![]() Conceptually, we're just trying to think about what is the common ratio approach? What is the ratio betweenĬonsecutive terms? The N plus one term and the Nth term as N gets really, really, Conceptually, let me copy and paste this. ![]() Let's take the limit as NĪpproaches infinity of this thing. Look at the behavior here, and if this is approaching some actual values as N approaches infinity, well, it would make conceptual sense that we could kind of think of that as the limit of our common ratio. The behavior as our N's "get really, really, really large as the limit, as our N's, go to infinity." So what if we were to This ratio, I guess youĬould say, is a function of N so this doesn't seem too useful, but what if I were to say, "O.K., well look, with any of these series "we really care about Term divided by the Nth term, it's changing depending on what my N is. "Hey, wait, look, this isn'tĪ fixed common ratio here." The ratio between consecutive terms here, when I took the N plus oneth One to the tenth power over N plus one times N to the tenth. To this N right over here, and why is that useful? Because this N factorialĬancels with this N factorial, and we're left with N plus You're not use to seeing order of operations with factorials, but this factorial only applies Well, if we do a little algebra here N plus one factorial, andįactorial algebra is always fun. Or dividing by anything, is equivalent to multiplyingīy it's reciprocal, So let's just multiply by the reciprocal, so times N factorial over N to the tenth. Plus one to the tenth power over N plus one factorial andĭivide that by the Nth term, so let's divide that by N to So let's take the N plus oneth term, which is going to be N So let's see if there'sĪ common ratio here. Well, let's see if we canĬome up with a common ratio. This actually converges? And maybe we can use a littleīit of our intuition here. Grows very, very quickly, probably it grows much faster than even a high degree polynomial like this, or a high degree term like this, but how do we prove that it converges? We could definitely the diverges test to show that this does not diverge, but how do we prove that N to the tenth power over N factorial, and factorial we know also Whether a series like this, so starting at N equals five to infinity of, let's say, N to the tenth power. Let's say that we want to look at a, we want to figure out Now, with that out of the way for review, let's tackle something a That, even though this is an infinite sum, it willĬonverge to a finite value. On and on and on and on until it makes sense ![]() To go down by that common, or it's going to be multipliedīy that common ratio, and it's going to decrease Of R is less than one then each term here is going ![]() Logical sense that, look, if that absolute value We've proven it as well, but it also makes And if the absolute value of R is greater than or equal to one, then the series diverges. If the absolute value of the common ratio is less than one, then the series converges. And what's interestingĪbout this is we've proven to ourselves, in the videosĪbout geometric series, that if the common ratio, If it is not review, I encourage you to watch the videos on geometric series. When you go from one term to another you are just multiplying, youĪre just multiplying by R, and this is all review. To R, or R to the first power, and you see that here. That same thing with the N, that's just going to be equal Well, this is just going to be equal to R. Which is the ratio between consecutive terms, is going There are a few things we'veĪlready thought about here. Plus R to the K plus one plus R to the K plus two and keep going on and on and on forever. Infinity of R to the N, which would be R to the K Infinite geometric series starting at N equals K to We already have a lot of experience with the geometric series. Let me know if that didn't fully help and I can try explaining differently If you'd like to see it on a smaller scale, try (100x^2 + 100x)/(x^3) it starts with the numerator being larger, but eventually the denominator is bigger. if the denominator keeps getting bigger than the numerator than eventually it will equal 0. What does that mean? well, a large denominator makes the fraction get closer and closer to 0. the numerator may start out bigger, but as you head toward infinity, the larger exponent will always make a bigger term, so the denominator will get bigger than the numerator. Nowwhat happens as n gets bigger and bigger? n^11 will always be bigger than an^9 + bn^8 +. the n^10 in the numerator and denominator kinda cancel out, or at least they will be the same number no matter what n is, so we only need to worry about the rest. You could maybe look at it as n^10 + n^11. ![]() + z where a through z are some real numbers. So, the numerator is a massive polynomial, but the largest term is n^10, so it will be n^10 + an^9 +bn^8 +. ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |