Leonard Euler has made many contributions to math, and one of them is a formula for arctan. To see and investigate Euler's formula, I used GeoGebra's

*An Equation of Euler*by timteachesmath (Link). There are 5 sets of examples of the formula in each of the 3 options in the GeoGebra exploration. Below is my explanation of the equation in relation to the GeoGebra activity.

Option 1

First, at T=0, there is a square with a diagonal line (Figure 1). As I move T along the slider, the green square becomes longer to create a rectangle with the original square. I also get a blue square and blue rectangle that has been rotated. The reason the blue square/rectangle gets rotated is so the two corners of the blue rectangle remain touching the corners of the green rectangle.

Figure 1: Option 1 - T=0 |

When T=1, the formula appears (Figure 2). Essentially, it's saying the angles b and c will add up to equal angle d. Since tangent of an angle = opposite/adjacent, the first arctan corresponds to angle b (green line to the red diagonal) in the green rectangle, since tan (b) = 1/3. The second arctan corresponds to angle c (from blue line to red line) in the blue rectangle, since tan (c) = 1/2. The final arctan corresponds to the angle d (from green line to blue line) in the green square, since tan (d) =1/1.

Figure 2: Option 1 - T=1 |

Option 2

Again, Option 2 starts with the same square as Option 1 when T=0. When T=1 though, the first square gets divided into fourths, there are 3 large green squares and 1 large rotated blue square. Again, the two corners of the blue square remain on the two corners of the green rectangle. Now, the equation asserts that angle b of the green rectangle and angle c of the blue square add up to angle d of the green rectangle comprised of the 2 smaller squares (Figure 3). This is because tan(b)=1/3 , tan (c ) = 1/1, and tan (d) = 2/1.

Figure 3: Option 2 - T=1 |

Option 3

Here, Option 3 starts with 2 green squares stacked vertically at T=0 (Figure 4).

Figure 4: Option 3 - T=0 |

At T=1, the 2 green squares expand to 6 (3 columns and 2 rows). There are 5 blue rotated squares, and again the 2 corners of the bigger blue rectangle remain on the two corners of the bigger green rectangle (Figure 5). Because we have two rows of squares the numerator of the first arctan is 2. In other words, tan (b) = 2/3 (the big green rectangle), tan (c ) = 1/5 (the blue rectangle), and tan (d) = 1/1 (the green square).

Figure 5: Option 3 - T=1 |

Wrap-Up

The key for this equation is keeping the two opposite corners of the blue square/rectangle inline with the expanding green square/rectangles. This makes the blue squares rotate and expand with the addition of the green squares. This creates a blue diagonal that goes through the green squares and rectangles. Thus, the red and blue lines divide the green rectangle in different ways to make the equation possible.

There were two other interesting things I noticed. The first arctan term was the answer for the next formula in Option 1, and all the numerators and denominators are numbers in the Fibonacci Sequence in all the options. Also, the Fibonacci Sequence was in reverse order in Option 1. As the blue rectangle rotates and expands with the green rectangle, a relationship between the green squares and rectangles and the blue squares and rectangles is maintained. So, the equation only works when the original square (T=0) can divide the green rectangle equally and the blue rectangle can be divided equally by a different size square.

Update (11/9/2015): Eureka! The rotated blue rectangle creates a parallelogram with the green rectangle. The red diagonal splits the parallelogram in half, creating two congruent triangles. We can also see the blue lines that create the sides of the parallelogram as two congruent transversals that cut the green rectangle. So, the properties of parallel lines and parallelograms is why Euler's arctan formula works.

Screen shots are of timteachesmath's

*An Equation of Euler*on GeoGebra. I added the angles b, c, and d. I do not have any affiliation with GeoGebra or timteachesmath, and neither has endorsed my work. I am a college student.

Some nice exposition here, and I love Tim Cieplowski's stuff. He's a prof at Bowling Green. 5Cs: ++.

ReplyDeleteLove that you came back to it with your Eureka.