WEBVTT mathematics/math-analysis/selhorst-jones
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Hi--welcome back to Educator.com.
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Today, we are going to talk about geometric sequences and series.
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The other specific kind of sequence we will look at in this course is the geometric sequence,
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a sequence where we multiply by a constant number for each step.
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In the previous lesson, we looked at the arithmetic sequence, which is where you add by a constant number.
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Now, we are looking at geometric, where you multiply by a constant number every step.
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Just like arithmetic sequences, geometric sequences commonly appear in real life.
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Since geometric sequences are based on ratios, since we are always multiplying by the same thing,
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and ratios occur a lot in the world, they give us a way to describe a wide variety of things.
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In this lesson, we will begin by going over what a geometric sequence is, and how we can talk about them in general.
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Then, we will look into formulas for geometric series to make adding up a bunch of terms really easy and fast; let's go!
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We will start with a definition: a sequence is **geometric** if every term in the sequence can be given
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by multiplying the previous term by some constant number, r.
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a<font size="-6">n</font> is equal to r times a<font size="-6">n - 1</font>; that is, some term is equal to the previous term, multiplied by r.
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We call r the common ratio, because we can express it as a<font size="-6">n</font> divided by a<font size="-6">n - 1</font>; that is, some term divided by the previous term.
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And so, we have a ratio in the way that we are building it.
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Here are two examples of geometric sequences: 3, 6, 12, 24...continuing on; 4/5, -4/25, 4/125, -4/625...continuing on.
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They are geometric, because each step multiplies by the same number.
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For example, in this one, every step we go forward, we are multiplying by 2: 3 to 6--times 2; 6 to 12--times 2; 12 to 24--times 2.
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And this is going to continue on forever, as long as we keep going with that sequence.
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Over here, 4/5 and -4/25...it is not quite as simple, but it is basically the same thing, multiplying by -1/5.
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That is how we get from 4/5 to -4/25, if we are going to multiply.
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To get to 4/125, once again, we multiply by -1/5; to get from 4/125 to -4/625, we multiply by -1/5.
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And this is going to keep going, every time we keep stepping.
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Every step is multiplying by the same number: we are multiplying by the same number each step.
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The number can be anything: it can positive; it can be larger than one; it can be less than one; it can be negative.
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It doesn't matter, so long as it is always the same value for every step.
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The definition of a geometric sequence is based on the recursive relation a<font size="-6">n</font> = r(a<font size="-6">n - 1</font>).
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That is, every term is equal to the previous term, multiplied by r.
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How can we turn this into a formula for the general term, where we don't have to know what the previous term is--
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we can just say, "I want to know the nth term," plug it into a formula, and out will come the nth term?
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Remember: a recursive relation needs an initial term.
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So, while this relation that defines a geometric sequence is useful, we still need a little bit more.
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We need this initial term to know where we start--what our very first term is--because previous to that...there is nothing previous.
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So, we just have to state that as one specific term.
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Since we don't know its value yet, we will just leave it as a₁, our first term.
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From a<font size="-6">n</font> = r(a<font size="-6">n - 1</font>), we see that a₁ relates to later terms as: a₂ will be equal to r(a₁).
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The second term will be equal to the first term, times r; this will continue on.
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The third term, a₃, will be equal to r times a₂.
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But we just showed that a₂ is equal to r(a₁), so we can plug that in there;
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and we will have r times r(a₁), so we end up getting r²(a₁) = a₃.
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We can continue on here; a₄ is going to be equal to r times a₃.
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But we just figured out that a₃ is equal to r² times a₁.
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So, we replace a₃, and we end up having r times r² times a₁;
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so now we have that r³(a₁) = a₄.
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And we will see that this pattern will just keep going like this.
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So, we have that a₁ is equal to a₁ (there is no big surprise there).
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a₂ is equal to r times a₁; a₃ is equal to r² times a₁.
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a₄ is equal to r³ times a₁; and we see that the pattern is just going to keep going like this.
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The nth term is n - 1 steps away from a₁; it is n - 1 steps to get from a₁ to n.
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If you start on the first stepping-stone, and you go to the nth stepping-stone, you have to take n - 1 steps forward.
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We start at a₁; to get to the a<font size="-6">n</font>, we have to go n - 1 steps forward.
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Since every step means multiplying by r, that means we have multiplied by r n - 1 times, which is r raised to the n - 1 power.
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Thus, we have that a<font size="-6">n</font> = r^n - 1(a₁).
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So, to find the formula for the general term of a geometric sequence, we only need to figure out what the first term is, and the common ratio.
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As soon as we figure out a₁ and our value for r, we have figured out what the general term is, what the a<font size="-6">n</font> term is--pretty great.
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What if we want to find the nth partial sum of a geometric sequence (that is, adding up
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the first n terms of the sequence, a₁ + a₂ + a₃ up until + a<font size="-6">n</font>)?
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Well, we could just add it all up by hand for small values of n.
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If it was n = 2, so it was just a₁ and then a₂, it is probably not that hard to just figure it out by hand.
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If it was n = 3, we could probably do it by hand; if it was n = 10, n = 100, n = 1000,
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this gets really, really tiresome, really quickly, as the value of n gets larger and larger.
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So, how can we create a formula--how can we just have some formula where we can plug some stuff in,
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and we will immediately know what that nth partial sum is?
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Well, let's do two things: first, let's give the sum a name.
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So, we will call our nth partial sum s<font size="-6">n</font>, the sum for the nth partial sum.
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Second, let's use the form for the nth term of a geometric sequence.
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Remember: we just figured out that the general form for a<font size="-6">n</font> is r^n - 1(a₁).
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We can use this general term to put of the terms in this series into a format that will involve a₁.
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So, we have s<font size="-6">n</font> = a₁ + a₂ (which is r times a₁) + a₃
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(which is now r² times a₁), up until we get to + a<font size="-6">n</font> (which is now r^n - 1(a₁)).
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Great; but at the moment, we can't do anything with just this.
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This isn't quite enough information; we can't combine the various r's, because they all have different exponents.
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r to the 0, what is effectively here; r¹; r²; r³; up until r^n - 1--
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they don't talk the same language, because they don't have the same exponent.
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So, since they can't really communicate with each other, we can't pull them out all at once.
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We could pull out all of the a₁'s, but then we would still be left with all of these different kinds of r's.
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So, we don't really have a good thing that we can do right now.
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What we want, what we are really looking for, is a way to somehow get rid of having so many things to add up.
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We want fewer things to add up; if we only had a few things to add up and compute, it would be easy for us to calculate these values.
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So, that is what we want to figure out how to do.
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This is the really clever part; this is basically the part of the magic trick where suddenly we pull the rabbit out of the hat.
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And so, you might see this and think, "How would I figure this out?"
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And it is a little bit confusing at first; but just like, as you study magic more and more (if you were to study magic),
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you would eventually realize, "Oh, that is how they got the rabbit into the hat," or "this is how the trick works"--
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as you work with math more and more, you will be able to see, "Oh, that is how we can make these sorts of things."
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So, don't be worried by the fact that you would not be able to think of this immediately.
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The people who made this proof at first didn't think of it immediately.
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They thought about it for a while; they figured out different things, and maybe they tried something that did not work;
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and eventually they stumbled on something and said, "Oh, if I do this, it works,"
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and they were able to come up with this really easy, cool, clever way to do it.
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But it is not something that you just have immediately.
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It is something that you have to think about, until eventually you can "pull your own rabbit out of a hat."
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But first, you have to get the rabbit into the hat.
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Anyway, here is the clever part: what is it?
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We have s<font size="-6">n</font> = a₁ + r(a₁) + r²(a₁) + ... + r^n - 1 (a₁).
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The really cool trick that we do is say, "What if we multiplied r times s<font size="-6">n</font>?"
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Well, that would end up distributing to everything in here, since we are multiplying it on both sides of the equation.
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We would have r(a₁) + r²(a₁) + r³(a₁)...
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all the way until we get up to + r^n(a₁).
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Now, notice: we now have matching stuff going on.
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Here is r times a₁; here is r times a₁; there is a connection here.
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Here is r² times a₁; here is r² times a₁; there is a connection there.
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Here is r³ times a₁; and somewhere in the next spot in the decimals is r³ times a₁, as well.
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Here is r^n - 1 times a₁; and in the next back spot in the decimals, here is r^n - 1 times a₁.
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We have all of this matching going on; well, with this idea, since we have all of this matching,
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we can subtract this equation, the rs<font size="-6">n</font> equation, this one right here,
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from the s<font size="-6">n</font> equation by using elimination.
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From when we talked about systems of linear equations: if we have two equations, we can subtract, and we can add them together.
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We are just using elimination to do this.
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So, we have our s<font size="-6">n</font> equation here, and then we subtract by minus r times s<font size="-6">n</font>.
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So now, we have this matching pattern going on: r times a₁ matches to -r times a₁; they cancel each other out.
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r² times a₁ matches to minus r² times a₁; they cancel each other out.
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Everything in the dots here cancels out with all of the negatives in the dots here.
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We finally get to r^n - 1 times a₁ in our top equation,
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which cancels out with minus r^n - 1 times a₁ in our bottom equation.
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The only things that end up being left over are minus r^n times a₁ and +a₁ here.
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We get s<font size="-6">n</font> -rs<font size="-6">n</font> is equal to a₁ - r^n(a₁).
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So, through this immense cleverness (and once again, this isn't something that you would be expected to just know immediately,
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and be able to figure out really easily--this is the part that takes the really long thinking.
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This is the really clever part; this is what takes hours of thought, by just thinking, "I wonder if there is a clever way to do this."
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And eventually, you end up stumbling on it.
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So, through immense cleverness, we have shown that s<font size="-6">n</font> - rs<font size="-6">n</font> = a₁ - r^n(a₁).
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That is pretty great: we have gotten this from what we just figured out.
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Now, our original goal was to find the value of the nth partial sum, which was s<font size="-6">n</font>.
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Using the above, we can now solve for s<font size="-6">n</font>: we just pull out the s<font size="-6">n</font>,
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so we have s<font size="-6">n</font> times 1 minus r; we can also pull out the a₁ over here on the right side.
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We divide both sides by 1 - r, and we get that the nth partial sum is equal to
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the first term, a₁, times the fraction 1 - r^n over 1 - r.
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So, we now have a formula to find the value of any finite geometric series at all, really easily.
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All that we need to know is the first term, a₁; the common ratio, r,
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which shows up on the top and the bottom; and how many terms are being added together total (the n exponent on our top ratio).
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It is pretty great--a really, really powerful formula that lets us do a lot of what would be very tedious, very slow, difficult addition, just like that.
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But what if the series was not finite?
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So far, we have only talked about if we are doing a finite sum of the sequence.
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But what if we had an infinite sequence, and we wanted to add up infinitely many of the terms, so we kept adding terms forever and ever and ever?
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To understand this better, let's consider some geometric sequences and what happens as we take partial sums using more and more and more terms.
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First, we will look at 3, 9, 27, 81, 243...so it is times 3 each time; it is a geometric sequence.
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So, our first partial sum, s₁, would be just the 3.
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Our next partial sum would be s₂, so we add on the 9, as well; we get 12.
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Our next partial sum would be s₃; we add on the 27, as well; we get 39.
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The next partial sum is s₄; we add on the 81; we get 120.
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The next partial sum: we add on 243; we get 363; and this is just going to keep going on in this pattern.
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It is going to keep adding more and more and more.
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We look at this, and we notice that, as the partial sums use more and more terms, it continues to grow at this really fast rate.
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In fact, the rate is going to get faster and faster and faster as we add more and more terms.
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We see its rate of growth increasing as it goes to larger sums.
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So, if we add terms to the series forever, it is not going to really get to anything.
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It is going to blow out to infinity; it is going to just "blast off" to infinity.
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There is no stopping this thing; it is not going to give us a single value.
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It never stops growing; we say that such an infinite series, one that never stops growing, that doesn't go to a single value, "diverges."
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As we add more and more terms, it continues to change forever.
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It diverges from giving us a single, nice, clean value, because it instead just blows off to infinity.
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It keeps moving around on us; it doesn't stay still; it doesn't go to something; it just goes off, so this would be a divergent series.
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On the other hand, we could consider another partial sum from this below geometric sequence: 1, 1/2, 1/4, 1/8, 1/16...
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What we are doing each time here is dividing by 2, or multiplying by 1/2; so we see that this is a geometric sequence.
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Our first partial sum would be s₁ = 1; we just add in that first term.
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The next one, s₂, would be 1.5, because we added 1/2, so we are at 1.5.
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The next one, s₃: we add in 1/4; that is 0.25 added in; that becomes 1.75.
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The next one, s₄: we add in an 8; that becomes 1.875.
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The next one, s₅: we add in 1/16; that becomes 1.9375.
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And it would continue in this way; but we notice that it is not really growing the same way.
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This time, the sums are continuing to grow, but the rate of growth is slowing down with each step.
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It is not increasing out like the previous one (it was blowing out somewhere; it was becoming really, really big).
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But with this one, we see it settling down; as we add more and more terms, it is going to a specific value.
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It is going to 2; the infinite sum, this infinite series, is going to a very specific value; it is working its way towards 2.
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If you keep adding more, you will see even more, as it gets to 1.99, 1.999, 1.9999999...
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As you keep adding more and more terms, you will see that it is really just working its way to a single value.
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It is slowing down as it gets to 2.
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In this case, we say that such an infinite series converges; it is converging on a specific value.
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As we add more and more terms, it works its way towards a single value.
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There is this single value that it is working towards.
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From the two examples we have seen, we see that whether a series converges or diverges is based on the common ratio of its underlying sequence.
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If the common ratio is large, it causes the sequence to always grow.
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It keeps growing, because that ratio keeps multiplying it to get larger and larger and larger, moving around.
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Specifically, if the absolute value of r is greater than or equal to 1, the partial sums will always be changing,
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because the size of our terms never shrinks down; so the series will diverge.
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On the other hand, if the common ratio is small, it causes the sequence to shrink down.
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If we have a small common ratio, it is going to make it smaller and smaller and smaller with every term we work on.
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Since it gets smaller and smaller and smaller, we have that if the absolute value of r is less than 1,
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the rate of change for the series will slowly disappear to nothing, because every time we go to the next term,
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since r is less than 1, it makes it smaller; and then it makes it smaller; and then it makes it smaller, and makes it smaller,
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and makes it smaller, and makes it smaller; so every time it is getting smaller.
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So, every time, the rate of growth is going down to less and less and less.
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And so, over the long term, over that infinite number of terms, it ends up converging to a single value.
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So, as a general rule, an infinite geometric series will converge if and only if the absolute value of r is less than 1.
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So, the absolute value of r being less than 1 means that the infinite geometric series converges.
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If the infinite geometric series converges, then the absolute value of r must be less than 1; they are equivalent things for a geometric sequence.
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All right: assuming that the absolute value of r is less than 1 for a geometric sequence, how can we figure out a formula for its corresponding infinite geometric series?
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Well, we already figured out a formula that is true for any finite geometric series.
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Remember: s<font size="-6">n</font>, the nth partial sum of any finite geometric series,
00:16:46.700 --> 00:16:51.000
is a₁ times (1 - r^n)/(1 - r).
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We just figured out that formula; that is pretty cool.
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Not only that, but we also know that, as we look at partial sums containing more and more terms,
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as a partial sum has more and more terms, they have to be growing closer and closer to the value that the infinite series will converge to.
00:17:06.900 --> 00:17:12.400
As we put in more and more terms into our partial sum, it has to be getting closer to this value that it is going to converge to.
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Think about why that is: if the partial sums were not getting closer to a specific value,
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if they were moving around away from the specific value,
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then it couldn't be converging to that, because it would always be changing around.
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If it is going to converge to a single value, it has to be working its way towards it.
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If it is working its way towards it, it must always be getting closer to the thing.
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If it wasn't always getting closer, if it was sometimes jumping away, it wouldn't be working its way towards it; it would be going somewhere else.
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Since we know that it is converging, we know that it must be working its way, as we have more and more terms in our partial sum.
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As we put in more terms in our partial sum, we will be closer to the value that we are converging on.
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As we add many more and more and more terms in our partial sum, we are going to be closer to the thing that we are converging to.
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What we are asking ourselves is, "As we have really large values for n, what value are we getting close to?"
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What happens to the formula we figured out, our nth partial sum formula,
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s<font size="-6">n</font> = a₁ times (1 - r^n)/(1 - r), as the number of terms we have, our n, goes off to infinity?
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As the value for the number of terms we have, our n, becomes infinitely large, what will happen to this formula?
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Whatever happens to this formula is what we have to be converging to, because of the argument
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that we just talked about, about how it has to be getting closer as we put in more and more terms.
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Notice: the only term on the right side affected directly by the n is r^n; there is no other term that directly has an n connected to it.
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We can ask ourselves what happens to r^n as our n grows very large.
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Also, remember: r is less than 1; the absolute value of r has to be less than 1.
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These two things combine as we ask ourselves...as n goes to infinity, and we are looking at our r^n here,
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since the absolute value of r is less than 1, we have that as n goes to infinity, r^n has to go to 0.
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So, it is going to shrink down to 0 as n grows infinitely large.
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Why is that the case? Well, since the absolute value of r is less than 1, every step has to make it smaller.
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For example, if we look at .9 raised to the 100, we get that that is less than .0001.
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Why is this occurring? Well, since the absolute value of r is less than 1, we know that it is this fractional thing--
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that effectively, every time we iterate it, every time we hit a term with this common ratio, it takes a little bite out of it.
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Whether it is .1 or 1/2 or 3/5 or 922/1000, it is going to take a bite out of whatever the term that it is being multiplied against is.
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As it takes infinitely many bites, since it is always shrinking it down, it means that it is always working its way towards this value of 0.
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Infinitely many bites away gets us to having nothing.
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Therefore, because r^n is going to 0 as n goes to infinity, we have this part right here shrinking down to a 0 in our formula.
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So, we get the following formula for an infinite geometric series: the infinite sum is equal to a₁ times 1/(1 - r),
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assuming that the absolute value of r is less than 1.
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If the absolute value of r is greater than or equal to 1, we couldn't even talk about this in the first place,
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because our series would be diverging, because it would always be growing and changing around on us.
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But if we have that the absolute value of r is less than 1, all we need to know is our first term, a₁, and the rate that it is growing at.
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And we work it out through this formula, and we know what it will converge to over the long run.
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All right, cool--we are ready for some examples.
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The first one: Show that the sequence below is geometric; then give a formula for the general term (that is, the a<font size="-6">n</font>, the nth term).
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We have 7, 35, 175, 875...so what we want to ask ourselves is, "What number are we multiplying by each time?"
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How do we get from 7 to 35? Well, we multiply by 5.
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Let's check and make sure that that works: 35 to 175--yes, if we use a calculator (or do it in our heads,
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or write it out by hand), we realize that, yes, we can multiply by 5 to get from there to there.
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The same thing: 175 to 875: we multiply by 5; so we see that this is continuing.
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Yes, it is geometric; that checks out.
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Now, we want to give a formula for the general term.
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We talked, in the lesson, about how a<font size="-6">n</font> is equal to (let me write it the way we had it last time):
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r, the rate that we are increasing at, to the n - 1, times a₁.
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What is our a₁? Well, a₁ is equal to 7, because it is the first term.
00:21:25.800 --> 00:21:30.600
What is our r? r is equal to 5, since it is the number we are multiplying each time.
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So, r = 5; r = 5; a₁ = 7; so a<font size="-6">n</font> is equal to 5^n - 1 times 7.
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And if we wanted to check this out, we could do a really quick check.
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We could plug in...let's look at a₂; that would be equal to 5^2 - 1 times 7,
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so 5 to the 1 times 7; 5 times 7 is 35; we check that against what our second term was.
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And indeed, that checks out; so it looks like we have our answer--there is our answer.
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All right, the next example: Find the value of the finite geometric series below.
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Notice: this does have an end--we stop at 3072, so it is not an infinite one.
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If it were infinite, it would go out to infinity, so we wouldn't actually be able to find a value.
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All right, how do we figure this out?
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The first thing: what is the rate that we are increasing at?
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To get from 3 to 6, we multiply by 2; to get from 6 to 12, we multiply by 2; so at this point, we realize that r equals 2.
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What is the value a₁? That is 3.
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What we are looking for, remember: the formula we are going to do is: the nth partial sum, s<font size="-6">n</font>,
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is equal to a₁ times 1 minus the rate, raised to the nth power, divided by 1 minus the rate.
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So, the only thing we have to figure out, that is left, is what our n is; n = ?.
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How many value are we going to be at?
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Here we are at a₁, but this is a<font size="-6">?</font>; this is a<font size="-6">n</font>, right over here.
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What would that have to be? Well, we could set this up, using the formula that we talked about before, our general formula for the general term.
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a<font size="-6">n</font> is equal to r^n - 1 times a₁.
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We plug in our a<font size="-6">n</font>; that is 3072; 3072 =...our rate is 2, raised to the n - 1, times...a₁ is 3.
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So, we divide both sides by 3, because we are looking to figure out our n.
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Divide both sides by 3, and we get 1024 = 2^n - 1.
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Now, you might just know immediately that 1024 is the tenth power of 2: 2^10 = 1024.
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So, we would see that it is 10 steps to get forward; we would multiply 10 steps forward, so that would mean that our n is equal to 11.
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We have figured out that to get 2^10...that is 10 steps; we multiplied all of them on the 3; we started at 3
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as our first stepping-stone; we stepped forward 10 times...so the first stepping-stone, plus 10 steps forward,
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means a total of 11 stepping-stones; so we have n = 11.
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However, alternatively, we could just figure this equation right here out by using the work that we did with logarithms long ago in this class.
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1024 = 2^n - 1...well, what we can do is just take the log of both sides: log(1024) = log(2^n - 1).
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One of the properties of logs is that we can pull down exponents; that is why this is so useful.
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We have n - 1 times log(2); log(1024) over log(2) equals n - 1.
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log(1024)/log(2) is equal to 10; it equals n - 1, which tells us that n equals 11.
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So, you could work this out just through raw algebra and using logarithms;
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or you could work it out, if you recognized 2^10 as equal to 1024, if you just kept dividing 1024 with a calculator
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until you saw how many steps it was; either way will end up getting us this value, that n equals 11.
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All right, great; now we have everything that we need for our sum: s<font size="-6">n</font>, the nth partial sum
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(in this case, the 11th partial sum of what this sequence would be) is going to be equal to...
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a₁ is 3 (our first term), times 1 minus our rate (it multiplied by 2 on each one of them),
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raised to the 11th power (because the number of terms we have total is 11), divided by 1 minus the rate (2, once again).
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We work this out: 3 times 1 - 2^11 is -2047; 1 - 2 is -1, so the -1 cancels out with the negative on top.
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We have positive 2047 times 3; and we end up getting 6141; that is what we get once we add up all of those terms.
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Great; the third example: Find the value of the below sum.
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The sum is in sigma notation: from i = 0 up until 10 of 500 times 1/2 raised to the i minus 3 to the i divided by 64.
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Previously, when we talked about sigma notation, series notation, we talked about how summations have properties
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where we can separate some of the things in sigma notation, and we can pull out constants; great.
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So, let's start by separating this: we can write this as Σ...it still has the same upper limit; the limits will not change...
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the same index and lower limit of 500, times 1/2 to the i...minus...so what we are doing is separating around this subtraction...
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minus...the series...same limits...of 3 to the i over 64.
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So, we can separate, based on addition and subtraction, into two separate series.
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Furthermore, we can also pull out constants and just multiply the whole thing.
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We have 500 times the series; 10i equals 0, 1/2 to the i, minus...we pull out the 1/64, since that is just a constant, as well...
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on the series, 10i equals 0; 3 to the i...cool.
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So, at this point, we can now use our formulas.
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Remember: our formula was that the nth sum is equal to the first term, times 1 minus r^n, over 1 - r.
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So, each one of these is going to end up having different rates.
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But they are going to end up having the same n.
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Our n is going to be based off of going from 0, our starting index, all the way up to 10, our ending index.
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What is our n if we go from 0 to 10?
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Remember: we have to count that first step, so 0 up until 10 isn't 10; it is 11.
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1 to 10 is 10, so 0 to 10 must be one more, 11; so we have n = 11.
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All right, back into this: we have 500 times...let's use that formula...the series from i = 0 to 10 of 1/2 to the i...
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well, what would it be for our a₁--what would be the first term?
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Well, if we plugged in i = 0 on (1/2)^i, (1/2)⁰, or anything raised to the 0, is just 1; so our a₁ is 1 there.
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Times 1 minus...what is the rate? Well, we are multiplying by 1/2 each time, because it is 1/2 with an exponent on it.
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So, 1/2 is our rate, raised to the n; our n that we figured out was 11, divided by 1 minus the same rate, 1/2.
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Great; minus...over here, 1/64: we apply that formula again; so, the series from i = 0 to 10 of 3^i:
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what is our a₁--what is the first term?
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Well, 3 to the 0, because our starting index is 0 once again--the first term of this series would be 3 to the 0, because it is the first thing that would show up.
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3 to the 0 is just 1 again; 1 minus...what is our rate? Our rate is 3, because we multiply by 3,
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successively, for each iteration, because it is an exponent.
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3 to the...what is our number of terms? 11; n = 11...one minus three.
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Great; now we can work through a calculator to figure out what these things are.
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We will save the incredibly boring actual doing of the arithmetic, but it shouldn't be too difficult.
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This will become 500 times 2047, over 1024, minus 1 over 64, times 88573.
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We can combine these things; actually, let's distribute our constants first.
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2047...well, we will distribute just the multiplication; they are all constant numbers now.
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255 thousand on our left, 875, divided by 256, minus 88573 over 64;
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we have to put the right side on a common denominator, but once again, I will spare working out every single aspect.
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This would simplify to -98417 divided by 256; and there is our answer, exactly precise.
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If we wanted to, we could have approximated to decimals at some point, and we would get -384.44, approximately.
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Notice: probably back here, when we were working at these steps, where we have these giant fractions,
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there is a good chance, if you are working with a scientific calculator or a graphing calculator, that it doesn't show it perfectly in fractions.
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You would end up getting fractional numbers; and for the most part, most teachers and textbooks would be fine with giving an answer in decimals instead.
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So, either one of these two answers--the specific exact fraction, -98417/256, or approximately -384.44--both of them are perfectly fine.
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The fourth example: Many lessons ago, when we first learned about exponential functions,
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we had a story where a clever mathematician asked for one grain of rice on the first square of a chessboard,
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then two on the second square, four on the third square, eight on the fourth square, 16, 32, 64, 128...
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he keeps asking for doubling amounts of grains of rice.
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So, if he had actually been paid on all of the squares, just as he was initially promised, how many grains of rice would he have total?
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So, what we have here is a finite geometric series, because it is doubling each time.
00:31:31.600 --> 00:31:39.900
What is our first term? Well, a₁ was on the very first square; he had one grain, so our first term is going to be 1.
00:31:39.900 --> 00:31:41.900
What is the rate that it increases at each time?
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It goes 1, 2, 4, 8, 16...so it is doubling each time; so that means that the rate is multiplied by 2.
00:31:48.400 --> 00:31:52.900
And what is the number of terms total? If we are going from the first square of a chessboard to...
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a chessboard is 8 x 8, so the total squares on an 8 x 8 chessboard...we have 8 by 8, so that means we have 64 squares total.
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So, that is going to be 64 times that this is going to happen; we have 64 terms here.
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If we are going to figure this out, the nth partial sum, the 64th partial sum,
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is going to be the first term, a₁, times 1 minus the rate, 2, raised to the number of terms, 64, over 1 minus the rate, 2.
00:32:28.000 --> 00:32:41.800
We work this out: we end up getting 1 - 2^64 over -1, which is the same thing as negative 1 - 2^64,
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which we will end up seeing comes out to be very positive.
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This is our precise answer for the number of grains; but let's see what that is approximately.
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That comes out to be approximately 1.845x10^19 grains of rice.
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That is a big, incredibly huge number of grains of rice.
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It might be a little bit hard to see just how many grains of rice that is: 10^19--we are not used to working with scientific notation that large.
00:33:08.800 --> 00:33:11.300
So, let's turn it into some words that we might understand.
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That would be about the same as 18 quintillion (that is a quintillion: it goes million, billion, trillion, quadrillion, quintillion) grains of rice.
00:33:23.100 --> 00:33:28.400
And we are, once again, not really used to working with numbers as incredibly large as the quintillion scale.
00:33:28.400 --> 00:33:41.700
So, let's write that as 18 billion billion grains of rice; that is an incredible number of grains of rice--that is so many grains of rice!
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How much rice is that? That is about more than the rice that has ever been produced by humanity, over all of humanity's time existing!
00:33:50.400 --> 00:33:56.500
It might be approximately on the same scale as the amount of rice humanity has ever created.
00:33:56.500 --> 00:34:00.800
I am talking about all of humanity ever creating this over the whole history of the world.
00:34:00.800 --> 00:34:06.200
All of the rice that has ever been made...18 billion billion grains...that is probably somewhere on the same scale.
00:34:06.200 --> 00:34:09.300
It might be a little bit more; it might be a little bit less; but that is the sort of scale.
00:34:09.300 --> 00:34:13.900
All of the rice that humanity has ever created: we get to very, very large numbers very quickly
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with these exponential functions when we are working in geometric stuff; that is something to keep in mind.
00:34:18.400 --> 00:34:24.700
18 billion billion grains is as much rice as humanity has ever made--pretty amazing.
00:34:24.700 --> 00:34:28.100
The final example: a super ball is dropped from a height of 3 meters.
00:34:28.100 --> 00:34:31.500
On every bounce, it bounces 4/5 of the previous height.
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If allowed to bounce forever, what is the total back-and-forth distance that the ball travels over?
00:34:35.900 --> 00:34:38.800
Notice that it is total, so it is up and down.
00:34:38.800 --> 00:34:41.600
Let's draw a picture to help ourselves see what is going on here.
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We start...we have this ball; we drop the ball, and it falls three meters.
00:34:47.100 --> 00:34:52.100
Then, it bounces; it is a super ball, so it bounces up, and it will go up by how much?
00:34:52.100 --> 00:34:59.000
It will go up by 3 times 4/5, because it goes up to 4/5 of its previous height on every bounce.
00:34:59.000 --> 00:35:01.700
Now, it is going to fall down; well, what height did it just fall down from?
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It is going to fall down from the same thing, 3 times 4/5.
00:35:05.900 --> 00:35:08.400
Then, it is going to bounce up again; it is going to bounce up by how much?
00:35:08.400 --> 00:35:13.600
Well, it is going to bounce up to 3 times 4/5 times 4/5, because it is 4/5 of the previous height it came from.
00:35:13.600 --> 00:35:18.000
3 times 4/5 times 4/5...well, we could just write that as 3 times (4/5)².
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Then, it is going to fall down, once again, from that same height; so it is 3 times (4/5)².
00:35:24.000 --> 00:35:28.800
And then, it is going to just continue on in this manner of going a little bit less each time:
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3 times (4/5)³, and then down again: 3 times (4/5)³, and just continuing on in this manner forever.
00:35:40.600 --> 00:35:44.100
We want to figure out the total back-and-forth distance that the ball travels over.
00:35:44.100 --> 00:35:47.500
Notice: what we have here is ups and downs.
00:35:47.500 --> 00:35:57.800
So, I am going to go with calculating the ups first: how much is from this up plus this up plus this up, going on forever and ever and ever and ever?
00:35:57.800 --> 00:36:02.200
Notice: what that means we are dealing with is an infinite sum, because we are saying,
00:36:02.200 --> 00:36:06.800
"If it were to continue bouncing forever, what would be the amount of distance it would travel over?"
00:36:06.800 --> 00:36:16.000
That is equal to the first value, times 1 over 1 minus the rate; that was what we figured out that the infinite sum comes out to be.
00:36:16.000 --> 00:36:18.900
It is the first value, times 1 over 1 minus the rate.
00:36:18.900 --> 00:36:34.000
All right, in this case, for our up bounces, all of the ups are going to end up being...our first value
00:36:34.000 --> 00:36:38.500
is a₁, so our first value is this one right here, 3 times 4/5.
00:36:38.500 --> 00:36:42.900
We might be tempted to think that it is 3 meters, but remember: this is all of the ups--we are looking at the ups first.
00:36:42.900 --> 00:36:46.800
And we will see why at the end--why I chose to look at the ups first.
00:36:46.800 --> 00:36:53.200
So, all ups is going to be...3 times 4/5 is our first value that we end up having occur.
00:36:53.200 --> 00:37:00.800
And then, that is going to be times 1 over 1 minus...what is the rate? 4/5, because it is 4/5 on every bounce.
00:37:00.800 --> 00:37:06.700
It goes up to 4/5 of its previous height, so the rate for the next one would be 4/5 of that, and 4/5 of that, and 4/5 of that.
00:37:06.700 --> 00:37:10.400
Notice: there was one other requirement on being able to use this.
00:37:10.400 --> 00:37:13.100
The absolute value of our rate must be less than 1.
00:37:13.100 --> 00:37:16.700
But since it is 4/5, that ends up checking out; OK.
00:37:16.700 --> 00:37:31.200
Keep going with this: that is 3 times 4/5; let's write it as 12/5; 12/5 times 1 over...1 - 4/5 is 1/5;
00:37:31.200 --> 00:37:39.000
well, 1 divided by 1/5 is 12/5 times...1 over 1/5 is 5; so we have 5 times 1/5 on the bottom;
00:37:39.000 --> 00:37:47.200
that cancels out; so we have 12 meters total of bouncing for our up values.
00:37:47.200 --> 00:37:52.800
OK, but that is only part of it; now we also have to figure out what all of the down values are--
00:37:52.800 --> 00:37:58.800
what is the down value here? the down value here? the down value here?...going out this way infinitely.
00:37:58.800 --> 00:38:12.000
Well, we also have this part right here; but notice: all of these green things end up having a matching up value.
00:38:12.000 --> 00:38:17.100
They each match up: the purple and the green, the purple ups and the green downs; they all match up.
00:38:17.100 --> 00:38:22.800
The only one who is out of the normal case of matching up is that red first drop down.
00:38:22.800 --> 00:38:31.000
So, what that means is that we can match all of our downs; they have a matching 12 meters,
00:38:31.000 --> 00:38:38.700
because it matches all of the ups; but then we just have to add on the initial 3-meter drop at the beginning.
00:38:38.700 --> 00:38:41.400
So, how many downs do we have in total?
00:38:41.400 --> 00:38:47.400
Well, that 3-meter drop, plus the matched value (because it matches to the 12-meter value for all of the ups),
00:38:47.400 --> 00:39:05.300
12 meters, plus 3: we get 15 meters, so the total number is going to be equal to the 12 from our ups,
00:39:05.300 --> 00:39:13.800
plus the 15 from our downs; 12 + 15 means 27 meters in total.
00:39:13.800 --> 00:39:15.600
Great; there we are--we finished that one.
00:39:15.600 --> 00:39:18.300
All right, now we have a pretty good understanding of how sequences work.
00:39:18.300 --> 00:39:20.700
We have worked through geometric sequences and arithmetic sequences.
00:39:20.700 --> 00:39:25.000
We understand how series work; we have a pretty good idea of how sequences work; great.
00:39:25.000 --> 00:39:27.000
All right, we will see you at Educator.com later--goodbye!