A Controlled Spiral
Suppose a particle moves in a right-angled left spiral on an x y -grid. That is, it moves a distance D 1 ( x ) in a straight line, stops, makes a right-angled turn to it's "left", travels a distance D 2 ( x ) in a straight line, stops, makes a right angled turn to its "left", travels a distance D 3 ( x ) in a straight line and continues in this fashion forever.
If D n ( x ) = ( n − 1 ) ! x n − 1 for n ≥ 1 , and if x = 2 0 1 5 , then find the magnitude of the straight line distance between the particle's starting and finishing points.
The answer is 1.00.
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2 solutions
Moderator note:
Very subtle function of D n ( x ) . Beautiful question. And as always, elegant solution Brian.
Great Question!
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Thanks, Julian. It's a beautiful and somewhat surprising result. :)
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I totally agree, this is a great question!
I would suggest that you post a second part to this question, in which you ask for the floor of the natural log of the total distance covered by the particle in its entire journey. (It is a neat answer)
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@Pranshu Gaba – Thanks! Yes, I was thinking of posting a couple of follow-up questions, with your suggestion being one of them. :)
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@Brian Charlesworth – Arashiro and I are going to post a collab set of problems like these too, just a tad more complicated.
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@Jake Lai – Yes, I know. Trevor has asked me to create some problems for the set, which is why I created this problem and another one titled "Right up and away". I'll be creating some more as well. :)
Awesome question
Beautiful Question !
However , I formed Expression of final co-ordintes , But I don't know why I foolishly choose
X
i
=
(
e
(
x
)
+
e
(
−
x
)
)
/
2
... Hence got answer as infinity ... And I miss chance of solving this Beutifull problem , Even after reaching to last step ! I'am dissappointed due to my such careless mistake !
I found out the final coordinates as ( 2 e i x + e − i x , 2 i e i x − e − i x ) , only to realise what it meant. You can imagine the look on my face when realization struck :). Anyways, I am having a hard time digesting the answer. Could anyone try to justify? Great Problem.
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The odd-numbered moves of the particle are in the x -direction, alternating between being in the positive and negative directions. Thus the distance traveled in the x -direction is
1 − 2 ! x 2 + 4 ! x 4 − 6 ! x 6 + . . . . ,
which we recognize as being the series representation for cos ( x ) for any real x .
The even-numbered moves of the particle are in the y -direction, alternating between being in the positive and negative directions. Thus the distance traveled in the y -direction is
x − 3 ! x 3 + 5 ! x 5 − 7 ! x 7 + . . . . ,
which we recognize as being the series representation for sin ( x ) for any real x .
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Yes, actually that is what I was saying. I found out the final point as ( 2 e i x + e − i x , 2 i e i x − e − i x ) which IS nothing but ( cos ( x ) , sin ( x ) ) (the i is iota). Only then did i realize that it was indeed the series representation for cos ( x ) and sin ( x ) (which is the realization I mentioned) , calculating the answer to be 1. What I meant was that I couldn't believe that the answer could be 1.
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@Aakarshit Uppal – Oh. o.k., sorry, I see what you meant now. Yes, I agree, it is a very surprising result that for any real X the solution is 1 . I had to check that
lim x → ∞ x ! 2 0 1 5 x = 0
to convince myself that the "spiral" would converge to some point, and then trust the math that this point was a distance 1 from the starting point.
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@Brian Charlesworth – Oh, it's ok sir. Yes, i also tried to visualize it that way. At some value of x, x ! 2 0 1 5 x will reach its maximum value, from where the spiral will start "going back" to the starting point as x increases, and eventually it would converge to a point at unit distance from the starting point.
I found out the distance to be n → ∞ lim ( D n − D n − 2 ) 2 + ( D n − 1 ) 2 and got the answer as 0 . I generalized this by trying to find the distance between starting and finishing point by putting values for n as n=1,2,3,4,5 . Please correct me
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Since lim n → ∞ D n = 0 it would be expected that your expression would return an answer of 0 . However, I think that you may have misinterpreted the problem, since what is being asked for here is the distance between the starting point and the finishing point after making the infinite series of moves of lengths D n as described. Perhaps my response to Aakarshit Uppal's comment will help clarify things. I'm sorry if the wording of the problem was not clear enough. :(
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Oops....there was a mistake in my calculation when I took a look at my solution... Actually, the distance should be n → ∞ lim ( D n − D n − 2 + D n − 4 − . . . . . ) 2 + ( D n − 1 − D n − 3 + D n − 5 − . . . . ) 2 And there is no flaw in the wording of the problem, there was a flaw in my solution...
Notice that:
x-displacement : D 1 − D 3 + D 5 − D 7 = ∑ n = 0 ∞ ( − 1 ) n D 2 n + 1 = ∑ n = 0 ∞ ( 2 n ) ! ( − 1 ) n x 2 n = c o s x
y-displacement : D 2 − D 4 + D 6 − D 8 = ∑ n = 0 ∞ ( − 1 ) n D 2 n + 2 = ∑ n = 0 ∞ ( 2 n + 1 ) ! ( − 1 ) n x 2 n + 1 = s i n x
thus, by pythagorean theorem, total displacement will be c o s 2 x + s i n 2 x = 1
Without loss of generality, let the starting position of the particle be the origin, and let its first move be in the positive x -direction. Then the distance traveled in the positive x -direction over the course of the infinite series of movements is
k = 0 ∑ ∞ ( 2 k ) ! ( − 1 ) k X 2 k = cos ( X ) ,
and the distance traveled in the positive y -direction will be
k = 0 ∑ ∞ ( 2 k + 1 ) ! ( − 1 ) k X 2 k + 1 = sin ( X ) .
Thus for any real X , the distance between the starting and finishing points is
sin 2 ( X ) + cos 2 ( X ) = 1 .