Series Articles - Wyzant Tutor Blogshttps://www.wyzant.com/resources/blogs/seriesThis is an aggregate of all of the Series articles in Wyzant.com's Tutors' Blogs. Wyzant.com is your source for tutors and students.Tue, 19 Jan 2021 03:54:23 -0600https://www.wyzant.com/images/logos/wyzant-logo.pngSeries Articles - Wyzant Tutor Blogshttps://www.wyzant.com/resources/blogs/serieshttps://www.wyzant.com/resources/blogs/series277085https://www.wyzant.com/resources/blogs/277085/mathematical_journeys_carl_gauss_and_the_sum_of_an_arithmetic_seriesEllen S.https://www.wyzant.com/resources/users/view/75479140Mathematical Journeys: Carl Gauss and the Sum of an Arithmetic Series<div>There's a famous (and probably apocryphal) story about the mathematician Carl Friedrich Gauss that goes something like this:<br /><br />Gauss was 9 years old, and sitting in his math class. He was a genius even at this young age, and as such was incredibly bored in his class and would always goof off and get into trouble. One day his teacher wanted to punish him for goofing off, and told him that if he was so smart, why didn't he go sit in the corner and add up all the integers from 1 to 100? Gauss went and sat in the corner, but didn't pick up his pencil. The teacher confronted him, saying “Carl! Why aren't you working? I suppose you've figured it out already, have you?” Gauss responded with “Yes – it's 5,050.” The teacher didn't believe him and spent the next ten minutes or so adding everything up by hand, only to find that Gauss was right! <br /><br />So how did Gauss find the answer so fast? What did he see that his teacher didn't? The answer is simple, really – it's all about pattern recognition. Let's look at the problem more closely.<br /><br />1 + 2 + 3 + 4 + 5 +...+ 95 + 96 + 97 + 98 + 99 + 100 = ?<br /><br />Now it's true that adding all that up by hand would take forever, but we don't really need to add it all up by hand. Look at this series from each end simultaneously instead of just left to right. You'll see that we can think of this series as a set of pairs of numbers, each of which adds up to 101:<br /><br /><span class="orange">1</span> + <span class="greenText">2</span> + 3 + 4 + 5 +...+ 95 + 96 + 97 + 98 +<span class="greenText"> 99</span> + <span class="orange">100</span> = ?<br /><br />1 + 100 = 101<br />2 + 99 = 101<br />3 + 98 = 101<br />4 + 97 = 101<br />5 + 96 = 101<br /><br />and so on right through to the middle, where:<br /><br />48 + 53 = 101<br />49 + 52 = 101<br />50 + 51 = 101<br /><br />So we've got a total of 50 pairs, each of which adds up to 101. Since all our chunks are the same size, we can take a shortcut and simply multiply the size of the pair (101) by the number of pairs (50). Which is really easy to do in your head, since it's just (100 x 50) plus (1 x 50), or 5000 + 50 = 5,050.<br /><br />This reasoning can actually be extrapolated to work with any arithmetic series, and in fact is how we get the formula for the sum of an arithmetic series. Check it out:<br /><br />Each pair added up to the same number, so we could actually use the mathematical expression for any one of those pairs in our formula. Since the first and last term are the ones most often known, let's use those. Remember, the last term in the series is written as a<sub>n</sub>, where n is the number of terms in the series. So that 101 will be represented by:<br /><br />The first term (a<sub>1</sub>) + the last term (a<sub>n</sub>), or (a<sub>1</sub> + a<sub>n</sub>)<br /><br />Check out the first term in each of our pairs above. They range from 1 to 50, so there are 50 terms – exactly half the number of terms in the series. Which makes sense, since we're pairing up the terms and that by definition gives us half as many pairs as terms. <br /><br />So 50 in our example is represented by one-half of n, or n/2. <br /><br />And what are we doing with these two bits of information? Multiplying them together. So our final formula would be:<br /><br />Σ = (n/2)(a1 + an)<br /><br />Now, sometimes you see this formula written as n(<span style="text-decoration: underline;">a1 + an</span>), <br /> 2<br /><br />or the number of terms times the average of the first and last terms. Practically that is exactly the same thing as the way we wrote it first, it's just written a little differently. But they both have the same value, so if it's easier for you to remember it as the average times the number of terms, do so.<br /><br />This formula works for any arithmetic series, so the next time you come up against one, remember Gauss and his pairs of terms!</div>Tue, 08 Jul 2014 08:21:06 -05002014-07-08T08:21:06-05:00246872https://www.wyzant.com/resources/blogs/246872/complete_the_seriesJohn M.https://www.wyzant.com/resources/users/view/82890510Complete the Series<div>On I.Q. tests and in other places, one is often confronted with problems of the form: “What’s the next element in the following series: 1, 4, 9, …”</div>
<div><br />Technically, such questions have no right answer, because there are a multitude of ways to generate the initial elements of the series, and each way can produce a different result for how the series should continue. What is being sought is the generator for the series that is somehow the simplest, cleverest, most obvious, or most elegant. There is an esthetic at work in determining the preferred solution.</div>
<div><br />For example, for the series above, an obvious answer is 16, because the initial elements of the series are the squares of the first three natural numbers, and so the obvious way to continue the series is with the squares of the subsequent natural numbers. The series is given by s<sub>n</sub> = n<sup>2</sup>.</div>
<div><br />A similar type of question involves a mapping between a series of expressions and values. In that case, the preferred answer involves the simplest or most elegant way to interpret the expression to generate the value to which it is mapped. For example, Alberto Garcia-Luis Valencia recently posted the following problem on the LinkedIn professional social network:</div>
<div> </div>
<div>“Solve if u r a genius !<br />11 x 11 = 4<br />22 x 22 = 16<br />33 x 33 = ???”</div>
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<div>Since there are only two initial data points, the latitude for choosing a generator for this series is extremely wide. Consider the following possible ways to express the “x” operator as a function:</div>
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<div>1. X<sub>1</sub>(a,b) = dsum(ab)<br />2. X<sub>2</sub>(a,b) = dsum(a)dsum(b)<br />3. X<sub>3</sub>(a,a) = 12a/11 – 8<br />4. X<sub>4</sub>(aa,aa) = 12a – 8</div>
<div> </div>
<div>In X<sub>1</sub> and X<sub>2</sub>, dsum(x) is the sum of the digits in the base 10 representation of the number x. I.e., dsum(11)=2. X<sub>3</sub> takes advantage of the fact that in our series the numbers being operated on are always the same, and X<sub>4</sub> takes advantage of the fact that in our series the numbers being operated on always have repeated digits.</div>
<div><br />Note that all four versions of X generate the first two terms of the series. However, they produce different predictions for the next entry. The predictions are:</div>
<div><br />1. X<sub>1</sub>: 33 x 33 = 18 (because 33 * 33 = 1089, and the sum of these digits is 18)<br />2. X<sub>2</sub>: 33 x 33 = 36 (because the digits of 33 sum to 6, and 6*6 = 36)<br />3. X<sub>3</sub>: 33 x 33 = 28 (because 12*33/11 – 8 = 36 – 8 = 28)<br />4. X<sub>4</sub>: 33 x 33 = 28 (because 12*3 –8 = 36 – 8 = 28)</div>
<div><br />Which of these is to be preferred? Or, is there another solution that is preferable to all of them?</div>
<div><br />The posting on the website lead to thousands of comment replies, in which people offered different answers for the next entry. By far the most popular answer was 36, possibly indicating that X<sub>2</sub> is the generator that most people find simplest and most elegant. However, most people did not indicate how they arrived at their answer, so we can’t know what generator they had in mind. A great many people came up with other answers, including 81, 64, 256, etc. What patterns did they see in the first two entries that would lead them to these conclusions?</div>
<div><br />Personally, I preferred X<sub>1</sub>. It is shorter and simpler than X<sub>2</sub>, and does not involve any arbitrary constants. It could be argued that X<sub>3</sub> or X<sub>4</sub> is simpler than either X<sub>1</sub> or X<sub>2</sub>, as it does not involve the rather non-standard function dsum, and what could be simpler than a linear extrapolation (which is what they are)? X<sub>1</sub> and X<sub>2</sub> can be expressed in English as “the digit sum of the product” and the “product of the sums of the digits.” Expressed this way, it’s harder to see X<sub>1</sub> as necessarily simpler than X<sub>2</sub>. X<sub>1</sub> and X<sub>2</sub> also exhibit a certain degree of cleverness in noticing that summing the digits can play a role.</div>
<div><br />However, there is another way to think about this problem. We tend to see 11, 22 and 33 as two-digit decimal representations of numbers. Another way to see them is as the concatenation of one-digit numbers, and to consider that concatenation can represent the addition operator. We usually treat concatenation to mean multiplication, so this is “thinking outside the box.”</div>
<div><br />Then X<sub>5</sub>(a,b,c,d) = (a+b)*(c+d). X5(a,a,a,a) = (a+a)*(a+a) = (2a)<sup>2</sup> or 4a<sup>2</sup>.</div>
<div><br />This is mathematically equivalent to X<sub>2</sub>, but doesn’t require invoking the strange dsum function, and has the esthetically pleasing expression is English: Two a’s times two a’s (i.e., two 1’s times two 1’s = 4, two 2’s times two 2’s = 16, two 3’s times two 3’s = 36).</div>
<div><br />Thus, I will concede that this is the simpler, more elegant, and more clever solution, and I understand why so many people chose 36 as the answer. Stubbornly, though, I still like my original answer.</div>Sun, 01 Dec 2013 14:25:56 -06002013-12-01T14:25:56-06:00