# Limit of sum of areas of infinite amount of triangles

I apologize for the possible incorrect use of math terms since English is not my native language and I'm not a mathematician, but this issue came to my mind about a month ago and I was unable to solve it, so I will appreciate any help.

Let length of a line segment $$L$$ be $$1$$. Also define variable $$r$$ (ratio) that can be any real number on the interval $$(0;1)$$.

Let us put the vertical line segment with length $$L$$ starting from the point $$(0;0)$$ on the orthogonal coordinate system; the other point of this line segment is $$(0;L)$$. Put the next line segment with the following rules:

1. Starting point should be located on the X axis, let us assume it as a point $$(X(n);0)$$, where $$X(n)>X(n-1)$$;
2. Lets treat the previous line segment like a vector, multiply it by $$r$$. The end point of this vector is the end point of the new line segment.

Here is the example which displays $$n = 6$$ triangles built with the $$L = 1$$ and $$r = 0.8$$.

Initially I tried to solve the following tasks:

1. Find the function $$f(L, r, n)$$ which will return the sum of the areas of $$n$$ triangles giving the length of the line segment is $$L$$ and a ratio is $$r$$;
2. Define the limit of the $$f(L, r, n)$$ with $$L = 1$$; $$r \to 1$$ and $$n \to \infty$$;
3. Assuming that these issues were solved before, what is the correct way to call this task and where I can find the information about it?

Here is what I have discovered so far.

First, let us find one of the angles of the $$n$$-th triangle. Define $$\beta(n)$$ as an angle between $$(n-1)$$-th and $$n$$-th line segment; $$\alpha(n)$$ as an angle between X axis and the $$n$$-th line segment. For the sake of simplicity let us use $$\alpha_N$$ and $$\beta_N$$ instead of a function form.

Since the first triangle is the right triangle, $$\sin(\alpha_1)$$ is defined by the known relations:

$$\sin(\alpha_1) = \frac {r * L}{L} = r$$

Using the law of sines, investigate the second triangle.

$$\frac {\sin(\alpha_2)}{L*r} = \frac{\sin(\pi - alpha(1))}{L};$$ $$\sin(\alpha_2) = \sin(\pi - \alpha_1) * r = \sin(\alpha_1) * r$$

Since there is no dependency from the right triangle in this formula, we can generalize the result: $$\sin(\alpha_n) = \sin(\alpha_{n-1})*r$$

or for the calculation simplicity sake: $$\alpha_n = \arcsin(r^n)$$

Knowing that the sum of the angles of the triangle is $$\pi$$, we get the following: $$\pi = \alpha_n + \beta_n + (\pi - \alpha_{n-1});$$ $$0 = \alpha_n + \beta_n - \alpha_{n-1};$$ $$\beta_n = \alpha_{n-1} - \alpha_n$$

Find the area of the $$n$$-th triangle with the following formula:

$$S(n) = \frac 1 2 * L * L* r * \sin(\beta_n) = \frac 1 2 * L^2 * r * \sin(\beta_n)$$

Such formula is acceptable for the calculations, but we can represent it in a different way. $$\sin(\beta_n) = \sin(\alpha_{n-1} - \alpha_n) = \sin\alpha_{n-1}*\cos\alpha_n - \sin\alpha_n*\cos\alpha_{n-1}=$$ $$= r^{n-1}*\sqrt{1-r^{2n}} - r^n*\sqrt{1-r^{2n-2}}$$

Since that is the solution for the first question, I had tried to solve the second, but with no avail.

$$Sum(L,r) = \frac 1 2 *L^2 \lim_{r \to 1, n \to \infty} (r * \sum_{n=1}^\infty \sin(\beta_n))$$

I also had tried to change the way of area calculation to the sum of the trapezoids, but it wasn't successful as well.

$$S_t(n) = \frac {L*r^{n-1} + L*r^n} 2 * \cos\alpha_n*(1-r)*L$$

I was unable to apply any known to me technique (such as L'Hôpital's rule, Taylor series investigation) to reach the solution, so I resorted to approximate solution.

I have managed to calculate the result of the function $$f$$ with the $$L = 1$$; $$r = 0.999999$$; $$n = 100000000$$: $$f(1,0.999999,10^8) = 0.7853973776669734$$ The length of the whole construct was approximately equal to $$100.6931$$, knowing that the side of the $$n$$-th triangle on the X axis is: $$B(n) = L*r*(\frac {\cos\alpha_n} r - \cos\alpha_{n-1}))$$

The result is more or less close to the $$\frac \pi 4 (0.7853981633974483)$$, which was surprising. I had tried to apply this knowledge (the infinite sum of $$\sin\beta_n$$ should approach to $$\frac \pi 2$$; knowledge that $$\int_0^{+\infty}\frac {dx}{(1+x)*\sqrt{x}} = \pi$$), but was unable to do so.

So here is the final composite question:

1. Are there any errors in my calculations?
2. Is there a way to solve this without resorting to approximations?
3. Is this sum really approaches to the $$\frac \pi 4$$?
4. How can I define the curve which is shaped by these triangles?
5. Assuming that these issues were solved before, what is the correct way to call this task and where I can find the information about it?

Every point on the limiting curve has a distance 1 to the $$x$$-axis along the tangent line at that point. Intuitively, this should be plausible.

More formally, let $$x=f(y)$$ be the equation for the curve. The equation for the tangent line at $$(f(y_0),y_0)$$ is $$x-f(y_0)=f'(y_0)(y-y_0).$$ This meets the $$x$$-axis at $$(-y_0f'(y_0)+f(y_0),0)$$.

The condition requires that the distance from $$(f(y_0),y_0)$$ to $$(-y_0f'(y_0)+f(y_0),0)$$ equals 1, i.e.: $$\sqrt{[f(y_0)-(-y_0f'(y_0)+f(y_0))]^2+(y_0-0)^2}=y_0\sqrt{f'(y_0)^2+1}=1$$ $$f'(y)=-\frac{\sqrt{1-y^2}}{y}\tag1,$$ where the minus sign comes from the fact that $$x$$ decreases as $$y$$ increases.

Your construction is actually a numerical approximation to the solution of this differential equation, where at every step, you move a distance $$1-r$$ in the direction given by Eq. (1) evaluated at the last value of $$y$$. To see this, let the coordinate of the $$n^\text{th}$$ point $$P_n$$ be $$(x_n,y_n)$$. The two points on the $$x$$-axis that is a distance 1 away from $$P_n$$ are $$(x_n\pm\sqrt{1-y_n^2},0)$$, as you can verify. Your construction takes the positive sign. The line from $$P_n$$ to $$(x_n+\sqrt{1-y_n^2},0)$$ has equation $$x-x_n=-\frac{\sqrt{1-y_n^2}}{y_n}(y-y_n)$$ which is exactly the one given by Eq. (1). (I don't yet have a formal proof of its convergence, but I believe it's possible.)

Now, Eq. (1) can be solved by an integration using the substitution $$u=\sqrt{1-y^2}$$. The result is $$f(y)=-\sqrt{1-y^2}+\frac{1}{2}\ln\frac{1+\sqrt{1-y^2}}{1-\sqrt{1-y^2}}.$$

The area can then be calculated by $$\int_0^1f(y)dy$$, which can be shown to be equal to $$\pi/4$$ (again by a substitution $$u=\sqrt{1-y^2}$$).

• Thank you for your answer. Could you please elaborate the first part of the solution (before integration) a bit thoroughly? I'm failing to move further after indicating the point on X axis. – Nick Semianiuk Aug 19 '19 at 18:01
• Sure, see the edited answer – Tipping Octopus Aug 21 '19 at 11:03

One can also express the area (as noted in the question) as a sum of trapezoids, obtained by dropping perpendicular lines from $$E$$, $$G$$, $$I$$, $$K$$, ... to the $$x$$-axis. The vertical bases of $$k$$-th trapezoid measure $$r^k$$ and $$r^{k+1}$$, while its height is $$(1-r)\sqrt{1-r^{2k}}$$. Hence total area $$S$$ can also be computed from: $$S=\sum_{k=1}^\infty{1\over2}(r^k+r^{k+1})(1-r)\sqrt{1-r^{2k}}= {1\over2}(1-r^2)\sum_{k=0}^\infty r^k\sqrt{1-r^{2k}}$$ (we can start the last sum with $$k=0$$ because the $$0$$-th term vanishes).

I tried to get a closed expression for this series with Mathematica, to no avail. We can however use Taylor's expansion $$\sqrt{1+x}=\sum_{n=0}^\infty\binom{1/2}{n}x^n$$ to write: $$S={1\over2}(1-r^2)\sum_{k=0}^\infty r^k\sum_{n=0}^\infty\binom{1/2}{n}(-1)^n r^{2nk} ={1\over2}(1-r^2)\sum_{n=0}^\infty\binom{1/2}{n}(-1)^n\sum_{k=0}^\infty r^{(2n+1)k}$$ and the last series can be summed (for $$r<1$$) to give: $$S={1\over2}(1-r^2)\sum_{n=0}^\infty\binom{1/2}{n}{(-1)^n\over 1-r^{2n+1}}.$$ We can now factor and simplify a term $$1-r$$ both in $$1-r^2$$ and in $$1-r^{2n+1}$$: $$S={1\over2}(1+r)\sum_{n=0}^\infty\binom{1/2}{n}{(-1)^n\over 1+r+r^2+\dots+r^{2n}}.$$ The advantage is that we can now carry out the limit $$r\to1$$ just substituting $$r=1$$ in this formula, to get: $$\lim_{r\to1}S=\sum_{n=0}^\infty\binom{1/2}{n}{(-1)^n\over {2n+1}}.$$ This is a simpler series and Mathematica evaluates it to $$\pi/4$$, thus confirming the given conjecture.

EDIT.

With the help of OEIS I found that the above series is Maclaurin expansion of $$x\sqrt{x^2+1} +{1\over2} \ln\big(x+\sqrt{x^2+1}\big)$$ (which is the arc length of Archimedes' spiral), computed for $$x=i$$ and divided by $$i$$.

• Thank you for your answer, it was easy to understand. It is quite interesting that the next to final part of the solution is similar with Leibniz formula . And it seems that OEIS has literally everything, I'm quite surprised that it has such rare sequences. – Nick Semianiuk Aug 19 '19 at 18:10