Mathematics of the Renaissance

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Mathematics of the Renaissance
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The 15th and 16th centuries in Europe are often
referred to as the Renaissance period. The word
renaissance means rebirth and describes the
renewed interest in intellectual pursuits which
characterized the period. This interest focused
on literature, architecture, sculpture, painting,
and reviving the Greek and Roman cultures of the
past. Advances in science and mathematics came
along with a growing appreciation for learning in
general.
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In the 15th century Europeans began to develop
the art of navigation and to travel to distant
continents, bringing their culture with them. By
the end of the 16th century, Jesuit schools and
been established all over the world. As a result
European mathematics was taught and studied
everywhere, and eventually became the dominant
form of mathematics worldwide.
As European sailors began to travel to other
continents, solving the technical problems of
navigation became increasingly important.
Long-range navigation depended on astronomy and
understanding of the geometry of the sphere.
Astrology was also an important part of the
culture of this period, and making star chars
also depended on having a good grip on
trigonometry. For these reasons trigonometry was
one of the major themes of Renaissance
mathematics.
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An important work of Renaissance trigonometry
came from Johann Müller who worked at the
University of Vienna. Müller is more commonly
known by his Latinized name, Regiomontanus. In
the book On Triangles, Regiomontanus summarized
trigonometry in an understandable way, and his
text became the standard means by which students
learned the subject.
Most of On Triangles concerns ways of finding all
three sides and angles of a given triangle.
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Proposition 11.1 on page 304 is an example of a
result contained in On Triangles.
Rule 11.1 gives a procedure for doing what
Regiomantus claimed he could do in Proposition
11.1.
One of the homework problems asks you to apply
this procedure to a triangle with sides 3 and 4,
and base 5.
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The Renaissance also saw a growing interest in
arithmetic and algebra. With the rise of the
merchant class, more people found that they
needed to be able to compute. Since algebra was
thought of as a kind of generalized arithmetic,
it was natural for scholars to move from
arithmetic to algebra as they went deeper into
their studies.
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Nicholas Chuquet was a physician who lived in
Paris and made contributions to mathematics. His
book Triparty contains several techniques for
solving algebraic problems such as the rule of
three and double false position. The rule of
mean numbers which appears as Rule 11.2 on page
308 was his own original contribution.
One of the homework problems asks you to explain
why this rule works.
Chuquet used this rule to solve Problem 11.1.
One of the homework problems asks you to use this
method to find solutions to other equations.
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Section 11.2.2 describes how algebraic notation
as we use it today was developed during the
Renaissance. Until this point in the history of
mathematics, algebraic problems were either
presented verbally or described using a
combination of words and abbreviations (in the
manner of Diophantus). The fact that it took so
long to develop algebraic notation is a testament
to its conceptual difficulty. Many students have
a hard time making sense of the symbols used in
algebra. It took mathematicians centuries
(millennia, actually) of working verbally with
algebra until the concepts were familiar enough
to be able to meaningfully describe with abstract
labels.
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Another important advance in algebra made during
the Renaissance period was the general solution
of certain cubic equations. A cubic equation has
the form ax3 bx2 cx d 0. Since
Renaissance mathematicians were still not
comfortable using negative numbers, they would
write a cubic equations such as x3 2x 1 0
(with a 1, b 0, c 2 and d 1) as x3 2x
1.
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Renaissance scholars were typically supported by
rich patrons and had to earn their keep by
defeating other scholars in public competitions.
A series of such competitions prompted the study
and solution of cubic equations.
In 1535 the Italian Niccolo Fontana, better known
as Tartaglia, was challenged by Antonio Maria
Fiore. Each posed thirty questions to be
answered by the other. Tartaglia gave a variety
of challenges, but those posed by Fiore were all
cubic equations of the form x3 px q.
Tartaglia discovered a general method that
allowed him to solve all thirty problems and win
the competition.
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If x3 6x 20, then what is x?
Tartaglia probably wouldnt have told you. Until
he met Giralamo Cardano, he never told anybody.
He wanted to keep his method secret so that he
could win competitions with it. Cardano had a
friend who was from the same town as Tartaglia,
and this friend told Cardano about Tartaglias
victory in the competition. Cardano was
intrigued by the idea of solving cubic equations
using a general method because many prominent
mathematicians believed it was impossible.
Cardano went to visit Tartaglia and eventually
persuaded him to share his secret method. He
then promised never to share the method with
anyone else.
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Ludovico Ferrari worked for Cardano as a valet,
but showed promise as a scholar. Cardano taught
him several subjects, including mathematics. In
particular he taught Tartaglias secret method
for solving certain cubic equations. Ferrari
used this knowledge to solve the fourth degree
equation x4 6x2 30
60x, a result which Cardano wanted to include in
an algebra book he was writing. Unfortunately
Cardano had promised not to reveal Tartaglia
secret, and the solution to this equation
required the solution of a cubic equation in the
process. To get around the promise, Tartaglia
located the papers in which Fiore (the scholar
who had challenged Tartaglia in the first place)
had written his method for solving cubic
equations. Cardano presented Fiores methods in
his algebra book, but credited Tartaglia for
having independently discovered them.
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The methods we now use for solving cubic
equations are labeled with Cardanos name. He
published them in The Great Art, but he wasnt
the one who invented them.
Our book describes Cardanos methods for solving
three types of cubic equations. The types are
distinguished by which of the coefficients in the
equation ax3 bx2 cx d 0 are negative or
zero.
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Rule 11.3 on page 316 gives Cardanos method for
solving equations of the form x3 px q (cube
and things equal the number) in the way it
appeared in The Great Art.
Cardano wrote his method in words whereas we
would use algebraic notation. Lets translate
Cardanos method into algebraic notation.
Lets use our formula to solve x3 6x 20.
Example 11.2 gives Cardanos more verbal way of
solving the equation.
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The way Cardano solved the equation x3 px q
was to write q as a difference of cubes (q u3
v3) and p as three times the product of the sides
of the cubes (p 3uv). When he did this he
could use geometry to prove that x was the
difference in the sides (x u v ).
One of the homework problems asks you to verify
that if you have an equation of the form x3 px
q where q u3 v3 and p 3uv, then x u v
is a solution to the equation. What does this
mean you should do?
Rules 11.4 and 11.5 give Cardanos method for
solving other types of cubic equations. To use
them, first translate them into algebraic
notation like we did with Rule 11.3. Lets do
Rule 11.4 now.
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What do the equations solved by Rules 11.3, 11.4,
and 11.5 all have in common?
They dont have an x2 term. Cardano solved cubic
equations that did have a quadratic term by first
converting them to cubic equations that didnt.
The equation x3 6x2 10 can be converted into
a cubic equation with no quadratic term by making
the substitution x y 2. What happens
when we make this substitution, and why did we
choose y 2?
How do we obtain an answer to the original
equation from the new one?
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One of the homework problems asks you to show how
the equation x3 6x2 100 can be converted into
the equation y3 12y 116 using the
substitution x y 2.
Exercise 6 on page 321 asks you to first find a
substitution that will turn a given cubic
equations with a quadratic term into an equation
without a quadratic term, then solve the
equation. Lets do part (a).
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An argument started between Cardano and Tartaglia
because Cardano wanted to publish the Ferraris
solution to the fourth degree equation x4 6x2
36 60x. (Page 318 reproduces this solution.)
The solution involves transforming the equation
so that both sides are perfect squares. Ferrari
did this so that he could take the square root of
both sides and be left with a quadratic equation
which he could solve. Along the way a cubic
equation also had to be solved. For this reason
Cardano needed to publish Tartaglias secret
method which he had promised never to reveal.
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Cardanos methods often led to solutions that
required taking the square root of a negative
number. He called such square roots impossible
roots. He observed that when he ignored the
mental tortures which resulted from
contemplating impossible roots, and blindly
performed arithmetic with them, then the
impossible roots did yield correct answers in the
end.
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Cardano was concerned about general solutions to
cubic equations, not because they had practical
value, but rather because he was attracted by the
challenge of doing something so many considered
difficult or even impossible. If he needed the
solution to a cubic equation for a practical
purpose, he could have used one of the various
known methods for finding approximate solutions
to cubic equations. Such approximate solutions
could be made accurate as desired. Cardano
included two methods for obtaining approximate
solutions to cubic equations in The Great Art.
One method is now called the method of secants
and appears on page 319. It is based on the
technique of double false position.