Stacking with harmonic series - Revision history

Alexandk at 15:26, 21 January 2022

2022-01-21T15:26:29Z

Alexandk at 15:25, 21 January 2022

2022-01-21T15:25:34Z

Alexandk at 15:25, 21 January 2022

2022-01-21T15:25:11Z

Alexandk at 02:25, 16 January 2022

2022-01-16T02:25:42Z

Alexandk at 02:21, 16 January 2022

2022-01-16T02:21:52Z

Alexandk at 02:16, 16 January 2022

2022-01-16T02:16:00Z

Alexandk: Alexandk moved page Harmonic series to Stacking with harmonic series without leaving a redirect

2022-01-16T01:57:04Z

Alexandk moved page Harmonic series to Stacking with harmonic series without leaving a redirect

Alexandk at 00:56, 16 January 2022

2022-01-16T00:56:18Z

Alexandk at 00:54, 16 January 2022

2022-01-16T00:54:05Z

Alexandk at 00:37, 16 January 2022

2022-01-16T00:37:29Z

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 How ''long'' can dominoes (or blocks, tiles etc.) stack be? Can it be 50 cm? 1m? 1 km? Let’s take a look:
 [[File:Domino.png|alt=Domino stacks|none|thumb|777x777px|Domino stacks]]
-For each of the stacks above we need to make sure that it does not collapse.  Suppose that we have already built a stack of <math> n </math> dominoes. We then take this stack and place it on top of the next domino, shifting the stack over as much as we can without tipping it over. What is the maximal length of this shift? If we shift dominoes by <nowiki><math> x </math></nowiki> then the total extra weight we add to the left is proportional to <nowiki><math> (n-1) \cdot x </math></nowiki>, and the same weight is removed from the right. We need to balance it with the weight of the portion of the lowest domino in the stack which remains at the right (rests entirely on top of the domino in the base). Assuming that the length of each domino is <math> 1, </math> this weight is proportional to <math> 1-x. </math>  Therefore we must have <math> (1-x) - (n-1)x = (n-1)x +x </math> which implies <math> x = 1/2n. </math>
+For each of the stacks above we need to make sure that it does not collapse.  Suppose that we have already built a stack of <math> n </math> dominoes. We then take this stack and place it on top of the next domino, shifting the stack over as much as we can without tipping it over. What is the maximal length of this shift? If we shift dominoes by <math> x </math> then the total extra weight we add to the left is proportional to <math> (n-1) \cdot x </math>, and the same weight is removed from the right. We need to balance it with the weight of the portion of the lowest domino in the stack which remains at the right (rests entirely on top of the domino in the base). Assuming that the length of each domino is <math> 1, </math> this weight is proportional to <math> 1-x. </math>  Therefore we must have <math> (1-x) - (n-1)x = (n-1)x +x </math> which implies <math> x = 1/2n. </math>
 Continuing in the same way for <math>  n </math>  dominos, the first overhanging by 1/2, the second by 1/4, the third by 1/6,..., the <math>  n </math>th by <math>  1/2n </math>, we get the total length of <math>  1/2 +1/4+1/6+\dots +1/2n = \frac{1}{2}(1+1/2+1/3+ \dots +1/n)</math>. Since the harmonic series diverges, we can get ''any'' length as soon as we have enough dominos!
 Note. In practical terms, it may take huge number of dominoes to build a stack of a modest length. This is due to the fact that the harmonic series diverges very slowly and, for any <math> n </math>, we have <math> \ln (n +1) < 1+1/2 + \dots + 1/n < \ln n +1. </math> Suppose, for example, that each domino has length 5 cm. Then, to built a stack of length 1 meter we need <math> n </math> such that <math> (\ln n + 1)\cdot 0.05 \ge 2 \</math> and therefore <math> \ln n \ge 39 </math> . This forces <math> n </math> to be close to <math> 10^{17}. </math> If the thickness of each domino is 5 mm, then  the height of the stack will be <math> 5 \cdot 10^{10} </math> km, which is comparable with the size of the Solar system (and exceeds the distance from Sun to Pluto).

@@ Line 1: / Line 1: @@
 How ''long'' can dominoes (or blocks, tiles etc.) stack be? Can it be 50 cm? 1m? 1 km? Let’s take a look:
 [[File:Domino.png|alt=Domino stacks|none|thumb|777x777px|Domino stacks]]
-For each of the stacks above we need to make sure that it does not collapse.  Suppose that we have already built a stack of <math> n </math></nowiki> dominoes. We then take this stack and place it on top of the next domino, shifting the stack over as much as we can without tipping it over. What is the maximal length of this shift? If we shift dominoes by <nowiki><math> x </math></nowiki> then the total extra weight we add to the left is proportional to <nowiki><math> (n-1) \cdot x </math></nowiki>, and the same weight is removed from the right. We need to balance it with the weight of the portion of the lowest domino in the stack which remains at the right (rests entirely on top of the domino in the base). Assuming that the length of each domino is <math> 1, </math> this weight is proportional to <math> 1-x. </math>  Therefore we must have <math> (1-x) - (n-1)x = (n-1)x +x </math> which implies <math> x = 1/2n. </math>
+For each of the stacks above we need to make sure that it does not collapse.  Suppose that we have already built a stack of <math> n </math> dominoes. We then take this stack and place it on top of the next domino, shifting the stack over as much as we can without tipping it over. What is the maximal length of this shift? If we shift dominoes by <nowiki><math> x </math></nowiki> then the total extra weight we add to the left is proportional to <nowiki><math> (n-1) \cdot x </math></nowiki>, and the same weight is removed from the right. We need to balance it with the weight of the portion of the lowest domino in the stack which remains at the right (rests entirely on top of the domino in the base). Assuming that the length of each domino is <math> 1, </math> this weight is proportional to <math> 1-x. </math>  Therefore we must have <math> (1-x) - (n-1)x = (n-1)x +x </math> which implies <math> x = 1/2n. </math>
 Continuing in the same way for <math>  n </math>  dominos, the first overhanging by 1/2, the second by 1/4, the third by 1/6,..., the <math>  n </math>th by <math>  1/2n </math>, we get the total length of <math>  1/2 +1/4+1/6+\dots +1/2n = \frac{1}{2}(1+1/2+1/3+ \dots +1/n)</math>. Since the harmonic series diverges, we can get ''any'' length as soon as we have enough dominos!
 Note. In practical terms, it may take huge number of dominoes to build a stack of a modest length. This is due to the fact that the harmonic series diverges very slowly and, for any <math> n </math>, we have <math> \ln (n +1) < 1+1/2 + \dots + 1/n < \ln n +1. </math> Suppose, for example, that each domino has length 5 cm. Then, to built a stack of length 1 meter we need <math> n </math> such that <math> (\ln n + 1)\cdot 0.05 \ge 2 \</math> and therefore <math> \ln n \ge 39 </math> . This forces <math> n </math> to be close to <math> 10^{17}. </math> If the thickness of each domino is 5 mm, then  the height of the stack will be <math> 5 \cdot 10^{10} </math> km, which is comparable with the size of the Solar system (and exceeds the distance from Sun to Pluto).

@@ Line 1: / Line 1: @@
 How ''long'' can dominoes (or blocks, tiles etc.) stack be? Can it be 50 cm? 1m? 1 km? Let’s take a look:
 [[File:Domino.png|alt=Domino stacks|none|thumb|777x777px|Domino stacks]]
-For each of the stacks above we need to make sure that it does not collapse.  Suppose that we have already built a stack of <nowiki><math> n </math></nowiki> dominoes. We then take this stack and place it on top of the next domino, shifting the stack over as much as we can without tipping it over. What is the maximal length of this shift? If we shift dominoes by <nowiki><math> x </math></nowiki> then the total extra weight we add to the left is proportional to <nowiki><math> (n-1) \cdot x </math></nowiki>, and the same weight is removed from the right. We need to balance it with the weight of the portion of the lowest domino in the stack which remains at the right (rests entirely on top of the domino in the base). Assuming that the length of each domino is <math> 1, </math> this weight is proportional to <math> 1-x. </math>  Therefore we must have <math> (1-x) - (n-1)x = (n-1)x +x </math> which implies <math> x = 1/2n. </math>
+For each of the stacks above we need to make sure that it does not collapse.  Suppose that we have already built a stack of <math> n </math></nowiki> dominoes. We then take this stack and place it on top of the next domino, shifting the stack over as much as we can without tipping it over. What is the maximal length of this shift? If we shift dominoes by <nowiki><math> x </math></nowiki> then the total extra weight we add to the left is proportional to <nowiki><math> (n-1) \cdot x </math></nowiki>, and the same weight is removed from the right. We need to balance it with the weight of the portion of the lowest domino in the stack which remains at the right (rests entirely on top of the domino in the base). Assuming that the length of each domino is <math> 1, </math> this weight is proportional to <math> 1-x. </math>  Therefore we must have <math> (1-x) - (n-1)x = (n-1)x +x </math> which implies <math> x = 1/2n. </math>
 Continuing in the same way for <math>  n </math>  dominos, the first overhanging by 1/2, the second by 1/4, the third by 1/6,..., the <math>  n </math>th by <math>  1/2n </math>, we get the total length of <math>  1/2 +1/4+1/6+\dots +1/2n = \frac{1}{2}(1+1/2+1/3+ \dots +1/n)</math>. Since the harmonic series diverges, we can get ''any'' length as soon as we have enough dominos!
 Note. In practical terms, it may take huge number of dominoes to build a stack of a modest length. This is due to the fact that the harmonic series diverges very slowly and, for any <math> n </math>, we have <math> \ln (n +1) < 1+1/2 + \dots + 1/n < \ln n +1. </math> Suppose, for example, that each domino has length 5 cm. Then, to built a stack of length 1 meter we need <math> n </math> such that <math> (\ln n + 1)\cdot 0.05 \ge 2 \</math> and therefore <math> \ln n \ge 39 </math> . This forces <math> n </math> to be close to <math> 10^{17}. </math> If the thickness of each domino is 5 mm, then  the height of the stack will be <math> 5 \cdot 10^{10} </math> km, which is comparable with the size of the Solar system (and exceeds the distance from Sun to Pluto).

← Older revision		Revision as of 02:25, 16 January 2022
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	Continuing in the same way for <math> n </math> dominos, the first overhanging by 1/2, the second by 1/4, the third by 1/6,..., the <math> n </math>th by <math> 1/2n </math>, we get the total length of <math> 1/2 +1/4+1/6+\dots +1/2n = \frac{1}{2}(1+1/2+1/3+ \dots +1/n)</math>. Since the harmonic series diverges, we can get ''any'' length as soon as we have enough dominos!		Continuing in the same way for <math> n </math> dominos, the first overhanging by 1/2, the second by 1/4, the third by 1/6,..., the <math> n </math>th by <math> 1/2n </math>, we get the total length of <math> 1/2 +1/4+1/6+\dots +1/2n = \frac{1}{2}(1+1/2+1/3+ \dots +1/n)</math>. Since the harmonic series diverges, we can get ''any'' length as soon as we have enough dominos!

−	Note. In practical terms, it may take huge number of dominoes to build a stack of a modest length. This is due to the fact that the harmonic series diverges very slowly and, for any <math> n </math>, we have <math> \ln (n +1) < 1+1/2 + \dots + 1/n < \ln n +1. </math> Suppose, for example, that each domino has length 5 cm. Then, to built a stack of length 1 meter we need <math> n </math> such that <math> (\ln n + 1)\cdot 0.05 \ge 2 \</math> and therefore <math> \ln n \ge 39 </math> . This forces <math> n </math> to be close to <math> 10^{17}. </math> If the thickness of each domino is 5 mm, then the height of the stack will be <math> 5 \cdot 10^{10} </math> km, which is comparable with the size of the Solar system (and ~~more than~~ the distance from Sun to Pluto).	+	Note. In practical terms, it may take huge number of dominoes to build a stack of a modest length. This is due to the fact that the harmonic series diverges very slowly and, for any <math> n </math>, we have <math> \ln (n +1) < 1+1/2 + \dots + 1/n < \ln n +1. </math> Suppose, for example, that each domino has length 5 cm. Then, to built a stack of length 1 meter we need <math> n </math> such that <math> (\ln n + 1)\cdot 0.05 \ge 2 \</math> and therefore <math> \ln n \ge 39 </math> . This forces <math> n </math> to be close to <math> 10^{17}. </math> If the thickness of each domino is 5 mm, then the height of the stack will be <math> 5 \cdot 10^{10} </math> km, which is comparable with the size of the Solar system (and exceeds the distance from Sun to Pluto).

← Older revision		Revision as of 02:21, 16 January 2022
Line 4:		Line 4:
	Continuing in the same way for <math> n </math> dominos, the first overhanging by 1/2, the second by 1/4, the third by 1/6,..., the <math> n </math>th by <math> 1/2n </math>, we get the total length of <math> 1/2 +1/4+1/6+\dots +1/2n = \frac{1}{2}(1+1/2+1/3+ \dots +1/n)</math>. Since the harmonic series diverges, we can get ''any'' length as soon as we have enough dominos!		Continuing in the same way for <math> n </math> dominos, the first overhanging by 1/2, the second by 1/4, the third by 1/6,..., the <math> n </math>th by <math> 1/2n </math>, we get the total length of <math> 1/2 +1/4+1/6+\dots +1/2n = \frac{1}{2}(1+1/2+1/3+ \dots +1/n)</math>. Since the harmonic series diverges, we can get ''any'' length as soon as we have enough dominos!

−	Note. In practical terms, it may take huge number of dominoes to build a stack of a modest length. This is due to the fact that the harmonic series diverges very slowly and, for any <math> n </math>, we have <math> \ln (n +1) < 1+1/2 + \dots + 1/n < \ln n +1. </math> Suppose, for example, that each domino has length 5 cm. Then, to built a stack of length 1 meter we need <math> n </math> such that <math> (\ln n + 1)\cdot 0.05 \ge 2 \</math> and therefore <math> \ln n \ge 39 </math> . This forces <math> n </math> to be close to <math> 10^{17}. </math> If the thickness of each domino is 5 mm, then the height of the stack will be <math> 5 \cdot 10^10 </math> km, which is comparable with the size of the Solar system (and more than the distance from Sun to Pluto).	+	Note. In practical terms, it may take huge number of dominoes to build a stack of a modest length. This is due to the fact that the harmonic series diverges very slowly and, for any <math> n </math>, we have <math> \ln (n +1) < 1+1/2 + \dots + 1/n < \ln n +1. </math> Suppose, for example, that each domino has length 5 cm. Then, to built a stack of length 1 meter we need <math> n </math> such that <math> (\ln n + 1)\cdot 0.05 \ge 2 \</math> and therefore <math> \ln n \ge 39 </math> . This forces <math> n </math> to be close to <math> 10^{17}. </math> If the thickness of each domino is 5 mm, then the height of the stack will be <math> 5 \cdot 10^{10} </math> km, which is comparable with the size of the Solar system (and more than the distance from Sun to Pluto).

← Older revision	Revision as of 01:57, 16 January 2022
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