Consider a circle with a starting circumference of C 0 = 1 0 0 units. Suppose you start on a point on that circle and proceed to walk along it until you take your first step on or past your starting position. Given the rules below:
which of the following gives the best (i.e. most accurate) approximation to the number of steps needed to complete the task?
Note:
If you think that the task will never be completed, then answer
∞
. Also, the
⌊
x
⌋
is the usual floor function.
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The radius of the circle is 100, so its circumference is 2 π ⋅ 1 0 0 = 2 0 0 π . This means the first step covers 2 0 0 π 1 of the total distance, the second step covers 2 ⋅ 2 0 0 π 1 , and so on. Your numbers are off by a factor of 2 π .
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Indeed! Thanks a lot of the remark. I have updated the problem accordingly.
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Why change it? you defined the original circumference as 100 unite, not the radius. Ed Gray
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@Edwin Gray – This problem was created 2 weeks ago and he originally said radius instead of circumference as the 7th word
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@Dylan Wedel – They should have it say the last date it was modified with a short desc of it when hovered so discussions like this don't come up.
It says "circle with a starting circumference of 100 units" so you are wrong in that regard.
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This problem was created 2 weeks ago and he originally said radius instead of circumference
I strongly dissent, the probability of hit a predetermined point inside the circle is 1/ divided by infinite no matter if circle increase in size. Remenber the steps can be taking in any direction it not the case of a dotted circle.
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I think you misunderstand the problem. You are walking in one direction along the edge of the circle, not randomly inside of it.
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It maybe the case but the reading says " Suppose you start on a point on that circle and proceed to walk along it until you go back to your starting position." it should say on the edge of the circle or better on the circumference and dismiss the word circle at all.
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@Mara Jares – I don't agree. A circle is a one-dimensional object only made up of its circumference (so to speak). Unlike a disk, which includes the inside. Stating that you walk on/along the circle hence leaves only one option as to what path can to be taken. I will try to be more careful with my wording next time though. Thanks for the comment.
The question asks the number of steps to return to the starting point, i.e. hitting the exact point where the walk began. This requires H n = 1 0 0 , which does not have a solution since H n is always a non-integer for n > 1 . So, the correct answer should have been ∞ .
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I agree! For 1 unit travelled, you create 100 additional units to travel.
That is, unless i completely missed what is meant here?
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That's not the case, but I agree, the question wasn't stated very well. When they said the circle is scaled, it's scaled about your current position. So if you were 1 unit in, and then the circle is scaled to 2(100), you are then 2 units in, and 198 units out. Notice, you're not creating 100 more units to travel for each step, that's 99. If you try more numbers, you'll see the distance necessary to travel becomes smaller and smaller (in fact, the distance is first multiplied by 2/1, then 3/2, 4/3, 5/4, etc, which converges towards 1, thus making the trip possible). Once it goes below 1, you can simply start making your way to the end without creating for steps to walk. Hopefully that makes sense.
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@Kevin Tong – That's a good explanation
@Kevin Tong – I am not certain to agree. By definition of scaling, the distance between any two points is scaled by the appropriate factor (assuming a "homogenous" scaling, which is the case here). So you could scale about any point you want (even outside the circle). Usually, one should think of it about the center for symmetry/intuition though.
Another way I look at this (without referring to coordinates) is that a circle is a one-dimensional manifold. Up to a point on the circle, it is topologically equivalent to a line. Now, when we talk about scaling a line, it should be clear geometrically that it is like "stretching" the line. There is no "reference point" from which it is scaled.
I found my "mistake", although I still think it is very ambiguous. I didn't think about that somehow my steps scale as well. The problem this way is HIGHLY unrealistic. Each step should be scaling only the circumference of the shape I am walking, not space behind me (formulated badly). I understand that they wanted to trick people into thinking of infinity, but that's not a very brilliant thing to do.
you should know,if (lim n →∞),Hn is a divergent series(p series,p=1). it means that there must be a value of n exist(the answer is that it is close to e^100) makes the value of Hn equal to 100.So you can complete this circle. Just like, lim n → ∞,Tn=1/2+1/4+1/8+.....1/(2^n).Tn equals 1. 100/100,you can also complete this circle.That's my opinion.
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H n cannot be ever an integer for n ≥ 2 . So your argument is not correct. Also, if you plot the values of H n against n it is going to be a staircase type function with lots of gaps in its values, and all the integers reside in those gaps.
I foolishly answered that it was not possible.. however realised later that distance covered in my first step would double too.. however my math deserted me right here !!😊
Before beginning the explanation, I want to point out that the question wasn't stated very clearly, although it can be easily understood nonetheless. Let's first consider what it means to scale the circle. In order to scale it, we must have a point to scale it about (not necessarily the center). If we were to scale it at any point other than our current position, we would end up inside of the circle since the surface would be scaled away from where we were (unless we were scaled it with). Thus, we can reasonably conclude that we are scaling the circle about our current position. Now we can begin the problem.
Let's start off observing what happens each time we take a step. After the first step before the scaling, we are 1 unit in, 99 units to go. After scaling, both our distance traveled and distance left are scaled by a factor of 1 2 . Thus, we have 198 units left to go by that point. Since we only care about the distance left to go, we only need to pay attention to that. The main problem when solving this is the factor by which we scale the circle by each time. Since we're adding 100 units to the circumference of the circle, the factor we multiply by is C C + 1 0 0 , which, as C gets larger, converges to 1. Thus, we now know the answer cannot be infinity since the factor converges.
Now let's setup an equation of what this would look like: After the first step and scaling, we have 1 2 ( 1 0 0 − 1 ) ; After the second step and scaling, we have 2 3 ( 1 2 ( 1 0 0 − 1 ) − 1 ) ; After the third step and scaling, we have 3 4 ( 2 3 ( 1 2 ( 1 0 0 − 1 ) − 1 ) − 1 ) . Notice, when we expand this, the fractions being multiplied to 1 0 0 − 1 = 9 9 cancel out to simply n + 1 and the − 1 s are multiplied by a partial sum of the harmonic series multiplied by n + 1 ) . In other words, the distance left to travel can be modeled by the formula 9 9 ( n + 1 ) − ( n + 1 ) k = 2 ∑ n k 1 = ( n + 1 ) ( 9 9 − k = 2 ∑ n k 1 ) We want to find when this is equal to zero, thus, we can divide n + 1 on both sides, and we end up with 9 9 − k = 2 ∑ n k 1 = 0 ⟹ 1 0 0 − H n = 0 ⟹ H n = 1 0 0
Now, we want to find the nth harmonic series such that it is greater than or equal to 100. To do this, notice that the harmonic series can be modeled by the function f ( x ) = ⌊ x 1 ⌋ . Now, notice x + 1 1 < f ( x ) ≤ x 1 . Thus, we know that the sum of the harmonic series is between the integrals of the two other functions, namely, ∫ 1 n x + 1 1 d x < ∫ 1 n f ( x ) d x ≤ ∫ 1 n x 1 d x ⟹ ln ( n + 1 ) − ln ( 2 ) < H n ≤ ln ( n ) ⟹ ln ( n + 1 ) − ln ( 2 ) < 1 0 0 ≤ ln ( n ) ⟹ n ≈ ⌊ e 1 0 0 ⌋
Can anyone explain the answer to me. If I neglect the fact thats this is a circel (which should not matter i presume) and see the problem as a 100 meter long track having a finish-flag at the end. It seems to me that every time i take a single 1-meter step someone plants the flag 100 meters farther away; adding more meters to the track than i made up for. And this happens 'every single 1-meter step i take!' The steps needed to finish a track that forever grows more than i travel could not logically converge to anything other than infinity to me. Please enlighten me !!!
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That's because the flag isn't put 100 meters farther from you, but from your starting point, which is also being put farther from you with each step.
After your first step, you've covered 1 % of the track. The track then "scales up", which essentially means the road in front of you lengthens, but so does the road behind you. So for your second step you'll still be positioned at the 1 % mark on the track. Continuing in this fashion, your second step takes you 2 1 % forward on the track, your third - 3 1 % and so on. Since the harmonic series diverges, this should convince you you will indeed at some point reach the 1 0 0 % mark, and hence complete the race.
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Aha. That did it. I was even wondering why they didn't just write 'add' instead of that long-winded scaling formula. I thank you in slight embarrasment sir.
In a very similar fashion of other solution posted, I assumed that goal was to cover a 2π path but in our mission our steps are reduced following the sequence 1/2 ; 1/3 ; 1/4 and so on. Since our initial step is 2π/100 to achieve our goal we must perform N steps so
2π/100 ( 1+1/2+ 1/3+ …. 1/N) = 2π; simplifying ( 1+1/2+ 1/3+ …. 1/N) = 100
So N will be the term of the Harmonic Series which balance the above equation. It is well known since Euler for large number the Harmonic Series can be approximated by Ln (N) plus γ the Euler Constant. Since the answer is a multiple choice the closest is indeed e^100
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Relevant wiki: Harmonic Series
Let us note that, after the first step, before the circle's circumference is doubled, we have completed 1/100 of the required distance. After the radius is doubled, we note that this percentage does NOT change! Indeed, the new distance is then 200 units, but the already completed distance has been scaled accordingly as well and is of 2 units! Therefore, their ratio stays invariant! Of course, for any subsequent steps, this quantity will remain invariant. In other words, after the first step, we have completed 1/100th of the distance, irrespective of how the circle is scaled!
Now, the second step contributes to half what the first step was worth, that is: 1/200, or (1/2)/100. Again, by the argument above, this is scaling invariant. And so, after the second step, we will have completed: 1/100 + (1/2)/100 of the task.
It is now easy to see where this is going and in general, after the nth step, we will have done: 1 0 0 1 + 1 0 0 1 / 2 + 1 0 0 1 / 3 + ⋯ + 1 0 0 1 / n ≡ 1 0 0 H n where H n is the familiar harmonic series: H n = 1 ∑ n k 1
The question is then, what is the smallest n such that 1 0 0 H n ≥ 1 or H n ≥ 1 0 0 ?
One quick way to go about this is using the fact that H n ≈ lo g ( n ) and therefore n ≈ e 1 0 0 .
A more formal approach is to note the inequality : lo g ( n + 1 ) < H n ≤ 1 + lo g ( n ) Picking H n ≈ 1 0 0 , we find that n ∈ ( e 9 9 , e 1 0 0 − 1 ] and again we see that n ≈ e 1 0 0 is the best approximation.