Building Blocks Problem Related to Harmonic Series

Transcription

1 TMME, vol3, o, p.76 Buildig Blocks Problem Related to Harmoic Series Yutaka Nishiyama Osaka Uiversity of Ecoomics, Japa Abstract: I this discussio I give a eplaatio of the divergece ad covergece of ifiite series through the buildig blocks problem ad at the same time I touch o the fact that mathematics is ot just about maipulatig complicated umerical formulas but also a field i which logical ways of thought are leart. I emphasize that i order to overcome the aversio of uiversity studets to mathematics, teachers must pour their eergies ito developig study materials take from topics relevat to the studets. Keywords: harmoic series, ceter of gravity, covergece ad divergece, logarithmic fuctios, Mathematical educatio. Is it possible to stagger buildig blocks by more tha the width of oe block? There has bee a lot of publicity about how youg people avoid ad are "allergic" to mathematics. The goal of mathematics is ot difficult umerical formulas but a mathematical way of lookig at ad thikig about thigs ad I would like to preset oe eample of this. Let us thik about the buildig blocks problem i Figure. There are a few buildig blocks stacked up, ad the problem is whether or ot it is possible to stack them i such a way that the positios of the bottom block ad the top block are horizotally separated by more tha the width of oe block. Most people asked this questio would immediately aswer that it is ot possible. I woder if this tedecy to come to a coclusio before eve attemptig to thik about whether somethig is possible or ot is a reflectio of the digital age. Sometimes it is possible, sometimes it is ot possible, additioally sometimes we do ot kow. But they hate vague aswers very much. This is ot magic or a trick, I promise that a solutio certaily eists. If a perso is told to stack buildig blocks i a staggered way, he or she will stagger them uiformly. But if they are staggered uiformly they will fall dow every time. I woder if this tedecy to stack the blocks uiformly is also a maifestatio of digital thikig. We will ot obtai a solutio immediately. Let us start by lookig at the case of two blocks. It is ituitively obvious that the distace they ca be staggered is / of the width of the blocks. So the problem is the third block. Let us hold the third block i our right had ad thik about this problem. Most people would try to stack this block o top of the other two but the they The Motaa Mathematics Ethusiast, ISSN 55-30, Vol. 3, o., pp The Motaa Coucil of Teachers of Mathematics

2 TMME, vol3, o, p.77 always fall dow. If your approach does ot work it is importat to abado it, ad you should search for a alterative approach. To be able to do this, it is ecessary to chage your way of thikig about the problem. Figure : Is it possible to stagger buildig blocks by more tha the width of oe block? Figure : / Stagger. Calculatig the ceter of gravity This is the problem of calculatig the ceter of gravity. Rather tha thikig about this problem with a pe ad paper, it is surprisigly fast to use buildig blocks ad look for the aswer through trial ad error. Here I will give you a hit. Are buildig blocks best stacked o top of each other? You will probably be perpleed by this hit. That is because of the fied precoceptio that it is because we stack them o top of each other that they are buildig blocks. But buildig blocks should ot be stacked o top of each other; they should be slid uder each other. If the third buildig block is placed at the bottom, ad we gradually stagger the first two buildig blocks o top of the third buildig block while maitaiig the relatioship betwee the first two buildig blocks as it was, we fid that we ca stagger the top two blocks by /. I the same way, the fourth block ca be placed uder the other three ad staggered by /6, ad the fifth block ca be placed uder the other four ad staggered by /8. If we add /, /, /6 ad /8 the sum is greater tha. I other words, we have stacked the blocks i such a way that the positio of the top block is horizotally separated from that of the bottom block by more tha the width of oe block.

3 TMME, vol3, o, p.78 While referrig to Figures, 3, ad, let us cofirm the above approach as a ceter of gravity calculatio usig umerical formulas. First of all let us thik about buildig block ad buildig block. It is clear that we ca oly stagger them by / of the width of a block (Figure ). 3 Figure 3: / Stagger 3 6 Figure : /6 Stagger Net we are goig to put buildig block 3 uder the first two blocks so let us thik about the ceter of gravity of buildig blocks ad together (Figure 3, left). Because buildig block 3 ca be staggered up to the ceter of gravity, I will obtai the momet, callig the stagger distace. Momet is the product of weight ad arm legth so the momet of buildig block (rotated clockwise) is, the momet of buildig block (rotated ati-clockwise) is ( ) ad because these two values are equal, = ( ) Solvig this equatio, we show that = /. I other words, the stagger distace for buildig

4 TMME, vol3, o, p.79 block 3 is / (Figure 3, right). Net we are goig to place buildig block so let us thik about the ceter of gravity of buildig blocks, ad 3 together. Let us obtai this ceter of gravity from the combiatio of the ceter of gravity of buildig blocks ad together ad the ceter of gravity of buildig block 3 (Figure, left). Takig buildig blocks ad together gives a weight of. The momet of buildig blocks ad (rotated clockwise) is, ad the momet of buildig block 3 (rotated ati-clockwise) is ( ) ad because these two values are equal, = ( ) Solvig this equatio, we show that = / 6. I other words, the stagger distace for buildig block is /6 (Figure, right). Let us obtai the geeral result for the ceter of gravity of buildig blocks. As this is determied by the ceter of gravity of ( ) buildig blocks plus the ceter of gravity of oe buildig block, ( ) = ( ), therefore =. 3 5 Figure 5: -Block Stagger Rearragig this equatio we ca see that if the stagger positio is as follows,,,, the the buildig blocks ca be stacked so that they will ot fall dow. Whe the progressio produced by reciprocal umbers is a arithmetic progressio, it is called a harmoic

5 TMME, vol3, o, p.80 progressio. For eample,, /, /3, ad, /3, /5, are harmoic progressios. Harmoic progressios are said to have bee used i the study of harmoies theory by the Pythagorea School i aciet Greece ad the ame of harmoic progressios is derived from it. Harmoic series are the totals of harmoic progressios so we ca also write: = ( ) 6 3 So ow let us calculate the value of this series. = 0.5, = 0.5, 0.67, = 0.5 therefore > 8 So we ow kow that if we have 5 buildig blocks we ca stagger them by more tha the width of oe block (Figure 5). 3. Covergece ad divergece I high school ad uiversity differetial ad itegral calculus tetbooks there are chapters o progressios ad series. I those chapters the followig eercise ivariably appears: 3 is diverget, ad 3 is coverget. Whe goes to ifiity, there are iterestig eercises i which sometimes eve if the geeral term of the progressio coverges to 0 the ifiite series diverges. Covergece ad divergece ca be approimately kow by performig itegratio as follows: d = [log ] therefore d = [ ].

6 TMME, vol3, o, p.8 Figure 6: y = Figure 7: y = The first of these two equatios is i log order ad diverges (Figure 6), ad the secod of these two equatios coverges (Figure 7). Geerally, ifiite series of the form ( p > 0) diverge if p ad coverge if p >. Furthermore, it is kow p that coverges to π. Furthermore, whether or ot coverges is determied by 6 the Cauchy covergece criteria for the progressio. The sum of the first terms of the progressio a, a,, a, is defied as S = a a a. As for the ecessary ad sufficiet coditio for the series a to be coverget, if we make N sufficietly large compared to ay give positive umberε, for all ad m where m > > N it ca be show that: S m S = a a am < ε. Assumig that S =, the o matter how big we make, 3 S S = > = terms So the Cauchy covergece criteria are ot met. Therefore is diverget. Let us look at

7 this more closely. If we take the umber of terms S S = S S = > 3 = S S = > S = S S S S S S 8 = S 8 S > So we ca see that the series diverges. TMME, vol3, o, p.8 as powers of like this:,, 8,..., the S. Divergece i log order I have eplaied that the harmoic series diverges to ifiity but let us look closely at how quickly diverge. I used a persoal computer to calculate the value of, the total stagger distace. The results were as follows: Whe = =.07 > Whe = 3 =.036 > Whe = 7 = 3.00 > 3 So the series does diverge to ifiity but at a etremely slow speed. If we ow compare with the itegratio of the fuctio iequality as follows: 3 From the fact that d < d < (log ) (log ) < <. So we kow that whe = [log ] = log, we ca show that diverges i log order., y = we ca establish a

8 TMME, vol3, o, p.83 Figure 8: -Block Stagger Because oe more etra buildig block is ecessary at the bottom, the umber of buildig blocks ecessary is actually. Oly five buildig blocks () are sufficiet to stagger the pile of buildig blocks by the width of oe buildig block, but 3 buildig blocks (3) are ecessary to stagger the pile by the width of two buildig blocks ad 8 buildig blocks (7) are ecessary to stagger the pile by the width of three buildig blocks. Figure 8 shows a stack of 3 buildig blocks but i practice it is impossible to stack up 3 blocks accurately staggered i this way. This is just a theoretical discussio. Figure 9 is a graph showig the fuctio y = ad y = log, the fuctio resultig from the itegratio of y =. If we rotate the log fuctio 90 degrees clockwise ad reverse it horizotally it becomes the buildig block stackig problem i Figure 8. I will leave it to you to cofirm this.

9 TMME, vol3, o, p.8 Figure 9: Log fuctio That completes the proof. I have show that the harmoic series describes the solutio to the buildig blocks problem. If we solve these kids of problems, mathematics should be more ejoyable I thik. Mathematics i high school ad uiversity progressively becomes more distat from reality ad sometimes studets come close to losig sight of them. At times like that the studet must ot forget to apply the problems to reality. The buildig blocks problem is the problem of the calculatio of the ceter of gravity; it also ivolves harmoic series ad is etremely mathematical. If we limit ourselves to just solvig the problem, we do ot eed to use complicated umerical formulas. The importat thigs are to employ logical ways of thought ad to have the ability to chage your way of thikig. Iroically, uiversity studets doig sciece subjects caot solve this buildig blocks problem. They ca prove with umerical formulas that harmoic series diverge to ifiity, but they caot solve the real world problem of the buildig blocks. This is a blid spot i moder educatio. I leaed about the buildig blocks problem from a 958 work by George Gamow. He was both a researcher ad educator ad it appears that he was of the opiio that the studets will ot get ecited about mathematics if the teacher is ot ecited about it. If you would like to cofirm the solutio to the buildig blocks problem but you do ot have ay buildig blocks at had, you could try doig it with te volumes of a ecyclopedia or te video tapes. Refereces: Gamow, G & Ster, M. (958). Puzzle-Math, Vikig Press Ic. Nishiyama, Y. (00). Buildig Blocks ad Harmoic Series, Osaka Keidai Roshu, 50 (), 95-0.