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Use generating functions to solve simultaneously the following recurrences:@n+1an + bn + Cn,n > 1,bn+1 = 4n _ Cn,n 2 1,Cn+l = 4n _ bn,n > 1,he initial conditi...

Question

Use generating functions to solve simultaneously the following recurrences:@n+1an + bn + Cn,n > 1,bn+1 = 4n _ Cn,n 2 1,Cn+l = 4n _ bn,n > 1,he initial conditions @1 = b1 = C1 = 1.

Use generating functions to solve simultaneously the following recurrences: @n+1 an + bn + Cn,n > 1, bn+1 = 4n _ Cn,n 2 1, Cn+l = 4n _ bn,n > 1, he initial conditions @1 = b1 = C1 = 1.



Answers

Use generating functions to solve the recurrence relation
$a_{k}=3 a_{k-1}+2$ with the initial condition $a_{0}=1$

Gearbox is equal to one plus some. Okay, from one to infinity in K ham's extra power. Okay, which is equal to one plus, um Okay. From one team in the tea. And here we replace a I three times a month. One us or to K minus one. I'm just okay, this is equal to one plus reacts some and from one t o ky a k minus one. Access to a free man. Swan us ex im some hay from 1 to 20. Well, that's the off a menace. One This part is ICO too. Jia vice on this part is equal to you. One over one, minus flax. So what's my excuse? And we have gearbox Zico, too. One class acts or one minus. Relax. You wanted by one minus reacts you can simply by this term this is You can chew one over one minus wax. We can ride this. You go to Psalm a three year old or to K excuse. Okay, so Okay. It's equal to war is about Kay

Were given or occurrence relation with initial conditions were fast to use. Generating functions to solve this relation. Their occurrence relation is a K is equal to four A K minus one minus four and K minus to plus case weird. Within your conditions, a zero equals two and a one equals five. Suppose that G of X is the generating function for sequence a K then G of X is equal to some from K equals zero to infinity of a K x to the king. This is equal to a zero plus a one x plus the some from K equals two to infinity of a piece of K extra decay by the recurrence relation. This is equal to a zero plus a one x plus to some from K equals two to infinity of for a K minus one plus ar minus for a K minus two plus case. Weird. Thanks to the King, this is equal to a zero plus a one x plus four times this some from K equals two to infinity of a K minus one Next to decay minus four times the some from k equals to to infinity uh, a k minus two extra decay plus some from K equals two to infinity of case, where extra decay This is equal to a zero plus a one X plus four times and then substituting K plus one for K they had This is the some from cake one in fact, on a straight this way, four times x times the some from K equals one to infinity of a K X to the K minus four times X squared tens. This some from K equals zero to infinity, a k x to the king. And plus, Now we substitute t equals K minus two. So we get Plus, Lawrence did. Teak was K minus two simply right. This as some from K equals zero to infinity of case where next to the K minus and then zero. This is simply zero times 10 Casey, the one we have one squared is one times X to the first since his minus X. This reduces to a zero plus a one x plus four x times. We have that. This is equal to G of X minus when K equals zero, which is Cynthia zero minus four x weird times G of X and plus some from cake zero to infinity of case. Where next to the K? No. In order to find Martin's Seiss form for this, some consider that K squared. This is going to be equal to the sum from K equals zero to infinity of que times K minus one over to times two. This is K Times K minus one and we're going to subtract from that or add to it K minus one. And so now we have que teams came as one of her two times two plus K minus one. This is going to be K plus one times K minus one. The difference of squares This is K squared minus one. Yes, we want to add a plus one. So now a case where? A minus one plus one, which is going to be simply k squared? That's how we get from K squared to this expression. Times X decay minus X. This is equal to zero plus a one x plus four x g of X minus for a zero x minus four x we're g of X and this becomes plus some from K equals zero to infinity. This is going to be que times K minus 1/2. Well, this is que factorial Hoover K minus one factorial K minus two factorial times two factorial. So this is really the some of K Choose to times two plus in K minus one. This is simply K minus one factorial our choose one plus one. We don't to write it this way we can keep It is K minus one mrs to a zero plus anyone minus four a zero X plus or X g of X minus four X squared G of X plus two times the sum from cake Will zero to infinity of K Choose to X to the K plus the some from K equals zero to infinity Minute made a mistake here. So going back were it? This is K equals one to infinity. This is simply from K equals one to infinity. Yes, this is from Kati Rolls one to infinity of K Choose two times two. So this last expression here is from K equals one to infinity of K Choose to next to the K plus some from K equals one to infinity uh, K minus one extra decay plus the some from K equals one to infinity of next to the K minus X, and this is equal to a zero plus a one minus for a zero x plus four x g of X minus four X squared G of X plus two. And then this binomial coefficient tells us that this is going to be I made a mistake. That should be This is the same as instead of K trees, too. I can write this as que choose K minus two. We have by a formula from the book that this expression it's going to be after I felt factor on next squared. This is equal to the sum from K equals zero to infinity of Cape List to choose K X to the K plus This is going to be came I just one extra decay. This is the same as one over one minus x squared. We can also factor on next. So we get went over X minus one, squared just the same and then add to that whenever one minus sex. And so we yet is equal to a zero plus a one in minus four a zero x plus four x g of X minus four X squared G of X and this is going to be we have plus two over one minus X cute plus 1/1 minus X squared plus 1/1 minus X is equal to zero plus a one minus for a zero x plus four x g of X minus four X squared G of X. This is equal to plus at least coming. Denominator is one minus x cubed, so we have one minus X squared plus one minus x and are we have to plus one minus X plus one minus X squared. This reduces to X squared minus three X plus four Oliver one minus execute and so we can solve for G of X. So we have that attracting for extra vex and adding for X squared g of X to both sides, we get one minus four x plus four X squared g of X is equal to this is a zero plus a one minus four a zero x plus X squared minus three x plus four over one minus x cubed. And so we have that G of X is equal to we plug in a zero a one where a zero is two and a one is five. We have to unless five minus eight is negative. Three x plus x squared minus three x to the fourth over. One minus X cubed while divided by one minus for X plus four X squared, and this reduces to three x to the fourth minus 11. Execute plus 16 X squared minus 12 x all over. This is going to be one minus. X cubed times one minus two X squared and in order to find the terms for the sequence a k who need to use partial fraction decomposition. So we have that say this is equal to hey over one minus X plus B over one minus X squared plus C over one minus x cubed Plus do you over one minus two x plus you over one minus two x squared. And to solve this system of equations, we'll use a computer algebra system. Since the system of five equations and five variables take a very long time and we get that a is equal to 29. B equals nine c equals two d equals negative 1 13 over to and easy quote of 45/2 and plugging in these values, we get expression of g of X in terms of these sums. And so we have that G of X is going to be equal to after doing this and converting these sums. Using the formula from the book, we find that sequence member a K is given by 29 plus nine times K plus one plus two times two plus K over K minus 133 over to to the K plus 45/2 K plus one to decay. This is our answer.

Shell backs you go to war. Plus while thanks plus some crave from true between unity a que ext Okay, then replace a k i a k minus one us two times a a minus two last. True to our okay, X k, this is you call Jew War class while blacks plus x times some Okay, um, true to you. T ache a minus one thanks to a A minus one last chu ex player, Some from true three Unity A k minus two x a month too. Plus, this one is equal to a lack squire over one minus two ax. This is equal to or plus 12 bucks. Last X cams. Relax. Minus four. Plus Xu Ex lawyer jukebox. Plus, we'll explain. We're one minus two ax. We can simplify this equation on behalf gearbox you called true or minus? Well, explain over one. Ministry likes times one minus X minus. Chu X squared and buy a copy leader. We have. This is equal to Chu. Third time swan over one, minus two. Ex lawyer plus 38 over nine. Arms one over one, minus two x minus eight over 91 open one plus x. So we're half a k. It's equal. Chu True. Third time's a swan Chew it is. Okay. Class 38 over nine. It's true. Okay, Minus it's over. Nine times. Make you want to, okay?

For this problem. We have been given eight recursive relationships, um, with initial conditions for different sequences. Now the Rikers Asian method is a really nice, compact, way too put down sequences, unfortunately, defined, like the 100th sequence of the 100 determine the sequence requires us to know all of the other previous 99. So sometimes it's nice to be able to write these sequences without having to use, um, recursive elements. So we're going to take recursive definitions of sequences and change them so that they're not recursive. So let's take a look at the first one and see how this works. We have eight to the end equals three times a to the N minus one. OK, so that's our death recursive definition. And we're also told that a sub zero equals two. So we're going to do for this problem and all of the subsequent ones. In this example, we're gonna find maybe the first maybe up through a sub four and see if we can find a pattern that emerges that we could use to define the sequence. So a sip one means I take three times my prior once said to be three times to How about a sick to? Well, that's going to be three times the prior term. It tends to Well, a sub three will be three times the prior term, which is three times three times to okay, and you can hopefully see the pattern emerging here. In each case, I have three to the same power as that as the base You A sub four has three to the fourth. A sub three has three to the third so we can take the sequence and say the length term of my sequence will be three raised to that end power times two. So, while the recursive definition is nice, our new definition we confined the 1000th term and Onley the 1000th term and not have to find all of the ones leading up to it. So it is convenient to be able to rewrite thes. Okay, be a recursive definition, says a Saban equals a sub end minus one plus two. We're going to start a sub zero equals three. Well, again, let's do the first few here and see if we can see a pattern emerge. Okay, so a sub one we take the prior term plus two. Step two, we take the prior term plus two a sub three We take the prior term plus two. Sometimes it's useful to actually do the math here to do the addition or multiplication. If possible, I'm gonna leave it out spread out like this because it is rather than just seeing three and five and seven and nine. This really shows how the numbers we're getting put together, and I think it makes it a little easier to see the pattern. Okay, let's just do one more. That's the prior term plus two. So in this case, if I look at what the pattern is, in each case I have a three in the beginnings of three plus and then I have I'm adding to and I'm adding one more to each term. So I'm adding to times. And so for the A sub three, I've got three twos for Ace of four. I have four twos. So here is my definition for this sequence. Okay, Next, I have a sub en equals a sub end minus one plus m. And in this case, we're starting with a sub zero equals one. So let's put out a few terms here. Okay. Ace of one. I take the prior term plus n okay, it's up to I take the prior term, plus n prior term plus n prior term plus n. Okay, So what pattern do we see emerging here? Well, in each case, it z pyramid ing. So I have I have this plus one just in the front here. So I think whatever I do, I'm gonna say one plus and then I've got a pyramid. I've got a sub one is one ace up to is one plus to a sub +31231234 Well, we do have a formula for adding the first end. Integers It is n times n plus one divided by two. Right. If you recall why this works, I'll take a step four is my example. I have if I add the first and the last I have, that's and and then plus one. So that's four. And one is five. And every pairing is going to equal that same amount, and I'm gonna have n over two of those pairs. So and plus one times and over two pairs. So this is my sequence. Are by the definition for this sequence, not using a recursive definition. Okay, on to D. De says a seven equals a sub n minus one plus two n plus three. And I'm going to start with a subzero equaling four to a set of 3/8 of four. Okay, so let's see what we get first, a sub one we take the prior term plus two times one plus three. Hey, it's up to we take the prior term plus two times on and plus three ace of three prior term plus, well, that two times two. Let's just kind of put that together here. Plus two times three plus three. Let's just do one more. Unfortunately, I'm not really seeing a solid pattern come out. There's the prior term, plus two times and plus three. Yeah, well, lots of fours and twos and threes, but nothing really is jumping out at me. So let's try another tack. Let's actually add them up and see what we get. Hey, I'm just change a different color here, so I've got four for the first one. Next I have nine. Then I have 16. Okay, Now I see a pattern by adding these. I can see that I'm going up. These are all perfect squares, and I'll just put this one in blue as well, so I can You can really see that pattern now. This is not and squared, though. Put Edward and zero. It's two squared when it is one. It's three squared, so it looks like I'm taking too, plus n, and that's what I'm squaring. So when a sub n 81 end is four, it's and plus two or six squared when end is three to more than that were five squared. So there's that one. So sometimes it's easier to see it when you write it all out sometimes adding, it shows the pattern best. Hey, let's do e a. Saban equals two times a sub end minus one minus one. Hey, and we're going to start with a subzero equally one. Okay, so let's begin a sip One. I take two times the prior one minus one. Well, that equals one. That's exactly where I just started. A sub to I'll take two times the prior one minus one equals one. I'm not going anywhere is of 32 times the prior one, minus one this is not changing. So for in this case, my sequence is one. Every single term is going to be one. It doesn't matter which and I'm at. So let's look a f f says a Saban equals three times a sub n minus one plus one. And I'm gonna start with a sub zero equaling one. So let's do a few here. Yeah, take three times the prior three times one plus one, right ace, up to three times the prior plus one. Well, I'm okay. Let's just do one more of these. Take three times the prior plus one. Well, let's take a look at this one here. I want to get rid of these parentheses. This gives me three squared plus three plus one when I go to a step three and I multiplied by three again. Now I have three cubed plus three squared, plus three plus one. What about a set of four? If I multiply that by three, I get three to the fourth plus three to the third plus three squared plus three. And then I do plus one. So my pattern here these air all descending powers of 33 to the end plus three to the end, minus one plus three. The end minus two, all the way down to three to the zero. So how do we write this as an actual definition? I could see the pattern, but what do we right? Well, I'm going to show you the solution, but we also talk about how we get to it, because there is a way, Ah, formula to write this. Just like there was a formula to write the some of the first and integers. There is a formula to write the some off the first and powers off some given base. So I'm just gonna write the formula first. But then we're gonna talk about how you can actually find it if you didn't know this. Okay, so what it is is you take the base to the N plus one, power minus one over to. So this is the actual sequence for this. But how did you come about something like that? That's not when we use a whole lot. So I'm gonna put this in blue, just what stands out, and I'm going to write down what we saw here, so h the end, Um, and I'm just going to use a generic p to show that it doesn't have to just be based three. It could be any base. So I'm gonna use to use based P. So I have p to the end, plus p to the n minus one plus p to the N minus two, all the way down toe one or p to the zero. Okay. Actually, I'm just gonna Right, this is one. Now let's take this and I'm gonna multiply both sides by p so that gives me P to the end. Plus one, Pete to the N P to the n minus one, all the way down to pee. Okay, so I just multiply every term. Both sides by P. Okay, well, let's take a look at that. Right hand side E, this is a seven. All I need to do to make it a sub end is to add a plus one. And as long as I add that a minus one, I haven't changed that side. So what I have over here is I have p to the end. Plus, I'm sorry. Sorry. Not not Peter. Then. My my apologies. I said the right thing. I wrote the wrong thing. It's a to the end because this whole piece here is a Sabet A This whole thing is a sub en going to get my subscription superscripts messed up here. Okay, Because this when I added that plus one, that's a sub subtracted one. So I haven't changed anything, so I have peach, the end plus one minus one. Okay. Sorry about that. Subscript and superscript mixed up. Hopefully this is making sense. Now I want to solve for a sub ends. I'm gonna subtract a sub in from both sides. So on the left hand side, I have p minus one times a PSA. Ben, if I factor that a sub and out on the right hand side, I have p to the end, plus one minus one. And now I'm going to divide. So I have a over here, a sub n equals p to the end, plus one minus one over p minus one. So that is for any pne base. That's my answer. So, in this case, PS three. So that in other words, I have three to the end, plus one minus one on top, and the bottom is going to be three minus one or two. So hopefully that helps. You might have known this one already. If not, hopefully this will help if you see it in the future. Okay, we have two more to go. Let's look a TGI Yeah, G is a Saban equals n times a sub and minus one. And we're starting with a sub zero equals five. Go down and look at a few entries here. Just get myself a little more of room. Okay? So a subzero was five. A sub one is n this case one times my prior term. A sub two is en times my prior term, three times my prior term, four times my prior term. Okay, well, this right here looks a whole lot like a factorial. The only difference is I'm multiplying it by five. So a Sabet is gonna be five times and factorial, okay. And last sequence H A Saban equals to en times a sub end minus one where a sub zero equals one. Okay, So same process. We're going to see if we can find a pattern. A sub one. It is, too. Times n times my prior term. Okay. And all those ones in there. I'm just gonna simplify that to To ace of To To Times n times my Prior. Okay, it's up three, two times n times my prior It's a 42 times n times my prior to 3 to 2 to Okay, Make sure you can read that. So let's see what we have here. We have a lot of two is going on. We also. Every time we're multiplying by the next image, you're up. So this is looking very much like a factorial. Four times, three times to three times to, um two. Okay, so so those are actually let me keep in the same general spot. Okay, so it looks like in each of these I have I have a factorial going on. But what else do I have? So I've underlined the factorial piece. Hey, here I have a two to choose. Three twos, four choose. So it looks like I also have to to the end going on, because every one of these terms I've got one more factor. Two plus the factorial piece. So there is. The definition of this sequence is well


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