Polynomials that no one can solve! Supriya Pisolkar IISER Pune - - PowerPoint PPT Presentation

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polynomials that no one can solve

Polynomials that no one can solve! Supriya Pisolkar IISER Pune - - PowerPoint PPT Presentation

Feb 19, 2024 280 likes •1.24k views

Polynomials that no one can solve! Supriya Pisolkar IISER Pune April 16, 2017 S. Pisolkar (IISER Pune) Polynomials that no one can solve! April 16, 2017 1 / 23 What are polynomials? S. Pisolkar (IISER Pune) Polynomials that no one can

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Polynomials that no one can solve!

Supriya Pisolkar

IISER Pune

April 16, 2017

S. Pisolkar (IISER Pune)

Polynomials that no one can solve! April 16, 2017 1 / 23

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What are polynomials?

S. Pisolkar (IISER Pune)

Polynomials that no one can solve! April 16, 2017 2 / 23

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What are polynomials?

X + 1

S. Pisolkar (IISER Pune)

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What are polynomials?

X + 1 X 2 + 2

S. Pisolkar (IISER Pune)

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What are polynomials?

X + 1 X 2 + 2 3X 3 + 2X 2 − 5

S. Pisolkar (IISER Pune)

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What are polynomials?

X + 1 X 2 + 2 3X 3 + 2X 2 − 5 In general a polynomial can be expressed as anX n + an−1X n−1 + · · · + a1X + a0

S. Pisolkar (IISER Pune)

Polynomials that no one can solve! April 16, 2017 2 / 23

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What are polynomials?

X + 1 X 2 + 2 3X 3 + 2X 2 − 5 In general a polynomial can be expressed as anX n + an−1X n−1 + · · · + a1X + a0 where ai are some numbers.

S. Pisolkar (IISER Pune)

Polynomials that no one can solve! April 16, 2017 2 / 23

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What are polynomials?

X + 1 X 2 + 2 3X 3 + 2X 2 − 5 In general a polynomial can be expressed as anX n + an−1X n−1 + · · · + a1X + a0 where ai are some numbers.

S. Pisolkar (IISER Pune)

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Degree of a polynomial

The highest integer power n appearing in anX n + an−1X n−1 · · · + a2X 2 + a1X + a0 is called the degree of a polynomial.

S. Pisolkar (IISER Pune)

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Degree of a polynomial

The highest integer power n appearing in anX n + an−1X n−1 · · · + a2X 2 + a1X + a0 is called the degree of a polynomial. Type Example Degree Linear X + 1 1 Quadratic X 2 + 2X + 1 2 Cubic X 3 + 2X 3 Quartic 2X 4 + x3 + 2 4 Quintic X 5 + 1 5

S. Pisolkar (IISER Pune)

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History of polynomials

S. Pisolkar (IISER Pune)

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History of polynomials

Egyptians and Babylonians (∼ 4000 years ago ) :

S. Pisolkar (IISER Pune)

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History of polynomials

Egyptians and Babylonians (∼ 4000 years ago ) : Problem: Given a specific area, they were unable to calculate lengths of the sides of certain shapes, and without these lengths they were unable to design floor plan for their kingdom.

S. Pisolkar (IISER Pune)

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Solving polynomials

S. Pisolkar (IISER Pune)

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Solving polynomials

Solving a polynomial p(X) means finding numbers which when substituted in place of X give zero.

S. Pisolkar (IISER Pune)

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Solving polynomials

Solving a polynomial p(X) means finding numbers which when substituted in place of X give zero. This is also called finding roots of a polynomial p(X).

S. Pisolkar (IISER Pune)

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Solving polynomials

Solving a polynomial p(X) means finding numbers which when substituted in place of X give zero. This is also called finding roots of a polynomial p(X). p(X) = X + 1,

S. Pisolkar (IISER Pune)

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Solving polynomials

Solving a polynomial p(X) means finding numbers which when substituted in place of X give zero. This is also called finding roots of a polynomial p(X). p(X) = X + 1, If we put X = −1,

S. Pisolkar (IISER Pune)

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Solving polynomials

Solving a polynomial p(X) means finding numbers which when substituted in place of X give zero. This is also called finding roots of a polynomial p(X). p(X) = X + 1, If we put X = −1, p(−1) = (−1) + 1 = 0

S. Pisolkar (IISER Pune)

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Solving polynomials

Solving a polynomial p(X) means finding numbers which when substituted in place of X give zero. This is also called finding roots of a polynomial p(X). p(X) = X + 1, If we put X = −1, p(−1) = (−1) + 1 = 0 So −1 is a root of X + 1.

S. Pisolkar (IISER Pune)

Polynomials that no one can solve! April 16, 2017 5 / 23

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Solving polynomials

Solving a polynomial p(X) means finding numbers which when substituted in place of X give zero. This is also called finding roots of a polynomial p(X). p(X) = X + 1, If we put X = −1, p(−1) = (−1) + 1 = 0 So −1 is a root of X + 1. Let p(X) = (X − 5)2 = (X − 5) · (X − 5),

S. Pisolkar (IISER Pune)

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Solving polynomials

Solving a polynomial p(X) means finding numbers which when substituted in place of X give zero. This is also called finding roots of a polynomial p(X). p(X) = X + 1, If we put X = −1, p(−1) = (−1) + 1 = 0 So −1 is a root of X + 1. Let p(X) = (X − 5)2 = (X − 5) · (X − 5), Substitute, X = 5

S. Pisolkar (IISER Pune)

Polynomials that no one can solve! April 16, 2017 5 / 23

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Solving polynomials

Solving a polynomial p(X) means finding numbers which when substituted in place of X give zero. This is also called finding roots of a polynomial p(X). p(X) = X + 1, If we put X = −1, p(−1) = (−1) + 1 = 0 So −1 is a root of X + 1. Let p(X) = (X − 5)2 = (X − 5) · (X − 5), Substitute, X = 5 p(5) = (5 − 5)2 = 0, So, 5 is a root of (X − 5)2.

S. Pisolkar (IISER Pune)

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Square root of a number

S. Pisolkar (IISER Pune)

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Square root of a number

For a positive integer a, there are two numbers √a and −√a such that (√a)2 = a and (−√a)2 = a

S. Pisolkar (IISER Pune)

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Square root of a number

For a positive integer a, there are two numbers √a and −√a such that (√a)2 = a and (−√a)2 = a So, these two numbers are roots of the polynomial X 2 = a

S. Pisolkar (IISER Pune)

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Square root of a number

For a positive integer a, there are two numbers √a and −√a such that (√a)2 = a and (−√a)2 = a So, these two numbers are roots of the polynomial X 2 = a They are called square roots of a.

S. Pisolkar (IISER Pune)

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Square root of a number

For a positive integer a, there are two numbers √a and −√a such that (√a)2 = a and (−√a)2 = a So, these two numbers are roots of the polynomial X 2 = a They are called square roots of a. (4)2 = 16 so, we write √ 16 = 4. The sign √· is called a radical sign.

S. Pisolkar (IISER Pune)

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Solving a general Quadratic polynomials

S. Pisolkar (IISER Pune)

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Solving a general Quadratic polynomials

Consider a simple quadratic polynomial p(X) = X 2 + 5X + 6.

S. Pisolkar (IISER Pune)

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Solving a general Quadratic polynomials

Consider a simple quadratic polynomial p(X) = X 2 + 5X + 6. How to find a root of this equation?

S. Pisolkar (IISER Pune)

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Solving a general Quadratic polynomials

Consider a simple quadratic polynomial p(X) = X 2 + 5X + 6. How to find a root of this equation? Nearly 1400 years ago Brahmagupta, an Indian mathematician, gave the first explicit root of p(X).

S. Pisolkar (IISER Pune)

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Solving a general Quadratic polynomials

Consider a simple quadratic polynomial p(X) = X 2 + 5X + 6. How to find a root of this equation? Nearly 1400 years ago Brahmagupta, an Indian mathematician, gave the first explicit root of p(X).

S. Pisolkar (IISER Pune)

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Roots of a quadratic polynomials

In general, for a quadratic equation aX 2 + bX + c, there are two roots;

S. Pisolkar (IISER Pune)

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Roots of a quadratic polynomials

In general, for a quadratic equation aX 2 + bX + c, there are two roots; Muhammad ibn Musa al-Khwarizmi (Baghdad, 1100 years ago), inspired by Brahmagupta, developed a set of formulas that worked for x = −b + ( √ b2 − 4ac) 2a , x = −b − ( √ b2 − 4ac) 2a

S. Pisolkar (IISER Pune)

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Solving Cubic polynomials

S. Pisolkar (IISER Pune)

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Solving Cubic polynomials

Gerolamo Cardano ( Italy, 600 years ago ) first published the formula for roots of cubic polynomials

Figure: Gerolamo Cardano (1501-1576)

S. Pisolkar (IISER Pune)

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Solving Cubic polynomials

S. Pisolkar (IISER Pune)

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Solving Cubic polynomials

Consider a cubic equation aX 3 + bX 2 + cX + d = 0 Formula for a root of this polynomial is given by

S. Pisolkar (IISER Pune)

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Quartic polynomial

S. Pisolkar (IISER Pune)

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Quartic polynomial

What happens to quartic polynomials?

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Quartic polynomial

What happens to quartic polynomials? There is a analogous formula for finding roots given by Lodovico Ferrari (1545), a student of Cardano but it is much worse and I won’t even try to write it down here!!

S. Pisolkar (IISER Pune)

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Mysterious Quintic polynomials

S. Pisolkar (IISER Pune)

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Mysterious Quintic polynomials

For next nearly 400 years people tried to solve quintic polynomials!

S. Pisolkar (IISER Pune)

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Mysterious Quintic polynomials

For next nearly 400 years people tried to solve quintic polynomials! Then came Evariste Galois ( who was only 20 years old at that time), and proved that a general quintic polynomial need not have a root that can be expressed by radicals.

S. Pisolkar (IISER Pune)

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Mysterious Quintic polynomials

For next nearly 400 years people tried to solve quintic polynomials! Then came Evariste Galois ( who was only 20 years old at that time), and proved that a general quintic polynomial need not have a root that can be expressed by radicals. Allegedly he was in love with a woman engaged to an artillery officer. Galois challenged the officer in order to win her affection. In the nights leading up to the duel, he did write many letters to his friends/colleagues. The most significant of these had the ideas which are foundation of Galois Theory which brought to light non-solvability of a general quintic.

S. Pisolkar (IISER Pune)

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Who am ‘i’ ?

S. Pisolkar (IISER Pune)

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Who am ‘i’ ?

Question

Is it true that every polynomial will always have a solution somewhere?

S. Pisolkar (IISER Pune)

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Who am ‘i’ ?

Question

Is it true that every polynomial will always have a solution somewhere? A root of X 2 = 1 which is i = √−1 does not belong to the set of real numbers.

S. Pisolkar (IISER Pune)

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Who am ‘i’ ?

Question

Is it true that every polynomial will always have a solution somewhere? A root of X 2 = 1 which is i = √−1 does not belong to the set of real numbers. Thus we need to extend the set of real numbers by introducing, or ‘adjoining’, i.

S. Pisolkar (IISER Pune)

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Who am ‘i’ ?

Question

Is it true that every polynomial will always have a solution somewhere? A root of X 2 = 1 which is i = √−1 does not belong to the set of real numbers. Thus we need to extend the set of real numbers by introducing, or ‘adjoining’, i.

S. Pisolkar (IISER Pune)

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Complex Numbers C

S. Pisolkar (IISER Pune)

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Complex Numbers C

Now lets ‘adjoin’ i = √−1 to the set of reals.

S. Pisolkar (IISER Pune)

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Complex Numbers C

Now lets ‘adjoin’ i = √−1 to the set of reals. R(i) := {a + bi} where a, b are real numbers and i = √−1.

S. Pisolkar (IISER Pune)

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Complex Numbers C

Now lets ‘adjoin’ i = √−1 to the set of reals. R(i) := {a + bi} where a, b are real numbers and i = √−1. This is called the set set of complex numbers.

S. Pisolkar (IISER Pune)

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Complex Numbers C

Now lets ‘adjoin’ i = √−1 to the set of reals. R(i) := {a + bi} where a, b are real numbers and i = √−1. This is called the set set of complex numbers. Example: −1 + 3i,

S. Pisolkar (IISER Pune)

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Complex Numbers C

Now lets ‘adjoin’ i = √−1 to the set of reals. R(i) := {a + bi} where a, b are real numbers and i = √−1. This is called the set set of complex numbers. Example: −1 + 3i, 2 + i, −3i

S. Pisolkar (IISER Pune)

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Complex Numbers C

Now lets ‘adjoin’ i = √−1 to the set of reals. R(i) := {a + bi} where a, b are real numbers and i = √−1. This is called the set set of complex numbers. Example: −1 + 3i, 2 + i, −3i

S. Pisolkar (IISER Pune)

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Complex Numbers C

Now lets ‘adjoin’ i = √−1 to the set of reals. R(i) := {a + bi} where a, b are real numbers and i = √−1. This is called the set set of complex numbers. Example: −1 + 3i, 2 + i, −3i

S. Pisolkar (IISER Pune)

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Fundamental theorem of Algebra

S. Pisolkar (IISER Pune)

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Fundamental theorem of Algebra

By using complex numbers, Carl Friedrich Gauss( Germany), proved a following remarkable result.

S. Pisolkar (IISER Pune)

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Fundamental theorem of Algebra

By using complex numbers, Carl Friedrich Gauss( Germany), proved a following remarkable result.

Figure: Carl Friedrich Gauss (1777-1855)

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Fundamental theorem of Algebra

By using complex numbers, Carl Friedrich Gauss( Germany), proved a following remarkable result.

Figure: Carl Friedrich Gauss (1777-1855)

‘Every polynomial X n + an−1X n−1 + · · · + a1X + a0 of degree n > 0 where a′

is are complex numbers, has a root in complex numbers’.

S. Pisolkar (IISER Pune)

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Sets

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Sets

Consider the set of integers : {· · · , −2, −1, 0, 1, 2, · · · }

S. Pisolkar (IISER Pune)

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Sets

Consider the set of integers : {· · · , −2, −1, 0, 1, 2, · · · } Take two integers m and n, then m + n is again a integer.

S. Pisolkar (IISER Pune)

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Sets

Consider the set of integers : {· · · , −2, −1, 0, 1, 2, · · · } Take two integers m and n, then m + n is again a integer. There is a special element called zero 0 such that m + 0 = m.

S. Pisolkar (IISER Pune)

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Sets

Consider the set of integers : {· · · , −2, −1, 0, 1, 2, · · · } Take two integers m and n, then m + n is again a integer. There is a special element called zero 0 such that m + 0 = m. Also, there is an integer −m such that m + (−m) = 0.

S. Pisolkar (IISER Pune)

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Groups

A group is a set G, with an operation +, such that:

S. Pisolkar (IISER Pune)

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Groups

A group is a set G, with an operation +, such that: (1) a, b are in G then a + b is in G.

S. Pisolkar (IISER Pune)

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Groups

A group is a set G, with an operation +, such that: (1) a, b are in G then a + b is in G. (2) G contains a special element called identity e.

S. Pisolkar (IISER Pune)

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Groups

A group is a set G, with an operation +, such that: (1) a, b are in G then a + b is in G. (2) G contains a special element called identity e. (3) G contains inverses i.e. if a is in G then there is an element −a in G such that a + (−a) = e.

S. Pisolkar (IISER Pune)

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Groups

A group is a set G, with an operation +, such that: (1) a, b are in G then a + b is in G. (2) G contains a special element called identity e. (3) G contains inverses i.e. if a is in G then there is an element −a in G such that a + (−a) = e. Example: the set of integer numbers

S. Pisolkar (IISER Pune)

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Fields

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Fields

The set of rational numbers, Q := {m n where m and n(= 0) are integers}

S. Pisolkar (IISER Pune)

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Fields

The set of rational numbers, Q := {m n where m and n(= 0) are integers} Then Q is a group.

S. Pisolkar (IISER Pune)

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Fields

The set of rational numbers, Q := {m n where m and n(= 0) are integers} Then Q is a group. There is one more operation on Q which is multiplication. Also, every non-zero rational number has a inverse.

S. Pisolkar (IISER Pune)

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Fields

The set of rational numbers, Q := {m n where m and n(= 0) are integers} Then Q is a group. There is one more operation on Q which is multiplication. Also, every non-zero rational number has a inverse. A set with two operations + and ’·’ where every non-zero element has a inverse is called a field.

S. Pisolkar (IISER Pune)

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Fields

The set of rational numbers, Q := {m n where m and n(= 0) are integers} Then Q is a group. There is one more operation on Q which is multiplication. Also, every non-zero rational number has a inverse. A set with two operations + and ’·’ where every non-zero element has a inverse is called a field. Example: Q, R, C.

S. Pisolkar (IISER Pune)

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Polynomials and groups

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Polynomials and groups

Consider a polynomial p(X) which has no roots in Q.

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Polynomials and groups

Consider a polynomial p(X) which has no roots in Q. Example: X 2 + 1 has no root in Q.

S. Pisolkar (IISER Pune)

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Polynomials and groups

Consider a polynomial p(X) which has no roots in Q. Example: X 2 + 1 has no root in Q. One constructs a field Q(α) which contains a root α of p(X)

S. Pisolkar (IISER Pune)

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Polynomials and groups

Consider a polynomial p(X) which has no roots in Q. Example: X 2 + 1 has no root in Q. One constructs a field Q(α) which contains a root α of p(X) Example: Q(i) contains a root i of X 2 + 1.

S. Pisolkar (IISER Pune)

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Polynomials and groups

Consider a polynomial p(X) which has no roots in Q. Example: X 2 + 1 has no root in Q. One constructs a field Q(α) which contains a root α of p(X) Example: Q(i) contains a root i of X 2 + 1. When we ‘adjoin’ all roots of p(X) to Q, then that ‘bigger’ field is called the splitting field of that polynomial.

S. Pisolkar (IISER Pune)

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Polynomials and groups

Consider a polynomial p(X) which has no roots in Q. Example: X 2 + 1 has no root in Q. One constructs a field Q(α) which contains a root α of p(X) Example: Q(i) contains a root i of X 2 + 1. When we ‘adjoin’ all roots of p(X) to Q, then that ‘bigger’ field is called the splitting field of that polynomial.

S. Pisolkar (IISER Pune)

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Galois theory

Galois showed that, for a polynomial p(X) , there is a way to associate; p(X) → splitting field → Galois group

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SLIDE 88

Galois theory

Galois showed that, for a polynomial p(X) , there is a way to associate; p(X) → splitting field → Galois group The association is such that to study roots of a polynomial it is enough to study the corresponding group and vice-versa.

S. Pisolkar (IISER Pune)

Polynomials that no one can solve! April 16, 2017 20 / 23

slide-89

SLIDE 89

Galois theory

Galois showed that, for a polynomial p(X) , there is a way to associate; p(X) → splitting field → Galois group The association is such that to study roots of a polynomial it is enough to study the corresponding group and vice-versa. He proved that roots of a general quintic polynomial can not be expressed in terms of radicals whenever the associated group is ‘non-solvable’.

S. Pisolkar (IISER Pune)

Polynomials that no one can solve! April 16, 2017 20 / 23

slide-90

SLIDE 90

Summary

S. Pisolkar (IISER Pune)

Polynomials that no one can solve! April 16, 2017 21 / 23

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SLIDE 91

Summary

There are some mysterious polynomials of degree 5 and higher whose solutions can not be expressed by using radicals i.e. √·,

3

√·,

4

√·

S. Pisolkar (IISER Pune)

Polynomials that no one can solve! April 16, 2017 21 / 23

slide-92

SLIDE 92

Summary

There are some mysterious polynomials of degree 5 and higher whose solutions can not be expressed by using radicals i.e. √·,

3

√·,

4

√· You can think of it as if, some of the functions are missing from your calculator.!

S. Pisolkar (IISER Pune)

Polynomials that no one can solve! April 16, 2017 21 / 23

slide-93

SLIDE 93

Summary

There are some mysterious polynomials of degree 5 and higher whose solutions can not be expressed by using radicals i.e. √·,

3

√·,

4

√· You can think of it as if, some of the functions are missing from your calculator.! One such polynomial is X 5 − 4X + 2.

S. Pisolkar (IISER Pune)

Polynomials that no one can solve! April 16, 2017 21 / 23

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SLIDE 94

Timeline

3700 years ago Egyptians made a table. 3600 years ago Pythagoras worked with integers and rational numbers. 2500 years ago Babylonians solved some quadratic equations. 1100 years ago Al-Khwarizmi proved a formula for roots of a quadratic. 600 years ago Cardano gave a formula for finding roots of a cubic. 600 year ago Ferrari proved a formula for roots of quartic polynomials. 217 years ago Gauss proved Fundamental theorem of Algebra. 200 years ago Galois proved non-solvability of some quintic polynomials.

S. Pisolkar (IISER Pune)

Polynomials that no one can solve! April 16, 2017 22 / 23

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SLIDE 95

Thank You!

S. Pisolkar (IISER Pune)

Polynomials that no one can solve! April 16, 2017 23 / 23