Draft:Original research/MacLaurin series

A MacLaurin series is a Taylor series that has a term at (0,0).

A Taylor series is a representation of a function as an infinite sum of terms that are calculated from the values of the function's derivatives at a single point.^[1][2][3]

Notation: let the symbol ${\displaystyle \Delta }$ represent difference in a variable.

Notation: let the symbol ${\displaystyle d}$ represent an infinitesimal difference in a variable.

Notation: let the symbol ${\displaystyle \partial }$ represent an infinitesimal difference in one input to a function of more than one input.

Let

${\displaystyle y=f(x)}$

be a function where values of ${\displaystyle x}$ may be any real number and values resulting in ${\displaystyle y}$ are also any real number.

${\displaystyle \Delta x}$ is a small finite difference in ${\displaystyle x}$ which when put into the function ${\displaystyle f(x)}$ produces a ${\displaystyle \Delta y}$.

These small differences can be manipulated with the operations of arithmetic: addition (${\displaystyle +}$), subtraction (${\displaystyle -}$), multiplication (${\displaystyle *}$), and division (${\displaystyle /}$).

${\displaystyle \Delta y=f(x+\Delta x)-f(x)}$

Dividing ${\displaystyle \Delta y}$ by ${\displaystyle \Delta x}$ and taking the limit as ${\displaystyle \Delta x}$ → 0, produces the slope of a line tangent to f(x) at the point x.

For example,

${\displaystyle f(x)=x^{2}}$

${\displaystyle f(x+\Delta x)=(x+\Delta x)^{2}=x^{2}+2x\Delta x+(\Delta x)^{2}}$

${\displaystyle \Delta y/\Delta x=(x^{2}+2x\Delta x+(\Delta x)^{2}-x^{2})/\Delta x}$

${\displaystyle \Delta y/\Delta x=2x+\Delta x}$

as ${\displaystyle \Delta x}$ and${\displaystyle \Delta y}$ go towards zero,

${\displaystyle dy/dx=2x+dx=limit_{\Delta x\to 0}{f(x+\Delta x)-f(x) \over \Delta x}=2x.}$

This ratio is called the derivative.

Let

${\displaystyle y=f(x,z)}$

then

${\displaystyle \partial y=\partial f(x,z)=\partial f(x,z)\partial x+\partial f(x,z)\partial z}$

${\displaystyle \partial y/\partial x=\partial f(x,z)}$

where z is held constant and

${\displaystyle \partial y/\partial z=\partial f(x,z)}$

where x is held contstant.

Notation: let the symbol ${\displaystyle \nabla }$ be the gradient, i.e., derivatives for multivariable functions.

${\displaystyle \nabla f(x,z)=\partial y=\partial f(x,z)=\partial f(x,z)\partial x+\partial f(x,z)\partial z.}$

For

${\displaystyle \Delta x*\Delta y=[f(x+\Delta x)-f(x)]*\Delta x}$

the area under the curve shown in the diagram at right is the light purple rectangle plus the dark purple rectangle in the top figure

${\displaystyle \Delta x*\Delta y+f(x)*\Delta x=f(x+\Delta x)*\Delta x.}$

Any particular individual rectangle for a sum of rectangular areas is

${\displaystyle f(x_{i}+\Delta x_{i})*\Delta x_{i}.}$

The approximate area under the curve is the sum ${\displaystyle \sum }$ of all the individual (i) areas from i = 0 to as many as the area needed (n):

${\displaystyle \sum _{i=0}^{n}f(x_{i}+\Delta x_{i})*\Delta x_{i}.}$

Notation: let the symbol ${\displaystyle \int }$ represent the integral.

${\displaystyle limit_{\Delta x\to 0}\sum _{i=0}^{n}f(x_{i}+\Delta x_{i})*\Delta x_{i}=\int f(x)dx.}$

This can be within a finite interval [a,b]

${\displaystyle \int _{a}^{b}f(x)\;dx}$

when i = 0 the integral is evaluated at ${\displaystyle a}$ and i = n the integral is evaluated at ${\displaystyle b}$. Or, an indefinite integral (without notation on the integral symbol) as n goes to infinity and i = 0 is the integral evaluated at x = 0.

Def. a branch of mathematics that deals with the finding and properties ... of infinitesimal differences is called a calculus.

Calculus focuses on limits, functions, derivatives, integrals, and infinite series.

"Although calculus (in the sense of analysis) is usually synonymous with infinitesimal calculus, not all historical formulations have relied on infinitesimals (infinitely small numbers that are nevertheless not zero)."^[4]

Taylor Series:

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

where fⁿ refers to the number (n) of derivatives taken.

A MacLaurin series of a function ƒ(x) for which a derivative may be taken of the function or any of its derivatives at 0 is the power series

${\displaystyle f(0)+{\frac {f'(0)}{1!}}(x)+{\frac {f''(0)}{2!}}(x)^{2}+{\frac {f^{(3)}(0)}{3!}}(x)^{3}+\cdots .}$

which can be written in the more compact sigma, or summation, notation as

${\displaystyle \sum _{n=0}^{\infty }{\frac {f^{(n)}(0)}{n!}}\,(x)^{n}}$

where n! denotes the factorial of n and ƒ ⁽ⁿ⁾(0) denotes the nth derivative of ƒ evaluated at the point 0. The derivative of order zero ƒ is defined to be ƒ itself and (x)⁰ and 0! are both defined to be 1.

Taylor series is defined as

${\displaystyle \sum _{n=0}^{\infty }{\frac {f^{(n)}(t)}{n!}}\,(x-t)^{n}}$

The MacLaurin series occurs when t=0

${\displaystyle \sum _{n=0}^{\infty }{\frac {f^{(n)}(0)}{n!}}\,(x)^{n}}$

The derivatives are

${\displaystyle {y}'=e^{x}}$

${\displaystyle {y}''=e^{x}}$

${\displaystyle {y}'''=e^{x}}$

.

.

.

Development of MacLaurin series for ${\displaystyle e^{x}}$

${\displaystyle e^{x}={\frac {1}{0!}}+{\frac {1}{1!}}x+{\frac {1}{2!}}x^{2}+{\frac {1}{3!}}x^{3}\ldots }$

${\displaystyle e^{x}=1+x+{\frac {1}{2}}x^{2}+{\frac {1}{6}}x^{3}\ldots }$

Explicit form can be written as

${\displaystyle e^{x}=\sum _{n=0}^{\infty }{\frac {1}{n!}}\,(x)^{n}}$

The natural logarithm (with base e) has Maclaurin series

${\displaystyle {\begin{aligned}\log(1-x)&=-\sum _{n=1}^{\infty }{\frac {x^{n}}{n}}=-x-{\frac {x^{2}}{2}}-{\frac {x^{3}}{3}}-\cdots ,\\\log(1+x)&=\sum _{n=1}^{\infty }(-1)^{n+1}{\frac {x^{n}}{n}}=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-\cdots .\end{aligned}}}$

They converge for ${\displaystyle |x|<1}$.

${\displaystyle y=sin(t)}$

${\displaystyle y'=cos(t)}$

${\displaystyle y''=-sin(t)}$

${\displaystyle y'''=-cos(t)}$

${\displaystyle y^{iv}=sin(t)}$

${\displaystyle y^{v}=cos(t)}$

.

.

.

Development of MacLaurin series for ${\displaystyle sin(x)}$

${\displaystyle sin(x)={\frac {0}{0!}}+{\frac {1}{1!}}x-{\frac {0}{2!}}x^{2}-{\frac {1}{3!}}x^{3}+{\frac {0}{4!}}x^{4}+{\frac {1}{5!}}x^{5}...}$

${\displaystyle sin(x)=1x-{\frac {1}{6}}x^{3}+{\frac {1}{120}}x^{5}...}$

Explicit form can be written as

${\displaystyle sin(x)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)!}}\,(x)^{(2n+1)}}$

Development of MacLaurin series for ${\displaystyle cos(x)}$

${\displaystyle y=cos(t)}$

${\displaystyle y'=-sin(t)}$

${\displaystyle y''=-cos(t)}$

${\displaystyle y'''=sin(t)}$

${\displaystyle y^{iv}=cos(t)}$

.

.

.

${\displaystyle cos(x)={\frac {1}{0!}}+{\frac {0}{1!}}x-{\frac {1}{2!}}x^{2}+{\frac {0}{3!}}x^{3}+{\frac {1}{4!}}x^{4}\ldots }$

${\displaystyle cos(x)=1-{\frac {1}{2}}x^{2}+{\frac {1}{24}}x^{4}...}$

Explicit form can be written as

${\displaystyle cos(x)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n)!}}\,(x)^{2n}}$

Recalling Euler's Formula:

${\displaystyle e^{i\omega x}=\cos \omega x+i\sin \omega x\!}$

Recall the Taylor Series from above for ${\displaystyle e^{x}}$ at :${\displaystyle t=0}$ (also called the MacLaurin series)

${\displaystyle e^{x}=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}\!}$

By replacing x with ${\displaystyle i\omega x}$, the Taylor series for ${\displaystyle e^{i\omega x}}$ can be found:

${\displaystyle e^{i\omega x}=\sum _{n=0}^{\infty }{\frac {(i\omega x)^{n}}{n!}}\!}$

even powers of n = 2k:

${\displaystyle i^{2k}=(i^{2})^{k}=(-1)^{k}\!}$

odd powers of n = 2k+1:

${\displaystyle i^{2k+1}=(i^{2})^{k}i=(-1)^{k}i\!}$

For ${\displaystyle i\omega x}$:

${\displaystyle e^{i\omega x}=\sum _{n=0}^{\infty }{\frac {i^{n}(\omega x)^{n}}{n!}}=\sum _{k=0}^{\infty }{\frac {i^{2k}(\omega x)^{2k}}{(2k)!}}+\sum _{k=0}^{\infty }{\frac {i^{2k+1}(\omega x)^{2k+1}}{(2k+1)!}}\!}$

Using the two previous equations:

${\displaystyle e^{i\omega x}=\sum _{k=0}^{\infty }{\frac {(-1)^{k}(\omega x)^{2k}}{(2k)!}}+i\sum _{k=0}^{\infty }{\frac {(-1)^{k}(\omega x)^{2k+1}}{(2k+1)!}}\!}$

${\displaystyle \Rightarrow e^{i\omega x}=\cos(\omega x)+i\sin(\omega x)\!}$

Therefore, the first part of the equation is equal to the Taylor series for cosine, and the second part is equal to the Taylor series for sine as follows:

${\displaystyle \cos(\omega x)=\sum _{k=0}^{\infty }{\frac {(-1)^{k}(\omega x)^{2k}}{(2k)!}}\!}$

${\displaystyle \sin(\omega x)=\sum _{k=0}^{\infty }{\frac {(-1)^{k}(\omega x)^{2k+1}}{(2k+1)!}}\!}$

${\displaystyle \displaystyle f(x)={\frac {1}{(1-x)^{a}}}}$

Table for Maclaurin Series
${\displaystyle f(x)={\frac {1}{(1-x)^{a}}}}$	${\displaystyle f(0)={\frac {1}{(1-0)^{a}}}=1}$
${\displaystyle f'(x)={\frac {a}{(1-x)^{a+1}}}}$	${\displaystyle f'(0)={\frac {a}{(1-0)^{a+1}}}=a}$
${\displaystyle f''(x)={\frac {a(a+1)}{(1-x)^{a+2}}}}$	${\displaystyle f''(0)={\frac {a(a+1)}{(1-0)^{a+2}}}=a(a+1)}$
${\displaystyle f^{(3)}(x)={\frac {a(a+1)(a+2)}{(1-x)^{a+3}}}}$	${\displaystyle f^{(3)}(0)={\frac {a(a+1)(a+2)}{(1-0)^{a+3}}}=a(a+1)(a+2)}$
And so on..	..

Rewriting the Maclaurin series expansion,

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

Substituting the values from the table, we get

${\displaystyle \displaystyle {\frac {1}{(1-x)^{a}}}=1+{\frac {a}{1!}}(x)+{\frac {a(a+1)}{2!}}(x)^{2}+{\frac {a(a+1)(a+2)}{3!}}(x)^{3}+\cdots .}$

${\displaystyle \displaystyle {\frac {1}{(1-x)^{a}}}=\sum _{n=0}^{\infty }(a)_{k}{\frac {x^{k}}{n!}}}$

Using

${\displaystyle (a)_{0}:=1}$

${\displaystyle (a)_{k}:=a(a+1)(a+2)\cdots (a+k-1)}$

We can represent

${\displaystyle \sum _{n=0}^{\infty }{\frac {(a)_{k}\,(b)_{k}}{(c)_{k}}}\,{\frac {x^{k}}{n!}}=F(a,b;c;x)}$

${\displaystyle \sum _{n=0}^{\infty }{\frac {(a)_{k}(b)_{k}}{(b)_{k}}}{\frac {x^{k}}{n!}}=F(a,b;b;x)}$

The binomial series is the power series

${\displaystyle (1+x)^{\alpha }=\sum _{n=0}^{\infty }{\binom {\alpha }{n}}x^{n}}$

whose coefficients are the generalized binomial coefficients

${\displaystyle {\binom {\alpha }{n}}=\prod _{k=1}^{n}{\frac {\alpha -k+1}{k}}={\frac {\alpha (\alpha -1)\cdots (\alpha -n+1)}{n!}}.}$

(If n = 0, this product is an empty product and has value 1.) It converges for ${\displaystyle |x|<1}$ for any real or complex number α.

When α = −1, this is essentially the infinite geometric series mentioned in the previous section. The special cases α = 1/2 and α = −1/2 give the square root function and its inverse:

${\displaystyle {\begin{aligned}(1+x)^{\frac {1}{2}}&=1+{\tfrac {1}{2}}x-{\tfrac {1}{8}}x^{2}+{\tfrac {1}{16}}x^{3}-{\tfrac {5}{128}}x^{4}+{\tfrac {7}{256}}x^{5}-\ldots ,\\(1+x)^{-{\frac {1}{2}}}&=1-{\tfrac {1}{2}}x+{\tfrac {3}{8}}x^{2}-{\tfrac {5}{16}}x^{3}+{\tfrac {35}{128}}x^{4}-{\tfrac {63}{256}}x^{5}+\ldots .\end{aligned}}}$

When only the linear term is retained, this simplifies to the binomial approximation.

We have the function

${\displaystyle \displaystyle arctan(1+x)}$

Expand ${\displaystyle \displaystyle arctan(1+x)}$

Table for Maclaurin Series
${\displaystyle f(x)=arctan(1+x)}$	${\displaystyle f(0)=arctan(1+0)={\frac {\pi }{4}}}$
${\displaystyle f'(x)={\frac {1}{1+(1+x)^{2}}}}$	${\displaystyle f'(0)={\frac {1}{1+(1+0)^{2}}}={\frac {1}{2}}}$
${\displaystyle f''(x)={\frac {-2(x+1)}{\left[1+(1+x)^{2}\right]^{2}}}}$	${\displaystyle f''(0)={\frac {-2(0+1)}{\left[1+(1+0)^{2}\right]^{2}}}={\frac {-1}{2}}}$
${\displaystyle f^{(3)}(x)={\frac {2(3x^{2}+6x+2)}{\left[1+(1+x)^{2}\right]^{3}}}}$	${\displaystyle f^{(3)}(0)={\frac {2(3(0)^{2}+6(0)+2)}{\left[1+(1+0)^{2}\right]^{3}}}={\frac {1}{2}}}$
And so on..	..

Rewriting the Maclaurin series expansion,

${\displaystyle \displaystyle arctan(1+x)=f(0)+{\frac {f'(0)}{1!}}(x-0)+{\frac {f''(0)}{2!}}(x-0)^{2}+{\frac {f^{(3)}(0)}{3!}}(x-0)^{3}+\cdots .}$

Substituting the values from the table, we get

${\displaystyle \displaystyle arctan(1+x)={\frac {\pi }{4}}+{\frac {1}{2*1!}}(x)-{\frac {1}{2*2!}}(x)^{2}+{\frac {1}{2*3!}}(x)^{3}+\cdots .}$

${\displaystyle arctan(x)}$

Expanding ${\displaystyle arctan(x)}$ using Maclaurin's series

Table for Maclaurin Series
${\displaystyle f(x)=arctan(x)}$	${\displaystyle f(0)=arctan(0)=0}$
${\displaystyle f'(x)={\frac {1}{1+x^{2}}}}$	${\displaystyle f'(0)={\frac {1}{1+(0)^{2}}}=1}$
${\displaystyle f''(x)={\frac {-2x}{\left[1+x^{2}\right]^{2}}}}$	${\displaystyle f''(0)={\frac {-2(0)}{\left[1+(0)^{2}\right]^{2}}}=0}$
${\displaystyle f^{(3)}(x)={\frac {6x^{2}-2}{\left[1+x^{2}\right]^{3}}}}$	${\displaystyle f^{(3)}(0)={\frac {6(0)^{2}-2}{\left[1+(0)^{2}\right]^{3}}}=-2}$
${\displaystyle f^{(4)}(x)={\frac {-24x(x^{2}-1)}{\left[1+x^{2}\right]^{4}}}}$	${\displaystyle f^{(4)}(0)={\frac {-24(0)((0)^{2}-1)}{\left[1+(0)^{2}\right]^{4}}}=0}$
${\displaystyle f^{(5)}(x)={\frac {24(5x^{4}-10x^{2}+1)}{\left[1+x^{2}\right]^{5}}}}$	${\displaystyle f^{(5)}(x)={\frac {24(5(0)^{4}-10(0)^{2}+1)}{\left[1+(0)^{2}\right]^{5}}}=24}$
And so on..	..

Rewriting the Maclaurin series expansion,

${\displaystyle \displaystyle arctan(x)=f(0)+{\frac {f'(0)}{1!}}(x-0)+{\frac {f''(0)}{2!}}(x-0)^{2}+{\frac {f^{(3)}(0)}{3!}}(x-0)^{3}+\cdots .}$

Substituting the values from the table, we get

${\displaystyle \displaystyle arctan(x)=0+{\frac {1}{1!}}(x)+{\frac {0}{2!}}(x)^{2}+{\frac {-2}{3!}}(x)^{3}+{\frac {0}{4!}}(x)^{4}+{\frac {24}{5!}}(x)^{5}\cdots .}$

${\displaystyle {\begin{aligned}\sin x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)!}}x^{2n+1}&&=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-\cdots &&{\text{for all }}x\\[6pt]\cos x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n)!}}x^{2n}&&=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-\cdots &&{\text{for all }}x\\[6pt]\tan x&=\sum _{n=1}^{\infty }{\frac {B_{2n}(-4)^{n}\left(1-4^{n}\right)}{(2n)!}}x^{2n-1}&&=x+{\frac {x^{3}}{3}}+{\frac {2x^{5}}{15}}+\cdots &&{\text{for }}|x|<{\frac {\pi }{2}}\\[6pt]\sec x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}E_{2n}}{(2n)!}}x^{2n}&&=1+{\frac {x^{2}}{2}}+{\frac {5x^{4}}{24}}+\cdots &&{\text{for }}|x|<{\frac {\pi }{2}}\\[6pt]\arcsin x&=\sum _{n=0}^{\infty }{\frac {(2n)!}{4^{n}(n!)^{2}(2n+1)}}x^{2n+1}&&=x+{\frac {x^{3}}{6}}+{\frac {3x^{5}}{40}}+\cdots &&{\text{for }}|x|\leq 1\\[6pt]\arccos x&={\frac {\pi }{2}}-\arcsin x\\&={\frac {\pi }{2}}-\sum _{n=0}^{\infty }{\frac {(2n)!}{4^{n}(n!)^{2}(2n+1)}}x^{2n+1}&&={\frac {\pi }{2}}-x-{\frac {x^{3}}{6}}-{\frac {3x^{5}}{40}}-\cdots &&{\text{for }}|x|\leq 1\\[6pt]\arctan x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}x^{2n+1}&&=x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-\cdots &&{\text{for }}|x|\leq 1,\ x\neq \pm i\end{aligned}}}$

All angles are expressed in radians. The numbers B_k appearing in the expansions of tan x are the Bernoulli numbers. The E_k in the expansion of sec x are Euler numbers.

The "performance of a Markov system under different operating strategies [can be estimated] by observing the behavior of the system under the [strategy of having] a Maclaurin series for the performance measures of [the] Markov chains."^[5]

1. Any non-convergent function can be represented by a MacLaurin series.

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