Estimating Sum of Series as the Midpoint is a much better Approximation than the Partial Sum by @mes on 3Speak
In this video, I show that modifying our earlier Theorem 2 remainder estimate by adding the partial sum to both sides of the inequality provides a much better approximation than simply using the pa...
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theorem 3, so if we add the partial sum s n to each side of the inequalities in theorem 2, so recall theorem 2 over here, so if we add s n to both of these, and notice here, so number rn is equal to s minus s n, if you add s n to it, what we end up having is this will be s, and plus s and there, plus s and there. So what we end up having is, yeah, so we'll end up having as s n. plus the integral from the s n plus 1 of infinity f x d x then this can be less than or equal to s as opposed to rn this can be less than or equal to now we have an s and plus plus like this integral from n to infinity of f x d x like that and i'll just box this And, and then now just for a clarity, I'm just going to write even further. So this is because, because SN plus RN, that's just the partial sum plus the remainder of the partial sum after, they're relative to the total sum, this just equals to S. So yes, that's just SN plus RN, equals to S like that. Just fascinating how they stuff. So yes, the inequalities in theorem 3 give a lower bound and an upper bound for S, the total or the exact sum. They provided more accurate approximation to the sum of the series than the partial sum S and us. Yes, so now we have a lower and upper bound like that instead of just saying S-N. So we know we add the remainders to them, or the possible values of the remainder. or the limits of them. So now let's look at example 6, and this continues further from example 5. So it states, use this new theorem, so theorem 3 with n equals 10 to estimate the sum of the following series. There's a summation from n equals 1 to infinity of 1 over n cubed. It's the exact same as the one we were dealing with in example 5, so 1 over n-cube, so exact same one. So we used the values from before, and the formulas we derived from it as well. Yes, so the inequalities in theorem 3 is they become, so inequalities, we're just going to plug the sense, we're using S-10, I know, we're using the nth term, I mean the 10th term for n, so S-10 equals 2, so we have S-10 plus, now the integral from, this is going to be from 11, so this is n plus 1, infinity of fx, dx, and now our fx would be 1 over x cubed, so it's going to be 11, and it's going to be infinity, it's going to be 1 over x cubed, 1 over x cubed, d x, and this is going to be less than or equal to s, and this can be less than or equal to, and now we have s-10 as well, plus, plus now we have s-10 infinity, 1 over x cubed d x, like that. And now, yeah, from example 5, we already know that, well, the formula for this one here, let's go back to example 5, where we derive here, so the integral from n to infinity of fx, dx, where fx equals 1 over x cube, it's going to be 1 over 2 and squared. So we could replace that whole thing with 1 over 2 and 2, 1.2 and squared, so we get S10 plus Now we're going to have a 1 over 2, now the N is going to be 11, 11 squared like that, this is going to be less than equal to S, and which is going to be less than equal to S 10, plus, and now what we have is a 1 over 2, now that n is 10, squared like that, so yes. And now using S-10, or the 10th partial sum, as being 1.1975, and then 3-2, note the extra 3-2 is just a more fine-tuned calculation in my calculus book, or my calculus book uses, which my one-note automatic calculator cuts off. And again, remember, this is from example 5, where the calculus book had it here, but then there's added now, for example, 6, an extra 3-2. Just scroll up to it. This is example 5. recall that this was for part a 1 .1.1975, which is the 10th partial sum is approximation. Yeah, this is just the sum of all these terms, and this is S10 right here, and my calculus book cuts off here, and, oh, I mean, my calculations cut off here, my one note, but also calculus book doesn't add 3 to there. But anyways, they added for the... Example 6. So going further, so using S-10 as that, to put it in here, plus this term, and then it's going to be in between over here, then plug it in here as well as 10, plus that, this term right here, 1 over 2 times 10 squared, which is 1 over 200, as before, we get this right here. So we get the sum is between 1 .201-664 and 1.202. 532, and here's calculation check. My one note, so we could just have, erases just for completeness. So 1.197532 plus here 1, plus 1 over 1 divided by 2 times 11 squared, equals automatic calculates, exactly this number, and this way here, 1.97532 plus 1 over 2 times 10 to power of 2 or squared. Erase this. Automatic calculations, absolutely amazing how this has an automatic calculator built in while you write notes. So going further, so if we approximate S, you notice these are exactly the same thing then. So if we approximate S by the midpoint of this interval, so the summation approximated by the midpoint of this, these are the limits, so this is the lower bound. that's the upper bound, let's just take the midpoint or the average, then the error is at most half the length of the interval. So yes, the full interval is the full error possibility, but if we go exactly in the middle, so we're going to have error of at most half, because it could be lower bound and upper bound or set in motion right there. So the interval length is, if you subtract these, 1.202532, so that's upper bound minus lower bound, 1.2. 01664, we get this value, and you can see exactly that equals that. So 0.003038, but we're going to take the midpoint so that we could split this in half. So the midpoint or average value is 1 .201664 plus 1.202-532, divide this by 2. Or you could just divide this by 2 and add it to any, add this to this or subtract from this. And then it's equals to 1.20298, and you could round up, my Congress book does, to 1.2021. Thus, the max error is half the interval length. So we get, just divide this by 2 .00868 divided by 2. uh yeah and also i i meant here to divide this by two and then you get added to this or subtract from this anyways so the max error length is 0 0 00868 divided by 2 equals to 0 .000 434 completeness check it again equals, and there it is, automatic calculation there, and then it's less, and note that this is less than, there's 3 zeros, and there's a 4, that's less than 0.0005, which, recall an example 5, we needed 32 terms to ensure accuracy of 0.005, so this one, note, here, we're just having 10 terms, so that's fascinating stuff, so what we end up having is the series, so the series, like this, infinity n equals 1 of 1 over n cubed is approximately equal to 1 .2021, so that's approximately that, and with error, error less than 0.005, so yes, fascinating, fascinating stuff. So yes, if we compare example 6 with example 5, we see that the improved estimate in theorem 3 can be much better than the estimate SN is approximately equal to partial sum of. I mean, S is approximately equal to the partial sum SM. So to make the air smaller than 0 .005, we had to use 32 terms if we were just using SN, in example 5, but only 10 terms in example 6, because we found the air bounce up and bottom, and we took the middle, so that means we could just split that in half the actual. Yeah, error length inside, we get this 0 .005, just using 10 terms, so without having to actually calculate, or needing 32 terms to get approximate S right there. So it's fascinating.