Power series
Power series or polynomial with infinitely many terms
The sum of a power series is a function
Maclaurin and Taylor series

Representing polynomial using Maclaurin's and Taylor's formula, examples
Power series or polynomial with infinitely many terms
A real power series in x around the origin (or centered at the origin) is a series of functions of the form
and the power series around a given point x = x0 (or centered at x0) is a series of the form
where the coefficients an are fixed real numbers and x is a real variable.
A power series with real coefficients is said to be real or complex according as both x and x0 are real or complex numbers.
Therefore, the nth partial sum of a power series is a polynomial of degree n,
The sum of a power series is a function
The sum of a power series is a function
the domain of which is the set of those values of x for which the series converges to the value of the function.
Maclaurin and Taylor series
Consider the polynomial function
f (x) = anxn + an - 1xn - - 1 + · · · a3x3 + a2x2 + a1x + a0.
If we write the value of the function and the values of its successive derivatives, at the origin, then
f (0) = a0,     f '(0) = 1· a1,     f ''(0) = 1· 2a2,     f '''(0) = 1· 2· 3a3,  . . .  ,  f (n)(0) = n!an
 so we get the coefficients;
Then, the polynomial f (x) with infinitely many terms, written as the power series
 and
where 0! = 1,   f (0)(x0) =  f (x0) and  f (n)(x0) is the nth derivative of  f at x0,
represents an infinitely differentiable function and is called Maclaurin series and Taylor series respectively.
Representing polynomial using Maclaurin's and Taylor's formula
If given is an n-th degree polynomial
Pn (x) = an xn + an - 1xn -1 + · · ·a2 x2 + a1 x + a0    then,
Pn(0) = a0Pn' (0) = a1Pn''(0) = 2! a2Pn'''(0) = 3! a3, . . . ,  Pn(n)(0) = n! an  and  Pn(n + 1) = 0
so, the coefficients of the polynomial
therefore, applying Maclaurin's formula, every polynomial can be written as
since Pn(n + 1) = 0, the remainder vanishes.
Example:  Represent the quintic  y = 2x5 + 3x4 - 5x3 + 8x2 - 9x + 1 using Maclaurin's formula.
Solution:  Let write all successive derivatives of the given quintic function and evaluate them at the origin,
y' (x) = 10x4 + 12x3 - 15x2 + 16x - 9,           y' (0) = - 9
y'' (x) = 40x3 + 36x2 - 30x + 16,                    y'' (0) = 16
y''' (x) = 120x2 + 72x - 30                            y''' (0) = - 30
yIV (x) = 240x + 72                                       yIV (0) = 72
yV (x) = 240,                                                   yV (0) = 240
yVI = 0  and the last term of the polynomial  a0 y(0) = P5(0) = 1,
then substitute obtained values into Maclaurin's formula
Example:  Represent the quartic  yx4 - 4x3 + 4x2 + x - 4 at x0 = 2 using Taylor's formula.
Solution:  Let write all successive derivatives of the given quartic and evaluate them at x0 = 2,
y' (x) = 4x3 - 12x2 + 8x + 1,              y' (2) = 1
y'' (x) = 12x2 - 24x + 8,                    y'' (2) = 8
y''' (x) = 24x - 24                            y''' (2) = 24
yIV (x) = 24                                     yIV = 24
yV (x) = 0     and the last term,   a0 y(2) = P4(2) = - 2,
then substitute obtained values into Taylor's formula
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