a surprising sum of arctangents -...

A Surprising Sum of Arctangents

Jared Ruiz

Youngstown State University

June 30, 2013

Jared Ruiz (Youngstown State University) A Surprising Sum of Arctangents June 30, 2013 1 / 1

Acknowledgements

The following researchers and organizations need to be thanked:

Dr. Jacek Fabrykowski

John Hoffman (Dr. J. Douglas Faires and Dr. Barbara T. FairesEndowment)

W. Ryan Livingston (Dr. J. Douglas Faires and Dr. Barbara T. FairesEndowment)

I was personally funded by the McNair Scholars Program inassociation with the University of Akron.

Support for all was given by the Center for Undergraduate Research,Youngstown State University.


There is a “well-known” trigonometric identity:

Identity

arctan 1 + arctan 2 + arctan 3 = π

This appears in [1] by Michael W. Ecker

Thus the question arises:Are there other formulas which can give us any multiple of π?


There is a “well-known” trigonometric identity:

Identity


This appears in [1] by Michael W. EckerThus the question arises:Are there other formulas which can give us any multiple of π?


Theorem

For every positive integer n, there exists s > 0 and distinct positiveintegers k1, k2, . . . , ks such that

s∑j=1

tj arctan kj = nπ

for suitable tj ∈ {−1, 1}.

Note

For all a ∈ R,arctan(−a) = − arctan a


Lemma 1

For a ∈ R, a 6= 0,

arctan a =

{− arctan(1a ) + π

2 ; a > 0

− arctan(1a )− π2 ; a < 0

Proof: Define the function f (a) = arctan a + arctan(1a ).

f ′(a) =1

1 + a2+

1

1 + 1a2

(−1

a2

)=

1

1 + a2− 1

1 + a2= 0

If a > 0, let a =√

3. Then we have

arctan√

3 + arctan1√3

=π

3+π

6=π

2

If a < 0, let a = −√

3. Then

arctan(−√

3)

+ arctan−1√

3= −

(arctan

√3 + arctan

1√3

)= −π

2

2


Lemma 2

Lemma (Incorrect)

For a, b ∈ R, ab 6= 1

arctan a + arctan b = arctan

(a + b

1− ab

)

We can easily check this with a numeric example:

arctan 2 + arctan 3 = 2.356194490

6= − .7853981635 = arctan(−1)

This is listed wrong in:

Calculus, 5th ed. by James Stewart (He does mentionarctan a + arctan b ∈ (−π

2 ,π2 ))

Calculus, 2nd ed. by J. Douglas Faires and Barbara T. Faires (Theysay |ab| 6= 1)


Lemma 2

Lemma (Incorrect)

For a, b ∈ R, ab 6= 1


(a + b

1− ab

)We can easily check this with a numeric example:

arctan 2 + arctan 3 = 2.356194490

6= − .7853981635 = arctan(−1)



2 ,π2 ))



Lemma 2

Lemma (Incorrect)

For a, b ∈ R, ab 6= 1


(a + b

1− ab


arctan 2 + arctan 3

= 2.356194490

6= − .7853981635 = arctan(−1)



2 ,π2 ))



Lemma 2

Lemma (Incorrect)

For a, b ∈ R, ab 6= 1


(a + b

1− ab


arctan 2 + arctan 3 = 2.356194490

6= − .7853981635 = arctan(−1)



2 ,π2 ))



Lemma 2

Lemma (Incorrect)

For a, b ∈ R, ab 6= 1


(a + b

1− ab


arctan 2 + arctan 3 = 2.356194490

6= − .7853981635 =

arctan(−1)



2 ,π2 ))



Lemma 2

Lemma (Incorrect)

For a, b ∈ R, ab 6= 1


(a + b

1− ab


arctan 2 + arctan 3 = 2.356194490

6=

− .7853981635 = arctan(−1)



2 ,π2 ))



Lemma 2

Lemma (Incorrect)

For a, b ∈ R, ab 6= 1


(a + b

1− ab


arctan 2 + arctan 3 = 2.356194490 6= − .7853981635 = arctan(−1)



2 ,π2 ))



Lemma (Incorrect)

For a, b ∈ R, ab 6= 1


(a + b

1− ab


arctan 2 + arctan 3 = 2.356194490 6= − .7853981635 = arctan(−1)



2 ,π2 ))



Lemma 2

For a, b ∈ R, ab 6= 1

arctan a + arctan b =

arctan a+b

1−ab ; ab < 1

arctan a+b1−ab + π; ab > 1, a, b > 0

arctan a+b1−ab − π; ab > 1, a, b < 0

Proof: If ab ∈ [−1, 1] then arctan a + arctan b ∈(−π

2 ,π2

).

Since

tan(arctan a + arctan b) =tan(arctan a) + tan(arctan b)

1− tan(arctan a) tan(arctan b)

arctan a + arctan b = arctana + b

1− ab


Lemma 2

arctan a + arctan b ={

arctan a+b1−ab ; ab < 1

Assume ab < 1.If 0 < a < 1, and 0 < b < 1 then we are done.WLOG, let a > 1. Then 0 < 1

a < 1, and 0 < b < 1.

arctan a + arctan b = − arctan

(1

a

)+π

2+ arctan b

= arctan

(−1 + ab

a + b

)+π

2

Now −1+aba+b < 0, since 0 < ab < 1. So by Lemma 1

arctan

(−1 + ab

a + b

)+π

2= − arctan

(a + b

−1 + ab

)− π

2+π

2

= arctana + b

1− ab


Lemma 2



Assume ab < 1.

If 0 < a < 1, and 0 < b < 1 then we are done.WLOG, let a > 1. Then 0 < 1

a < 1, and 0 < b < 1.


(1

a

)+π

2+ arctan b

= arctan

(−1 + ab

a + b

)+π

2


arctan

(−1 + ab

a + b

)+π

2= − arctan

(a + b

−1 + ab

)− π

2+π

2

= arctana + b

1− ab


Lemma 2



Assume ab < 1.If 0 < a < 1, and 0 < b < 1 then we are done.

WLOG, let a > 1. Then 0 < 1a < 1, and 0 < b < 1.


(1

a

)+π

2+ arctan b

= arctan

(−1 + ab

a + b

)+π

2


arctan

(−1 + ab

a + b

)+π

2= − arctan

(a + b

−1 + ab

)− π

2+π

2

= arctana + b

1− ab


Lemma 2




a < 1, and 0 < b < 1.


(1

a

)+π

2+ arctan b

= arctan

(−1 + ab

a + b

)+π

2


arctan

(−1 + ab

a + b

)+π

2= − arctan

(a + b

−1 + ab

)− π

2+π

2

= arctana + b

1− ab


Lemma 2




a < 1, and 0 < b < 1.


(1

a

)+π

2+ arctan b

= arctan

(−1 + ab

a + b

)+π

2

Now −1+aba+b < 0, since 0 < ab < 1.

So by Lemma 1

arctan

(−1 + ab

a + b

)+π

2= − arctan

(a + b

−1 + ab

)− π

2+π

2

= arctana + b

1− ab


Lemma 2




a < 1, and 0 < b < 1.


(1

a

)+π

2+ arctan b

= arctan

(−1 + ab

a + b

)+π

2


arctan

(−1 + ab

a + b

)+π

2= − arctan

(a + b

−1 + ab

)− π

2+π

2

= arctana + b

1− ab


Lemma 2


{arctan a+b

1−ab + π; ab > 1, a, b > 0


Assume ab > 1.

Let a, b > 0. Then 0 < 1a ·

1b < 1

arctan a + arctan b = − arctan1

a− arctan

1

b+ π

= arctana + b

1− ab+ π

Now let a, b < 0. Again we have 0 < 1a ·

1b < 1


a− arctan

1

b− π

= arctana + b

1− ab− π 2


Lemma 2


{arctan a+b

1−ab + π; ab > 1, a, b > 0


Assume ab > 1.Let a, b > 0. Then 0 < 1

a ·1b < 1


a− arctan

1

b+ π

= arctana + b

1− ab+ π


1b < 1


a− arctan

1

b− π

= arctana + b

1− ab− π 2


Lemma 2


{arctan a+b

1−ab + π; ab > 1, a, b > 0



a ·1b < 1


a− arctan

1

b+ π

= arctana + b

1− ab+ π


1b < 1


a− arctan

1

b− π

= arctana + b

1− ab− π

2


Lemma 2


{arctan a+b

1−ab + π; ab > 1, a, b > 0



a ·1b < 1


a− arctan

1

b+ π

= arctana + b

1− ab+ π


1b < 1


a− arctan

1

b− π

= arctana + b

1− ab− π 2


Corollary


Proof: By Lemma 2,

arctan 1 + arctan 2 + arctan 3 = (arctan 1 + arctan 2) + arctan 3

= arctan

(3

−1

)+ π + arctan 3

= − arctan 3 + π + arctan 3

= π

2

This is a new proof of this identity!


Lemma 3

For a ∈ R,arctan a+arctan(1−a)+arctan(2−a+a2)+arctan(3−3a+4a2−2a3+a4) =π.

Proof: For all a ∈ R, a(1− a) < 1.

arctan a + arctan(1− a) = arctan

(1

1− a + a2

)

Now for all a ∈ R,2− a + a2

1− a + a2> 1.

Then adding arctan(

11−a+a2

)+ arctan(2− a + a2) gives


Lemma 3

arctan a+arctan(1−a)+arctan(2−a+a2)+arctan(3−3a+4a2−2a3+a4) =π.

arctan

(1

1−a+a2+ 2− a + a2

1− 2−a+a2

1−a+a2

)+ π = arctan

1 + (2−a+a2)(1−a+a2)1−a+a2

(1−a+a2)−(2−a+a2)1−a+a2

+ π

= arctan

(3− 3a + 4a2 − 2a3 + a4

−1

)+ π

Now adding

arctan

(3− 3a + 4a2 − 2a3 + a4

−1

)+π+arctan(3−3a+4a2−2a3+a4) = π

2


Lemma 4

Let a ∈ R, a < 0. Then

|a| < 1− a < 2− a + a2 < 3− 3a + 4a2 − 2a3 + a4

Proof: Since a < 0,|a| < 1− a < 2− a + a2

For the last inequality,

2− a + a2 < 3− 3a + 4a2 − 2a3 + a4

0 < 1− 2a + 3a2 − 2a3 + a4

2


Recall

Lemma 3

For a ∈ R,arctan a+arctan(1−a)+arctan(2−a+a2)+arctan(3−3a+4a2−2a3+a4) =π.

Lemma 4

Let a ∈ R, a < 0. Then

|a| < 1− a < 2− a + a2 < 3− 3a + 4a2 − 2a3 + a4


Theorem

For all n > 0, there exists s > 0 and distinct positive integers k1, k2, . . . , kssuch that

s∑j=1

tj arctan kj = nπ


Proof: Let a0 < 0, and let k1 = |a0|. Then

k2 = 1− a0, k3 = 2− a0 + a20, and k4 = 3− 3a0 + 4a20 − 2a30 + a40

We have by Lemma 3

−

arctan

k1

+ arctan k2 + arctan k3 + arctan k4 = π (1)

Let a1 = −(k4 + 1), and let k5 = |a1|.

−

arctan

k5


(??) + (??) = 2π. This can be continued indefinitely. 2


Theorem


s∑j=1

tj arctan kj = nπ



k2 = 1− a0, k3 = 2− a0 + a20, and k4 = 3− 3a0 + 4a20 − 2a30 + a40

We have by Lemma 3

−

arctan a0

k1


Let a1 = −(k4 + 1), and let k5 = |a1|.

−

arctan

k5




Theorem


s∑j=1

tj arctan kj = nπ



k2 = 1− a0, k3 = 2− a0 + a20, and k4 = 3− 3a0 + 4a20 − 2a30 + a40

We have by Lemma 3

− arctan k1 + arctan k2 + arctan k3 + arctan k4 = π (1)

Let a1 = −(k4 + 1), and let k5 = |a1|.

−

arctan

k5




Theorem


s∑j=1

tj arctan kj = nπ



k2 = 1− a0, k3 = 2− a0 + a20, and k4 = 3− 3a0 + 4a20 − 2a30 + a40

We have by Lemma 3


Let a1 = −(k4 + 1), and let k5 = |a1|.

−

arctan a1

k5




Theorem


s∑j=1

tj arctan kj = nπ



k2 = 1− a0, k3 = 2− a0 + a20, and k4 = 3− 3a0 + 4a20 − 2a30 + a40

We have by Lemma 3


Let a1 = −(k4 + 1), and let k5 = |a1|.




Theorem


s∑j=1

tj arctan kj = nπ



k2 = 1− a0, k3 = 2− a0 + a20, and k4 = 3− 3a0 + 4a20 − 2a30 + a40

We have by Lemma 3


Let a1 = −(k4 + 1), and let k5 = |a1|.


(??) + (??) = 2π.

This can be continued indefinitely. 2


Theorem


s∑j=1

tj arctan kj = nπ



k2 = 1− a0, k3 = 2− a0 + a20, and k4 = 3− 3a0 + 4a20 − 2a30 + a40

We have by Lemma 3


Let a1 = −(k4 + 1), and let k5 = |a1|.


(??) + (??) = 2π. This can be continued indefinitely.

2


Theorem


s∑j=1

tj arctan kj = nπ



k2 = 1− a0, k3 = 2− a0 + a20, and k4 = 3− 3a0 + 4a20 − 2a30 + a40

We have by Lemma 3


Let a1 = −(k4 + 1), and let k5 = |a1|.




Example

Let a0 = −2, and k1 = 2.

Then

− arctan 2 + arctan 3 + arctan 8 + arctan 57 = π (3)

Now a1 = −58, and k5 = 58 (so t5 = −1). Then


Then (??) + (??) = 2π. Etc.


Example

Let a0 = −2, and k1 = 2. Then




Then (??) + (??) = 2π. Etc.


Example



Now a1 = −58, and k5 = 58 (so t5 = −1).

Then


Then (??) + (??) = 2π. Etc.


Example





Then (??) + (??) = 2π. Etc.


Example





Then (??) + (??) = 2π.

Etc.


Example





Then (??) + (??) = 2π. Etc.


References

1 Ecker, Michael W. An arctangent triangle: puzzling overarctan1+arctan2+arctan3=π. Mathematics and ComputerEducation, 2003.

2 Edwards, C. Henry and David E. Penney. Calculus, Ed. 6. PrenticeHall, 2002: p. 476.

3 Faires, J. Douglas and Barbara T. Faires. Calculus, Ed. 2. RandomHouse, 1988: p. 404.

4 Stewart, James. Calculus, Ed. 5. Brooks/Cole (A division ofThomson Learning), 2003: p. 485.


Where Does Lemma 3 Come From?

Let a, b ∈ Z, and a + b = 1.

Then arctan a + arctan b = arctana + b

1− abNow

arctana + b

1− ab+ arctan c = arctan

a + b + c − abc

1− ab − ac − bc

We want 1− ab − ac − bc to equal 1.So c = −ab = −a + a2. We add 2 to c so that

a + b

1− abc > 1

Then we add arctana + b + c − abc

1− ab − ac − bc+ arctan d and so on.



Let a, b ∈ Z, and a + b = 1. Then arctan a + arctan b = arctana + b

1− ab

Now

arctana + b


a + b + c − abc



a + b

1− abc > 1






1− abNow

arctana + b


a + b + c − abc



a + b

1− abc > 1






1− abNow

arctana + b


a + b + c − abc


We want 1− ab − ac − bc to equal 1.

So c = −ab = −a + a2. We add 2 to c so that

a + b

1− abc > 1






1− abNow

arctana + b


a + b + c − abc



a + b

1− abc > 1




a surprising sum of arctangents -...

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