Step on it

Calculus Level 5

Suppose a particle, starting at the origin of a standard x y xy -grid, moves in a step-like manner, first going in a straight line right, (i.e., in the positive x x -direction), then up, (i.e., in the positive y y -direction), then right, up, and so on ad infinitum, such that the n n th move has length d n = ( ln ⁡ ( 2 ) ) n − 1 ( n − 1 ) ! . d_{n} = \dfrac{(\ln(2))^{n-1}}{(n - 1)!}.

If the (magnitude of the) displacement between the starting and finishing points of the particle is a b , \dfrac{\sqrt{a}}{b}, where a a and b b are positive integers with a a square-free, then find a + b . a + b.


The answer is 38.

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Brian Charlesworth by Brian Charlesworth , 55, Canada

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1 solution

The distance the particle travels in the positive x x -direction is

∑ k = 0 ∞ ( ln ⁡ ( 2 ) ) 2 k ( 2 k ) ! = cosh ⁡ ( ln ⁡ ( 2 ) ) = 2 + 1 2 2 = 5 4 . \displaystyle\sum_{k=0}^{\infty} \dfrac{(\ln(2))^{2k}}{(2k)!} = \cosh(\ln(2)) = \dfrac{2 + \frac{1}{2}}{2} = \dfrac{5}{4}.

The distance the particle travels in the positive y y -direction is

∑ k = 0 ∞ ( ln ⁡ ( 2 ) ) 2 k + 1 ( 2 k + 1 ) ! = sinh ⁡ ( ln ⁡ ( 2 ) ) = 2 − 1 2 2 = 3 4 . \displaystyle\sum_{k=0}^{\infty} \dfrac{(\ln(2))^{2k + 1}}{(2k + 1)!} = \sinh(\ln(2)) = \dfrac{2 - \frac{1}{2}}{2} = \dfrac{3}{4}.

Thus the magnitude of the distance is ( 5 4 ) 2 + ( 3 4 ) 2 = 34 4 , \sqrt{\left(\dfrac{5}{4}\right)^{2} + \left(\dfrac{3}{4}\right)^{2}} = \dfrac{\sqrt{34}}{4}, and so a + b = 34 + 4 = 38 . a + b = 34 + 4 = \boxed{38}.

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we can also solve simultaneous equations for x and y where X + Y = e l n 2 X + Y = e^{ln2} and X − Y = e − l n 2 X - Y = e^{-ln2}

Abhinav Raichur - 6 years, 1 month ago

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Oh, right. Nice short-cut. :)

Brian Charlesworth - 6 years, 1 month ago

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sir I have observed that majority of your problems have the mention of hyperbolic functions in some way ....... Is there some special affinity you have towards hyperbolicS?? { just askin :) }

Abhinav Raichur - 6 years, 1 month ago

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@Abhinav Raichur – Well, I've posted about 140 problems, and at most a dozen or so have made use of hyperbolic functions, so I guess you've just seen more of those ones than the rest I've posted. That said, they are beautiful functions and thus deserve some attention every once in a while. :)

Brian Charlesworth - 6 years, 1 month ago

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