leonhard euler tercentennary — 2007mkondra/papers/euler07.pdfalgebra: 7% other: 3% leonhard euler...
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Leonhard Euler Tercentennary —2007
Dr. Margo Kondratieva and Andrew Stewart
Leonhard Euler Tercentennary — 2007 – p. 1/47
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Overview
Leonhard Euler Tercentennary — 2007 – p. 2/47
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Overview
Introduction
Leonhard Euler Tercentennary — 2007 – p. 2/47
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Overview
Introduction
Timeline
Leonhard Euler Tercentennary — 2007 – p. 2/47
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Overview
Introduction
Timeline
Polyhedral Problem
Leonhard Euler Tercentennary — 2007 – p. 2/47
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Overview
Introduction
Timeline
Polyhedral Problem
Basel Problem
Leonhard Euler Tercentennary — 2007 – p. 2/47
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Overview
Introduction
Timeline
Polyhedral Problem
Basel Problem
Conclusion
Leonhard Euler Tercentennary — 2007 – p. 2/47
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Introduction
Some numbers about Euler
Leonhard Euler Tercentennary — 2007 – p. 3/47
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Introduction
Some numbers about Euler
76 years
Leonhard Euler Tercentennary — 2007 – p. 3/47
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Introduction
Some numbers about Euler
76 years
13 children
Leonhard Euler Tercentennary — 2007 – p. 3/47
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Introduction
Some numbers about Euler
76 years
13 children
886 articles
Leonhard Euler Tercentennary — 2007 – p. 3/47
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Introduction
Some numbers about Euler
76 years
13 children
886 articles
72 · 600
Leonhard Euler Tercentennary — 2007 – p. 3/47
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Introduction
Some numbers about Euler
76 years
13 children
886 articles
72 · 600 = 43200 pages
Leonhard Euler Tercentennary — 2007 – p. 3/47
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Introduction
Opera Omnia
Leonhard Euler Tercentennary — 2007 – p. 4/47
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Introduction
Opera Omnia
Pure Math (29 volumes)
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Introduction
Opera Omnia
Pure Math (29 volumes)
Mechanics and Astronomy (31 volumes)
Leonhard Euler Tercentennary — 2007 – p. 4/47
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Introduction
Opera Omnia
Pure Math (29 volumes)
Mechanics and Astronomy (31 volumes)
Physics and Miscellaneous (12 volumes)
Leonhard Euler Tercentennary — 2007 – p. 4/47
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Introduction
In Math
Leonhard Euler Tercentennary — 2007 – p. 5/47
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Introduction
In Math
Analysis: 60%
Leonhard Euler Tercentennary — 2007 – p. 5/47
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Introduction
In Math
Analysis: 60%
Geometry: 17%
Leonhard Euler Tercentennary — 2007 – p. 5/47
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Introduction
In Math
Analysis: 60%
Geometry: 17%
Number Theory: 13%
Leonhard Euler Tercentennary — 2007 – p. 5/47
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Introduction
In Math
Analysis: 60%
Geometry: 17%
Number Theory: 13%
Algebra: 7%
Leonhard Euler Tercentennary — 2007 – p. 5/47
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Introduction
In Math
Analysis: 60%
Geometry: 17%
Number Theory: 13%
Algebra: 7%
Other: 3%
Leonhard Euler Tercentennary — 2007 – p. 5/47
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Introduction
Some of Euler’s Topics
Here are some simple, but important, results of Euler:
Leonhard Euler Tercentennary — 2007 – p. 6/47
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Introduction
Some of Euler’s Topics
Here are some simple, but important, results of Euler:
Logarithms: the formula
loga(b) =logc(b)
logc(a)
Leonhard Euler Tercentennary — 2007 – p. 6/47
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Introduction
Some of Euler’s Topics
Here are some simple, but important, results of Euler:
Logarithms: the formula
loga(b) =logc(b)
logc(a)=
ln(b)
ln(a)
Leonhard Euler Tercentennary — 2007 – p. 6/47
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Introduction
Some of Euler’s Topics
Here are some simple, but important, results of Euler:
Logarithms: the formula
loga(b) =logc(b)
logc(a)=
ln(b)
ln(a)
The symbol i to denote√−1
Leonhard Euler Tercentennary — 2007 – p. 6/47
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Introduction
Some of Euler’s Topics
Here are some simple, but important, results of Euler:
Logarithms: the formula
loga(b) =logc(b)
logc(a)=
ln(b)
ln(a)
The symbol i to denote√−1
The base of the natural number, 2.719281828459 . . ., andthe symbol e to denote 2.719281828459 . . .
Leonhard Euler Tercentennary — 2007 – p. 6/47
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Introduction
Some of Euler’s Topics
Here are some simple, but important, results of Euler:
Logarithms: the formula
loga(b) =logc(b)
logc(a)=
ln(b)
ln(a)
The symbol i to denote√−1
The base of the natural number, 2.719281828459 . . ., andthe symbol e to denote 2.719281828459 . . .
Euler’s Identity:eiπ + 1 = 0
Leonhard Euler Tercentennary — 2007 – p. 6/47
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Introduction
Some of Euler’s Topics
Here are some simple, but important, results of Euler:
Logarithms: the formula
loga(b) =logc(b)
logc(a)=
ln(b)
ln(a)
The symbol i to denote√−1
The base of the natural number, 2.719281828459 . . ., andthe symbol e to denote 2.719281828459 . . .
Euler’s Identity:eiπ + 1 = 0
(Note that this last formula contains e, i, π, 0, and 1, whichare five of the most important symbols in mathematics.)
Leonhard Euler Tercentennary — 2007 – p. 6/47
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Timeline
1707: Euler was born.
Leonhard Euler Tercentennary — 2007 – p. 7/47
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Timeline
1707: Euler was born.1720: University of Basel
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Timeline
1707: Euler was born.1720: University of Basel1723: Master’s Degree (talk in Latin about the comparisonof philosophies of DeCartes and Newton)
Leonhard Euler Tercentennary — 2007 – p. 8/47
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Timeline
1707: Euler was born.1720: University of Basel1723: Master’s Degree (talk in Latin about the comparisonof philosophies of DeCartes and Newton)
Leonhard Euler Tercentennary — 2007 – p. 8/47
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Timeline
1707: Euler was born.1720: University of Basel1723: Master’s Degree
Leonhard Euler Tercentennary — 2007 – p. 9/47
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Timeline
1707: Euler was born.1720: University of Basel1723: Master’s Degree1726: First Paper — Analysis of the placement of masts ona sailing ship.
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Timeline
1707: Euler was born.1720: University of Basel1723: Master’s Degree1726: First Paper — Analysis of the placement of masts ona sailing ship. (Receives an award from Paris Academy ofScience)
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Timeline
1707: Euler was born.1720: University of Basel1723: Master’s Degree1726: First Paper — Analysis of the placement of masts ona sailing ship.1727: Moves to St. Petersburg, Russia
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Timeline
1707: Euler was born.1720: University of Basel1723: Master’s Degree1726: First Paper — Analysis of the placement of masts ona sailing ship.1727: Moves to St. Petersburg, Russia1730: Professor of Physics
Leonhard Euler Tercentennary — 2007 – p. 10/47
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Timeline
1707: Euler was born.1720: University of Basel1723: Master’s Degree1726: First Paper — Analysis of the placement of masts ona sailing ship.1727: Moves to St. Petersburg, Russia1730: Professor of Physics1733: Chair in Pure Mathematics
Leonhard Euler Tercentennary — 2007 – p. 11/47
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Timeline
1707: Euler was born.1720: University of Basel1723: Master’s Degree1726: First Paper — Analysis of the placement of masts ona sailing ship.1727: Moves to St. Petersburg, Russia1730: Professor of Physics1733: Chair in Pure Mathematics
Duties: Scientific consultant to the government,prepares maps, advises Russian Navy, tests designs forengines
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Timeline
1707: Euler was born.1720: University of Basel1723: Master’s Degree1726: First Paper — Analysis of the placement of masts ona sailing ship.1727: Moves to St. Petersburg, Russia1730: Professor of Physics1733: Chair in Pure Mathematics1734: Marries Katharina Gsell
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Timeline
1707: Euler was born.1720: University of Basel1723: Master’s Degree1726: First Paper — Analysis of the placement of masts ona sailing ship.1727: Moves to St. Petersburg, Russia1730: Professor of Physics1733: Chair in Pure Mathematics1734: Marries Katharina Gsell1736: First book — Mechanica
Leonhard Euler Tercentennary — 2007 – p. 11/47
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Timeline
1707: Euler was born.1720: University of Basel1723: Master’s Degree1726: First Paper — Analysis of the placement of masts ona sailing ship.1727: Moves to St. Petersburg, Russia1730: Professor of Physics1733: Chair in Pure Mathematics1734: Marries Katharina Gsell1736: First book — Mechanica1741: Moves to Berlin
Leonhard Euler Tercentennary — 2007 – p. 11/47
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Timeline
1748: Introductio in Analysin
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Timeline
1748: Introductio in Analysin1755: Books on Differential Calculus
Leonhard Euler Tercentennary — 2007 – p. 13/47
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Timeline
1748: Introductio in Analysin1755: Books on Differential Calculus1755: Letters to Princess
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Timeline
1748: Introductio in Analysin1755: Books on Differential Calculus1755: Letters to Princess(over 300 in total)
Leonhard Euler Tercentennary — 2007 – p. 13/47
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Timeline
1748: Introductio in Analysin1755: Books on Differential Calculus1755: Letters to Princess1766: Returns to St. Petersburg
Leonhard Euler Tercentennary — 2007 – p. 13/47
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Timeline
1748: Introductio in Analysin1755: Books on Differential Calculus1755: Letters to Princess1766: Returns to St. Petersburg1768: Book on Integral Calculus
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Timeline
1748: Introductio in Analysin1755: Books on Differential Calculus1755: Letters to Princess1766: Returns to St. Petersburg1768: Book on Integral Calculus1772: Treatise on the motion of the moon
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Timeline
1748: Introductio in Analysin1755: Books on Differential Calculus1755: Letters to Princess1766: Returns to St. Petersburg1768: Book on Integral Calculus1772: Treatise on the motion of the moon1773: Katharina died
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Timeline
1748: Introductio in Analysin1755: Books on Differential Calculus1755: Letters to Princess1766: Returns to St. Petersburg1768: Book on Integral Calculus1772: Treatise on the motion of the moon1773: Katharina died1776: Remarries (his wife’s sister)
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Timeline
1748: Introductio in Analysin1755: Books on Differential Calculus1755: Letters to Princess1766: Returns to St. Petersburg1768: Book on Integral Calculus1772: Treatise on the motion of the moon1773: Katharina died1776: Remarries (his wife’s sister)1783: Died
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Timeline
Gottfried Leibniz (1646-1716)
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Timeline
Jacob Bernoulli (1654-1705)
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Timeline
Johann Bernoulli (1667-1748)
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Timeline
Christian Goldbach (1690-1764)
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Timeline
Daniel Bernoulli (1700-1782)
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Timeline
Leonhard Euler
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Timeline
Peter the Great (1672)-(1725)
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Timeline
Catherine the Great (1729-1796)
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Timeline
Frederick the Great (1712-1786)
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Timeline
Joseph Louis Lagrange(1736-1813)
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Timeline
St. Petersburg Academy of Science(1724)
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Euler’s Formula for Polyhedra
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Euler’s Formula for Polyhedra
While Euler was studying polyhedra, he noticed arelationship between the number of edges, vertices andfaces of a polyhedra.
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Euler’s Formula for Polyhedra
While Euler was studying polyhedra, he noticed arelationship between the number of edges, vertices andfaces of a polyhedra.
Let V be the number of vertices.
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Euler’s Formula for Polyhedra
While Euler was studying polyhedra, he noticed arelationship between the number of edges, vertices andfaces of a polyhedra.
Let V be the number of vertices.
Let E be the number of edges.
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Euler’s Formula for Polyhedra
While Euler was studying polyhedra, he noticed arelationship between the number of edges, vertices andfaces of a polyhedra.
Let V be the number of vertices.
Let E be the number of edges.
Let F be the number of faces.
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Euler’s Formula for Polyhedra
While Euler was studying polyhedra, he noticed arelationship between the number of edges, vertices andfaces of a polyhedra.
Let V be the number of vertices.
Let E be the number of edges.
Let F be the number of faces.
Then we haveV − E + F = 2.
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Euler’s Formula for Polyhedra
Auxilary Problem: Suppose we have n points inside atriangle that are connected to form triangles (triangulation).How many triangles are there?
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Euler’s Formula for Polyhedra
Auxilary Problem: Suppose we have n points inside atriangle that are connected to form triangles (triangulation).How many triangles are there?
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Euler’s Formula for Polyhedra
Suppose there are N interior triangles. We will find N .
a
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Euler’s Formula for Polyhedra
Suppose there are N interior triangles. We will find N .
Sum the angles of the interior triangles in two ways.
a
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Euler’s Formula for Polyhedra
Suppose there are N interior triangles. We will find N .
Sum the angles of the interior triangles in two ways.
First way: N triangles; sum of angles isa Nπ.
aRemember that π radians is the same as 180◦.
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Euler’s Formula for Polyhedra
Suppose there are N interior triangles. We will find N .
Sum the angles of the interior triangles in two ways.
First way: N triangles; sum of angles isa Nπ.
Second way: Interior points + exterior triangle;
aRemember that π radians is the same as 180◦.
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Euler’s Formula for Polyhedra
Suppose there are N interior triangles. We will find N .
Sum the angles of the interior triangles in two ways.
First way: N triangles; sum of angles isa Nπ.
Second way: Interior points + exterior triangle; sum ofangles is 2nπ
aRemember that π radians is the same as 180◦.
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Euler’s Formula for Polyhedra
Suppose there are N interior triangles. We will find N .
Sum the angles of the interior triangles in two ways.
First way: N triangles; sum of angles isa Nπ.
Second way: Interior points + exterior triangle; sum ofangles is 2nπ + π
aRemember that π radians is the same as 180◦.
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Euler’s Formula for Polyhedra
Suppose there are N interior triangles. We will find N .
Sum the angles of the interior triangles in two ways.
First way: N triangles; sum of angles isa Nπ.
Second way: Interior points + exterior triangle; sum ofangles is 2nπ + π = (2n + 1)π.
aRemember that π radians is the same as 180◦.
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Euler’s Formula for Polyhedra
Suppose there are N interior triangles. We will find N .
Sum the angles of the interior triangles in two ways.
First way: N triangles; sum of angles isa Nπ.
Second way: Interior points + exterior triangle; sum ofangles is 2nπ + π = (2n + 1)π.
Hence, we have Nπ = (2n + 1)π, or
N = 2n + 1.
aRemember that π radians is the same as 180◦.
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Euler’s Formula for Polyhedra
Suppose there are N interior triangles. We will find N .
Sum the angles of the interior triangles in two ways.
First way: N triangles; sum of angles isa Nπ.
Second way: Interior points + exterior triangle; sum ofangles is 2nπ + π = (2n + 1)π.
Hence, we have Nπ = (2n + 1)π, or
N = 2n + 1.
Thus, if we have n interior points, we will have 2n + 1triangles.
aRemember that π radians is the same as 180◦.
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Euler’s Formula for Polyhedra
Now, suppose we have a polyhedra that can be projectedonto the plane as in the problem above. Then we have
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Euler’s Formula for Polyhedra
Now, suppose we have a polyhedra that can be projectedonto the plane as in the problem above. Then we have
V = 3 + n
F = N + 1 = 2(n + 1)
E =3F
2= 3(n + 1)
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Euler’s Formula for Polyhedra
Now, suppose we have a polyhedra that can be projectedonto the plane as in the problem above. Then we have
V = 3 + n
F = N + 1 = 2(n + 1)
E =3F
2= 3(n + 1)
Then we see that
V − E + F =
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Euler’s Formula for Polyhedra
Now, suppose we have a polyhedra that can be projectedonto the plane as in the problem above. Then we have
V = 3 + n
F = N + 1 = 2(n + 1)
E =3F
2= 3(n + 1)
Then we see that
V − E + F = (3 + n)
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Euler’s Formula for Polyhedra
Now, suppose we have a polyhedra that can be projectedonto the plane as in the problem above. Then we have
V = 3 + n
F = N + 1 = 2(n + 1)
E =3F
2= 3(n + 1)
Then we see that
V − E + F = (3 + n) − (3n + 3)
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Euler’s Formula for Polyhedra
Now, suppose we have a polyhedra that can be projectedonto the plane as in the problem above. Then we have
V = 3 + n
F = N + 1 = 2(n + 1)
E =3F
2= 3(n + 1)
Then we see that
V − E + F = (3 + n) − (3n + 3) + (2n + 2)
Leonhard Euler Tercentennary — 2007 – p. 31/47
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Euler’s Formula for Polyhedra
Now, suppose we have a polyhedra that can be projectedonto the plane as in the problem above. Then we have
V = 3 + n
F = N + 1 = 2(n + 1)
E =3F
2= 3(n + 1)
Then we see that
V − E + F = (3 + n) − (3n + 3) + (2n + 2) = 2
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Euler’s Formula for Polyhedra
Now, suppose we have a polyhedra that can be projectedonto the plane as in the problem above. Then we have
V = 3 + n
F = N + 1 = 2(n + 1)
E =3F
2= 3(n + 1)
Then we see that
V − E + F = (3 + n) − (3n + 3) + (2n + 2) = 2,
which is Euler’s formula!
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Euler’s Formula for Polyhedra
We have proven Euler’s Formula for polyhedra with onlytriangular faces. We can prove it in general, but we mustrecall a useful fact.
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Euler’s Formula for Polyhedra
We have proven Euler’s Formula for polyhedra with onlytriangular faces. We can prove it in general, but we mustrecall a useful fact.Theorem: For n greater than 3, the sum of the interior angles of apolygon with n sides will be
(n − 2)π.
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Euler’s Formula for Polyhedra
We have proven Euler’s Formula for polyhedra with onlytriangular faces. We can prove it in general, but we mustrecall a useful fact.Theorem: For n greater than 3, the sum of the interior angles of apolygon with n sides will be
(n − 2)π.
For example, for a triangle, n = 3, and the sum of the angles is(3 − 1)π = π.
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Euler’s Formula for Polyhedra
We have proven Euler’s Formula for polyhedra with onlytriangular faces. We can prove it in general, but we mustrecall a useful fact.Theorem: For n greater than 3, the sum of the interior angles of apolygon with n sides will be
(n − 2)π.
For example, for a triangle, n = 3, and the sum of the angles is(3 − 1)π = π.For a square, n = 4, and the sum of the angles is (4 − 2)π = 2π.
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Euler’s Formula for Polyhedra
Now assume that we can project any polyhedron onto theplane so that none of the projections of the edges cross.We will prove Euler’s Formula for the general case.
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Euler’s Formula for Polyhedra
Now assume that we can project any polyhedron onto theplane so that none of the projections of the edges cross.We will prove Euler’s Formula for the general case.
Let the exterior polygon have n0 edges.
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Euler’s Formula for Polyhedra
Now assume that we can project any polyhedron onto theplane so that none of the projections of the edges cross.We will prove Euler’s Formula for the general case.
Let the exterior polygon have n0 edges.
Let there be m polygons on the projection.
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Euler’s Formula for Polyhedra
Now assume that we can project any polyhedron onto theplane so that none of the projections of the edges cross.We will prove Euler’s Formula for the general case.
Let the exterior polygon have n0 edges.
Let there be m polygons on the projection. (This meansthat there were m + 1 faces on the polyhedron, wherethe extra face is the exterior polygon in the projection.)
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Euler’s Formula for Polyhedra
Now assume that we can project any polyhedron onto theplane so that none of the projections of the edges cross.We will prove Euler’s Formula for the general case.
Let the exterior polygon have n0 edges.
Let there be m polygons on the projection.
Let ni be the number of edges of the ith polygon.
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Euler’s Formula for Polyhedra
Now assume that we can project any polyhedron onto theplane so that none of the projections of the edges cross.We will prove Euler’s Formula for the general case.
Let the exterior polygon have n0 edges.
Let there be m polygons on the projection.
Let ni be the number of edges of the ith polygon.
Then we have
n1 + n2 + . . . + nm
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Euler’s Formula for Polyhedra
Now assume that we can project any polyhedron onto theplane so that none of the projections of the edges cross.We will prove Euler’s Formula for the general case.
Let the exterior polygon have n0 edges.
Let there be m polygons on the projection.
Let ni be the number of edges of the ith polygon.
Then we have
n1 + n2 + . . . + nm + n0
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Euler’s Formula for Polyhedra
Now assume that we can project any polyhedron onto theplane so that none of the projections of the edges cross.We will prove Euler’s Formula for the general case.
Let the exterior polygon have n0 edges.
Let there be m polygons on the projection.
Let ni be the number of edges of the ith polygon.
Then we have
n1 + n2 + . . . + nm + n0 = 2E.
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Euler’s Formula for Polyhedra
Again, we sum the interior angles of interior polygons of theprojection in two ways:
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Euler’s Formula for Polyhedra
Again, we sum the interior angles of interior polygons of theprojection in two ways:
Interior polygons:
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Euler’s Formula for Polyhedra
Again, we sum the interior angles of interior polygons of theprojection in two ways:
Interior polygons:
(n1 − 2)π + (n2 − 2)π + . . . + (nm − 2)π
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Euler’s Formula for Polyhedra
Again, we sum the interior angles of interior polygons of theprojection in two ways:
Interior polygons:
(n1 − 2)π + (n2 − 2)π + . . . + (nm − 2)π
= (n1 + n2 + . . . + nm)π −m
︷ ︸︸ ︷
(2 + 2 + . . . + 2) π
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Euler’s Formula for Polyhedra
Again, we sum the interior angles of interior polygons of theprojection in two ways:
Interior polygons:
(n1 − 2)π + (n2 − 2)π + . . . + (nm − 2)π
= (n1 + n2 + . . . + nm)π −m
︷ ︸︸ ︷
(2 + 2 + . . . + 2) π
+(n0 − 2)π − (n0 − 2)π
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Euler’s Formula for Polyhedra
Again, we sum the interior angles of interior polygons of theprojection in two ways:
Interior polygons:
(n1 − 2)π + (n2 − 2)π + . . . + (nm − 2)π
= (n1 + n2 + . . . + nm)π −m
︷ ︸︸ ︷
(2 + 2 + . . . + 2) π
+(n0 − 2)π − (n0 − 2)π
= (n1 + n2 + . . . + nm + n0)π − 2(m + 1)π − (n0 − 2)π
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Euler’s Formula for Polyhedra
Again, we sum the interior angles of interior polygons of theprojection in two ways:
Interior polygons:
(n1 − 2)π + (n2 − 2)π + . . . + (nm − 2)π
= (n1 + n2 + . . . + nm)π −m
︷ ︸︸ ︷
(2 + 2 + . . . + 2) π
+(n0 − 2)π − (n0 − 2)π
= (n1 + n2 + . . . + nm + n0)π − 2(m + 1)π − (n0 − 2)π
= 2Eπ − 2Fπ − (n0 − 2)π
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Euler’s Formula for Polyhedra
Again, we sum the interior angles of interior polygons of theprojection in two ways:
Interior polygons:
2Eπ − 2Fπ − (n0 − 2)π
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Euler’s Formula for Polyhedra
Again, we sum the interior angles of interior polygons of theprojection in two ways:
Interior polygons:
2Eπ − 2Fπ − (n0 − 2)π
Interior points + exterior polygon:
Leonhard Euler Tercentennary — 2007 – p. 35/47
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Euler’s Formula for Polyhedra
Again, we sum the interior angles of interior polygons of theprojection in two ways:
Interior polygons:
2Eπ − 2Fπ − (n0 − 2)π
Interior points + exterior polygon:
(V − n0)2π
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Euler’s Formula for Polyhedra
Again, we sum the interior angles of interior polygons of theprojection in two ways:
Interior polygons:
2Eπ − 2Fπ − (n0 − 2)π
Interior points + exterior polygon:
(V − n0)2π + (n0 − 2)π
Leonhard Euler Tercentennary — 2007 – p. 35/47
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Euler’s Formula for Polyhedra
Again, we sum the interior angles of interior polygons of theprojection in two ways:
Interior polygons:
2Eπ − 2Fπ − (n0 − 2)π
Interior points + exterior polygon:
(V − n0)2π + (n0 − 2)π
= 2V π − n0π − 2π
Leonhard Euler Tercentennary — 2007 – p. 35/47
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Euler’s Formula for Polyhedra
Again, we sum the interior angles of interior polygons of theprojection in two ways:
Interior polygons:
2Eπ − 2Fπ − (n0 − 2)π
Interior points + exterior polygon:
(V − n0)2π + (n0 − 2)π
= 2V π − n0π − 2π + 2π − 2π
Leonhard Euler Tercentennary — 2007 – p. 35/47
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Euler’s Formula for Polyhedra
Again, we sum the interior angles of interior polygons of theprojection in two ways:
Interior polygons:
2Eπ − 2Fπ − (n0 − 2)π
Interior points + exterior polygon:
(V − n0)2π + (n0 − 2)π
= 2V π − n0π − 2π + 2π − 2π
= 2V π − (n0 − 2)π − 4π
Leonhard Euler Tercentennary — 2007 – p. 35/47
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Euler’s Formula for Polyhedra
Again, we sum the interior angles of interior polygons of theprojection in two ways:
Interior polygons:
2Eπ − 2Fπ − (n0 − 2)π
Interior points + exterior polygon:
2V π − (n0 − 2)π − 4π
Leonhard Euler Tercentennary — 2007 – p. 36/47
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Euler’s Formula for Polyhedra
Again, we sum the interior angles of interior polygons of theprojection in two ways:
Interior polygons:
2Eπ − 2Fπ − (n0 − 2)π
Interior points + exterior polygon:
2V π − (n0 − 2)π − 4π
So
2Eπ − 2Fπ − (n0 − 2)π = 2V π − (n0 − 2)π − 4π
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Euler’s Formula for Polyhedra
Again, we sum the interior angles of interior polygons of theprojection in two ways:
Interior polygons:
2Eπ − 2Fπ − (n0 − 2)π
Interior points + exterior polygon:
2V π − (n0 − 2)π − 4π
So
2Eπ − 2Fπ − (n0 − 2)π = 2V π − (n0 − 2)π − 4π
⇒ E − V + F = 2.
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The Basel Problem
Here we attempt to solve the problem of finding the sum
1 +1
4+
1
9+ . . . +
1
n2+ . . . .
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The Basel Problem
Here we attempt to solve the problem of finding the sum
1 +1
4+
1
9+ . . . +
1
n2+ . . . .
This was known as the Basel problem — mathematiciansbefore Euler had tremendous difficulties trying to solve it.
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The Basel Problem
Here we attempt to solve the problem of finding the sum
1 +1
4+
1
9+ . . . +
1
n2+ . . . .
This was known as the Basel problem — mathematiciansbefore Euler had tremendous difficulties trying to solve it. Itwas introduced by the Italian Pietro Mengoli in 1644, andJacob Bernoulli popularized it in 1689.
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The Basel Problem
Here we attempt to solve the problem of finding the sum
1 +1
4+
1
9+ . . . +
1
n2+ . . . .
This was known as the Basel problem — mathematiciansbefore Euler had tremendous difficulties trying to solve it. Itwas introduced by the Italian Pietro Mengoli in 1644, andJacob Bernoulli popularized it in 1689. The Bernoulli familyspend so much time solving it while residing in Basel,Switzerland, that it became known as the Basel problem.
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The Basel Problem
Here we attempt to solve the problem of finding the sum
1 +1
4+
1
9+ . . . +
1
n2+ . . . .
This was known as the Basel problem — mathematiciansbefore Euler had tremendous difficulties trying to solve it. Itwas introduced by the Italian Pietro Mengoli in 1644, andJacob Bernoulli popularized it in 1689. The Bernoulli familyspend so much time solving it while residing in Basel,Switzerland, that it became known as the Basel problem.When Euler solved the problem in 1735, the exact value ofthe sum was a genuine surprise.
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The Basel Problem
First we give a little explanation of the term infinite series. Ifwe ignore all the terms after the N th term, we obtain
1 +1
4+
1
9+ . . . +
1
N2.
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The Basel Problem
First we give a little explanation of the term infinite series. Ifwe ignore all the terms after the N th term, we obtain
1 +1
4+
1
9+ . . . +
1
N2.
We call this a partial sum.
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The Basel Problem
First we give a little explanation of the term infinite series. Ifwe ignore all the terms after the N th term, we obtain
1 +1
4+
1
9+ . . . +
1
N2.
We call this a partial sum. If the partial sums tend to acertain value S as N becomes large, then we say theinfinite series has a value S.
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The Basel Problem
First, we note that partial sums don’t always tend to acertain number.
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The Basel Problem
First, we note that partial sums don’t always tend to acertain number. For example, if we consider the simpleseries
1 + 1 + 1 + . . .
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The Basel Problem
First, we note that partial sums don’t always tend to acertain number. For example, if we consider the simpleseries
1 + 1 + 1 + . . . ,
its partial sums are
N︷ ︸︸ ︷
1 + 1 + . . . + 1 = N.
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The Basel Problem
First, we note that partial sums don’t always tend to acertain number. For example, if we consider the simpleseries
1 + 1 + 1 + . . . ,
its partial sums are
N︷ ︸︸ ︷
1 + 1 + . . . + 1 = N.
It is easy to see that as N grows, this partial sums go toinfinity.
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The Basel Problem
But if we look at the graph of the partial sums of our series,we obtain the following picture.
10 20 30 40 50
1.3
1.4
1.5
1.6
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The Basel Problem
But if we look at the graph of the partial sums of our series,we obtain the following picture.
10 20 30 40 50
1.3
1.4
1.5
1.6
From this picture, we suspect that the partial sums tend tosome number which is a little more than 1.6.
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The Basel Problem
Euler, in 1735, solved the Basel problem by using someingenious methods.
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The Basel Problem
Euler, in 1735, solved the Basel problem by using someingenious methods. He applied techniques used in finitecases to our infinite series.
Leonhard Euler Tercentennary — 2007 – p. 41/47
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The Basel Problem
Euler, in 1735, solved the Basel problem by using someingenious methods. He applied techniques used in finitecases to our infinite series. We will introduce the idea interms of a quadratic.
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The Basel Problem
Euler, in 1735, solved the Basel problem by using someingenious methods. He applied techniques used in finitecases to our infinite series. We will introduce the idea interms of a quadratic.
Consider a quadratic function f(x) = x2 + bx + c withroots p and q.
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The Basel Problem
Euler, in 1735, solved the Basel problem by using someingenious methods. He applied techniques used in finitecases to our infinite series. We will introduce the idea interms of a quadratic.
Consider a quadratic function f(x) = x2 + bx + c withroots p and q.
Then we can write f(x) = (x − p)(x − q).
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The Basel Problem
Euler, in 1735, solved the Basel problem by using someingenious methods. He applied techniques used in finitecases to our infinite series. We will introduce the idea interms of a quadratic.
Consider a quadratic function f(x) = x2 + bx + c withroots p and q.
Then we can write f(x) = (x − p)(x − q).
Expand f(x):
f(x) = x2 − (p + q)x + pq.
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The Basel Problem
Euler, in 1735, solved the Basel problem by using someingenious methods. He applied techniques used in finitecases to our infinite series. We will introduce the idea interms of a quadratic.
Consider a quadratic function f(x) = x2 + bx + c withroots p and q.
Then we can write f(x) = (x − p)(x − q).
Expand f(x):
f(x) = x2 − (p + q)x + pq.
Then (p + q) = −b.
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The Basel Problem
Euler used the same idea as we used for the quadratic —for an infinite polynomial!
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The Basel Problem
Euler used the same idea as we used for the quadratic —for an infinite polynomial! Consider the function
f(x) =sin(x)
x.
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The Basel Problem
Euler used the same idea as we used for the quadratic —for an infinite polynomial! Consider the function
f(x) =sin(x)
x.
Any root of f has to be a root of the numerator, sin(x).
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The Basel Problem
Euler used the same idea as we used for the quadratic —for an infinite polynomial! Consider the function
f(x) =sin(x)
x.
Any root of f has to be a root of the numerator, sin(x).The roots of sin(x) are
0,±π,±2π, . . . ,±nπ, . . . .
Near x = 0, sin(x) ≈ x, so f(x) ≈ 1, so x = 0 is not aroot.
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The Basel Problem
Euler used the same idea as we used for the quadratic —for an infinite polynomial! Consider the function
f(x) =sin(x)
x.
Any root of f has to be a root of the numerator, sin(x).The roots of sin(x) are
0,±π,±2π, . . . ,±nπ, . . . .
Near x = 0, sin(x) ≈ x, so f(x) ≈ 1, so x = 0 is not aroot.
Therefore, the roots of f(x) are
±π,±2π, . . . ,±nπ, . . . .
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The Basel Problem
Since the roots of f(x) are
±π,±2π, . . . ,±nπ, . . . ,
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The Basel Problem
Since the roots of f(x) are
±π,±2π, . . . ,±nπ, . . . ,
it is reasonable to assume that
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The Basel Problem
Since the roots of f(x) are
±π,±2π, . . . ,±nπ, . . . ,
it is reasonable to assume that
f(x) =(
1 −x
π
) (
1 +x
π
) (
1 −x
2π
)(
1 +x
2π
)
· · ·
(Instead of writing the factor (nπ − x), we write the factor(1 − x
nπ), which corresponds to the same root, nπ.)
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The Basel Problem
Since the roots of f(x) are
±π,±2π, . . . ,±nπ, . . . ,
it is reasonable to assume that
f(x) =
︷ ︸︸ ︷(
1 −x
π
) (
1 +x
π
) (
1 −x
2π
)(
1 +x
2π
)
· · ·
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The Basel Problem
Since the roots of f(x) are
±π,±2π, . . . ,±nπ, . . . ,
it is reasonable to assume that
f(x) =
︷ ︸︸ ︷(
1 −x
π
) (
1 +x
π
) ︷ ︸︸ ︷(
1 −x
2π
) (
1 +x
2π
)
· · ·
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The Basel Problem
Since the roots of f(x) are
±π,±2π, . . . ,±nπ, . . . ,
it is reasonable to assume that
f(x) =
︷ ︸︸ ︷(
1 −x
π
)(
1 +x
π
) ︷ ︸︸ ︷(
1 −x
2π
) (
1 +x
2π
)
· · ·
=
(
1 −x2
π2
)(
1 −x2
4π2
)
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The Basel Problem
Since the roots of f(x) are
±π,±2π, . . . ,±nπ, . . . ,
it is reasonable to assume that
f(x) =
︷ ︸︸ ︷(
1 −x
π
) (
1 +x
π
) ︷ ︸︸ ︷(
1 −x
2π
) (
1 +x
2π
)
· · ·
=
(
1 −x2
π2
) (
1 −x2
4π2
)(
1 −x2
9π2
)
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The Basel Problem
Since the roots of f(x) are
±π,±2π, . . . ,±nπ, . . . ,
it is reasonable to assume that
f(x) =
︷ ︸︸ ︷(
1 −x
π
) (
1 +x
π
) ︷ ︸︸ ︷(
1 −x
2π
) (
1 +x
2π
)
· · ·
=
(
1 −x2
π2
) (
1 −x2
4π2
)(
1 −x2
9π2
)(
1 −x2
16π2
)
· · ·
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The Basel Problem
Since the roots of f(x) are
±π,±2π, . . . ,±nπ, . . . ,
it is reasonable to assume that
f(x) =
︷ ︸︸ ︷(
1 −x
π
) (
1 +x
π
) ︷ ︸︸ ︷(
1 −x
2π
) (
1 +x
2π
)
· · ·
=
(
1 −x2
π2
) (
1 −x2
4π2
)(
1 −x2
9π2
)(
1 −x2
16π2
)
· · ·
Now if we expand f(x) as if it were a polynomial, we cansee that the coefficient of x2 would be
−1
π2
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The Basel Problem
Since the roots of f(x) are
±π,±2π, . . . ,±nπ, . . . ,
it is reasonable to assume that
f(x) =
︷ ︸︸ ︷(
1 −x
π
) (
1 +x
π
) ︷ ︸︸ ︷(
1 −x
2π
) (
1 +x
2π
)
· · ·
=
(
1 −x2
π2
) (
1 −x2
4π2
)(
1 −x2
9π2
)(
1 −x2
16π2
)
· · ·
Now if we expand f(x) as if it were a polynomial, we cansee that the coefficient of x2 would be
−1
π2−
1
4π2
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The Basel Problem
Since the roots of f(x) are
±π,±2π, . . . ,±nπ, . . . ,
it is reasonable to assume that
f(x) =
︷ ︸︸ ︷(
1 −x
π
) (
1 +x
π
) ︷ ︸︸ ︷(
1 −x
2π
) (
1 +x
2π
)
· · ·
=
(
1 −x2
π2
) (
1 −x2
4π2
)(
1 −x2
9π2
)(
1 −x2
16π2
)
· · ·
Now if we expand f(x) as if it were a polynomial, we cansee that the coefficient of x2 would be
−1
π2−
1
4π2−
1
9π2− . . . .
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The Basel Problem
Now Euler also knew (as you will in a few years!) that wecan also express the function f as
f(x) =sin(x)
x= 1 −
x2
3!+
x4
5!− . . . +
(−1)nx2n
(2n + 1)!+ . . .
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The Basel Problem
Now Euler also knew (as you will in a few years!) that wecan also express the function f as
f(x) =sin(x)
x= 1 −
x2
3!+
x4
5!− . . . +
(−1)nx2n
(2n + 1)!+ . . .
Therefore we have the infinite product from the last slidemust be equal to the infinite sum above.
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The Basel Problem
Now Euler also knew (as you will in a few years!) that wecan also express the function f as
f(x) =sin(x)
x= 1 −
x2
3!+
x4
5!− . . . +
(−1)nx2n
(2n + 1)!+ . . .
Therefore we have the infinite product from the last slidemust be equal to the infinite sum above.The coefficient of x2 in the infinite sum is −1
6.
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The Basel Problem
Now Euler also knew (as you will in a few years!) that wecan also express the function f as
f(x) =sin(x)
x= 1 −
x2
3!+
x4
5!− . . . +
(−1)nx2n
(2n + 1)!+ . . .
Therefore we have the infinite product from the last slidemust be equal to the infinite sum above.The coefficient of x2 in the infinite sum is −1
6. Equating the
coefficients of x2 gives
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The Basel Problem
Now Euler also knew (as you will in a few years!) that wecan also express the function f as
f(x) =sin(x)
x= 1 −
x2
3!+
x4
5!− . . . +
(−1)nx2n
(2n + 1)!+ . . .
Therefore we have the infinite product from the last slidemust be equal to the infinite sum above.The coefficient of x2 in the infinite sum is −1
6. Equating the
coefficients of x2 gives
− 1
π2 −1
4π2−
1
9π2− . . . = −
1
6
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The Basel Problem
Now Euler also knew (as you will in a few years!) that wecan also express the function f as
f(x) =sin(x)
x= 1 −
x2
3!+
x4
5!− . . . +
(−1)nx2n
(2n + 1)!+ . . .
Therefore we have the infinite product from the last slidemust be equal to the infinite sum above.The coefficient of x2 in the infinite sum is −1
6. Equating the
coefficients of x2 gives
− 1
π2 −1
4π2−
1
9π2− . . . = −
1
6
⇒ 1 − 1
4−
1
9− . . . =
π2
6.
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The Basel Problem
It is interesting to note that Euler was not yet aware of thesymbol π for the number 3.14159265. . ., and stated theresult as follows:
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The Basel Problem
It is interesting to note that Euler was not yet aware of thesymbol π for the number 3.14159265. . ., and stated theresult as follows:
“I have discovered for the sum of six of this series tobe equal to the square of the circumference of acircle whose diameter is 1.”
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The Basel Problem
It is interesting to note that Euler was not yet aware of thesymbol π for the number 3.14159265. . ., and stated theresult as follows:
“I have discovered for the sum of six of this series tobe equal to the square of the circumference of acircle whose diameter is 1.”
The series he is talking about is the series
1 +1
4+
1
9+ . . . ,
Leonhard Euler Tercentennary — 2007 – p. 46/47
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The Basel Problem
It is interesting to note that Euler was not yet aware of thesymbol π for the number 3.14159265. . ., and stated theresult as follows:
“I have discovered for the sum of six of this series tobe equal to the square of the circumference of acircle whose diameter is 1.”
The series he is talking about is the series
1 +1
4+
1
9+ . . . ,
and the circumference of a circle whose diameter is 1 is, ofcourse, the number π!
Leonhard Euler Tercentennary — 2007 – p. 46/47
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Conclusion
Leonhard Euler Tercentennary — 2007 – p. 47/47