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
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Acknowledgements
The following researchers and organizations need to be thanked:
Dr. Jacek Fabrykowski
John Hoffman (Dr. J. Douglas Faires and Dr. Barbara T. Faires
Endowment)
W. Ryan Livingston (Dr. J. Douglas Faires and Dr. Barbara T. Faires
Endowment)
I was personally funded by the McNair Scholars Program in
association with the University of Akron.
Support for all was given by the Center for Undergraduate Research,
Youngstown State University.
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
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There is a “well-known” trigonometric identity:
Identity
arctan 1 + arctan 2 + arctan 3 = π
This appears in [1] by Michael W. Ecker
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
3/1
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?
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
3/1
Theorem
For every positive integer n, there exists s > 0 and distinct positive
integers k1 , k2 , . . . , ks such that
s
X
tj arctan kj = nπ
j=1
for suitable tj ∈ {−1, 1}.
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
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Theorem
For every positive integer n, there exists s > 0 and distinct positive
integers k1 , k2 , . . . , ks such that
s
X
tj arctan kj = nπ
j=1
for suitable tj ∈ {−1, 1}.
Note
For all a ∈ R,
arctan(−a) = − arctan a
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
4/1
Lemma 1
For a ∈ R, a 6= 0,
(
− arctan( 1a ) + π2 ; a > 0
arctan a =
− arctan( 1a ) − π2 ; a < 0
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
5/1
Lemma 1
For a ∈ R, a 6= 0,
(
− arctan( 1a ) + π2 ; a > 0
arctan a =
− arctan( 1a ) − π2 ; a < 0
Proof: Define the function f (a) = arctan a + arctan( 1a ).
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
5/1
Lemma 1
For a ∈ R, a 6= 0,
(
− arctan( 1a ) + π2 ; a > 0
arctan a =
− arctan( 1a ) − π2 ; a < 0
Proof: Define the function f (a) = arctan a + arctan( 1a ).
−1
1
1
1
1
0
f (a) =
+
=
−
=0
1
2
2
2
1+a
1+a
1 + a2
1 + a2 a
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
5/1
Lemma 1
For a ∈ R, a 6= 0,
(
− arctan( 1a ) + π2 ; a > 0
arctan a =
− arctan( 1a ) − π2 ; a < 0
Proof: Define the function f (a) = arctan a + arctan( 1a ).
−1
1
1
1
1
0
f (a) =
+
=
−
=0
1
2
2
2
1+a
1+a
1 + a2
1 + a2 a
√
If a > 0, let a = 3. Then we have
√
1
π π
π
arctan 3 + arctan √ = + =
3
6
2
3
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
5/1
Lemma 1
For a ∈ R, a 6= 0,
(
− arctan( 1a ) + π2 ; a > 0
arctan a =
− arctan( 1a ) − π2 ; a < 0
Proof: Define the function f (a) = arctan a + arctan( 1a ).
−1
1
1
1
1
0
f (a) =
+
=
−
=0
1
2
2
2
1+a
1+a
1 + a2
1 + a2 a
√
If a > 0, let a = 3. Then we have
√
1
π π
π
arctan 3 + arctan √ = + =
3
6
2
3
√
If a < 0, let a = − 3. Then
√ √
−1
1
π
arctan − 3 + arctan √ = − arctan 3 + arctan √
=−
2
3
3
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
5/1
Lemma 1
For a ∈ R, a 6= 0,
(
− arctan( 1a ) + π2 ; a > 0
arctan a =
− arctan( 1a ) − π2 ; a < 0
Proof: Define the function f (a) = arctan a + arctan( 1a ).
−1
1
1
1
1
0
f (a) =
+
=
−
=0
1
2
2
2
1+a
1+a
1 + a2
1 + a2 a
√
If a > 0, let a = 3. Then we have
√
1
π π
π
arctan 3 + arctan √ = + =
3
6
2
3
√
If a < 0, let a = − 3. Then
√ √
−1
1
π
arctan − 3 + arctan √ = − arctan 3 + arctan √
=−
2
3
3
2
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
5/1
Lemma 2
For a, b ∈ R, ab 6= 1
arctan a + arctan b = arctan
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
a+b
1 − ab
June 30, 2013
6/1
Lemma 2
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:
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
6/1
Lemma 2
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
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
6/1
Lemma 2
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
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
6/1
Lemma 2
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
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
arctan(−1)
June 30, 2013
6/1
Lemma 2
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
Jared Ruiz (Youngstown State University)
− .7853981635 = arctan(−1)
A Surprising Sum of Arctangents
June 30, 2013
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Lemma 2
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)
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
6/1
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)
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
6/1
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 mention
arctan a + arctan b ∈ (− π2 , π2 ))
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
6/1
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 mention
arctan a + arctan b ∈ (− π2 , π2 ))
Calculus, 2nd ed. by J. Douglas Faires and Barbara T. Faires (They
say |ab| =
6 1)
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
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Lemma 2
For a, b ∈ R, ab 6= 1

a+b

arctan 1−ab ; ab < 1
a+b
arctan a + arctan b = arctan 1−ab
+ π; ab > 1, a, b > 0


a+b
arctan 1−ab − π; ab > 1, a, b < 0
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
7/1
Lemma 2
For a, b ∈ R, ab 6= 1

a+b

arctan 1−ab ; ab < 1
a+b
arctan a + arctan b = arctan 1−ab
+ π; ab > 1, a, b > 0


a+b
arctan 1−ab − π; ab > 1, a, b < 0
Proof: If ab ∈ [−1, 1] then arctan a + arctan b ∈
Since
tan(arctan a + arctan b) =
.
tan(arctan a) + tan(arctan b)
1 − tan(arctan a) tan(arctan b)
arctan a + arctan b = arctan
Jared Ruiz (Youngstown State University)
−π π
2 , 2
A Surprising Sum of Arctangents
a+b
1 − ab
June 30, 2013
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Lemma 2
n
a+b
arctan a + arctan b = arctan 1−ab
; ab < 1
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
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Lemma 2
n
a+b
arctan a + arctan b = arctan 1−ab
; ab < 1
Assume ab < 1.
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
8/1
Lemma 2
n
a+b
arctan a + arctan b = arctan 1−ab
; ab < 1
Assume ab < 1.
If 0 < a < 1, and 0 < b < 1 then we are done.
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
8/1
Lemma 2
n
a+b
arctan a + arctan b = arctan 1−ab
; ab < 1
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.
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
8/1
Lemma 2
n
a+b
arctan a + arctan b = arctan 1−ab
; ab < 1
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
π
arctan a + arctan b = − arctan
+ + arctan b
a
2
−1 + ab
π
= arctan
+
a+b
2
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
8/1
Lemma 2
n
a+b
arctan a + arctan b = arctan 1−ab
; ab < 1
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
π
arctan a + arctan b = − arctan
+ + arctan b
a
2
−1 + ab
π
= arctan
+
a+b
2
Now
−1+ab
a+b
< 0, since 0 < ab < 1.
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
8/1
Lemma 2
n
a+b
arctan a + arctan b = arctan 1−ab
; ab < 1
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
π
arctan a + arctan b = − arctan
+ + arctan b
a
2
−1 + ab
π
= arctan
+
a+b
2
Now
−1+ab
a+b
< 0, since 0 < ab < 1. So by Lemma 1
−1 + ab
π
a+b
π π
arctan
+
= − arctan
− +
a+b
2
−1 + ab
2
2
a+b
= arctan
1 − ab
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
8/1
Lemma 2
(
a+b
arctan 1−ab
+ π; ab > 1, a, b > 0
arctan a + arctan b =
a+b
arctan 1−ab − π; ab > 1, a, b < 0
Assume ab > 1.
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
9/1
Lemma 2
(
a+b
arctan 1−ab
+ π; ab > 1, a, b > 0
arctan a + arctan b =
a+b
arctan 1−ab − π; ab > 1, a, b < 0
Assume ab > 1.
Let a, b > 0. Then 0 <
Jared Ruiz (Youngstown State University)
1
a
·
1
b
<1
A Surprising Sum of Arctangents
June 30, 2013
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Lemma 2
(
a+b
arctan 1−ab
+ π; ab > 1, a, b > 0
arctan a + arctan b =
a+b
arctan 1−ab − π; ab > 1, a, b < 0
Assume ab > 1.
1
b
<1
arctan a + arctan b
=
Let a, b > 0. Then 0 <
1
a
·
=
Jared Ruiz (Youngstown State University)
1
1
− arctan + π
a
b
a+b
arctan
+π
1 − ab
− arctan
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June 30, 2013
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Lemma 2
(
a+b
arctan 1−ab
+ π; ab > 1, a, b > 0
arctan a + arctan b =
a+b
arctan 1−ab − π; ab > 1, a, b < 0
Assume ab > 1.
1
b
<1
arctan a + arctan b
=
Let a, b > 0. Then 0 <
1
a
·
=
1
1
− arctan + π
a
b
a+b
arctan
+π
1 − ab
− arctan
Now let a, b < 0. Again we have 0 <
Jared Ruiz (Youngstown State University)
1
a
·
A Surprising Sum of Arctangents
1
b
<1
June 30, 2013
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Lemma 2
(
a+b
arctan 1−ab
+ π; ab > 1, a, b > 0
arctan a + arctan b =
a+b
arctan 1−ab − π; ab > 1, a, b < 0
Assume ab > 1.
1
b
<1
arctan a + arctan b
=
Let a, b > 0. Then 0 <
1
a
·
=
1
1
− arctan + π
a
b
a+b
arctan
+π
1 − ab
− arctan
Now let a, b < 0. Again we have 0 <
arctan a + arctan b
Jared Ruiz (Youngstown State University)
=
1
a
·
1
b
<1
− arctan
A Surprising Sum of Arctangents
1
1
− arctan − π
a
b
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Lemma 2
(
a+b
arctan 1−ab
+ π; ab > 1, a, b > 0
arctan a + arctan b =
a+b
arctan 1−ab − π; ab > 1, a, b < 0
Assume ab > 1.
1
b
<1
arctan a + arctan b
=
Let a, b > 0. Then 0 <
1
a
·
=
1
1
− arctan + π
a
b
a+b
arctan
+π
1 − ab
− arctan
Now let a, b < 0. Again we have 0 <
arctan a + arctan b
=
=
Jared Ruiz (Youngstown State University)
1
a
·
1
b
<1
1
1
− arctan − π
a
b
a+b
−π
arctan
1 − ab
− arctan
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June 30, 2013
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Lemma 2
(
a+b
arctan 1−ab
+ π; ab > 1, a, b > 0
arctan a + arctan b =
a+b
arctan 1−ab − π; ab > 1, a, b < 0
Assume ab > 1.
1
b
<1
arctan a + arctan b
=
Let a, b > 0. Then 0 <
1
a
·
=
1
1
− arctan + π
a
b
a+b
arctan
+π
1 − ab
− arctan
Now let a, b < 0. Again we have 0 <
arctan a + arctan b
=
=
Jared Ruiz (Youngstown State University)
1
a
·
1
b
<1
1
1
− arctan − π
a
b
a+b
−π
arctan
1 − ab
− arctan
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June 30, 2013
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9/1
Corollary
arctan 1 + arctan 2 + arctan 3 = π
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
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Corollary
arctan 1 + arctan 2 + arctan 3 = π
Proof: By Lemma 2,
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
10 / 1
Corollary
arctan 1 + arctan 2 + arctan 3 = π
Proof: By Lemma 2,
arctan 1 + arctan 2 + arctan 3 = (arctan 1 + arctan 2) + arctan 3
3
= arctan
+ π + arctan 3
−1
= − arctan 3 + π + arctan 3
= π
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
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Corollary
arctan 1 + arctan 2 + arctan 3 = π
Proof: By Lemma 2,
arctan 1 + arctan 2 + arctan 3 = (arctan 1 + arctan 2) + arctan 3
3
= arctan
+ π + arctan 3
−1
= − arctan 3 + π + arctan 3
= π
2
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A Surprising Sum of Arctangents
June 30, 2013
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Corollary
arctan 1 + arctan 2 + arctan 3 = π
Proof: By Lemma 2,
arctan 1 + arctan 2 + arctan 3 = (arctan 1 + arctan 2) + arctan 3
3
= arctan
+ π + arctan 3
−1
= − arctan 3 + π + arctan 3
= π
2
This is a new proof of this identity!
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
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Lemma 3
For a ∈ R,
arctan a+arctan(1−a)+arctan(2−a+a2 )+arctan(3−3a+4a2 −2a3 +a4 ) =
π.
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
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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
2 − a + a2
> 1.
1 − a + a2
1
Then adding arctan 1−a+a
+ arctan(2 − a + a2 ) gives
2
Now for all a ∈ R,
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
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Lemma 3
arctan a+arctan(1−a)+arctan(2−a+a2 )+arctan(3−3a+4a2 −2a3 +a4 ) =
π.
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Lemma 3
arctan a+arctan(1−a)+arctan(2−a+a2 )+arctan(3−3a+4a2 −2a3 +a4 ) =
π.
rctan
1
1−a+a2
+ 2 − a + a2
1−
2−a+a2
1−a+a2

!
+ π = arctan 
= arctan
Jared Ruiz (Youngstown State University)

(2−a+a2 )(1−a+a2 )
1−a+a2
+π
(1−a+a2 )−(2−a+a2 )
1−a+a2
− 3a + 4a2 − 2a3 + a4
1+
3
A Surprising Sum of Arctangents
+π
−1
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Lemma 3
arctan a+arctan(1−a)+arctan(2−a+a2 )+arctan(3−3a+4a2 −2a3 +a4 ) =
π.
rctan
1
1−a+a2
+ 2 − a + a2
1−
2−a+a2
1−a+a2

!
+ π = arctan 
= arctan

(2−a+a2 )(1−a+a2 )
1−a+a2
+π
(1−a+a2 )−(2−a+a2 )
1−a+a2
− 3a + 4a2 − 2a3 + a4
1+
3
+π
−1
Now adding
3 − 3a + 4a2 − 2a3 + a4
arctan
+π +arctan(3−3a+4a2 −2a3 +a4 ) = π
−1
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A Surprising Sum of Arctangents
June 30, 2013
12 / 1
Lemma 3
arctan a+arctan(1−a)+arctan(2−a+a2 )+arctan(3−3a+4a2 −2a3 +a4 ) =
π.
rctan
1
1−a+a2
+ 2 − a + a2
1−
2−a+a2
1−a+a2

!
+ π = arctan 
= arctan

(2−a+a2 )(1−a+a2 )
1−a+a2
+π
(1−a+a2 )−(2−a+a2 )
1−a+a2
− 3a + 4a2 − 2a3 + a4
1+
3
+π
−1
Now adding
3 − 3a + 4a2 − 2a3 + a4
arctan
+π +arctan(3−3a+4a2 −2a3 +a4 ) = π
−1
2
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A Surprising Sum of Arctangents
June 30, 2013
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Lemma 4
Let a ∈ R, a < 0. Then
|a| < 1 − a < 2 − a + a2 < 3 − 3a + 4a2 − 2a3 + a4
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A Surprising Sum of Arctangents
June 30, 2013
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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
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
13 / 1
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
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
13 / 1
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
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
13 / 1
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
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
13 / 1
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
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
14 / 1
Theorem
For all n > 0, there exists s > 0 and distinct positive integers k1 , k2 , . . . , ks
such that
s
X
tj arctan kj = nπ
j=1
for suitable tj ∈ {−1, 1}.
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
15 / 1
Theorem
For all n > 0, there exists s > 0 and distinct positive integers k1 , k2 , . . . , ks
such that
s
X
tj arctan kj = nπ
j=1
for suitable tj ∈ {−1, 1}.
Proof: Let a0 < 0, and let k1 = |a0 |. Then
k2 = 1 − a0 ,
k3 = 2 − a0 + a02 , and k4 = 3 − 3a0 + 4a02 − 2a03 + a04
We have by Lemma 3
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
15 / 1
Theorem
For all n > 0, there exists s > 0 and distinct positive integers k1 , k2 , . . . , ks
such that
s
X
tj arctan kj = nπ
j=1
for suitable tj ∈ {−1, 1}.
Proof: Let a0 < 0, and let k1 = |a0 |. Then
k2 = 1 − a0 ,
k3 = 2 − a0 + a02 , and k4 = 3 − 3a0 + 4a02 − 2a03 + a04
We have by Lemma 3
arctan a0
Jared Ruiz (Youngstown State University)
+ arctan k2 + arctan k3 + arctan k4 = π
A Surprising Sum of Arctangents
June 30, 2013
(1)
15 / 1
Theorem
For all n > 0, there exists s > 0 and distinct positive integers k1 , k2 , . . . , ks
such that
s
X
tj arctan kj = nπ
j=1
for suitable tj ∈ {−1, 1}.
Proof: Let a0 < 0, and let k1 = |a0 |. Then
k2 = 1 − a0 ,
k3 = 2 − a0 + a02 , and k4 = 3 − 3a0 + 4a02 − 2a03 + a04
We have by Lemma 3
− arctan k1 + arctan k2 + arctan k3 + arctan k4 = π
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
(1)
15 / 1
Theorem
For all n > 0, there exists s > 0 and distinct positive integers k1 , k2 , . . . , ks
such that
s
X
tj arctan kj = nπ
j=1
for suitable tj ∈ {−1, 1}.
Proof: Let a0 < 0, and let k1 = |a0 |. Then
k2 = 1 − a0 ,
k3 = 2 − a0 + a02 , and k4 = 3 − 3a0 + 4a02 − 2a03 + a04
We have by Lemma 3
− arctan k1 + arctan k2 + arctan k3 + arctan k4 = π
(1)
Let a1 = −(k4 + 1), and let k5 = |a1 |.
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
15 / 1
Theorem
For all n > 0, there exists s > 0 and distinct positive integers k1 , k2 , . . . , ks
such that
s
X
tj arctan kj = nπ
j=1
for suitable tj ∈ {−1, 1}.
Proof: Let a0 < 0, and let k1 = |a0 |. Then
k2 = 1 − a0 ,
k3 = 2 − a0 + a02 , and k4 = 3 − 3a0 + 4a02 − 2a03 + a04
We have by Lemma 3
− arctan k1 + arctan k2 + arctan k3 + arctan k4 = π
(1)
Let a1 = −(k4 + 1), and let k5 = |a1 |.
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
15 / 1
Theorem
For all n > 0, there exists s > 0 and distinct positive integers k1 , k2 , . . . , ks
such that
s
X
tj arctan kj = nπ
j=1
for suitable tj ∈ {−1, 1}.
Proof: Let a0 < 0, and let k1 = |a0 |. Then
k2 = 1 − a0 ,
k3 = 2 − a0 + a02 , and k4 = 3 − 3a0 + 4a02 − 2a03 + a04
We have by Lemma 3
− arctan k1 + arctan k2 + arctan k3 + arctan k4 = π
(1)
Let a1 = −(k4 + 1), and let k5 = |a1 |.
arctan a1
Jared Ruiz (Youngstown State University)
+ arctan k6 + arctan k7 + arctan k8 = π
A Surprising Sum of Arctangents
June 30, 2013
(2)
15 / 1
Theorem
For all n > 0, there exists s > 0 and distinct positive integers k1 , k2 , . . . , ks
such that
s
X
tj arctan kj = nπ
j=1
for suitable tj ∈ {−1, 1}.
Proof: Let a0 < 0, and let k1 = |a0 |. Then
k2 = 1 − a0 ,
k3 = 2 − a0 + a02 , and k4 = 3 − 3a0 + 4a02 − 2a03 + a04
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 + arctan k6 + arctan k7 + arctan k8 = π
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
(2)
15 / 1
Theorem
For all n > 0, there exists s > 0 and distinct positive integers k1 , k2 , . . . , ks
such that
s
X
tj arctan kj = nπ
j=1
for suitable tj ∈ {−1, 1}.
Proof: Let a0 < 0, and let k1 = |a0 |. Then
k2 = 1 − a0 ,
k3 = 2 − a0 + a02 , and k4 = 3 − 3a0 + 4a02 − 2a03 + a04
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 + arctan k6 + arctan k7 + arctan k8 = π
(2)
(??) + (??) = 2π.
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
15 / 1
Theorem
For all n > 0, there exists s > 0 and distinct positive integers k1 , k2 , . . . , ks
such that
s
X
tj arctan kj = nπ
j=1
for suitable tj ∈ {−1, 1}.
Proof: Let a0 < 0, and let k1 = |a0 |. Then
k2 = 1 − a0 ,
k3 = 2 − a0 + a02 , and k4 = 3 − 3a0 + 4a02 − 2a03 + a04
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 + arctan k6 + arctan k7 + arctan k8 = π
(2)
(??) + (??) = 2π. This can be continued indefinitely.
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
15 / 1
Theorem
For all n > 0, there exists s > 0 and distinct positive integers k1 , k2 , . . . , ks
such that
s
X
tj arctan kj = nπ
j=1
for suitable tj ∈ {−1, 1}.
Proof: Let a0 < 0, and let k1 = |a0 |. Then
k2 = 1 − a0 ,
k3 = 2 − a0 + a02 , and k4 = 3 − 3a0 + 4a02 − 2a03 + a04
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 + arctan k6 + arctan k7 + arctan k8 = π
2
(??) + (??) = 2π. This can be continued indefinitely.
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
(2)
June 30, 2013
15 / 1
Example
Let a0 = −2, and k1 = 2.
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
16 / 1
Example
Let a0 = −2, and k1 = 2. Then
− arctan 2 + arctan 3 + arctan 8 + arctan 57 = π
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
(3)
16 / 1
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).
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
16 / 1
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
− arctan 58 + arctan 59 + arctan 3424 + arctan 11720353 = π
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
(4)
16 / 1
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
− arctan 58 + arctan 59 + arctan 3424 + arctan 11720353 = π
(4)
Then (??) + (??) = 2π.
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
16 / 1
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
− arctan 58 + arctan 59 + arctan 3424 + arctan 11720353 = π
(4)
Then (??) + (??) = 2π. Etc.
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
16 / 1
References
1
Ecker, Michael W. An arctangent triangle: puzzling over
arctan1+arctan2+arctan3=π. Mathematics and Computer
Education, 2003.
2
Edwards, C. Henry and David E. Penney. Calculus, Ed. 6. Prentice
Hall, 2002: p. 476.
3
Faires, J. Douglas and Barbara T. Faires. Calculus, Ed. 2. Random
House, 1988: p. 404.
4
Stewart, James. Calculus, Ed. 5. Brooks/Cole (A division of
Thomson Learning), 2003: p. 485.
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
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Where Does Lemma 3 Come From?
Let a, b ∈ Z, and a + b = 1.
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
18 / 1
Where Does Lemma 3 Come From?
Let a, b ∈ Z, and a + b = 1. Then arctan a + arctan b = arctan
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
a+b
1 − ab
June 30, 2013
18 / 1
Where Does Lemma 3 Come From?
Let a, b ∈ Z, and a + b = 1. Then arctan a + arctan b = arctan
Now
arctan
a+b
1 − ab
a+b
a + b + c − abc
+ arctan c = arctan
1 − ab
1 − ab − ac − bc
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
18 / 1
Where Does Lemma 3 Come From?
Let a, b ∈ Z, and a + b = 1. Then arctan a + arctan b = arctan
Now
a+b
1 − ab
a+b
a + b + c − abc
+ arctan c = arctan
1 − ab
1 − ab − ac − bc
We want 1 − ab − ac − bc to equal 1.
arctan
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
18 / 1
Where Does Lemma 3 Come From?
Let a, b ∈ Z, and a + b = 1. Then arctan a + arctan b = arctan
Now
a+b
1 − ab
a+b
a + b + c − abc
+ arctan c = arctan
1 − ab
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
arctan
a+b
c >1
1 − ab
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
18 / 1
Where Does Lemma 3 Come From?
Let a, b ∈ Z, and a + b = 1. Then arctan a + arctan b = arctan
Now
a+b
1 − ab
a+b
a + b + c − abc
+ arctan c = arctan
1 − ab
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
arctan
a+b
c >1
1 − ab
Then we add arctan
a + b + c − abc
+ arctan d and so on.
1 − ab − ac − bc
Jared Ruiz (Youngstown State University)
A Surprising Sum of Arctangents
June 30, 2013
18 / 1