hey what's up guys it's misra myiasis here and today i'd like to talk to you about the alternating series remainder so we're looking at air approximations and really what we're looking at is the error is what's left over when we approximate a suit when we approximate the sum of a series using only the first n terms so if you kind of remember when we're talking about Taylor polynomials and we estimated the value of a function using a Taylor polynomial and then we can create a Taylor series and that would give us the exact value well if we're only gonna take part of it so the only taking the Taylor polynomial part then we're gonna have some leftover right we're gonna have extra that we didn't account for it's not gonna be perfect gonna be an estimate how much how close are we what's the left over part what's the remainder part that's our error that's how much we're not hitting the actual value okay so for an alternating series it turns out that the error is gonna be less than or equal to whatever the value of the next term after the first n terms that we've used to estimate that series so let's take a look at a few examples and hopefully that'll make a little bit more sense looking at this so Clos what I'm gonna do for just hit the desk shake it it's not an earthquake guys are just this guy bang the desk here so let's do the first example let's say we're gonna approximate the sum of this series that I have here and we're gonna use the first six terms to do that so what we're gonna do is we're just gonna write out the first six terms so we're gonna say that's approximate equal to one plus we're going to go and plug in two in there we're gonna have negative one half right because it's 1 over 2 factorial plus 1 over 3 factorial minus 1 over 4 factorial plus 1 over 5 factorial minus 1 over 6 factorial so then what I'm gonna do is I'm gonna plug this here in a calculator to calculate that out and that's gonna give me point six three one nine I'm gonna put an extra I'm gonna put an extra value here just to show you what we're doing and remind you that I will remind you let me just get rid of this here buddy am i from my AP read remind you that we are gonna go four decimal places in in a in calculus three decimal three decimal places in calculus right so 632 all right find the upper bound for the remainder for the approximation from example one so this is this is my estimation now I have an amount of error there are a lot of values in there that I didn't take in consideration because remember this series is infinite so all all of that stuff the difference between all of that and what I have is going to be less than or equal to the value of my next term so all I'm gonna do is take my next term so I know by using this alternating series remainder that my remainder is going to be less than or equal to whatever the next term was which was 1 over 7 factorial and 1 over 7 factorial calculates out to 0.0001 9 8 so this is our upper bound for the remainder for our approximations this is our error so we're gonna go and find the upper and lower bounds for the actual sum so what does that mean the actual sum well I don't know what the actual sum is but I do know that the actual sum is going to be somewhere between zero point six three one nine four because that's what I calculated it to be i approximated it to be for the first six terms and that actual value let me go and write this series in here the actual value of that's gonna be between my lower bound of 0.6 3 1 9 4 bits which is what I got for the first six terms plus anything I have left over well what do I have left over at least as much as this value plus this value so I'm going to add those two together for those of you that have taken statistics ap statistics you can kind of relate this to something like a margin of error all right so we've got a lower bound and upper bound it's gonna be as low as this because that's what I estimated it to be for the first six terms and as high is this because that's the first six terms plus the error that I have using that sixth term so let's go ahead and do another example here let's suppose I want to approximate this the value of this series with an error that's less than point zero zero one well if I know that this is an alternating series and the remainder is going to the remainder is going to be less than or equal to the nth plus one term really what I got to do is I got to find which term is gonna give me a value that's less than point zero zero one once I find that term then I'm gonna take everything before that right because that's that's the term that I want to be to error less then so I'm gonna go ahead and write this guy out oops write a few a few of these guys out my first term is going to be one N equals zero I'm using zero plug zero in right plug one in and I'm going to get 1/2 then I'm gonna get one over four factorial and when over for Fecteau is still not small enough minus one over six factorial which I calculated that out to be point zero zero one three eight so that's still not less than point zero zero one so I think I'm going to need to go one more and one over eight factorial 1 over 8 factorial is actually zero point zero zero zero zero to four so that's plenty small the problem was that one over six factorial was point zero zero one three eight and that's still bigger than point zero zero one so this part is gonna be my air so I want my error to be less than this so I'm gonna go to this one I'm gonna not use that term and I'm gonna add up all of these alright so I'm gonna add all those up and I'm gonna get my approximation that is that has an error that's less than point zero zero one alright so let's go ahead and use an elementary series to actually see what that value is well negative one to the N over two to the N or 2n factorial this is the Sigma notation for we've covered this in my in the last unit this is an elementary series this is Sigma notation for cosine of one because there's no this is like X to the N right X to the N right here but X in this case is one so cosine of one is equal to you know approximately equal to here zero point five four zero three zero when I thrown in the calculator and notice here that we in fact are within point zero zero one of our calculation in example four so in general terms folks when we want to find the remainder the error of an alternating series that remainder is going to be less than or equal to the first one use term in that Taylor pollen in that Taylor series so the first unused term is going to be our is gonna be it's going to whatever that value is the air is gonna be less than or equal to that so you can write it like this that I have here highlighted it's gonna be less than or equal to F of the n plus one derivative X minus C if it's centered around C to the n plus one over n plus one factorial now that works for an alternating series what happens if the series doesn't alternate well we got to use a little bit of a different kind of a harder kind of a harder method it's not not as easy as this one and that's called the Lagrangian der or a Taylor's theorem remainder and I will cover lagravis remainder in another video catch you guys later good luck - ow