 We are at lecture 57 today, so I think my earlier prediction of getting close to 60 by the end of the class I think we're actually going to be very very close to that. Maybe we won't quite reach 60, but a couple things before we dive into the calculus we have one section to cover 8.9 the first part of it. I don't know that we'll deal with that in-depth It's stuff. We've already covered really in a lot of ways it deals with error estimates and what are some different ways we can truncate this infinite series and Have an idea about how bad our answer might be So we can just kind of reiterate those we won't do much with that But there are a couple of applications to physics that I think are worth looking at Here's some information that I got yesterday, so it's pretty up to the date some unemployment information So if you need a little extra motivation to Stay in school and get your degree This was for March 2009 If you have less than a high school diploma, so not you but anybody else that's watching 15.5% of those people were unemployed in March of this year If you are a high school graduate with no college, I guess that applies to one person in here I guess he had some college because he had a couple days there Those people in March ten percent of those people are unemployed So high school graduate no college a little bit better some college or an associate degree in In March 7.8% of those people were unemployed and A bachelor's degree or higher 4.3% of those people were unemployed So what what's the message of this data? Stay in school and get a degree is what it says to me, especially in this day and time where unemployment is kind of Overinflated over exaggerated. I think a meaningful college degree is very valuable at least that statistically that plays out Another thing before I had this written down yesterday and I just didn't address it This is probably the I guess my curse of mathematics that I like it so much and so I was fixing a Couch a love seat for my niece who's a student at Carolina Even though she's over there. We still kind of like her So one of the springs on the love seat was malfunctioning So I'm taking it all apart underneath and trying to figure out So I go online to figure out kind of how this clip looked Before it broke because I just had a piece of it. So Then it gives me all the specs of this clip and it's got all this in minus one and D and K And I should have brought that in but you know, I can't get away from it so I'm trying to fix a couch and Here's all this numerical Jumbo, you know, just tell me where I can buy the clip and tell me how to install it You know, I can feel like I can fix it so then to cap the weekend off the New York Yankees open their new Yankee Stadium one billion dollar stadium. So a thousand million dollar stadium So that's just outrageous for a sports arena So they had a little trouble with the Cleveland Indians and one of their anybody Yankee fan Do we have a Yankee fan in here? They had a little trouble didn't they with the Cleveland Indians So they did get it corrected, but one of their games it was 22 to 4 they got beat In this, you know, billion dollar stadium. So the I think it was the New York Post the next day had a Headline and the headline was 22 to 4 and then they put an exclamation mark. So where does my mind go? 22 to 4 factorial Well, if this was 4 factorial the Yankees would have actually won the game Right because it would have been 22 to 24 so they probably shouldn't have put that in the New York Post on the headline So it wasn't 22 to 4 factorial. It was 22 to 4 The other thing that is related So that's you know, it's just the curse I guess that you see this stuff and it's oh, it's not for you know With an exclamation mark, it's 4 factorial One of the problems on the test and I know you don't all have it back yet, but it's a problem. Hopefully that You remember is this one That's the hyperbolic cosine So you're probably probably seen that key on your calculator Cache it's sometimes referred to the Cache function it is not a trigonometric function pretty clearly. It's an exponential function But as some of you noted on your test those that I've graded thus far is that it When you're done with the series the power series for it It actually resembles in a lot of ways the power series for a cosine a regular trigonometric cosine function Some things that happen in this Derivative of e to the x's itself and that's going to be all the way down the page derivative of e to the negative x is itself with a Negative one tacked on so the first derivative is this and then when we get to second derivative What happens? It goes back to the original function, right? So the derivative of e to the x's itself derivative of negative one e to the negative x is back to positive and And this pattern continues when you evaluate this at zero the function at zero e to the zero plus e to the Negative zero That's one. There's another one. So there's two in the parentheses times one half Unfortunately, or I guess kind of fortunately in a way We don't have to mess with the term at all because when you put zero into the first derivative e to the zero minus e to the negative zero you get a zero here So you alternate zeros and ones. So as we saw in the cosine This happens as well is you lose every other term. So what does the Function that we're calling hyperbolic cosine What does the power series look like? There's the when n equals zero The original function at zero is one x to the zero over zero factorial So one right Is that an even like the cosine is that an even expression X to the zero zero factorial zeros even So the next term we lose because it's got a coefficient of zero. So we lose the what? Linear term the first degree term so we pick up again with the second degree term And what is it? It's also positive. That's where it's different, right? From the cosine the cosine the regular trigonometric cosine function Alternates signs. So does the sign by the way that that was a Kind of a common mistake on the one sign of x and then sign of the quantity x squared and the integral of sign of the quantity of x squared problem is that a Few of you a handful of you had that sign function as Non-alternating but sine and cosine are both alternating this one is not Over two factorial right The next one we skip then we go to the fourth So it happens to also be positive And I think I ask you to carry it to six right to n equals six So there's what it is. There are other terms to this if it's actually going to be an equation But we could truncate it by saying t6 is approximately the f of x And what is t6? What do you think is the derivative of the hyperbolic cosine? There you go hyperbolic sine well if the derivative of hyperbolic cosine is the hyperbolic sine What then is the hyperbolic sine? Isn't it this so not only are they the derivative of one another? Very convenient that they don't have some s i g n change involved They are the derivative of one another they are the integral of one another So you don't have to deal with this negative sign like we do with regular trig this function right here is the hyperbolic sine and This function up here is the hyperbolic cosine. They are in fact the derivative of one another. What do you think the hyperbolic? Sine is going to look like it's going to be the if this is the even function part of it, it's going to be the odd function part right because the Constance is going to be gone the linear term is going to be there And again we could truncate that but that's what the hyperbolic sine now hyperbolic Any guesses as to why that's part of this what a What is regular trig what's true about that? Well, they're both really exponential stays within the range of think one once Is this related to something geometrically and This is related to something else geometrically well The age hyperbolic is because it's related to a hyperbola The same way that the sine of x and cosine of x are related to a Unit circle right if this is the angle theta What are the coordinates of this point on the unit circle? So we call this circular trig a lot, but that's kind of the regular sine and cosine What's the x value of this point on a unit circle having rotated through? theta degrees or radians cosine theta Y value is Sine theta so the way that regular trig is related to a unit circle This kind of trig hyperbolic trig is related to points on a guess what? hyperbola and actually technically it'd be called a Unit hyperbola so it has to do with coordinates of points on a hyperbola Just the way regular trig deals with coordinates of points on a unit circle Somebody graph a hyperbolic cosine and then we'll also try to get a picture. Maybe it'll graph Cosh, I don't know if you can actually put that in to your calculator and it would graph that but I know it'll graph this What's it look like if you had to describe what it looked like? What other figure does it kind of resemble? Kind of looks parabolic doesn't it and it can you tell what point this is right here? It's one I think yes, it's got to be one so it is not a parabola, but it kind of looks Parabolic that's what hyperbolic cosine looks like if you were to plot points on this thing Hyperbolic sign this is sometimes called a cinch Okay, or hyperbolic sign it kind of gets old saying hyperbolic sign over and over What does it look like at zero you get zero right because this is zero Kind of looks a little bit like an x cubed function, so I think it looks something like like this That's what this guy looks like So they are kind of nice smooth continuous curves Can be dealt with a lot of times the same way that we study first second derivative Integrate them we just in this book. It just isn't addressed very often these two functions hyperbolic sign and hyperbolic cosine But I'm sure you've seen that key on your calculator If you have a scientific calculator that has a cosine button a lot of times the second function is the hyperbolic cosine Same thing with sine and hyperbolic sine, but that's what those two graphs look like One thing that you've probably seen a picture of or maybe have actually seen in person Is the arch in st. Louis? It doesn't look like this that'd be kind of a funny-looking arch, but it looks like that Flipped so the symmetric image of this is Actually, if you look at and this again is the kind of the curse of mathematics if you get a brochure For the arch of st. Louis on the back of it. It actually gives you the equation for the arch of st. Louis and It actually is not parabolic. It is it is a hyperbolic cosine is the main Function of that particular art so you may come across that in an engineering class As far as something structurally Probably pretty sound I would imagine is not this but the symmetric image of that So unfortunately, that's not addressed a whole lot in in our book All right now. Let's go into section 8.9 the first part of section 8.9 Talks about error estimates Let's just kind of briefly Mention what it is they address and I think we have addressed them somewhat sufficiently So they're talking about error estimates associated with Taylor polynomials which kind of includes Maclaurin as Well by the way if we say polynomial that means we're truncating the series We're not going to have all the terms all the way out to infinity that we only want the first five or six or seven of those How do we Estimate the error and by the way this also includes binomial series because binomial series are rooted in Taylor expansion, so they are also included in this category One that's not we haven't really addressed very much because we don't require in this class a graphing calculator My guess is that most of you have one and use it one of the ways that you can deal with error estimates is to actually graph the series itself and graph the truncated series the Taylor polynomial or the binomial series polynomial and For certain values you can kind of compare and contrast the actual Series with the truncated version so you could do that on graphing calculator Other ways that we have and we have done this if you have an alternating series That when I see that I see a series. It's a Taylor series, and it's an alternating series One of the first things I think about is that if I need to approximate I Know my upper bound for my error can be basically determined by the value of the next term, right? so we've dealt with errors on Alternating series you just wherever you stop it you look at the next term, and that's an upper bound for the error So we've addressed that I don't think we need to address it again Taylor's inequality is where we Looked at that capital M We had an M here. We had the N plus first power of X or X minus a depending on the function Over n plus 1 so it kind of is the next term With one slight adaptation and we are going to include this in the problem We're going to go through today If in fact we get that far M is the maximum value of the n plus first derivative on the function Somewhere in the interval Between X and a now. I don't know that a is less than X or greater than than X, but I'll just write it like that We can choose a value anywhere. We want between X and a or in fact X or a it's an upper bound for the error anyway So we want to know what the error is at its worst So we are going to use This in the problem we're going to look at today and the numbers are hideous I'll warn you because we're going to be dealing with the speed of light But That value actually ends up being in the denominator Which kind of helps the cause and makes it our error a whole lot less Then you might expect it to be All right, I don't remember if this is the first example in the book But it's probably one that you'll recognize so that's why I think it's good for us to look at this one I have another one from the list of problems ready That we probably won't get to today, but we can get to that tomorrow It's probably not as familiar, but we can still look at the equation that's handed to us Manipulated in the same way. We're going to manipulate this one that is kind of familiar to us and we don't have to be experts about what we're Manipulating to get it in the form of that we want it or to talk about the error associated with that So this has to do with physics and a branch of physics that deals with the theory of special relativity and Here's the first equation which May not look all that familiar to it kind of depends on who in here is a physics major or has used this in another math or math related class as an example Illustrating some physics so the mass of an object that's in motion. That's what M is is Related to the mass of the object at rest so M sub zero is the mass of the object at rest in this fashion V is the velocity of this object that's moving and C is the speed of light how many of you've used this equation in other courses before is that a common equation? okay we're going to Work with this and you maybe you can see how we can manipulate this a little bit Because we've got something that is in the form of kind of we have to adapt it a little bit one Plus I know that's not plus, but we can make it a plus x To a power so we're kind of Headed in this direction with parts of this problem Not only in the development of the equation But also in the error estimate we're going to use this guy right here So M is the mass of the object that's in motion M sub zero is the mass of the object at rest V is the velocity of the object C is the speed of light all right, we're going to use this equation to Simplify although it makes it look more complicated initially for a kinetic energy equation So somebody tell me what that equation says since we have some physicists in here that have had this stuff before Kinetic energy what it what is this? That's kind of the total energy and what would this be? The energy of the object at rest, okay So here's what's going to initially make it look more complicated But then we'll manipulate it a little bit and hopefully make it look a little simpler So here's what M is equal to We're going to put that in there for M and replace it and see what happens So that's M. We're going to multiply that by C squared So it looks worse temporarily Do you see any possible gains having made that substitution? What do they both have now? M sub zero C squared so that could be factored out front let's Bring that up from the denominator. That's all that's left right in the first term So we have a one minus V squared over C squared to what power? To the negative one half and Then when we factor out M sub zero C squared out of this one, what's left one? So let's take this thing and the binomial series expansion that we have done That is the same thing as one plus negative V squared over C squared To the negative a half What we have dealt with kind of generically is One plus X to the K Well our X value that we normally Occupy this position with is now going to be negative V squared over C squared and our K value is negative one half So let's take a little bit of time and energy to Expand this thing. We know we're not going to be able to find all terms But we want to keep a few of them around To show that this is in fact something that you have probably used before whether you're a physics student or not You've probably come across this final result that we're going to come up with All right, so this little piece of it. Let's expand it and then we'll throw it back in this position So one plus X to the K. What's the first term of that binomial series expansion? one That's going to kind of help the cause because it's going to get knocked out by this other one At the end. All right. What's the next term? Typically it would be what K times X For us that's going to be Everybody agree. That's K times X What's the next term? K times K minus one. What would that be Jacob? negative one-half times negative three-halves over Two factorial X. Well, we don't have X. We have something that's being replaced for X I don't like the way that's headed here. I'm running out of room So our X value is negative V squared over C squared and we want that squared. Is that right? So there's K K minus one over two factorial X well, we don't have X. We have negative V squared over C squared that quantity squared See if we can squeeze one more in here negative a half negative three-halves negative five-halves over three factorial Negative V squared over C squared cubed Does that work? So we're going to put those things in right here and let's see if we can simplify some of them as we go So we're going to replace one minus V squared over C squared to the negative one-half with this binomial series expansion. So we get a what? one-half V squared over C squared. Is that correct? How about the next term where we've got a negative times a negative And then we've got a negative V squared over C squared squared. So it's a positive again I think it's positive again And we've got a three over two four eight. Does that look right? Three-eighths So we've got negative V squared over C squared that quantity squared. So V to the fourth over C to the fourth and the last one let's see We've got three negatives, which is negative and a negative cubed which is also negative So it looks like the end result is again going to be positive In the numerator, we've got 15 Actually, let's knock the three out numerator and denominator So in the numerator, we've got five in the denominator two four eight sixteen So there are more terms. Those are the ones that we found and you'll see we're not going to need any more than that It's probably well more than what we need This one no should be positive Here's what I'm looking at Negative negative negative so their product is negative and then we've got this negative thing That's cubed which is also negative. So the end result should be positive This is Part of and with the three dots. It's actually technically all of one minus V squared over C squared to the negative one half What are we going to do with that? We just did that Now we're going to subtract one from that There's a one. There's a one. So that's gone Jacob Negative set up are are you supposed to subtract them because then why isn't the second term negative the second term? If I've written this down right this was K was negative a half and This is the X value and it's the X value to the first So it was negative. So there's a negative and there's a negative which should be positive Well, it looks that way Based on what we see thus far the next one's going to have four negatives And it's going to be the negative to the fourth. So it appears that all the terms are going to be positive Some of those terms way out to the right We're not going to care much about anyway for certain values of V Compared to what we know how large C is C is the speed of light. So we've got a whole bunch of terms We're not going to look at very many of them. And in fact We're going to stop right here Not class. I mean some of you're looking at your watch. Oh, yeah, that sounds good. I like that It's what I was waiting for not quite there yet if V is small Compared to C C is the speed of light, which is a large number. So for small values of V If even though this is V to the fourth, which takes this small value of V and raises it to the fourth, isn't it going to be completely outweighed by C? Which is way larger than V, the C value to the fourth And it's going to only get worse as we go out to the right So if V is small compared to C, C is not small Then these terms In a sense become somewhat meaningless Well, if we ignore these terms realizing that it's truncated now, we are going to have some error And that's part of what we'll do. I don't know if we'll get to that today I don't know if I should use equal. Let me put equal with the dot above it. So if we stop it Truncate the series There we end up with this and for small values of V This is going to be very very close to the full-blown Kinetic energy equation. By the way, what could we reduce here? C squared over itself. Does that look familiar? So that's not exactly kinetic energy, but isn't that the one that's probably used more often than not? I think this was actually kind of Discovered or published first by Newton But that's probably what you've used for kinetic energy more so than the first formula that we started with So it isn't exactly But it's very very close Especially for small values of V, and they don't have to be all that small. That's why I put small in quotation marks A lot of things are small compared to the speed of light So for small values of V This is what that formula for kinetic energy actually is So it's the result of a binomial series expansion Now let's take a look at the beginning of the error term. I don't know if we'll finish this but We haven't really used this a lot This Taylor's inequality which deals with the error. So we want the error associated with the first Taylor because we Chose the n equals zero term. We used it. That was the one We went on to the next one n equals one and we kept that one But then we got rid of all the rest of them. So this is the error associated with the first Taylor I know we didn't use Taylor. We used the binomial series, but that is the result of the Taylor Series So the upper bound for the error associated with the first Taylor polynomial would be This m value that we addressed earlier Because don't we go to the n plus first term so if this is one we want to go to the n plus one value the value of two for that So m is the maximum value of the second derivative Second derivative of what well, we've got this Function which is kind of an ugly function with v squared and c squared in it. Let's simplify it a little bit to this Probably would have been helpful if I would write that down, which it doesn't appears that I actually wrote it down So I am putting an x in there just to make the derivative process Go a little bit better Chandler make sure you tell them I said hello I want to go ahead and give my salutations there too. So this in the formula is what negative v squared Over c squared, right? Let's go ahead and find the derivative. This stuff is coefficients, right? What's the derivative of this thing? Wouldn't it be negative one half? one plus x to the negative three halves times the derivative of what's inside which is one and then the derivative of minus one which is Zero so there's the first derivative We need the second derivative So all this is lead coefficients. We bring that all along Now we need the derivative of one plus x to the negative three halves, which would be negative three halves One plus x to the negative five halves Times the derivative of what's inside which is one. So there's our second derivative So what do we have we have two negatives? Which is positive we have three over four M zero we have a c squared and we have a one plus x to the negative five halves which I'll just put that down in the denominator where it belongs So we want to maximize the value of the second derivative M sub zero we're not going to know that so that's going to remain in the problem everything else We should be able to sub in C is three times Ten to the eighth meters per second. Does that look right for C? Okay, I'm going to use that value in a while and we're also handed in this particular problem We want to know the error associated with Values of v they're small and they give us a relatively small value of v So this is given to us as a hundred So we want to know how far off this value is at its worst What's the upper bound for the error when we have a small velocity a hundred compared to the speed of light? Which is a whole lot larger than that and again, we don't care if the error is Positive or negative we just want to know how far off we might be at the worst Back in the what they call the Taylor's inequality. We stopped we put in n equals zero and We put in n equals one so we really use the first Taylor polynomial Here's the error associated with the first Taylor. So it's called r sub one because it's associated with t1 This was the value one and this was the value. What did we get there? one half V squared over c squared right and we kind of drew the little squiggly line We didn't take any of the rest of them. So that's why we use t1. So we want our one We want to go to the next value to the n plus first Power of x and the n plus first factorial and then this this is the maximum value of the n plus first derivative So that's why we're dealing with the second derivative because we stopped with t1 So there's the c value. We can plug that in We do know the x value and I called it x to make the derivative process simpler What was x temporarily representing? negative v squared over c squared actually Or no, oh wait. No second. Okay Never mind. We're okay. Yeah. No, it should be squared. Yeah Yes, c squared. Thank you Because we were just basically filling in what we know. We don't know m sub zero. So we're just carrying that through Three m sub zero. This is the c value squared. Correct that. Thank you and Down here. We want a four and our one plus x becomes one minus v squared over c squared that to the Five halves now. That's just the m part of this That's kind of the maximum value of the second derivative on this Interval let's go all the way back to the error and I this is probably we're probably going to be able to set this up And then we'll have to kind of finish the problem So here's our error. We need all of this So we need the m value We have that so that's m. We need x squared so what was x x was negative v squared over c squared With that squared kind of numerically kind of intensive here We push the buttons later It may take us from the end of class today to the beginning of class tomorrow to actually push all these buttons So we can get an error and all that is over what two factorial So we should have an answer that is not completely numerical because it's got a Mass of the object at rest in it, which we don't know but in terms of that number whatever it is If you have a few minutes to push these buttons in this order you'll you'll realize that those terms that we truncated are Virtually meaningless to the to the solution to this problem with a relatively small v value Compared to a pretty large c value This should be when all is said and done somebody can check the arithmetic But if we've got that level of accuracy Regardless of what that value is that's pretty good We've done really really well with a velocity of a hundred meters per second compared to the c value for the speed of light So there's the arithmetic that's supposed to get us that Everybody agree that's pretty good level of accuracy. I would take that All right, we're out of time today We will do probably one brief example you might want to bring your old tests Tomorrow especially if there's something you want to ask Specifically about a test question on one of the early tests that probably we could handle that tomorrow