# Building Blocks Problem Related to Harmonic Series

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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.

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