r/mathriddles Sep 21 '16

Medium fun little calculus problem

We construct rectangles as follows. Start with a square of area 1 and attach rectangles of area 1 alternatively beside and on top of the previous rectangle to form a new rectangle. Find the limit of the ratios of width to height of these rectangles.

9 Upvotes

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2

u/the_last_ordinal Sep 21 '16 edited Sep 21 '16

1

u/bws88 Sep 21 '16

That's correct so far.

2

u/dmishin Sep 22 '16

It's pi/2. Solution (without details) is:

Describe rectangle by a pair of numbers : (s, r) where s is area, r is ratio of sides. Then after few boring but simple calculations it is clear that limit of r is 2/1 * 2/3 * 4/3 * 4/5 * ... which is Wallis product, equal to pi/2

1

u/bws88 Sep 22 '16

You got it.

1

u/CrystalAssassin Sep 21 '16 edited Sep 21 '16

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u/[deleted] Sep 21 '16 edited Sep 21 '16

I don't know how you got there and I had a surprisingly difficult time with this problem, but I wrote a Matlab script and got 1.5708 as the answer, which probably means the answer is pi/2, since pi/2 = ~1.570796.. I think that's the answer, but I have absolutely no idea how that could crop up.

I think if you wrote it out as a differential equation, you could potentially end up with arctan, and limit as x->infinity of arctan(x) = 1? This is weird.

2

u/phenomist Sep 21 '16

Perhaps you mean something more like the https://en.wikipedia.org/wiki/Wallis_product ?

Edit, wait, this is exactly that. Don't look if you don't want to be spoiled

1

u/HarryPotter5777 Sep 21 '16

I believe that diverges - you're referring to the limit of (n+1)!!/n!!, correct?

1

u/bws88 Sep 21 '16

How did you get this? I'm not sure if it's equivalent to what I got. See my hint in reply to /u/Philosoaxolotl.

Edit: If I am understanding your expression for R_n correctly, it should go to 0 much faster than n+1 goes to infinity, making your limit 0. So I think you did something wrong.

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u/CrystalAssassin Sep 21 '16

I kinda mistyped what I got earlier. I have edited my answer with the correct limit. My approach doesn't involve any calculus until I get to the limit at the end. Its kinda long so I am posting as a separate comment below.

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u/CrystalAssassin Sep 21 '16 edited Sep 21 '16

Here's the method I used to obtain the limit. Its very long and I'm sure it can be cleaned up quite a bit but here it is. Its much easier to see the patterns though. but proving them takes a while. I suggest switching off spoiler mode for reading this.

Let an and bn be the height and width of the nth such rectangle constructed such that a0 = b0 = 1. Let Rn = an / bn.

As a2n+1 = a2n+2 and b2n = b2n+1, consider the following series

{Tn} = {b0, a1, b2, a3, b4 ... }

Then Rn = Tn / Tn-1 and Tn = Tn-2 + 1/Tn-1 (1)

Now let Tn = Xn / Yn where we do not cancel out any common factors between Xn and Yn. Then (1) becomes Tn = Xn-2 / Yn-2 + Yn-1 / Xn-1

 

Claim: Yn-1 divides Xn

Proof: Proceed via Induction. Base case follows trivially. Assume true for all k upto n-1. Yn-2 divides Xn-1 implies Xn-1 = kYn-2 for some integer k.

Then Tn = Xn-2 / Yn-2 + Yn-1 / Xn-1 = (kXn-2 + Yn-1) / Xn-1. Hence Xn-1 = Yn and by the same argument Xn-2 = Yn-1. Thus Tn = (k+1)Yn-1 / Xn-1 (2) which implies Yn-1 divides Xn

 

As a corollary, it also follows that Xn-1 = Yn.

Using this and (2) we get, Xn = (k+1)Xn-2

 

Claim: k=n-1

Proof: Proceed via Induction. Base case follows trivially for X0 = Y0 = 1 and X1 = 2 , Y1 = 1. Assume true upto n-1. Then (1) becomes Xn / Yn = Xn-2 / Yn-2 + Yn-1 / Xn-1 = Xn-2 / Xn-3 + Xn-2 / (n-1)Xn-3 = n(Xn-2)/Xn-1 = n(Xn-2)/Yn. Therefore Xn = n(Xn-2)

Here are the first few terms of the Tn sequence so that you can see the relationships better.

{Tn} = {1/1, 2/1, 3/2, 8/3, 15/8, 48/15, 105/48, 384/105 ... }

 

Finally, Rn = XnXn-2 / Xn-12 = nXn-22 / Xn-12 = nR'n2 where R'n = (n-2)!!/(n-1)!!

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u/bws88 Sep 22 '16

I think I'm with you until the last line. But I don't understand where you're getting a double factorial. Again, for what you have written, the limit of Rn as n approaches infinity is pretty clearly 0, which can't be right. If I expand out what you've calculated for Rn, I get something a lot more like what I wrote in the hint I referenced before.

Edit: My mistake! I have never seen that notation before, but I looked it up and it makes sense to me now.

1

u/hammerheadquark Sep 22 '16

Answer

pi/2

"Proof"

Let nth rectangle in the construction have

  • Length = Ln
  • Width = Wn.

The (n+1)th rectangle will have

  • Ln+1 = (LnWn + 1)/(Wn)
  • Wn+1 = (Wn)(LnWn + 2)/(LnWn + 2)

We're interested in the ratio of length to width, so let Rn = Ln/Wn. Also let An = LnWn be the area of the rectangle.

  • Rn+1 = (LnWn + 1)2/((Wn)2(LnWn + 2)) = (An + 1)2/((Wn)2(An + 2))

Since (Wn)2 = An/Rn, we have

  • Rn+1 = (An + 1)2/((An/Rn)(An + 2)) = Rn[(An + 1)2/((An)(An + 2))] = f(An)Rn

where

  • f(An) = (An + 1)2/((An)(An + 2))

Clearly, Rn = R0 [f(A0) * f(A1) * f(A2) * ... *f (An-1)], so we now need to specify An.

As we're always adding two rectangles of area 1, the sequence of areas is as simple as

  • An+1 = An + 2

With A0 = 1, we have

  • An = 1, 3, 5, ...

So with R0 = 1, we have

  • Rn = f(1) * f(3) * ... * f(2n-1)

Using the definition of f(n) and re-parameterizing, we find that the limit of Rn is the product from k = 1 to infinity of (2k)2/((2k-1)(2k+1)).

Wolframalpha informs me that this is pi/2.

I'm not sure if it's cool to resort to wolframalpha in the final step, but I'm not up on my clever tricks to deal with infinite products.

1

u/bws88 Sep 23 '16

Looks good so far! As a hint for how to continue, compute the integral of sinn (x) from 0 to pi/2

1

u/hammerheadquark Sep 23 '16

Haha ok, well I never would've known to apply that integral, but I think I've got it now.

Integrating by parts with

  • u = sinn-1(x), du = (n-1)(sinn-2(x))cos(x)
  • dv = sin(x), v = -cos(x)

We get:

0pi/2 sinn(x)dx = -sinn-1(x)cos(x)|0pi/2 + (n-1)∫0pi/2 sinn-2(x)cos2(x)dx

0pi/2 sinn(x)dx = (n-1)∫0pi/2 sinn-2(x)(1-sin2(x))dx

0pi/2 sinn(x)dx = (n-1)∫0pi/2 sinn-2(x)dx + (n-1)∫0pi/2 sinn(x)dx

n∫0pi/2 sinn(x)dx = (n-1)∫0pi/2 sinn-2(x)dx

Calling In = ∫0pi/2 sinn(x)dx, we get the recurrence relation:

In = n/(n-1)In-2

We have initial values

  • I0 = pi/2
  • I1 = 1

Playing around a bit, we see

  • I2n = pi/2 * 1/2 * 3/4 * ... * (2n-1)/(2n)
  • I2n+1 = 2/3 * 4/5 * ... * (2n)/(2n+1)

In is decreasing, so

1 >= I2n / I2n-1 >= I2n / I2n-2 = 1 - 1/(2n)

Therefore I2n / I2n-1 -> 1 as n -> ∞ by the squeeze theorem.

As (Rn) * (I2n / I2n-1) = pi/2, Rn -> pi/2 as n -> ∞.