In this video, I approximate the accuracy of the partial sums of a series that converges via the Integral Test by estimating the size of the remainder. This is done by comparing the areas of the re...
Transcript
estimating the sum of a series so suppose we have been able to use the integral test to show that a series a n is convergent and now we want to find an approximation to the sum of the series so yes we want to know more than just if it's convergent or not and then see if we can approximate the actual value so of course any partial partial sum as n is an approximation to s because the limit as n approaches infinity of the partial sum s of n equals s. But how good is such an approximation? To find out, we need to estimate the size of the remainder. Yes, and we can now write this down in formula form as rn, the remainder after n, after the nth term is equal to s, minus s of n. So that's the actual sum, minus the... a partial sum s n all the way there and this just equals to well it's going to be the terms a n plus one so after the nth term a n plus 1 a and plus 2 plus and then a plus 3 and so on like that so we're all adding all the rectangles as a show past or all the terms past the a and past the end term Yes, so the remainder RN is the error made when SN, or the partial sum S, n, the sum of the first N terms is used as an approximation to the total sum. And we use the same notation and ideas as in the integral test, assuming that F is decreasing on the interval from n to infinity, and comparing the areas of the rectangles with the area under Y equals F of X, 4x, is greater than N in the figure below, we see that. this one out. Yeah, we see like this, so if we have the x and y, so there is the y axis, and there's the x like that, this is the y. So let's say we have a function like this, drive like this, this is the y equals f of x. And let's say this is all the way, this is zero there, but let's say we have all the way to our nth term and it's all the way across here. This is our nth term. So we've summed up all before to the nth term, and now what we want to do is get the rectangles or the terms after the endth term. In other words, get the remainder. So what we'll do is I'll draw a point like here. This is going to be n plus 1, and this can be n plus 2, and so on. And so on, and so on. So let's draw like this. This part's going to go all the way across like this. I'm going to draw a rectangle from it, so now we have A, this is the term A, n plus 1, and now likewise over here, make it a bit closer like that, so we have this, and this is going to be our A, n plus 2, and so on, we're going to keep getting more and more like that, and these are all the terms after, yeah, all the terms after the nth term, we can put a dot, dot, dot, like that. Yes, so that's what we have over here is RN, let's write this down the remainder, RN is equal to A.N plus 1 plus A.N plus 2, plus dot, dot, dot, and this is going to be less than, or equal to, yeah, so there's the, it's going to be less than or equal to the area underneath, or the integral from n to infinity of fx, dx. like that, because in this case here, so n plus 1, we have the rectangle on the left soles can be less than or equal to it. And here, just move the things around, just boxed it up like that. So yes, that's for this case. But now let's look at the rectangles up on the other side. So similarly, we see from the following figure that, now we're going to use the rectangles on the other side of it. So let's just draw the X and Y coordinates again. So it's going to be like this. This is the x, there's the y, and we're going to have a function like this. Let's try this like that. I guess we have a function. This is our y of x, y equals f of x like that. And let's say we have our n is over here. There's our zero, so this is all the way up to the nth term. I'm just going to draw this one in black. And then we're going to go to the side like that. like that, let's go all the way down here, and I'll write this as a n, so that's the nth term of the series, or of the sequence, and now in red, this is, now we'll look at the remainder, that's going to be over here, n plus 1, so we're going to go all the way across like that, and this one's going to be our a n plus 1, and then likewise we're going to have an n plus 2 there, and it goes like that, and this is our a n plus 2, And this is always going to be, yes, it keeps going on and on, like that, and then just do one more. Dot, dot, dot, keeps going on. Notice this one is greater than the F of X curve there. Thus, what we have is the RN, the remainder, past the nth term, is going to be equal to A, n plus 1, and then going further, A, n plus 2. and then plus dot, dot, dot, but now instead of less than, it's going to be greater than or equal to the integral or the area underneath this curve from, this is going to be from, yeah, this is going to be from n plus once, we're starting from this side there, so n plus 1 up to infinity of f of x. DX, like that. Yes, in here, I've boxed that out. So now that we have a remainder like that. Yeah, remainder is greater than this N plus 1, from integral from N plus 1 to infinity of FX, DX. But likewise, it's less than, from this part, from n to infinity of FX, DX. So, so we approve the following error estimate. Theorem 2, remainder estimate for the integral test. So suppose F of K equals AK. Yeah, just defining that the function and AKR, they share the same function there, the series and this function, F of X or F of K, where F is a continuous positive decreasing function for X is greater than or equal to N, and the series, summation of A.N is convergent, so if it's convergent, and it's continued, the function F is continuous positive decreasing for the function. And if we have if, if we have the remainder RN equals to S, the actual sum minus partial sum S of N, then, so if that's what we have, then, what we end up having is the integral from n plus 1 up to infinity of F of X, like this, DX. This right here is going to be well to less than our RN. So RN is going to be greater than this, yeah, so the remainder is greater than on this side, greater than the area under the curve when we're dealing with like this, to the right of it, or above that function. So this is going to be less than or equal to RN, like that. But then this is going to be less than or equal to over here. This one is going to be the integral from n to infinity of fx, DX, like that. So those rectangles are smaller than. than that as well. And these recognize are exactly the same. They're just shifted based on, well, they're just, yeah, they're just shifted based on the curve there. So instead of this one, this can be shifted to the right, and so on. So this one's going to be like that. So it's going to be, so the error estimate, I mean, yes, so the remainder is going to be less than this, I mean, greater than this, but less than this right here. Yeah, less than this area of that curve. So, it's going to be the integral from n to infinity of f of x. DX like that and I'll box this out like this actually pretty fascinating if you take time to think about it. Alright, so now let's look at another example, and this is example 5, and it states, approximate the sum of the series. of this right here, summation 1 over n cubed by using the sum of the first 10 terms. So we just add those up. And then states estimate the error involved in this approximation. And then part B says how many terms are required to ensure that the sum is accurate to within 0.005, so 3-5s. Solution, before we get to A and B, in both parts A and B, we need to know the integral from n to infinity of FX DX. and with fx equals 1 over x cubed, which satisfies the conditions of the integral test we have. So yes, this is satisfied. This is just a P series. This is always decreasing, and it's positive, and it's continuous, because we have 1 over x cubed right there. It's always going to be positive for x is positive. and it's decreasing 1 divided by x cubed is also going to decrease in here, also cubing. The number's going to get larger and larger at the bottom, dividing by a large number is a small number. So what we have is, we need to take the integral of it, so we need to take n to infinity to get the upper limit here. So we're going to go from n to infinity to get the upper limit to see how accurate it is, and let's just keep going further here. So of fx. dX, this equals to integral from n to infinity of 1 over x cubed dx. And this is an improper integral, so we need to do the same thing, as always limit as t approaches infinity, and then I'm going to write this as n to t, and instead of writing 1 over x cube, I'm just going to write this as x to power of negative 3, just it's easier to solve for the integral. like this, the same thing, power function just put a negative, if you could flip it to the bottom or the top. So we have this here, so the integral, recall the integral is an anti-derivative, so in other words, to get this right here, before it used to be an X to the power of negative 2, or a higher power by 1, so this is going to be equal to, let's put the limit, as T approaches infinity, put this in bracket, we're going to have x minus 2, and to get rid of that, you know, the chain rule, or when you take the, or not necessarily the chain, just the power rule, when you take the derivative, you have to put the negative 2 down, so when I do, I'm going to divide this by negative 2, because we're taking the anti-derivative, like that. So that if you take a derivative of this one, you can put the negative 2 down, then minus a power, so then these negative 2 is cancel, around negative 3, like that. This is from n to t. Yeah, so what we end up having is, and then this is going to be limit. as T approaches infinity, and now what I'm going to do is I'm just going to move this at the bottom again, so we'll move this as, this is going to be as 1 over negative 2, put the T inside, T squared, just to get rid of that negative power, and then we're going to minus, and then the N inside, so that's going to be 1 over, and then it'll be negative, make this a positive like that, because the negatives cancel, so 2, n squared like that. So that's what we end up having. So what we end up having here, as T goes to infinity, this one right here, well, we're going to have a scenario that's 1 over infinity, is T squared, is T squared, is just infinity. Yeah, it's also going infinity, so then this can be 1 divided by infinity, it's going to 0. So what we end up having is, we have a term here, 0 plus this one here, there's no T terms, so that's all we have. So 1 over 2n squared. So that is the... Yeah, that is the integral from n to infinity. And this is the upper limit of that remainder. So, remainder is going to be less than, from integral from n to infinity of fx d x. So it's going to be less than this number here. So the, whatever, whatever. your partial sum that you have the it's going to be less than this error or this remainder part so 1 over 2 and squared I mean the remainder is going to be less than this it's going to be the larger and larger or more and more terms you get that and gets larger so then you're going to have smaller and smaller remainder there it's very interesting stuff So yes, that's a good way to check the calculation as well, because this should get smaller and smaller as you increase the number of terms, because the increased number of terms, you get more accurate sum, yeah, a more accurate sum, so then you're going to have that smaller and smaller remainder. So now let's look at part A, so now that we have that, so now that we have that, so a approximate the summation by using the first 10 terms, so summation 1 over n cubed, so we'll write right here is a, yeah, what I'm going to write is the summation from n equals to 1 to infinity, of 1 over m cubed. This is approximately equal to this first 10 terms. I assume it's approximates from my calculus book has this like that. It's approximately equal to the s tenth or the 10th partial sum, and this equals to 1 over 1 cubed plus 2 over cubed plus 1 over 3 cubed plus dot dot dot all the way until we get 1 plus 10 cubed. and this if you plug this new calculator you get it's approximately equal to and this is going to be 1.1975 and here's just a calculation check my one node has a built-in calculator which is pretty amazing so you just put this ear s 10 1 over 1 qute plus 1 over 2 cube all the way to all the way to over here 1 plus uh yeah plus 1 over 10 cubed so here's the calculation i'll delete this and write equal and then space and yes automatically calculates absolutely fascinating stuff here on bold that. All right, so yes, that is the correct one there. So according to the remainder estimate in 2, I mean theorem 2, so the theorem 2 is over here's theorem 2 remainder estimate. So according to this one here, this remainder is going to be less than or equal to, well, this figure there, an integral from n to infinity of fx-d-x, which is this number right here, so you've got to put a 10 inside there. Yeah, so what you end up having is, instead of n, we have r10, it's going to be less than or equal to, and it's going to be integral from 10 to infinity of 1 over x cubed d x. And this right here, that's the 10. So we already know that. That's just going to be this one here. This is from n to infinity of fx d x. So it's going to be 1 over 2 n squared. So 1 over 2 times 10. squared or cubed like that. Let's know, yeah, it is squared there, so that's this right here, 1 over 2 n squared like that. And this equals to, this is equals to 10 squared is 100, so we didn't have, end up having is 100 times 2, that's 1 over 200, like that. So the error is going to be less than, or the remainder is going to be less than 1 over 200. Yeah, so the size of the error is at most 1 over 200, which equals to 0.005, you can just do that as well. Automatic calculations. Amazing stuff here. So, yes, that's the error is at most that, which is pretty good for the first 10 terms. So 1.1975, so this is going to be a 5 in there somewhere. at most error. So that was the estimate they are involved in the approximation. And now part B says how many terms are required to ensure that the sum is accurate to within 0.005. So like a third zero there. So we want to see how many terms. So 10 terms, you can be within 0 .005, and now accuracy within 0.005 means that we have to find a value of n such that RN is less than 0 .005. So what'll do is, right, well, since, since we know that, yeah, that RN is less than or equal to, so yeah, let's scroll up again, so RN is less than or equal to the integral from S, from n to infinity of fx, dx, which that equals to 1 over n squared. So it's less than equal to n to infinity of 1 over n, yeah, 1 over fx, like that, and then this equals 2, we already solved it for our case, 1 over 2n squared. Yes, in other words, what we want, we want to be accuracy within that, so what we want is this, it has to be less than 0.005, so we want, yeah, the max error there, 1 over 2n squared, less than 0.0005, like that. So solving this inequality we get is, well, we can move the two over to the other side, first of all. So what you get is a 1 over n squared is less than, well, 2 times 0.005. And when we multiply that out, we just get a, well, this is 5 times 2 is 10, so we're just going to move it over here. We're going to have two zeros there. So this is equals to 0.001. and now this equals to just get rid of the decimal place to make it a fraction because we have a fraction here. So what we'll do is move this over one, two, three decimal places. In other words, we're going to have a, this is the same thing as writing 1 over 3 zeros here, the 1 over 1 ,000. So 1 with 3 zeros like that. So now what we have here is we can move this over to this side, and we can move over this to the other side. to get rid of the fractions. So what we end up having is a 1,000 is less than n squared. Or in other words, take the square root. So we're going to have a n is greater than, because we move the inequality across like this. So we have n is greater than, and then square root, number of squared could be plus or minus, but for practical use, in this case, it's a positive. So 1,000, like that. And then this one right here, if you were to put this in a calculator, you're going to get a 31.6, roughly, that figure there. But n is greater than it, yes, we need to be n is greater than it, and we're dealing with sequences in series, so it needs to be a whole number, so the nth term needs to be 32. So yes, calculation check as well here, you can just build in here. delete that equals space and there's the calculation. So, 1,000, the power of 1, 1 half, or square root of 1,000, is 31.6-28, and here's around it of 31.6, and we're, so we need, N is greater than that, so N is 32 terms to ensure accuracy to within 0.0 ,005.