WEBVTT mathematics/pre-calculus/selhorst-jones
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Hi--welcome back to Educator.com.
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Today, we are going to talk about an introduction to series.
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In the previous lesson, we introduced the idea of a sequence--that is, an ordered list of numbers.
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With this idea in mind, we can now discuss the concept of a series, summing up the terms of a sequence.
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Now, at first, this might seem a little bit series.
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It is just addition, and we have been doing addition since kindergarten; so why do we need to talk about addition in a special way?
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But consider how long it would take to add up 100 terms by hand.
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Could you imagine how much time it would take up if you had to add up 100 different things from a sequence?
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What if it were a thousand terms, or ten thousand terms?
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The study of series can make these seemingly colossal summation tasks, having to add up huge numbers of numbers, really, really easy.
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We can just turn this stuff into being some trivially easy task once we figure out how to talk about series in a deep way.
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Being able to easily add up lots of numbers has many applications.
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You will see it in science, engineering, economics, computer programming, advanced math, and many other fields.
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All of these things benefit greatly from the study of series,
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from being able to talk about adding up a whole bunch of numbers in an easy way,
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where we can talk about it in compact notation--all of these things are really important to a variety of fields.
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So, this is a great thing to study; let's start by defining a series.
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Given some sequence a₁, a₂, a₃, a₄...
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a **series** is the sum of the terms in the sequence; it is just adding up the terms,
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or maybe a portion of them--maybe not the entire sequence.
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But it is just adding up terms from the sequence.
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If the sequence is infinite, and we are adding up all of the terms from the sequence, we call it an infinite series.
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It adds all of the terms together; so I am saying that it keeps going forever.
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It is a₁ + a₂ + a₃ + a₄...going on forever...
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so plus a₅, a₆, a₇, a₈...forever and ever and ever.
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Since our sequence was infinite, we are just saying to keep adding them forever and ever and ever.
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That is an infinite series.
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On the other hand, we can talk about a sequence that is not infinite--or if we only wish to add up a finite number of its terms.
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We aren't going to add them forever; we are just adding up a finite portion of them--just a number where we can count how many are there.
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We call this a finite series; adding the first n terms of the sequence together is called the nth partial sum.
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If we add a₁ + a₂ + a₃, up until a₁,
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but notice: it just stop there--we don't go any farther past a<font size="-6">n</font>--then that is the nth partial sum,
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because we are adding in all of the terms, 1, 2, 3, up until we get to n; so it is the nth partial sum,
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because it is only a part of our infinite sequence.
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Consider if we had a sequence to find by the general term a<font size="-6">n</font> = n + 2^n over 3^n.
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That is 1 + 2 over 3, 2 + 2² over 3², 3 + 2³ over 3³,
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4 + 2⁴ over 3⁴...what if we wanted to talk about its thirtieth partial sum?
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In talking about its thirtieth partial sum, we will see very quickly why we need a special form of notation for series.
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Why is writing this stuff out by hand going to be a real pain?
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If we wanted to write this out, we would have to first show the pattern that occurs here.
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We have this pattern that created each one of these various terms.
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We would have to have that show up in our summation; otherwise, we wouldn't be able to realize what pattern is going on.
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So, we are going to at least need the first three terms--at least.
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Then, we could use an ellipsis, that ... , to say that the pattern continues on in this manner.
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We can not have to write a massive amount, because we can use the ellipsis to say that the pattern keeps going.
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But we are still going to have to show its stop at the thirtieth term.
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At a very minimum, we are going to have to write out the first three terms, the last term, and an ellipsis in the middle.
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We would have to write out (1 + 2)/3 + (2 + 2²)/3² + (3 + 2³)/3³ + ...
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+ (30 + 2^30)/3^30; that is the thirtieth partial sum, like we were talking about.
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But that is a lot of stuff to write out; that is a lot of writing for a fairly simple series.
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This isn't even that complicated; but if we had to write this on multiple lines, as we worked through a problem--
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if we had to keep talking about this over and over as we did steps in a problem--your wrist is going to hurt after doing one of these problems.
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And you are going to have to do a bunch of problems about summation.
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This is why we need some sort of notation: we need notation to make it easier to compactly describe a series.
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We don't have to write out this really, really long thing every time we want to talk about some summation from a sequence.
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We need some easy way to be able to do it in a short way.
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Enter sigma notation: to compactly describe sums, we use something called sigma notation.
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It is also sometimes called summation notation.
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Why is it called sigma notation?--because it uses the uppercase Greek letter sigma.
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Sigma is this right here; if you are drawing it by hand (this is a computer-drawn picture of sigma,
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made by a computer typesetting program), I would recommend writing it like this,
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where you have these little vertical hooks on either end; and then it is just kind of a capital M on its side.
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Just write it out like that; and that is great.
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If you are feeling lazy, you can end up just sort of writing it like that, as well.
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But it helps to write it like this, so that we can clearly see what it is.
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But if you are doing a lot of problems that way, don't worry if it ends up getting not quite absolutely perfect each time,
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because you will be able to recognize it from other things that are about to be seen.
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All right, let's see how we use sigma notation.
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Let's look at the anatomy of a series in sigma notation.
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The first thing to look at is the thing being summed: the terms from the sequence given by a<font size="-6">i</font>, this thing to the right of the sigma, are added together.
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This is what will be added together over and over with each step as we add up:
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the first thing, and then the second thing, and then the third thing, and so on.
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It might be a sequence, but it is much more often going to be an algebraic expression--something like 3 times i plus 10 or 2 to the i or i!,
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something that we could plug in a number and be able to churn out a value.
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You will often see algebraic expressions; but once in a while, you will see a sequence like a<font size="-6">i</font>.
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Next, the index of summation; i is the index--it increases by 1 for each step of summation.
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We do the first thing, and then we add the second thing in the next step;
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and then we add the next thing in the next step; and we add the next thing in the next step, and so on, and so forth.
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So, every time we step forward, the index will click up by 1.
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We start at some value, and then we go to the value one above that, then the value one above that, then the value one above that, and so on.
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So, every step, it will increase by 1.
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The index can be any symbol; but i is very common, and that is what we will end up using for the most part in this course.
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So, that is the thing that ends up changing; and notice how i will generally occur over here in the series, as well.
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That is saying that this is the thing that changes; this is what will change with each step--this little thing right here.
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The lower limit of summation: this is the first value used for the index.
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This is where the series starts; in this case, since i = 1, our first value would be plugging in a 1 for the i.
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The upper limit of summation: this is the last value used for the index, where the series ends.
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We would step up until we eventually got to some value n.
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All right, let's see it in action: we have this pictorial summary of how it works.
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This is a great picture; so check against this, if you get used confused in a later point; this is a great slide right here.
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Let's see it in action: we have sigma...it's i = 3 for the starting index, and then 7 is the upper limit, and we have 2i - 1 as the actual expression.
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The very first thing: we have the i right here, and we start at i = 3.
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So, the first value that we plug in for our i is 2 times 3, because our lower limit of summation was that 3.
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So, 2 times 3 minus 1 is the first thing.
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Then, we go on to our next step: plus...we increase our index by 1.
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Our index started at 3, so it goes from 3...plus 1...to 4; so 2 times 4, minus 1.
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Then, we go on to our next step; it will be 2 times...we were at 4 in the last step, so it increases to 5; so we are at 4 + 1 is 5.
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2 times 5 minus 1, plus...our index increases by 1 again, so 2 times 6 minus 1, plus...
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and now, finally, we have just hit our final, upper limit.
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So, at this point, we stop; this will be the last one, once we click up to this final upper limit for how high our index goes to.
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That is the last one in the series; so it is 2 times 7 minus 1.
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And at this point, we could simplify it out if we wanted to.
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We could get a value from this; but that is also a good way to see how it expands.
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And that is all we are looking for right now--seeing how it expands.
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But if you wanted to, you could simplify each one of these.
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And you could also combine them and be able to get a value for that series.
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So far, we have only seen how to use sigma notation for finite sums--that is to say, partial sums, not sums that are going on forever.
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Since they have all had some upper limit of summation, something on top of this sigma, they have all had to eventually stop.
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They have had some top limit to how far they step out.
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If we want to show an infinite series, one where the series' terms keep adding forever and ever,
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we simply put an infinity symbol on top of the sigma.
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This shows that the series has no upper limit, and instead continues on forever.
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So, if we had a sigma with i = 1 on the bottom, our index is i, and our lower limit is 1,
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and our upper limit is infinity, that says really "just keep going."
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It is not that we actually get to infinity and stop; we can't ever get to infinity.
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Infinity is just the idea of going on forever; so we start at 1, and then on to 2, and then on to 3, and then on to 4, and then on to 5, and then on to 6...
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And it just keeps going forever and ever.
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So we would have...since it is a<font size="-6">i</font>, we would have a₁ first, because it was a 1 for our lower limit,
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then a₂, and then + a₃, and then + a₄, and then + a₅...
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and it will just keep going on forever, because it is an infinite series that we are working with.
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If we wanted to see in specific, here is an example that uses specific numbers.
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We have some series with i = 1 on the bottom; it is an infinite series, because of the infinity on top.
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We have 1/2 to the i; so our first i is 1; that is 1/2 (is our first term, one-half), plus 1/2...
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our next step with the i will be now at a 2, so 1/2 to the 2, 1/4...and then our next step would be i at 3;
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so 1/2 to the 3; that is 8; at our next step, i will be at 4, so that is 1/2 to the 4, so 1/16.
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And then, i will be at 5, so it is 1/2 to the fifth, so 1/32; and that pattern will just keep going.
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We will keep adding them on forever and ever.
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Sometimes it is useful to re-index a series (or a sequence, sometimes).
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We might have a series that has the index begin at one value, but we want it to start at another value.
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For example, we might want a transformation like this one here, where we have, in our original sequence,
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i = 7 as our starting index, and it has some upper limit, and it has some expression that we are actually working with.
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And we want to transform it to k = 1; we want our starting index to be k = 1.
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And we are going to, of course, also need some upper limit and something here.
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Now, of course, we can't just change the number of the lower limit.
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If we just went and changed the lower limit, that would affect the whole series.
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We have to pay attention to how the thing being summed, and our upper limit, end up making the transformation, as well.
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They are going to change over the process; they are not going to be the same thing on the left side and the right side.
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They are going to end up becoming different things; otherwise we would have altered the whole thing by starting...
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we start at one starting place, and then we just change the starting place, and we don't do anything else;
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well, if you start in a different starting place, you are going to have different values coming out.
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So, since we want to end up having the same value come out of our two ways of talking about it,
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we have to alter every part of the sigma notation, so that we can get the index we want without changing the value that comes out of it.
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You are probably asking why we care; if we have it written in one way, why not leave it that way?
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Well, there are a bunch of reasons to do it.
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For example, if you are working on a proof, it can sometimes help to re-index it.
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But specifically to this course, a lot of formulas that we end up working with will be given in the form of something with an i = 1 on the bottom.
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So, since a lot of our formulas will have i = 1 on the bottom, we will have to start at this index of something = 1,
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some symbol = 1, to be able to work with some of the formulas we have.
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A lot of the formulas that you end up seeing are in this format.
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So, it is really helpful to be able to re-index, so we get to the format that we already have a formula for,
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as opposed to having to figure out an entirely new formula for some new, different index.
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So, how do we actually re-index?
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The most essential way is to expand the sigma notation into a written-out series,
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then see how we could rewrite the pattern with our chosen starting index in mind.
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So, for example, if we want to go from i = 7 to k = 1, we can just start by...
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a series is just a shorthand way of writing out something + something + something...+ something.
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And even that is a shorthand way of writing out every single term.
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So, we can expand it into that format; if it is i = 7 first, then we would have 3(7) - 16;
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the next thing would be + 3 times 1 step up to 8, minus 16; plus...it would continue in this form.
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And 3 times...our final upper limit is 22; so 3(22) - 16; great.
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If we want, we can figure out how to write this sigma over here, with just that in mind.
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But we can also simplify things to see if we can see another easier-to-see pattern.
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So, 3(7) would give us 21, minus 16, plus...3 times 8...16, 24...minus 16 + ... 3(22) is 6, minus 16...
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so, 21 - 16 is 5, plus 24 - 16 (that is 8), plus...plus 66 - 16 (that is 50).
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We look at this, and we might realize, "Oh, what is doing is going up by 3 each time."
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And that makes sense, since we started at 3i.
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It is going up at 3 each time; if that is the case, and we want to get this k = 1, then that means
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that we know that here, we are at k = 1; here we are at k = 2; here we are at k =...
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we will actually leave that as a question mark right now, because we don't quite know yet.
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We will talk about that in just a moment--what number we are at; we will see why.
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k = 1; if it is + 3 each time, we are going to want some 3k + some number.
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So, if we are at 5 here, then that would be 3k + 2, because 1 for k...3 times 1 plus 2 does get us 5, so that checks out.
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Here, with 3k + 2, it works, as well; so 3k + 2 is here; we plug in 2; 3(2) is 6, plus 2 is 8; that checks out, as well.
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So, it looks like it is going to be some 3k + 2 that will be here; we will have 3k + 2 for that part right there.
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Now, what is the upper limit of summation going to end up being?
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Well, if we are at 3k + 2, and it equals 50, we have to figure out what value would end up coming out here.
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So, k = ?; we could solve for this: 3k = 48; k...divided by 3 now on both sides...we get 16.
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So, we know that our upper limit is going to have to be 16.
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Another way of doing it is to notice what the difference is between our upper limit and our lower limit.
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In our original, we had 22 as our upper and 7 as our lower; so, 22 - 7 means 15.
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Oops, I'm sorry; not 15...oh, it is 15; 22 - 7 comes out to be 15.
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Over here, we know that we are going to end up having that difference of 15, as well.
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Since we start at k = 1, 1 plus that same 15 (because we are going to have to have the same number of steps,
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however we phrase it) comes out to be 16, which is the same thing we figured out that it has to be to work there.
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With that in mind, we now know that our lower limit with our index is going to be k = 1,
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because that is what we wanted to figure out in the first place.
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Our top, our upper limit of summation, will be 16; we have figured out two different ways that that has to be the case,
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because 16 - 1 is 15, which is the same as 22 - 17; or alternately, we can solve for it
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using the formation that we came up with for that general term.
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And then, 3k + 2 is what is actually going in that is being added on each term.
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And so, that is one way of doing it.
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We worked this out by expanding first; expanding is a very specific way to see it.
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It is a great thing to do if you get confused by the problem that you are working on.
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There is another way to do this, though; we can also re-index by thinking in terms of substitution.
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How do our old index and the new index we are creating relate to each other?
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We can think in terms of substitution: consider, once again, if we wanted to convert from i = 7, with all of the same thing, to k = 1.
00:17:00.100 --> 00:17:06.000
Well, since we have i = 7 and k = 1 both at their starting places, how is i related to k?
00:17:06.000 --> 00:17:11.300
Well, i is the same thing as k + 6; 1 + 6 gives us 7.
00:17:11.300 --> 00:17:16.000
So, we now have this relationship where i is equal to k + 6.
00:17:16.000 --> 00:17:23.600
With this in mind, we can substitute to figure out what the upper limit is, and then to figure out what the general term is.
00:17:23.600 --> 00:17:31.500
To figure out the upper limit, we know that i = 22 is what we are going to plug in, since that is its upper limit.
00:17:31.500 --> 00:17:42.400
And then, we are going to...over here, we know i = k + 6, so we plug in 22 for i: we have 22 = k + 6.
00:17:42.400 --> 00:17:51.800
Subtract 6 on both sides, and we get 16 = k for its upper limit.
00:17:51.800 --> 00:17:59.900
So, at this point, we can now write that our sigma is going to end up being...upper limit of 16;
00:17:59.900 --> 00:18:05.400
the lower starting index is at k = 1; now, what is it going to end up being?
00:18:05.400 --> 00:18:12.800
Well, here it was 3i - 16; so we have 3; what is our i in terms of k?
00:18:12.800 --> 00:18:19.700
Well, i is k + 6; and then, we continue on with the rest of it...minus 16.
00:18:19.700 --> 00:18:31.300
We have substituted out the i here for the k + 6 here, into what i was in our initial version for the sigma notation.
00:18:31.300 --> 00:18:34.600
So, at this point, we can expand this and work this out, and simplify it if we want.
00:18:34.600 --> 00:18:37.600
We could also just leave it as it is, and it would be just fine.
00:18:37.600 --> 00:18:41.600
The upper limit and lower limit and our index--they will all stay the same throughout.
00:18:41.600 --> 00:18:55.200
We work this out: 3 times k plus 6 gets 3k +...3 times 6 is 18; minus 16; and so, we get 16 for our upper limit; k = 1 for our index and lower limit.
00:18:55.200 --> 00:19:06.000
3k + 18 simplifies to 3k + 2, which is exactly what we had the first time when we did this through expanding it.
00:19:06.000 --> 00:19:08.400
So, substitution and expanding both work the same way.
00:19:08.400 --> 00:19:13.500
Substitution is probably a little bit faster and easier, but it is a little bit more complicated to see what is going on, to really understand.
00:19:13.500 --> 00:19:19.000
The important thing is to figure out how your i...whatever your initial symbol and the new symbol that you are changing to are...
00:19:19.000 --> 00:19:22.600
how they are related to each other; and you probably start with the lower limit,
00:19:22.600 --> 00:19:28.400
and then ask what your upper limit will have to be, to have that same relationship between old index and new index.
00:19:28.400 --> 00:19:37.100
And then, what is your expression going to end up being, if you just plug in your connection between the two indexes?
00:19:37.100 --> 00:19:41.100
All right, working with series, there are various properties that we can occasionally use to our advantage.
00:19:41.100 --> 00:19:46.900
Let's look at some properties of sums: below, let a<font size="-6">i</font> and b<font size="-6">i</font> be sequences.
00:19:46.900 --> 00:19:51.200
They are things that are liable to change; but c will simply always be a constant.
00:19:51.200 --> 00:19:57.400
Our first one is that the sum from i = 1 to n of c is equal to c times n.
00:19:57.400 --> 00:19:58.600
Remember: c is just a constant.
00:19:58.600 --> 00:20:06.200
Why is this the case? Well, remember: if we had the sum of i = 1 to n of c,
00:20:06.200 --> 00:20:10.200
well, c doesn't change any time, because it is a constant; it isn't affected by the index.
00:20:10.200 --> 00:20:16.500
So, it is just going to be c, plus c, plus c...plus c, that many times.
00:20:16.500 --> 00:20:18.800
If we have that, how many times did it show up?
00:20:18.800 --> 00:20:27.600
Well, we went from i = 1 up until n, so 1, 2, 3, 4...we count up to n; we have a total of n terms in there.
00:20:27.600 --> 00:20:32.700
If c adds to itself n times, then that is just c times n.
00:20:32.700 --> 00:20:38.800
So, that is equal to c times n; and that is where we get that property for how sums work.
00:20:38.800 --> 00:20:46.300
If it is just a constant being added through a series, we can just multiply it by the number of terms in that series.
00:20:46.300 --> 00:20:52.800
Next, we have the summation of some summation...and this will end up working for any summation limit;
00:20:52.800 --> 00:20:55.100
the only thing that we have to care about is the index.
00:20:55.100 --> 00:21:00.700
You can start with any lower limit and any upper limit, including an infinite sum; it is still going to end up working out the same.
00:21:00.700 --> 00:21:07.300
So, c times a<font size="-6">i</font>...we can do this where we pull out the constant, and we bring it out to the front.
00:21:07.300 --> 00:21:10.500
And it is c times the summation of a<font size="-6">i</font>.
00:21:10.500 --> 00:21:15.800
So, why is this the case? Well, notice: our first term would be c times a₁.
00:21:15.800 --> 00:21:22.900
Plus...our next term would be c times a₂; plus...our next term would be c times a₃.
00:21:22.900 --> 00:21:24.700
And it is going to continue on in this pattern.
00:21:24.700 --> 00:21:28.400
It might end; it might not end; we will just leave it as dots there.
00:21:28.400 --> 00:21:30.700
But notice: we have a c on each one of our terms.
00:21:30.700 --> 00:21:35.300
So, if we want, since we have a c here, a c here, a c here, and that thing is going to end up continuing
00:21:35.300 --> 00:21:39.000
(there is going to be a c on each one of these, because we see it is c times a<font size="-6">i</font>),
00:21:39.000 --> 00:21:49.700
we can pull out all of the c's; we pull them out, so it is c times a₁ + a₂ + a₃ + ....
00:21:49.700 --> 00:21:55.300
So, it is c multiplied against that entire series; it is not going to have any difference in how we do it.
00:21:55.300 --> 00:22:03.300
So, this entire series, here to here--we can think of it as c times a new series of just the a<font size="-6">i</font>'s changing.
00:22:03.300 --> 00:22:07.400
And that is how we end up getting the same thing here.
00:22:07.400 --> 00:22:15.500
The final idea: if we have addition, a<font size="-6">i</font> + b<font size="-6">i</font>, well, we can end up breaking that into two separate series, added together.
00:22:15.500 --> 00:22:20.500
And once again, this would end up working with any limit; so that is why we don't have a lower limit and don't have an upper limit--just an index.
00:22:20.500 --> 00:22:25.900
It is because this will work with any limits whatsoever, even an infinite series.
00:22:25.900 --> 00:22:33.100
We could write this...if it is series of i, a<font size="-6">i</font> + b<font size="-6">i</font>, well, that is going to be a₁ + b₁,
00:22:33.100 --> 00:22:40.600
plus a₂ + b₂, plus a₃ + b₃...
00:22:40.600 --> 00:22:42.700
And it is just going to continue on in that format.
00:22:42.700 --> 00:22:47.700
Well, order of addition doesn't matter; a₁ + b₁ + a₂ + b₂
00:22:47.700 --> 00:22:51.400
is the same thing as a₁ + a₂, then plus b₁ + b₂.
00:22:51.400 --> 00:23:01.100
So, what we do is order all of our a₁'s first; we will have all of the a₁'s in our series show up first.
00:23:01.100 --> 00:23:08.100
And then, we will add on our b₁'s: b₁ + b₂ + b₃...
00:23:08.100 --> 00:23:12.700
And that will also go on the same as it would have on its own.
00:23:12.700 --> 00:23:17.100
So, at that point, what we have done is just re-ordered how we are looking at the series, since we have expanded it,
00:23:17.100 --> 00:23:18.700
and now we have re-ordered the way we are looking at it.
00:23:18.700 --> 00:23:24.000
But we haven't changed anything; so at this point, we can just pull it into two separate series, i...
00:23:24.000 --> 00:23:30.300
so, we will have a<font size="-6">i</font> here, plus...and then the b<font size="-6">i</font>.
00:23:30.300 --> 00:23:33.400
There is the a portion of the series and the b portion of the series.
00:23:33.400 --> 00:23:39.400
So, we can either have them intermixed together, or we can spread them apart and work on each of them on its own.
00:23:39.400 --> 00:23:42.300
And that is how we end up having this final property.
00:23:42.300 --> 00:23:44.600
All right, we are ready for some examples.
00:23:44.600 --> 00:23:48.800
The first example here: Given the sequence below, find the third partial sum.
00:23:48.800 --> 00:23:52.900
If we are going to find the third partial sum, that is as easy as just adding up the first three terms.
00:23:52.900 --> 00:24:03.000
That is 0, 3, and 8; 0 + 3 + 8; we end up getting 11; it's done; it was easy.
00:24:03.000 --> 00:24:08.200
The next one: If we are looking for the seventh partial sum...well, so far, we only have 0, 3, 8, 15, 24...
00:24:08.200 --> 00:24:11.700
so we are going to have to figure out what the pattern will continue on to.
00:24:11.700 --> 00:24:14.800
Our first step is to figure out how this pattern ends up working.
00:24:14.800 --> 00:24:20.400
We look at this through addition: 0 to 3, 3 to 8, 8 to 15...well, the addition is changing each time.
00:24:20.400 --> 00:24:23.800
So, we could think in terms of some recursive relationship, but that is not really going to make it easier.
00:24:23.800 --> 00:24:26.600
Multiplication? Multiplication doesn't really work, either.
00:24:26.600 --> 00:24:35.800
But we look at this for a while, and we might realize that this looks kind of like 1, 4, 9, 16, 25...
00:24:35.800 --> 00:24:39.700
0, 3, 8, 15, 24...that is just - 1 on each one of the terms.
00:24:39.700 --> 00:24:46.200
So, we can think of this as being, in general, n² - 1.
00:24:46.200 --> 00:24:48.800
If that is the case, then we can figure out what the next terms are.
00:24:48.800 --> 00:24:57.300
If it is n² - 1, the next term to follow is going to end up being 36 - 1, because we are at 5 here; 5² - 1 is 24.
00:24:57.300 --> 00:25:01.800
So, the next will be 6² - 1; that is going to get us 35.
00:25:01.800 --> 00:25:07.500
The next one will be 7² - 1, 49 - 1, or 48.
00:25:07.500 --> 00:25:11.400
At this point, the seventh partial sum will be adding up these first seven terms.
00:25:11.400 --> 00:25:30.500
So, 0 + 3 + 8 + 15 + 24 + 35 + 48...we toss that all into a calculator, and we end up getting 133; there we go.
00:25:30.500 --> 00:25:35.800
The next one: Expand the below sigma notation into a series of terms; add them together using an ellipsis.
00:25:35.800 --> 00:25:38.900
We break it apart into the thing where we are not using sigma notation.
00:25:38.900 --> 00:25:42.600
We are just writing it as one term after another, with some pattern occurring.
00:25:42.600 --> 00:25:44.200
So, what would our very first term be?
00:25:44.200 --> 00:25:53.900
Our very first term would start at i = 3; so we plug in a 3 for our i first; that would be 3! over 7³.
00:25:53.900 --> 00:26:03.900
Plus...our next term would end up being...we click up one, from 3 to 4; that will be 4! over 7⁴.
00:26:03.900 --> 00:26:14.600
Plus...our next one: we click up another one to 5! over 7⁵, plus..., plus...we are going to end up finishing here at 15.
00:26:14.600 --> 00:26:19.500
We could also start with something before 15; for example, we could write out 14!,
00:26:19.500 --> 00:26:28.700
because that would be in the expansion: 7^14, plus 15! over 7^15.
00:26:28.700 --> 00:26:31.600
There we go; we have managed to write the whole thing out, using ellipsis.
00:26:31.600 --> 00:26:36.400
We have expanded the sigma notation into a series of terms.
00:26:36.400 --> 00:26:39.100
And we don't have to necessarily write out all of these.
00:26:39.100 --> 00:26:41.100
We could probably get away with not writing it.
00:26:41.100 --> 00:26:45.100
We could certainly get away with not writing this one right here, the 14!/7^14.
00:26:45.100 --> 00:26:48.900
And we might even be able to get away with not writing the 5!/7⁵.
00:26:48.900 --> 00:26:52.500
It helps us see the pattern, but it is not absolutely necessary.
00:26:52.500 --> 00:26:58.200
But by including more terms, sometimes the reason we want to expand things is so we can manipulate how the terms work.
00:26:58.200 --> 00:27:05.500
So, sometimes it is useful to include more things at the start and the end, so that we can end up seeing how things are interacting.
00:27:05.500 --> 00:27:07.500
It will sometimes allow things to work out better.
00:27:07.500 --> 00:27:10.800
And we will see that sometimes, when we are working in proofs.
00:27:10.800 --> 00:27:15.500
Example 3: Condense the sum below into a series expressed using sigma notation.
00:27:15.500 --> 00:27:17.400
Notice: it is an infinite series.
00:27:17.400 --> 00:27:23.400
The first thing: let's notice how these connect to each other; how do we get 1/2 - 1/4 + 1/8 - 1/16?
00:27:23.400 --> 00:27:27.000
Well, we can break this into a sequence, first, that just has a series going on.
00:27:27.000 --> 00:27:33.500
So, let's look at this underlying sequence; how do we get from 1/2 to -1/4? Well, we multiply by -1/2.
00:27:33.500 --> 00:27:38.200
How do we get from -1/4 to positive 1/8? We multiply by -1/2.
00:27:38.200 --> 00:27:41.300
How do we get the next one? We multiply by -1/2.
00:27:41.300 --> 00:27:45.100
So, with this in mind, how can we create a general term a<font size="-6">n</font>?
00:27:45.100 --> 00:27:52.500
That is going to be a<font size="-6">n</font> =...since we are multiplying by -1/2 each time, it will be -1/2 to the...
00:27:52.500 --> 00:27:56.900
is it going to be to the n? Well, here we are going to end up starting with something else, actually.
00:27:56.900 --> 00:28:07.400
It is n - 1, because here is n = 1; we want to not have anything at the beginning, which we could write as times 1/2... so -1/2 to the n - 1, times 1/2.
00:28:07.400 --> 00:28:18.700
Alternatively, another equivalent way to write this out: we could write this as -1 to the n - 1 over 2 to the n - 1, times 1/2.
00:28:18.700 --> 00:28:21.000
The 2's in the denominator will end up compacting together.
00:28:21.000 --> 00:28:29.700
And we could also write this as -1 to the n - 1, over 2 to the n.
00:28:29.700 --> 00:28:31.500
Both of these are just fine ways to write it.
00:28:31.500 --> 00:28:37.900
We will end up getting the same thing, whether we write it using this general term, or we end up writing it with this general term.
00:28:37.900 --> 00:28:39.700
We will end up getting the same series.
00:28:39.700 --> 00:28:44.500
What is our first term? We decided to set it at some n = 1; let's use an index of i,
00:28:44.500 --> 00:28:47.200
just because it is what we are used to, although we could use n.
00:28:47.200 --> 00:28:51.100
Sometimes, you might have a little bit of confusion, since we are used to talking about nth terms with n.
00:28:51.100 --> 00:28:54.600
So, it might be weird to use an index of n; but you will often see it used, as well.
00:28:54.600 --> 00:28:57.000
I like i, so I am going to use i.
00:28:57.000 --> 00:29:00.300
i = 1, because that is our first location; what do we go up to?
00:29:00.300 --> 00:29:04.700
Well, it is an infinite series, because it goes on forever, and we are never told that it stops.
00:29:04.700 --> 00:29:09.300
So, we use an infinity symbol on top; and then, let's do this one first.
00:29:09.300 --> 00:29:19.000
We could write this as -1/2 to the n - 1, times 1/2; and that whole thing has the series applied to it.
00:29:19.000 --> 00:29:30.100
Or equivalently, this would also be equal to the series--the same upper limit of going on forever, and the same starting index.
00:29:30.100 --> 00:29:35.000
We could also write this as -1 to the n - 1 over 2 to the n.
00:29:35.000 --> 00:29:39.700
Either way we end up working it out, both of these are just fine.
00:29:39.700 --> 00:29:43.400
They are going to give us the exact same answer; they are just two different ways to write it out.
00:29:43.400 --> 00:29:46.700
Depending on the specific problem, it might be useful to write it one way or the other.
00:29:46.700 --> 00:29:51.200
But any teacher, if this was just the question, should accept both of these.
00:29:51.200 --> 00:29:56.500
All right, calculate the value of sigma with a lower limit of 0 and an upper limit of 4.
00:29:56.500 --> 00:30:00.400
These are indexed where the expression is v! - 5v.
00:30:00.400 --> 00:30:04.900
The first thing we do is plug in v = 0 for our first 0.
00:30:04.900 --> 00:30:10.500
Our very first term is going to be 0! - 5(0).
00:30:10.500 --> 00:30:12.900
Our index is v, so that is the thing being swapped out.
00:30:12.900 --> 00:30:21.700
Plus...we step up our index to the next level, 0 up to 1 now; so 1! - 5(1).
00:30:21.700 --> 00:30:28.700
Plus...the next thing, stepping up from 1, is 2; so 2! - 5(2).
00:30:28.700 --> 00:30:41.300
Plus...the next thing, from 2 to 3: 3! - 5(3); and then, plus...3 to 4...4! - 5(4).
00:30:41.300 --> 00:30:45.000
Finally, at this point, we notice that that is our upper limit, so we stop.
00:30:45.000 --> 00:30:50.800
We step from our lower limit, 0, 1, 2, 3, 4...our upper limit, so that is where we stop.
00:30:50.800 --> 00:30:53.200
And we do each of the steps in between and add them all together.
00:30:53.200 --> 00:30:55.900
So now, it is just a matter of simplifying to actually get the value.
00:30:55.900 --> 00:31:01.100
What is 0!? Remember: 0! is simply defined to be equal to 1.
00:31:01.100 --> 00:31:08.300
All of the other numbers, factorial, are that number, times each of the positive integer numbers underneath it.
00:31:08.300 --> 00:31:11.800
But 0! is just simply defined to be 1.
00:31:11.800 --> 00:31:22.700
So, 0! gets us 1; minus...5 times 0 is 0, plus 1! is just 1 times itself; so 1 minus...5 times 1 is 5, plus 2!...
00:31:22.700 --> 00:31:33.100
2 times 1 is 2, minus 5(2) is 10, plus 3! is 3 times 2 times 1, so it is 6; minus 5(3) is 15;
00:31:33.100 --> 00:31:41.100
plus 4! is 4 times 3 times 2 times 1, 24; minus 5(4) is 20.
00:31:41.100 --> 00:31:52.300
We work this out; we have 1 here, plus -4, plus -8, plus -9, plus +4.
00:31:52.300 --> 00:31:56.900
We see that we have a +4 here and a -4 here; they cancel out.
00:31:56.900 --> 00:32:04.500
1 + -9 gets us -8, so we have -8 and -8, or -16, once we simplify the whole thing out.
00:32:04.500 --> 00:32:10.000
Great; the fifth example: Re-index the series below so that it starts at the index k = 1.
00:32:10.000 --> 00:32:17.000
We are looking to swap the index i = 5 out for k = 1, but have the exact same value come out of the series.
00:32:17.000 --> 00:32:23.300
Our sigma notation will change; the upper limit will end up changing, and the expression here will end up changing to some extent,
00:32:23.300 --> 00:32:27.500
so that we can achieve this without affecting the value of the series.
00:32:27.500 --> 00:32:31.400
We talked about two different ways to do this when we went through the lesson.
00:32:31.400 --> 00:32:37.500
The first way we will approach this is through expanding it.
00:32:37.500 --> 00:32:44.000
We will start by doing it through expanding it; and then later, we will look at doing it through substitution.
00:32:44.000 --> 00:32:49.500
There are two alternative ways; whichever one makes more sense to you, that is the one that you probably want to end up using for your own work.
00:32:49.500 --> 00:32:52.100
All right, we want to start by expanding this.
00:32:52.100 --> 00:33:01.400
If we are going to expand i = 5, 3^7 - i, that is going to be...our first thing would be 3^7 - 5, so 3².
00:33:01.400 --> 00:33:09.700
Plus...then the next would be 6 for our i; so 3^7 - 6 would be 3¹,
00:33:09.700 --> 00:33:17.000
plus...then at 7, we will have 3⁰, plus...plus...and then finally,
00:33:17.000 --> 00:33:23.700
when we get up to an upper limit of 20, 7 - 20 is -13, so it will be 3^-13.
00:33:23.700 --> 00:33:30.900
OK, that is what we see that we end up getting out of this: 3² + 3¹ + 3⁰...+ 3^-13.
00:33:30.900 --> 00:33:34.800
So, we see that the top steps down each time.
00:33:34.800 --> 00:33:37.500
We want it to figure out how we can end up showing this.
00:33:37.500 --> 00:33:46.400
We start with k = 1 for this first slot; use k = 2, k = 3...and here is k = we-don't-know-what-yet.
00:33:46.400 --> 00:33:54.100
We can actually figure out that it is going to have to be...20 - 5 is 15, so 1, our starting lower limit here, plus 15 would be 16.
00:33:54.100 --> 00:33:57.800
We know it will end up having it come out to 16; but let's work through it the other way.
00:33:57.800 --> 00:34:05.000
So, what we have here...it looks like we could write this as 3^3 - k.
00:34:05.000 --> 00:34:13.800
3^3 - 1 would get us 3²; the next one, 3^3 - 2, would get us 3¹; that works out.
00:34:13.800 --> 00:34:19.600
At k = 3 on our third term, we would have 3^3 - 3; it would be 3⁰; that ends up working out.
00:34:19.600 --> 00:34:29.600
So, if we are going to have 3^-13 = 3^3 - k, we are now going to know that -13 has to be equal to 3 - k.
00:34:29.600 --> 00:34:36.000
So, we have k = 16 as our upper limit, as well; great.
00:34:36.000 --> 00:34:39.600
That ends up making sense; we can also figure it out, once again, from:
00:34:39.600 --> 00:34:43.300
our starting upper limit was 20; our starting lower limit was 5.
00:34:43.300 --> 00:34:52.900
So, that means that there is a difference of 15; so if we have k = 1, then 1 + 15 has to come out to be 16, as well.
00:34:52.900 --> 00:34:55.400
So, that is going to be our upper limit; there are two ways of looking at it.
00:34:55.400 --> 00:34:57.500
Either way, just make sure that you are careful with this sort of thing.
00:34:57.500 --> 00:35:00.000
It is easy to get confused the first few times you work with it.
00:35:00.000 --> 00:35:03.300
So now, we know what our general term is; it is 3^3 - k.
00:35:03.300 --> 00:35:09.200
We know what our upper limit is, and we know what our index and starting term is, and lower limit.
00:35:09.200 --> 00:35:20.300
So, we can write the whole thing out: we have 16 as our upper limit; k = lower limit of 1; 3 to the 3 - k; great.
00:35:20.300 --> 00:35:25.100
So, that is our answer; and that is doing it through expanding as our method.
00:35:25.100 --> 00:35:28.300
Alternatively, we could do this through substitution.
00:35:28.300 --> 00:35:33.500
We could have also set this up by noticing that we have i = 5 here, and we want to start at k = 1.
00:35:33.500 --> 00:35:39.500
So, at the first place, we have i = 5; and then, at our second place, we have k = 1.
00:35:39.500 --> 00:35:46.600
So, if we have i = 5 here and k = 1 here, then we can see that i is equal to k + 4.
00:35:46.600 --> 00:35:53.800
So, if i is equal to k + 4, in general (this is just going to hold true in general), then what about our upper limit?
00:35:53.800 --> 00:35:59.700
Our upper limit: it is going to have to end up being the case that when i is equal to 20...what will our k end up being?
00:35:59.700 --> 00:36:07.000
We will have 20 = k + 4, which means 16 is equal to k for the upper limit.
00:36:07.000 --> 00:36:12.000
16 = k for our upper limit; now we just plug in through substitution.
00:36:12.000 --> 00:36:17.800
So, 16 is our upper limit; we just figured that out; k = 1 is our index and lower limit.
00:36:17.800 --> 00:36:26.100
So, it is going to be the same thing we started with, 3^7 - i, so 3 to the 7 minus...
00:36:26.100 --> 00:36:30.200
but we are not going to use i; we are going to use i = k + 4.
00:36:30.200 --> 00:36:34.300
So, we are going to swap out this i here for k + 4.
00:36:34.300 --> 00:36:39.000
We substitute that in, and we have k + 4.
00:36:39.000 --> 00:36:45.000
We can now work this out and simplify it, since it is the upper limit and lower limit and index...
00:36:45.000 --> 00:36:48.000
none of those will change; we are just simplifying what is on the inside.
00:36:48.000 --> 00:36:58.600
3 to the 7 - (k + 4), so minus k and minus 4...7 - 4 gets us 3, and minus k gets us 3 - k in the whole.
00:36:58.600 --> 00:37:04.400
So, we end up getting the exact same thing, either way we end up approaching this: through substitution or through expanding.
00:37:04.400 --> 00:37:08.800
We can end up re-indexing the series, so we can have it change at a different starting index.
00:37:08.800 --> 00:37:13.400
They both work fine; whichever one makes more sense to you, that is the one I would recommend using.
00:37:13.400 --> 00:37:19.700
All right, the final example: Calculate the value of the series below; use summation properties to make the math less tedious.
00:37:19.700 --> 00:37:32.100
We could just do this by hand; we could say, "All right, we have 3 times 1 minus 5, plus...plus...all the way up to 3 times 15 minus 5."
00:37:32.100 --> 00:37:38.300
OK, and then we could calculate what 3 times 1 minus 5 is, and what 3 times 2 minus 5 is, and what 3 times 3 minus 5 is.
00:37:38.300 --> 00:37:42.100
And we could do this by hand or through the calculator; but that is kind of a pain.
00:37:42.100 --> 00:37:45.200
That is going to be a lot of writing; that is going to be a lot of calculation.
00:37:45.200 --> 00:37:50.200
Luckily, there are some clever ways to use the summation properties that we learned early to make this at least...
00:37:50.200 --> 00:37:55.700
not easy...not fast (it will be easy), but less tedious, so we have to do less writing.
00:37:55.700 --> 00:38:00.400
How can we do this? Well, first, remember: we can split based on addition.
00:38:00.400 --> 00:38:10.400
We can see this as 3i + -5; and we can split this into i = 1; the limits to our summation will never end up changing,
00:38:10.400 --> 00:38:19.500
but we can split this into a 3i +...the same sum over here...15i = 1 on -5; cool.
00:38:19.500 --> 00:38:26.400
We also were allowed to pull out constants; we have this constant of a 3 here; we have this constant of a -1 here, effectively.
00:38:26.400 --> 00:38:38.700
We can pull those out front; we can now write this as 3 times the summation of 15, i = 1 of i, plus...
00:38:38.700 --> 00:38:50.500
Now, I am sorry...plus -1, so let's just write that as minus...summation...15...i = positive 1 (I got confused by the negative) on 5; great.
00:38:50.500 --> 00:38:57.700
So, 3 times...well, there are no cool properties that we have learned yet, although we will learn in the very next lesson,
00:38:57.700 --> 00:39:05.300
how to easily sum this one up; but we would have 1 plus 2 plus 3 plus...plus 15.
00:39:05.300 --> 00:39:11.200
And over here, we will have minus...well, we could do 5 + 5 + 5...but it is just going to show up 15 times.
00:39:11.200 --> 00:39:14.600
We have i = 1 to 15; we talked about this before.
00:39:14.600 --> 00:39:19.500
Since a constant is just showing up 15 times, that is going to be 15 times the constant.
00:39:19.500 --> 00:39:26.400
So, it will be 15 (the number of terms that are there) times the constant that is showing up over and over, so 15 times 5.
00:39:26.400 --> 00:39:31.900
We can now work out with a calculator 1 + 2 + 3 +...up until 15.
00:39:31.900 --> 00:39:49.200
That ends up working out to 120; so we have 3(120) - 15(5) is 75; 3 times 120 is 360, minus 75; that comes out to equal 285, and there is our answer.
00:39:49.200 --> 00:39:51.600
We still had to do a little bit of writing this out.
00:39:51.600 --> 00:39:59.600
The slowest part here is going to be adding by hand: 1 + 2 + 3 +...to 15, or by using a calculator.
00:39:59.600 --> 00:40:06.400
But either way, that is much better than having to multiply numbers, and then subtract 5, and then add that, each time, over and over.
00:40:06.400 --> 00:40:11.000
So, we can use these summation properties to split things up into ways that make them easier to work with.
00:40:11.000 --> 00:40:16.200
All right, cool; in the next lesson, we will end up looking at our first specific kind of sequence and series, arithmetic.
00:40:16.200 --> 00:40:21.000
And then later on, we will work with geometric, which will let us apply these ideas about series into one specific thing,
00:40:21.000 --> 00:40:24.400
where we can actually start creating some formulas to make things really easy and fast.
00:40:24.400 --> 00:40:27.000
All right, we will see you at Educator.com later--goodbye!