 In this video, let f of x equal the square root of x plus one and we're gonna find the Taylor polynomial, the fourth degree Taylor polynomial of this function centered at zero. So we're gonna grab the first couple of terms of its Maclaurin series. We wanna use this to approximate the value the square root of 0.9. So notice making a connection to the function right here that 0.9 is really just one minus 0.1 which is our function evaluated at negative 0.1. So that's gonna be the connection we want to make there. And so using the Taylor polynomial, we're gonna see that f of negative 0.1 is approximately the same thing as t4 of negative 0.1. We'll make that estimate and then we'll see how good that estimate's gonna be. Now, before we go into the details of the calculation, let's look at the graphs of these things for a little bit. So in all three of these graphs, the yellow curve right here will be our function y equals the square root of one plus x. This first reddish function, it's magenta, is our tangent line. It'll be the line at the point one comma, sorry, zero comma one. Like so, this is the linear approximation. You now see here in cyan, this bluish color. This is our tangent parabola which is doing a little bit better than the tangent line. Then in the next graph over here, you see this, the degree three Taylor polynomial. It's a cubic graph. It fits the curve even better. And then finally, this violet graph you see in the third one, it is going to be the degree four Taylor polynomial which we're gonna look for in just a second. And so graphically, you can see this thing does pretty good. Now, of course, you're gonna be really good near the point of tangency. Now, the farther you get away, the worse it does, but it does pretty good inside of these green lines right here. So where did these green lines come from? Now, one thing I wanna mention is that this function f of x right here, it could be written as one plus x to the one half power. And this helps us recognize that this is in fact a binomial series, a binomial series for which k equals one half. Now, for this binomial series, we can then very quickly see that the radius of convergence is one, that is our x value needs to sit between negative one and one, our interval of convergence. The function will equal its Maclaurin series when x is between negative one and one. Outside of that, the Maclaurin series diverges and therefore equality is not possible. So notice that between the values x equals negative one and x equals one, the fit between the function and the Taylor polynomial is really, really tight. But outside of that, you don't get any, you don't expect anything to happen here because only between negative one and one is the function equal to its Maclaurin series. And therefore only between negative one and one would we expect the Taylor polynomials to approximate this function. So we see that in this picture here. So like I said, our function is a binomial series. Therefore f of x, which again equals one plus x to the one half power, we can very quickly get its Maclaurin series where it equals zero to infinity. You're gonna get one half choose n times x to the n. You get this thing very, very, very quickly. And so remember this binomial coefficient has the following properties. That if you take k choose zero, you're always gonna equal one. If you take k choose one, it's always equal to k. And then more generally, if you're taking k choose n, this thing will look like k times k minus one times k minus two. And so you'll then continue on until you end up with k minus n minus one like this. Although this last few bits are typically written as k minus n plus one. And this all sits on top of n factorial, all right? And so we're gonna use this formula in the consideration for the above. Now we don't have to do all of the series. We just want the Taylor approximation degree four. So we are gonna say that this is approximately the same thing as t four of x. And so we just have to grab the first couple of coefficients. So like I said before, k zero is always one. So you're gonna get one plus k choose one is always k. So you're gonna get one half x. For the next ones, we need a little bit more to it perhaps. So you're gonna get k, which is one half times k minus one, which is actually negative one half over two factorial x squared. The next term will be one half, negative one half. And then you're gonna take the negative one half and subtract another one from a negative three halves. This will sit above three factorial times by x cubed. And then finally, you're gonna get a one half times a negative one half times a negative three halves times a negative five halves right there. And then this sits above four factorial. Don't forget your x to the fourth right there. Now let's try to simplify these things. The first couple of coefficients we already did, of course, so you get one plus one half x. For the next one, you do see there's a negative sign and so you'll get a negative right here. For the next one, so if you check all the twos, you have a two, you have a two. Two factorial is itself just a two. So this is gonna be one eighth, negative one eighth times x squared as the next term there. For three factorial, notice three factorial is three times two times one. The three on the bottom does cancel with the three on top. You have two negatives, which double negative actually makes it a positive. And then if you grab all the twos, we have a two, two, two on top, that's an eighth. There's a two on the bottom. That's gonna give us a sixteenth, so we get one positive, one sixteenth x cubed. And then finally, if you look at the last one, again, the four factorial becomes one times two times three times four. The three will cancel right here. You have a negative, negative, negative, so it's actually a net negative like so. And if we keep track of all the twos, you have a two, two, two, two. So there's four twos on the top. I should say four, one half's on the top. That's gonna give us a one sixteenth. And then you have a two and a four, that's an eight. So 16 combined with the eight that's in the denominator already. You're gonna end up with one twenty-eighth, or I should say one twenty-eighth on the bottom. And then the only other coefficient on the top that we haven't interacted with is the five. So you're gonna get five over one twenty-eight x to the fourth. And so this is our Taylor polynomial approximation of the function. So like we said earlier, the square root of point nine is a, this is equal to f of negative point one, which will be approximately the same thing as t four of negative point one. And so we have to shove this into the function above in all these different places for x. So we end up with one plus negative point one over two. Sorry, we're gonna get a minus negative point one squared over eight. And then we're going to get a positive negative one point one cubed over 16. And then finally minus, minus five times negative point one to the fourth over one twenty-eight, like so. All right. And so again, there's some tedious arithmetic that needs to go on here, but in the end, this thing would approximately become zero point nine four eight six, eight two six nine four, if I did all those correctly. So you get your estimate right here, that's good, that's good. What are we gonna do next? So the next thing is we have to estimate how good this thing is gonna be, which when you look at this thing, notice, well, how are we gonna estimate this thing? Well, since we estimated this thing using the Taylor polynomial, we're gonna use Taylor's inequality to help us compute the possible error, sort of worst case error here. And so the thing to remember with Taylor's inequality is the error is gonna be bounded above by m times the absolute value of x minus a to the n plus first all over n plus one factorial. And this is gonna happen for all x's, which are sufficiently close to a, that is the distance between x and a is d. And also remember that m is an upper bound for the n plus first derivative. Of the function on this interval as well. So what is the n plus first derivative here? Now remember our series we were using, we got about from the binomial series. So we kind of skipped all the derivative steps, step by step by step. And if you wanna see those, you can actually look in the lecture notes that are linked in the description of this video for a full detail here. But as it's a binomial series, the derivatives follow a basic pattern. We don't have to compute them step by step by step. The nth derivative of this binomial series, that is the nth derivative here, is gonna look like k choose n times one plus x raised to the k minus n power. k minus n, that's the pattern we got. And so when we plug in x equals zero in the center of this Maclaurin series, we end up with, I'm sorry, I got, let me back that up. Well, what we have here is fine, but this should be over-infectorial, excuse me. So the nth derivative over-infectorial is gonna equal this expression right here. This is the pattern we get when you do those derivatives after derivatives after derivatives after derivatives. So therefore if we choose n to equal four, we then have to look at the next piece going on here. And so the next piece to help us find this m value right here, m has to be bigger than the nth derivative. But then if we slap this n plus, if we slap this thing on the bottom right here, right, we see that m over n plus one factorial, it needs to be greater than or equal to this k choose n times one plus x to the k minus n power there. And so let's put in the specifics for our function. So we're gonna get one half choose n, sorry, n plus one. That's what I meant to put in there, n plus one. And then this one plus x, we're gonna raise that to, I guess I don't have to be so general here. The n is a four, right? So all of these should just be five. So let me kind of back up there and fix this. So m over five factorial, this needs to be bigger than the, this is gonna be bigger than one half choose five times one plus x raise to the one half minus five right there. So with that in mind, m is gonna be this value. Well, I guess we can just leave m over, m over 120 here is, we'll just move to that side. m is gonna be very equal to 120. And then when we do this one half choose five, like we saw before, we're gonna get one half times negative one half times negative three halves times negative five halves. And then one more, you should have five factors on the top, you're gonna get negative seven halves right there. And this will all sit above the five factorial, right? So you see that those ones are gonna cancel out. And so then what do we have? Now we're gonna have one plus x raise to the negative nine halves power, like so. And so how bad can this thing get in this interval in question? Let's try to clean this thing up a little bit because we're taking absolute values of all these things, all the negative signs don't really matter. If we keep track of the twos, you get one, two, three, four, five twos. So you're gonna get a 32 on the bottom, in which case then you get a three times a five times a seven, multiplying all those together should get 105 times one plus x to the negative nine halves. Now this function is a decreasing function on the interval in question here. And therefore the biggest that this can get in terms of absolute value will happen at the left end point. And so that's gonna happen at the point x equals negative point one, right? So our d value from before that we had this d value, we're taking the absolute value of negative point one. So how far can x, how big can this derivative get as x ranges between negative point one and point one? Well, because the function that this derivative here is decreasing, because it's decreasing, the biggest is gonna happen on the left at this negative point one. So we wanna insert that into this expression right here. And so when we do that, m needs to be greater than or equal to 105 over 32. And then when you plug in the negative point one, you're gonna get this point nine to the negative nine halves power like so. And so again, this thing would look like 105 over 32 times the square root of point nine to the negative nine. So I'm actually gonna put it in the denominator. That's gonna be raised to the ninth power. So we have somewhat of an issue right here, right? Because if we can compute the square root of point nine, why are we doing this whole exercise, right? So we're gonna have to think of a sort of a clever way of getting around this situation. And so what I'm gonna offer is the following. This right here is gonna be less than 105, 105 over 32 times the square root of point 81 to the ninth. Why is that a good choice? Well, because point, if you make the number point nine gets smaller, that actually makes the fraction get bigger. And now point 81 is actually a perfect square. And so this thing becomes 105 over 32 times, we're gonna get point nine again to the ninth. And so taking something to ninth power, although it can be kind of difficult, just use a calculator, this is just basic multiplication there. We're gonna end up with 105 divided by 32, 32 times, point nine to the ninth, you end up with point three, eight, seven, four, two, zero, four, eight, nine. That's gonna be even more difficult arithmetic here. I'm just gonna say that this is less than 105 over 32 times point three. That is this number right here, I'm just gonna replace it with something even smaller, point three, voila. And then that cleans up very dramatically. You're gonna end up with 175 over 16. So again, these are error bounds, these are inequalities, right? You can sometimes fudge the numbers a little bit to make life easier for you. So 175 over 16 is a much better value to use. And that way it doesn't actually require any sort of the reason, we don't actually have to compute a number that we haven't computed yet. So going back to our error, the error which is equal to the remainder of some four of x right here, this is gonna be bounded above by 175 over 16. Divided by five factor, which is 120 like we saw before. And then you're gonna take a point one here because that's as far away from the centers we're getting and we're gonna raise this to the fifth power. All right, a little bit tedious at this moment, the arithmetic might be, but as it just requires addition, multiplication, really just division of multiplication right here, these are four functions of arithmetic, well, two of those functions. This will be approximately 9.115 times 10 to the negative seventh. And so we see that this answer would be accurate to about six decimal places. Remember our calculation from before, where is it, aha, here it is. And so we can see what that will be accurate to about six decimal places, at least this much right here. And it turns out that the error we did with this article actually, it's a little bit better than what we estimated. The error bound tells us it's gonna be about, about nine times 10 to the negative seventh. But if we did a slightly more accurate calculation, we can see that this is actually a 6.041 times 10 to the negative seventh. That is our error bound gave us this, but the actual error was this illustrated in green right here. And so oftentimes our estimates can do much better than the error bound gives us. So when you're approximating a value using a Taylor polynomial, Taylor's inequality gives you the error bounds you wanna use to try to calculate worst case scenario, how bad your error is gonna be. And so this brings us to the end of lecture 48. And this actually brings us to the end of our lecture series calculus two here. For those who stuck around for this series, congratulations, give yourself a pat on the back. Calculus two is a very, very difficult course. I'll give you that. And so way to go making it to here. Now I do wanna mention that even though the series is gonna be officially over in the future, it's very possible that I actually might include some new videos that are related to calculus two that are sort of optional that students can look into if they are interested. So take a look for those things. You know, feel free to subscribe so you do get updates about those additional videos that might be added, not just for calculus two, but for other college level mathematics classes or questions you might have in the future. Feel free to hit the like button and hopefully I'll see you for next time, right? With the end of calculus two, I think the very natural thing to go to next was, well, we could go into differential equations. We did a little bit of those in our series. Calculus three is another way to continue on. We saw a little bit of that when we talked about polar coordinates and parametric curves. Calculus three is typically about finding, finding, working with multiple variables in your calculus problems. We did a little bit of probability in this course. So calculus based statistics class kind of makes sense. Or my very favorite, it would be linear algebra. In fact, we have a, this channel has a really good, not just video lectures, but also a free digital textbook based upon linear algebra. It's known as linear algebra done openly. And so if you are interested in learning kind of what's the next step in these, in your mathematical journeys, take a look at some of the links that hopefully you see on the screen right now. And I will hopefully see you in another, another lecture series sometime soon. Keep on calculating everyone and I'll hopefully see you next time. Bye.