When I look at the numbers in the series, the first thing I notice is that they're all powers of 2:
2^{0} + 2^{1} + 2^{2} + 2^{3} + 2^{4}
Notice how the exponents start at 0 and increase by 1 in each term. That's the behavior that the i variable has in our series. That makes me think that I can rewrite the series as
$$\sum_{i=0}^4 2^i$$Each series is unique so there's no formula that will always take you from a series to its Σ version. (It may not even be able to write a series that way, e.g. if the series is made up of random numbers.) There are a few things I always start off looking for.
- Look for numbers that increase by 1 in every term, e.g. the exponents in Example 4 went from 1 to 2 to 3, etc. When you see that, you can replace those numbers with i in your formula.
- Look for numbers that increase by a constant amount, e.g. 2, 5, 8, 11 starts at 2 and increases by 3 so you could write it as (2 + i).
- Look for numbers that increase by a constant exponent, e.g. 1, 3, 9, 27, 81, 243 are all powers of 3: 3^{0}, 3^{1}, 3^{2}, 3^{3}, 3^{4} so you would write them as 3i.
- Look for numbers that increase by a constant multiple, e.g. 5, 10, 15, 20, 25 are all multiples of 5: 1 · 5, 2 · 5, 3 · 5, 4 · 5, 5 ·5. You can replace the part that increases by 1 with i which makes the formula 5i.
- Look for factorials. This can be tricky to see because you probably aren't as familiar with factorials as you are with multiples so, if you get stuck, and the numbers seem to be increasing pretty quickly this is a good place to go. For example, 1, 1, 2, 6 is equal to 0!, 1!, 2!, 3! so, in a formula, it would become i!.
- Look for alternating signs, e.g. -1, +2, -3, +4, -5. You can write alternating signs by using (-1)^{i}. When i is an even number (-1)^{i} is equal to +1 but when i is an odd number, it's equal to -1. That means that I could write -1, +2, -3, +4, -5 as (-1)^{i} · i.