5

X 1 3 { # H [ 1 1 0 L 2 Ji 1 1 1 Ewt 3 O 1 I 1 1 3 1 [ 1 { L L 2 F 1 { 2 | 1 # { 1 [ L 1 1 L E 8 0 } F L 8 2 7 1 3 1...

Question

X 1 3 { # H [ 1 1 0 L 2 Ji 1 1 1 Ewt 3 O 1 I 1 1 3 1 [ 1 { L L 2 F 1 { 2 | 1 # { 1 [ L 1 1 L E 8 0 } F L 8 2 7 1 3 1

X 1 3 { # H [ 1 1 0 L 2 Ji 1 1 1 Ewt 3 O 1 I 1 1 3 1 [ 1 { L L 2 F 1 { 2 | 1 # { 1 [ L 1 1 L E 8 0 } F L 8 2 7 1 3 1



Answers

Fill in the entries in the following table $$\begin{array}{|c|c|c|c|c|}\hline x & f(x) & g(x)=f(x)+2 & h(x)=\frac{1}{2} f(x) & i(x)=3 f(x)-2 \\\hline-1 & -1 / 2 & & & \\\hline 0 & 1 & & & \\\hline 1 & 2 & & & \\\hline 2 & 6 & & & \\\hline 3 & 8 & & & \\\hline\end{array}$$

Matrix modification here. This time we got the zero matrix times F. Okay, If you don't already know the zero matrix times anything, we'll leave the zero and all of our spots. We'll just see why hand. Okay, so let's go through and do matrix modification. Row one, column one, zero times three. Zero times negative one. We know that's going to be zero. Okay, next spot in top right. Row one. Column one. Well, that's the same calculation. We just did we know that's going to give us zero. Okay, we can avoid a little bit of work there. Bottom left road to column one. Again, the same calculation we've already done because of how the numbers are falling here. The multiplying by zero in each spot is given a zero everywhere as well. Okay, last time, row two, column to again, pretty easy to see. That's gonna give us there. Okay? It's the same with real numbers at multiplying zero by anything, we'll give you zero. Just in this case it gives us the zero metrics rather than the real number zero.

Going on three times matrix F. So I've got matrix F right here, this is F. And we're just gonna modify it by three. So take each component and multiply it by three. I'll take one step in here. Make sure you put times three and all of these. Okay? If you want to do that step in your head, you can. But your result will be 99 negative three negative threes, just distributing that coefficient, distributing that number in front onto each element of the matrix.

All right for this problem. We want to confirm that the GMs or the ghost in hyper geometric Siri's With Alfa equals one half data equals one gamma equals three over to and using X Square two rather than X, that Gaussian hyper geometric function is the same thing as one half X to the power of negative one times the natural logarithms of one plus X, divided by one minus x. So let's first look at the left hand side, expand that out or figure out what that looks like. It's a power Siri's so that is going to be one plus the sum from n equals one up to infinity of we have factorial function one half times the factorial function on one divided by and factorial three half n r. Factorial function three half up to end times X The power of to n Now we know that's the factorial function on one upto end is actually the same thing as in factorial. So those cancel out So we're left with one plus the sum from n equals one up to infinity of one half, factorial on and divided by 3/2 factorial and x, the power of to win. So on the top and on the bottom, we'll have. I'm just rewriting some of the stuff here and equals one up to infinity. The top. We'll have one half times one half plus one. Those one half plus two dot, dot dot up to one half plus n minus one. So that end point up there, one half plus and minus one is going to be end minus one half. And on the bottom we have three over to times 3/2. That's one times three over to those two dot, dot dot, up to three over to plus n plus in les en minus one, then times extra power of tea. When Now I'll note. And also all note here three half plus and minus one. That's the same thing as and plus one half. And I'll note here that three over to is equal to one half plus one so we can actually rewrite this as one plus the sum that was a terrible Sigma or whatever from n equals one up to infinity of one half one half plus one, one half plus two that not up to end minus one half divided by one half, plus one times one half. So the three halfs plus one would be one half plus one plus one or one half plus two and so on up until we get so it's not plus times we'll have times Uh huh. And minus one half times and plus one half the end minus one half. There comes from the fact that in our product, the term before the end would be three halfs plus n minus two, which is an minus one half, but also the same. That is the same thing is what we end up with at top. So you can see now that a whole bunch of these or almost all of these are going to cancel out. So what we're left with and it's all multiplying X power to end. What we're left with is I'm from an equals one up to infinity of one half X, the power of to U N. Divided by and plus one half. Now we can do a little bit of rearranging here that end plus one half that's the same thing as to end plus one all divided by two. So that's equal to one plus the sum, and it was one up to infinity of one half extra power to n divided by two n plus one. Yes, do n Plus 1/2 two's cancel each other out, so we're left with that. Get one plus the sum and equals one up to Infinity X, the power of to N divided by two and plus one. So that's our left hand side. Now we can hop over and take a look at our right hand side. I'm just going thio, Grab this, Copy it, paste it into a new page here. That's that's what I wanted. Scuse me case into a new page. There we have one half extra power of negative one and lawn of one plus X so one thing that we can see here. That way we can make this the first step that when we have long of a over B, that's the same thing as long of a minus lawn of so weaken. First say that that's equal to one half X are negative one times lawn of one plus X minus lawn of one minus X. So now, recognizing that let's go, let's taken aside and look at what the Siri's expansions of lot of one plus X and lot of one minus X R. So we want and we'll be expanding them about X equals zero, so we'll have f zero plus F prime and zero divided by one times X plus half double prime at zero, divided by two times X squared, plus dot, dot dot so lawn of one plus zero is just going to be lawn of one in line of one is zero. And then, of course, we can represent the rest of that sum as some from n equals one to infinity, the inthe derivative of this case. Log or lawn Rather divided by n factorial x The power event. The reason why I write this this way is because it is actually possible to find a general expression for the month derivative of lawn as long as N is greater than or equal to one so and the ends derivative. Second here, the insta river tive of lawn of one plus X, and for the sake of time, I'm not going to go into exactly how you find this, but it's not excessively difficult to find the the anti derivative is negative. One. The power of n minus one kinds X plus one. The power of negative end times and minus one Factorial So and since we're evaluating this at zero, we'll have that lot of one. Plus X is equal to the sum from n equals one up to infinity. Since we're evaluating at zero this X plus one. Um, it's the power of negative end doesn't make any difference. It's just one divide or one to the power of negative end. It's always going to be one. So what we're left with is in the numerator will have negative one our end minus one times n minus one factorial divided by n factorial and X The power event now you can recognize here and minus one factorial divided by N factorial is the same thing as n minus one. Factorial divided by n times n minus one factorial so we can cancel those out. We're left with one over n So we get that it is some making from n equals one up to infinity of negative one to the power of end minus one and x the power of n divided by n. Next we take a look at lawn of one plus X Sorry lawn of one minus X. It's Siri's expansion, turns out very similarly except for where a lot of one plus X the anti derivative You have this negative one to the power of end minus one for lawn of one minus X. We have a to end minus one there instead. So the expression for it is going to be the some from n equals one up to infinity of negative one our of two and minus one and extra power of n divided by n. So if we look at lawn of one plus X minus one of one minus X, we get that's going to be some from n equals one up to infinity of negative one to the power of and minus one. That's X power end divided by n minus negative one are to end minus one times x the power of n over n which then simplify factoring out the exit power of en and the en. So we get its exit power of n over n times negative one the power of and minus one. But also we can recognize you have this negative multiplying this negative one to the power of two in minus one. So that's going to give us a plus. Negative one. The power of to n Now what? This ends up doing alright, actually, right out the end First few terms of the Siri's you'll have X times so negative one to the power of one minus one. That's going to be one less negative one hour of two n So one thing that we can recognize actually negative one Teoh a even power is always going to be positive. So this negative one to the power of to end can actually rewrite. What I'll do is I'll just copy and paste that over No, and in that for a second there so that negative one to the power of to N. You can change that to be just a plus one. So what this is going to end up doing is that we are going to be alternating between having a coefficient of to so we'd have if we have and even exponents on this negative one, which will happen whenever we have an odd N. Then we'll have one plus one times x power of and over n And whenever we have an even n, we'll have an odd exponents on the negative one. And so we'll have negative one plus ones will have zero. So what, we end up having that we get every odd term killed off. Does that make sense? Yes. Every no sorry. Every even term will be killed off. So we'll have the some from n equals one to infinity. And if we only want to look at the even terms Sorry. The odd terms or what will survive? Then we can write this as this extra power of Tuen minus one. Yeah. Sorry to tend X The power of two and minus one. Yeah, come on. Two times x, the power of to end minus one, divided by two and minus one. So next we still had this one half, uh, expert of negative one multiplying out front. So if you look at, we're flying by the one half. Expect of negative one a lot of one plus X minus Lawn of one minus X. That So the division by two will kill off that coefficient of two there, and the extra power of negative one is going to lower the power of this. So it's going to be the some from n equals one up to infinity of X, the power of to n minus not minus two. Sorry, not minus one. It's minus two, invited by two in minus one. But this to end minus two. That's going to be the same thing as to times n minus one. So we end up having it's the some from n equals one up to infinity of X power of to end or two times and months one divided by two n minus one. Now, now what we can dio is if we look at how this expands out the first term in this some when n equals one, that's going to be X to the power of zero. So one divided by two minus one. So the first time? Well, actually one. Then we'll have plus thes some from n equals two up to infinity. That's the power of to end minus one, divided by two end minus one. So what we can do now is we can change the index that we're using. I'm just going t o make a quick adjustment here, give myself some more space. So we can shift our index can say let MM equal n minus one. So for starting at and equals two are actually I'll do a quick thing. That means that n is equal to m plus one. So in equals when n equals two, that means that we have am equals one. And we get that we have the some or our expression is going to turn into the some from n equals or sorry am rather am equals one up to infinity and X to the power of to em Divided by now since n equals M plus one, we have two times M plus one minus one on the bottom. Now to save some space, just going to the right. Okay, so two times m plus one that's to m plus to minus one. And that is just going to equal to M minus one. So we get that it turns into one, plus the sum from M equals one up to Infinity X, the power of to M divided by two M minus one, and our index is completely arbitrary. We can call it whatever you want, so he can just say relabel relabel terrible out relabel Mm call it and again. So we have that one half X. The power of negative one lawn of one plus x, divided by one minus X, is equal to one plus thes some from n equals one to infinity X the power of to n divided by two n minus one If we go back Oh, one moment. Oh, yes, silly mistake I made to m plus two minus one. That should be to m plus one to m plus one. There we go. Yes. So it's a one plus the sum from n equals one up to infinity tends or of extra power to end provided by two and plus one, which is exactly what we found as the Siri's form of our Gaussian hyper geometric function. Thus it is equal to U F one half one three halves X squared.

They were multiplying matrices, E and F. Together here. Eight times F. We know this will give us a two by two matrix as a result. So. Top right element row one, column 11 times three plus three times negative one. Get zero there. Okay. Top right. Row one column to one times three was three times negative one. This is another zero there. Okay. Moving our way through. It just takes a bit of practice. These guys working through all the elements, say it to yourself in your head or out loud. If you want. Row one column, try road to column one. Two times three plus six times negative one. Okay, It's gonna give us another zero there. I guess. I think we're getting zero matrix out here, but we'll check with the last element Road to college too. Two times 36 times negative one. Another zero out there. Okay, so our result here is again zero metrics.


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