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## #251 2024-06-19 13:23:46

Jai Ganesh
Registered: 2005-06-28
Posts: 47,756

### Re: LaTeX - A Crash Course

Algebraic Identities - III

(a + b + c + d)^2

gives

is equal to

a^2 + b^2 + c^2 + d^2 + 2ab + 2ac + 2ad + 2bc + 2bd + 2cd

gives

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #252 2024-06-19 22:35:17

Jai Ganesh
Registered: 2005-06-28
Posts: 47,756

### Re: LaTeX - A Crash Course

Algebraic Identities - IV

(a + b - c)^2 = a^2 + b^2 + c^2 + 2ab - 2ac - 2bc
\text{gives this.}

.

(a - b + c)^2 = a^2 + b^2 + c^2 - 2ab + 2ac - 2bc
\text{gives this.}

(a - b - c)^2 = a^2 + b^2 + c^2 - 2ab - 2ac + 2bc
\text{gives this.}

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #253 2024-06-28 21:56:53

Jai Ganesh
Registered: 2005-06-28
Posts: 47,756

### Re: LaTeX - A Crash Course

Binomial Theorem - Part I

In elementary algebra, the binomial theorem (or binomial expansion) describes the algebraic expansion of powers of a binomial. According to the theorem, it is possible to expand the polynomial

(x + y)^n

gives

into a sum involving terms of the form axbyc, where the exponents b and c are nonnegative integers with b + c = n, and the coefficient a of each term is a specific positive integer depending on n and b. For example, for n = 4,

{\displaystyle (x+y)^{4}=x^{4}+4x^{3}y+6x^{2}y^{2}+4xy^{3}+y^{4}.}

gives

The coefficient a in the term of

ax^by^c

gives

is known as the binomial coefficient

{\displaystyle {\tbinom {n}{b}}}

gives

or

{\displaystyle {\tbinom {n}{c}}}

gives

(the two have the same value). These coefficients for varying n and b can be arranged to form Pascal's triangle. These numbers also occur in combinatorics, where

{\displaystyle {\tbinom {n}{b}}}

gives

gives the number of different combinations (i.e. subsets) of b elements that can be chosen from an n-element set. Therefore

{\displaystyle {\tbinom {n}{b}}}

gives

is usually pronounced as "n choose b".

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #254 2024-06-28 22:23:52

Jai Ganesh
Registered: 2005-06-28
Posts: 47,756

### Re: LaTeX - A Crash Course

Binomial Theorem - Part II

History

Special cases of the binomial theorem were known since at least the 4th century BC when Greek mathematician Euclid mentioned the special case of the binomial theorem for exponent n = 2. Greek mathematician Diophantus cubed various binomials, including

{\displaystyle x-1}

gives

. Indian mathematician Aryabhata's method for finding cube roots, from around 510 CE, suggests that he knew the binomial formula for exponent n=3.

Binomial coefficients, as combinatorial quantities expressing the number of ways of selecting k objects out of n without replacement, were of interest to ancient Indian mathematicians. The earliest known reference to this combinatorial problem is the Chandaḥśāstra by the Indian lyricist Pingala (c. 200 BC), which contains a method for its solution.  The commentator Halayudha from the 10th century AD explains this method.  By the 6th century AD, the Indian mathematicians probably knew how to express this as a quotient

{\textstyle {\frac {n!}{(n-k)!k!}}}

gives

, and a clear statement of this rule can be found in the 12th century text Lilavati by Bhaskara.

The first known formulation of the binomial theorem and the table of binomial coefficients appears in a work by Al-Karaji, quoted by Al-Samaw'al in his "al-Bahir". Al-Karaji described the triangular pattern of the binomial coefficients and also provided a mathematical proof of both the binomial theorem and Pascal's triangle, using an early form of mathematical induction. The Persian poet and mathematician Omar Khayyam was probably familiar with the formula to higher orders, although many of his mathematical works are lost. The binomial expansions of small degrees were known in the 13th century mathematical works of Yang Hui and also Chu Shih-Chieh. Yang Hui attributes the method to a much earlier 11th century text of Jia Xian, although those writings are now also lost.:

In 1544, Michael Stifel introduced the term "binomial coefficient" and showed how to use them to express

{\displaystyle (1+x)^{n}}

gives

in terms of

{\displaystyle (1+x)^{n-1}}

gives

, via "Pascal's triangle". Blaise Pascal studied the eponymous triangle comprehensively in his Traité du triangle arithmétique. However, the pattern of numbers was already known to the European mathematicians of the late Renaissance, including Stifel, Niccolò Fontana Tartaglia, and Simon Stevin.

Isaac Newton is generally credited with discovering the generalized binomial theorem, valid for any real exponent, in 1665. It was discovered independently in 1670 by James Gregory.

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #255 2024-06-28 23:41:18

Jai Ganesh
Registered: 2005-06-28
Posts: 47,756

### Re: LaTeX - A Crash Course

Binomial Theorem - Part III

Statement

According to the theorem, the expansion of any nonnegative integer power n of the binomial x + y is a sum of the form where

{\displaystyle {\tbinom {n}{k}}}

gives

is a positive integer known as a binomial coefficient.

This formula is also referred to as the binomial formula or the binomial identity. Using summation notation, it can be written more concisely as

{\displaystyle (x+y)^{n}=\sum _{k=0}^{n}{n \choose k}x^{n-k}y^{k}=\sum _{k=0}^{n}{n \choose k}x^{k}y^{n-k}.}

gives

The final expression follows from the previous one by the symmetry of x and y in the first expression, and by comparison it follows that the sequence of binomial coefficients in the formula is symmetrical,

{\textstyle {\binom {n}{k}}={\binom {n}{n-k}}.}

gives

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #256 2024-06-29 00:49:47

Jai Ganesh
Registered: 2005-06-28
Posts: 47,756

### Re: LaTeX - A Crash Course

Binomial Theorem - Part IV

According to the theorem, the expansion of any nonnegative integer power n of the binomial x + y is a sum of the form

{\displaystyle (x+y)^{n}={n \choose 0}x^{n}y^{0}+{n \choose 1}x^{n-1}y^{1}+{n \choose 2}x^{n-2}y^{2}+\cdots +{n \choose n-1}x^{1}y^{n-1}+{n \choose n}x^{0}y^{n},}

gives

where each

{\displaystyle {\binom {n}{k}}}

gives

is a positive integer known as a binomial coefficient, defined as

{\displaystyle {\binom {n}{k}}={\frac {n!}{k!\,(n-k)!}}={\frac {n(n-1)(n-2)\cdots (n-k+1)}{k(k-1)(k-2)\cdots 2\cdot 1}}.}

gives

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #257 2024-07-03 22:17:05

Jai Ganesh
Registered: 2005-06-28
Posts: 47,756

### Re: LaTeX - A Crash Course

Taylor Series - I

In mathematics, the Taylor series or Taylor expansion of a function is an infinite sum of terms that are expressed in terms of the function's derivatives at a single point. For most common functions, the function and the sum of its Taylor series are equal near this point. Taylor series are named after Brook Taylor, who introduced them in 1715. A Taylor series is also called a Maclaurin series when 0 is the point where the derivatives are considered, after Colin Maclaurin, who made extensive use of this special case of Taylor series in the 18th century.

The partial sum formed by the first n + 1 terms of a Taylor series is a polynomial of degree n that is called the nth Taylor polynomial of the function. Taylor polynomials are approximations of a function, which become generally more accurate as n increases. Taylor's theorem gives quantitative estimates on the error introduced by the use of such approximations. If the Taylor series of a function is convergent, its sum is the limit of the infinite sequence of the Taylor polynomials. A function may differ from the sum of its Taylor series, even if its Taylor series is convergent. A function is analytic at a point x if it is equal to the sum of its Taylor series in some open interval (or open disk in the complex plane) containing x. This implies that the function is analytic at every point of the interval (or disk).

Definition

The Taylor series of a real or complex-valued function f (x), that is infinitely differentiable at a real or complex number a, is the power series

{\displaystyle f(a)+{\frac {f'(a)}{1!}}(x-a)+{\frac {f''(a)}{2!}}(x-a)^{2}+{\frac {f'''(a)}{3!}}(x-a)^{3}+\cdots =\sum _{n=0}^{\infty }{\frac {f^{(n)}(a)}{n!}}(x-a)^{n}.}

gives

Here, n! denotes the factorial of n. The function

{f}^{(n)}(a)

gives

denotes the nth derivative of f evaluated at the point a. The derivative of order zero of f is defined to be f itself and

{(x - a)}^0

gives

and 0! are both defined to be 1. This series can be written by using sigma notation, as in the right side formula. With a = 0, the Maclaurin series takes the form:

{\displaystyle f(0)+{\frac {f'(0)}{1!}}x+{\frac {f''(0)}{2!}}x^{2}+{\frac {f'''(0)}{3!}}x^{3}+\cdots =\sum _{n=0}^{\infty }{\frac {f^{(n)}(0)}{n!}}x^{n}.}

gives

.

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #258 2024-07-05 00:43:53

Jai Ganesh
Registered: 2005-06-28
Posts: 47,756

### Re: LaTeX - A Crash Course

Taylor Series - II

The Taylor series of any polynomial is the polynomial itself.

The Maclaurin series of 1 - x is the geometric series

{\displaystyle 1+x+x^{2}+x^{3}+\cdots .}

gives

So, by substituting x for

\frac{1}{1 - x}

gives

, the Taylor series of
at a = 1 is

{\displaystyle 1-(x-1)+(x-1)^{2}-(x-1)^{3}+\cdots .}

gives

By integrating the above Maclaurin series, we find the Maclaurin series of ln(1 - x), where ln denotes the natural logarithm:

{\displaystyle -x-{\tfrac {1}{2}}x^{2}-{\tfrac {1}{3}}x^{3}-{\tfrac {1}{4}}x^{4}-\cdots .}

gives

The corresponding Taylor series of ln x at a = 1 is

{\displaystyle (x-1)-{\tfrac {1}{2}}(x-1)^{2}+{\tfrac {1}{3}}(x-1)^{3}-{\tfrac {1}{4}}(x-1)^{4}+\cdots ,}

gives

and more generally, the corresponding Taylor series of ln x at an arbitrary nonzero point a is:

{\displaystyle \ln a+{\frac {1}{a}}(x-a)-{\frac {1}{a^{2}}}{\frac {\left(x-a\right)^{2}}{2}}+\cdots .}

gives

.

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #259 2024-07-07 16:45:22

Jai Ganesh
Registered: 2005-06-28
Posts: 47,756

### Re: LaTeX - A Crash Course

Maclaurin series

The Maclaurin series of the exponential function

e^x

gives

is

{\displaystyle {\begin{aligned}\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}&={\frac {x^{0}}{0!}}+{\frac {x^{1}}{1!}}+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+{\frac {x^{4}}{4!}}+{\frac {x^{5}}{5!}}+\cdots \\&=1+x+{\frac {x^{2}}{2}}+{\frac {x^{3}}{6}}+{\frac {x^{4}}{24}}+{\frac {x^{5}}{120}}+\cdots .\end{aligned}}}

gives

The above expansion holds because the derivative of

e^x

gives

with respect to x is also

e^x,

gives

and

e^0

gives

equals 1. This leaves the terms

(x - 0)^n

gives

in the numerator and n! in the denominator of each term in the infinite sum.

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #260 2024-07-10 00:02:54

Jai Ganesh
Registered: 2005-06-28
Posts: 47,756

### Re: LaTeX - A Crash Course

Complex Number

Complex numbers are helpful in finding the square root of negative numbers. The concept of complex numbers was first referred to in the 1st century by a greek mathematician, Hero of Alexandria when he tried to find the square root of a negative number. But he merely changed the negative into positive and simply took the numeric root value. Further, the real identity of a complex number was defined in the 16th century by Italian mathematician Gerolamo Cardano, in the process of finding the negative roots of cubic and quadratic polynomial expressions.

Complex numbers have applications in many scientific research, signal processing, electromagnetism, fluid dynamics, quantum mechanics, and vibration analysis. Here we can understand the definition, terminology, visualization of complex numbers, properties, and operations of complex numbers.

What are Complex Numbers?

A complex number is the sum of a real number and an imaginary number. A complex number is of the form a + ib and is usually represented by z. Here both a and b are real numbers. The value 'a' is called the real part which is denoted by Re(z), and 'b' is called the imaginary part Im(z).  Also, ib is called an imaginary number.

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #261 2024-07-10 15:41:20

Jai Ganesh
Registered: 2005-06-28
Posts: 47,756

### Re: LaTeX - A Crash Course

i

gives

i^2 = -1

gives

i^3 = -i

gives

i^4 = 1

gives

i^5 = i

gives

.......

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #262 2024-07-10 16:03:56

Jai Ganesh
Registered: 2005-06-28
Posts: 47,756

### Re: LaTeX - A Crash Course

Value of i

It turns out that arithmetically,

i^2=-1

gives

. In other words, the value of i is the square root of −1. For now, don’t ask how! As we said earlier, the various different parts of the puzzle that is Complex Numbers will fall into place as you delve deeper into this subject.

At this point, just keep in mind that:-

a) i is a non-real number (it lies outside the Real set).

b) i represents one unit perpendicular to the Real direction.

c) yi represents y units perpendicular to the Real direction

d) x+iy represents the point (x, y)

e) i is the square root of -1, or,

i^2=-1

gives

.

When we say that i is a non-real number, we do not mean that i does not exist or is a figment of our imagination. We mean to say that i is non-real in the sense that it does not like in the Real set. However, it is a perfectly valid mathematical entity. i is also known by the name "iota".

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #263 2024-07-11 18:46:19

Jai Ganesh
Registered: 2005-06-28
Posts: 47,756

### Re: LaTeX - A Crash Course

Operation on Complex numbers

The various operations of addition, subtraction, multiplication, division of natural numbers can also be performed for complex numbers also. The details of the various arithmetic operations of complex numbers are as follows.

Th addition of complex numbers is similar to the addition of natural numbers. Here in complex numbers, the real part is added to the real part and the imaginary part is added to the imaginary part. For two complex numbers of the form

z_1 = a + id

gives

and

z_2 = c + id

gives

,
the sum of complex numbers

z_1 + z_2 = (a + c) + i(b + d)

gives

.

Closure Law: The sum of two complex numbers is also a complex number. For two complex numbers

z_1 \text {and} z_2

gives

,
the sum of

z_1 + z_2

gives

is also a complex number.

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #264 2024-07-11 19:30:18

Jai Ganesh
Registered: 2005-06-28
Posts: 47,756

### Re: LaTeX - A Crash Course

Operation of Complex Numbers - II

Commutative Law: For two complex numbers

z_1,z_2

gives

is

z_1 + z_2 = z_2 + z_1

gives

.

Associative Law: For the given three complex numbers

z_1, z_2, z_3

gives

we have

z_1 + (z_2 + z_3) = (z_1 + z_2) + z_3

gives

Additive Identity: For a complex number

z = a + ib

gives

, there exists

0 = 0 + i0

gives

, such that

z + 0 = 0 + z = 0

gives

.

Additive Inverse: For the complex number

z = a + ib

gives

, there exists a complex number

-z = -a -ib

gives

such that

z + (-z) = (-z) + z = 0

gives

Here -z is the additive inverse.

Subtraction of Complex Numbers

The subtraction of complex numbers follows a similar process of subtraction of natural numbers. Here for any two complex numbers, the subtraction is separately performed across the real part and then the subtraction is performed across the imaginary part. For the complex numbers

z_1 = a + ib, z_2 = c + id

gives

we have

z_1 - z_2  = (a - c) + i(b - d)

gives

.

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #265 2024-07-11 20:54:51

Jai Ganesh
Registered: 2005-06-28
Posts: 47,756

### Re: LaTeX - A Crash Course

Multiplication of Complex Numbers

The multiplication of complex numbers is slightly different from the multiplication of natural numbers. Here we need to use the formula of

i^2 = -1

gives

.

For the two complex numbers

z_1 = a + ib, z_2 = c + id

gives

the product is

z_1 \cdot z_2  = (ca - bd) + i(ad + bc)

gives

.

The multiplication of complex numbers is polar form is slightly different from the above mentioned form of multiplication. Here the absolute values of the two complex numbers are multiplied and their arguments are added to obtain the product of the complex numbers.

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #266 2024-07-11 22:43:33

Jai Ganesh
Registered: 2005-06-28
Posts: 47,756

### Re: LaTeX - A Crash Course

Multiplication of Complex Numbers - II

The product of two complex numbers is computed as follows:

{\displaystyle (a+bi)\cdot (c+di)=ac-bd+(ad+bc)i.}

gives

For example,

{\displaystyle (3+2i)(4-i)=3\cdot 4-(2\cdot (-1))+(3\cdot (-1)+2\cdot 4)i=14+5i.}

gives

In particular, this includes as a special case the fundamental formula

{\displaystyle i^{2}=i\cdot i=-1.}

gives

.

This formula distinguishes the complex number i from any real number, since the square of any (negative or positive) real number x always satisfies

{\displaystyle x^{2}\geq 0}

gives

.

This formula distinguishes the complex number i from any real number, since the square of any (negative or positive) real number x always satisfies

{\displaystyle x^{2}\geq 0}

gives

.

With this definition of multiplication and addition, familiar rules for the arithmetic of rational or real numbers continue to hold for complex numbers. More precisely, the distributive property, the commutative properties (of addition and multiplication) hold. Therefore, the complex numbers form an algebraic structure known as a field, the same way as the rational or real numbers do.

With this definition of multiplication and addition, familiar rules for the arithmetic of rational or real numbers continue to hold for complex numbers. More precisely, the distributive property, the commutative properties (of addition and multiplication) hold. Therefore, the complex numbers form an algebraic structure known as a field, the same way as the rational or real numbers do.

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #267 2024-07-12 00:00:27

Jai Ganesh
Registered: 2005-06-28
Posts: 47,756

### Re: LaTeX - A Crash Course

More on Complex numbers : I

For any complex number z = x + yi , the product

{\displaystyle z\cdot {\overline {z}}=(x+iy)(x-iy)=x^{2}+y^{2}}

gives

is a non-negative real number. This allows to define the absolute value (or modulus or magnitude) of z to be the square root

{\displaystyle |z|={\sqrt {x^{2}+y^{2}}}.}

gives

By Pythagoras' theorem,

{\displaystyle |z|}

gives

is the distance from the origin to the point representing the complex number z in the complex plane. In particular, the circle of radius one around the origin consists precisely of the numbers z such that

{\displaystyle |z|=1}.

gives

If

{\displaystyle z=x=x+0i}

gives

is a real number, then

{\displaystyle |z|=|x|}

gives

: its absolute value as a complex number and as a real number are equal.

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #268 2024-07-12 01:27:20

Jai Ganesh
Registered: 2005-06-28
Posts: 47,756

### Re: LaTeX - A Crash Course

More on Complex Numbers - II

Using the conjugate, the reciprocal of a nonzero complex number

{\displaystyle z=x+yi}

gives

can be computed to be

{\displaystyle {\frac {1}{z}}={\frac {\bar {z}}{z{\bar {z}}}}={\frac {\bar {z}}{|z|^{2}}}={\frac {x-yi}{x^{2}+y^{2}}}={\frac {x}{x^{2}+y^{2}}}-{\frac {y}{x^{2}+y^{2}}}i.}

gives

Using the conjugate, the reciprocal of a nonzero complex number

{\displaystyle z=x+yi}

gives

can be computed to be

{\displaystyle {\frac {1}{z}}={\frac {\bar {z}}{z{\bar {z}}}}={\frac {\bar {z}}{|z|^{2}}}={\frac {x-yi}{x^{2}+y^{2}}}={\frac {x}{x^{2}+y^{2}}}-{\frac {y}{x^{2}+y^{2}}}i.}

gives

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #269 2024-07-12 01:39:57

Jai Ganesh
Registered: 2005-06-28
Posts: 47,756

### Re: LaTeX - A Crash Course

More on Complex Numbers - III

More generally, the division of an arbitrary complex number

{\displaystyle w=u+vi}

gives

by a non-zero complex number

{\displaystyle z=x+yi}

gives

equals

{\displaystyle {\frac {w}{z}}={\frac {w{\bar {z}}}{|z|^{2}}}={\frac {(u+vi)(x-iy)}{x^{2}+y^{2}}}={\frac {ux+vy}{x^{2}+y^{2}}}+{\frac {vx-uy}{x^{2}+y^{2}}}i.}

gives

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #270 2024-07-12 13:27:32

Jai Ganesh
Registered: 2005-06-28
Posts: 47,756

### Re: LaTeX - A Crash Course

De Moivre's formula

In mathematics, de Moivre's formula (also known as de Moivre's theorem and de Moivre's identity) states that for any real number x and integer n it holds that

{\displaystyle {\big (}\cos x+i\sin x{\big )}^{n}=\cos nx+i\sin nx,}

gives

where
i is the imaginary unit

(i^2 = -1)

gives

.

The formula is named after Abraham de Moivre, although he never stated it in his works. The expression cos x + i sin x is sometimes abbreviated to cis x.

The formula is important because it connects complex numbers and trigonometry. By expanding the left hand side and then comparing the real and imaginary parts under the assumption that x is real, it is possible to derive useful expressions for cos nx and sin nx in terms of cos x and sin x.

As written, the formula is not valid for non-integer powers n. However, there are generalizations of this formula valid for other exponents. These can be used to give explicit expressions for the nth roots of unity, that is, complex numbers z such that

z^n = 1.

gives

Using the standard extensions of the sine and cosine functions to complex numbers, the formula is valid even when x is an arbitrary complex number.

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #271 2024-07-12 18:13:06

Jai Ganesh
Registered: 2005-06-28
Posts: 47,756

### Re: LaTeX - A Crash Course

De Moivre's formula - II

Example

For

{\displaystyle x=30^{\circ }}

gives

and

{\displaystyle n=2},

gives

,
de Moivre's formula asserts that

{\displaystyle \left(\cos(30^{\circ })+i\sin(30^{\circ })\right)^{2}=\cos(2\cdot 30^{\circ })+i\sin(2\cdot 30^{\circ }),}

gives

or equivalently that

{\displaystyle \left({\frac {\sqrt {3}}{2}}+{\frac {i}{2}}\right)^{2}={\frac {1}{2}}+{\frac {i{\sqrt {3}}}{2}}.}

gives

In this example, it is easy to check the validity of the equation by multiplying out the left side.

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #272 2024-07-12 20:13:04

Jai Ganesh
Registered: 2005-06-28
Posts: 47,756

### Re: LaTeX - A Crash Course

Relation to Euler's formula

De Moivre's formula is a precursor to Euler's formula

{\displaystyle e^{ix}=\cos x+i\sin x,}

gives

with x expressed in radians rather than degrees, which establishes the fundamental relationship between the trigonometric functions and the complex exponential function.

One can derive de Moivre's formula using Euler's formula and the exponential law for integer powers

{\displaystyle \left(e^{ix}\right)^{n}=e^{inx},}

gives

since Euler's formula implies that the left side is equal to

{\displaystyle \left(\cos x+i\sin x\right)^{n}}

gives

while the right side is equal to

{\displaystyle \cos nx+i\sin nx.}

gives

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #273 2024-07-12 23:09:41

Jai Ganesh
Registered: 2005-06-28
Posts: 47,756

### Re: LaTeX - A Crash Course

Powers and Roots

The n-th power of a complex number can be computed using de Moivre's formula, which is obtained by repeatedly applying the above formula for the product:

{\displaystyle z^{n}=\underbrace {z\cdot \dots \cdot z} _{n{\text{ factors}}}=(r(\cos \varphi +i\sin \varphi ))^{n}=r^{n}\,(\cos n\varphi +i\sin n\varphi ).}

gives

For example, the first few powers of the imaginary unit i are

{\displaystyle i,i^{2}=-1,i^{3}=-i,i^{4}=1,i^{5}=i,\dots }

gives

.

The n nth roots of a complex number z are given by

{\displaystyle z^{1/n}={\sqrt[{n}]{r}}\left(\cos \left({\frac {\varphi +2k\pi }{n}}\right)+i\sin \left({\frac {\varphi +2k\pi }{n}}\right)\right)}

gives

for

0 \leq k ≤ n - 1

gives

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #274 2024-07-13 00:28:07

Jai Ganesh
Registered: 2005-06-28
Posts: 47,756

### Re: LaTeX - A Crash Course

The History of this Theorem

In the 18th century complex numbers gained wider use, as it was noticed that formal manipulation of complex expressions could be used to simplify calculations involving trigonometric functions. For instance, in 1730 Abraham de Moivre noted that the identities relating trigonometric functions of an integer multiple of an angle to powers of trigonometric functions of that angle could be re-expressed by the following de Moivre's formula:

{\displaystyle (\cos \theta +i\sin \theta )^{n}=\cos n\theta +i\sin n\theta .}

gives

Euler's formula relates the complex exponential function of an imaginary argument, which can be thought of as describing uniform circular motion in the complex plane, to the cosine and sine functions, geometrically its projections onto the real and imaginary axes, respectively.

In 1748, Euler went further and obtained Euler's formula of complex analysis:

{\displaystyle e^{i\theta }=\cos \theta +i\sin \theta }

gives

by formally manipulating complex power series and observed that this formula could be used to reduce any trigonometric identity to much simpler exponential identities.

The idea of a complex number as a point in the complex plane was first described by Danish–Norwegian mathematician Caspar Wessel in 1799, although it had been anticipated as early as 1685 in Wallis's A Treatise of Algebra.

The English mathematician G.H. Hardy remarked that Gauss was the first mathematician to use complex numbers in "a really confident and scientific way" although mathematicians such as Norwegian Niels Henrik Abel and Carl Gustav Jacob Jacobi were necessarily using them routinely before Gauss published his 1831 treatise.

Augustin-Louis Cauchy and Bernhard Riemann together brought the fundamental ideas of complex analysis to a high state of completion, commencing around 1825 in Cauchy's case.

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #275 2024-07-13 14:06:07

Jai Ganesh
Registered: 2005-06-28
Posts: 47,756

### Re: LaTeX - A Crash Course

More on Complex Numbers

For any complex number z, with absolute value

{\displaystyle r=|z|}

gives

and argument

{\displaystyle \varphi }

gives

, the equation

{\displaystyle z=r(\cos \varphi +i\sin \varphi )}

gives

holds. This identity is referred to as the polar form of z. It is sometimes abbreviated as

{\textstyle z=r\operatorname {\mathrm {cis} } \varphi }.

gives

In electronics, one represents a phasor with amplitude r and phase

\varphi

gives

in angle notation:

{\displaystyle z=r\angle \varphi .}

gives

If two complex numbers are given in polar form, the product and division can be computed as

{\displaystyle z_{1}z_{2}=r_{1}r_{2}(\cos(\varphi _{1}+\varphi _{2})+i\sin(\varphi _{1}+\varphi _{2})).}

gives

,
if

{\displaystyle {\frac {z_{1}}{z_{2}}}={\frac {r_{1}}{r_{2}}}\left(\cos(\varphi _{1}-\varphi _{2})+i\sin(\varphi _{1}-\varphi _{2})\right),{\text{if }}z_{2}\neq 0.}

gives

(These are a consequence of the trigonometric identities for the sine and cosine function.) In other words, the absolute values are multiplied and the arguments are added to yield the polar form of the product. The picture at the right illustrates the multiplication of

{\displaystyle (2+i)(3+i)=5+5i.}

gives

Because the real and imaginary part of 5 + 5i are equal, the argument of that number is 45 degrees,
or

\frac{\pi}{4}}

gives

.

Thus, the formula

{\displaystyle {\frac {\pi }{4}}=\arctan \left({\frac {1}{2}}\right)+\arctan \left({\frac {1}{3}}\right)}

gives

holds. As the arctan function can be approximated highly efficiently, formulas like this – known as Machin-like formulas – are used for high-precision approximations of

\pi

gives

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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