Taylor Series Power

In mathematics, the Taylor series is a representation of a function as an infinite sum of terms calculated from the values of its derivatives at a single point. It is named after the English mathematician Brook Taylor. If the series is centered at zero, the series is also called a Maclaurin series, named after the Scottish mathematician Colin Maclaurin. It is common practice to use a finite number of terms of the series to approximate a function. The Taylor series may be regarded as the limit of the Taylor polynomials.

I like to share this taylor series expansions with you all through my article.

Definition of taylor series power:

Taylor series power:

Definition:   Taylor polynomial degree of function can be defined as the function of f. For an approximation of a function have a degree of polynomial. It can be differentiate for n times. The degree of Taylor polynomial for centered at a is

Pn(x) = f(a) + f′(a)(x - a) + (f″/ 2!)(x – a) ²+ …… + (f (n) (a)/ n!)(x - a)n

= ∑n k = 0 (f (k) (a) / k!) (x- a) k

Properties of taylor series power approximation:

In the interval (a + r, a - r), the series of converges for x and the result is equal to f(x) which function is analytic.
The power series representation can be used in the simplest form of Euler’s formula. It can be done in algebraic expressions
The result of expansions of Taylor series for cosine, sine and exponential functions are the one of fundamentals fields of harmonic analysis.
In singularity, the Taylor functions are sometimes cannot write as function.
The Taylor series is zero and also the function is not which is not analytic.


Example problem for Taylor series power:

Ex 1:   Solve: f(x) = (1 / (x + 1)), a = 1. Find the Taylor polynomial degree up to two.

Sol :    Given f(x) = (1 / x +1)          f (1) = 0.5

f’(x) = -1 / (x +1)2             f’ (1) = -0.25

f”(x) = 2 / (x +1)3             f’’(1) = 0.125

To find the Taylor polynomial:

General form of the Taylor polynomials is

Pn(x) = f (a) + f′ (a) (x - a) + (f″/ 2!)(x – a)2 + …… + (f (n) (a)/ n!)(x - a) n

Here we calculate up to third degree of Taylor polynomial

Pn(x)   = f (4) + f’ (4) (x – a) + f’’ (4) ((x – a)2/ 2!)

= 0.5 + (-0.25) (x - 1) + 0.125((x - 1)2 / 2!)

= 0.5 – 0.25x + 0.25 + (x2 – 2x + 1) (0.0625)

= 0.5 – 0.25x + 0.25 + 0.0625x2 – 0.125x + 0.0625

Pn(x) = 0.0625x2 – 0.375x +0.75.

Ex  2:  Solve: f(x) = (1 / (x - 3)) +1, a = 4. Find the Taylor polynomial degree up to three.

Sol :    Given f(x) = (1 / x - 3) +1          f (4) = 2

f’(x) = -1 / (x - 3)2             f’ (4) = -1

f”(x) = 2 / (x – 4)3             f’’(4) = 2

f’’’(x) = -6 / (x – 4)4          f’’’ (4) = -6

To find the Taylor polynomial:

General form of the Taylor polynomials is

Pn(x) = f (a) + f′ (a) (x - a) + (f″/ 2!)(x – a)2 + …… + (f (n) (a)/ n!)(x - a) n

Here we calculate up to third degree of Taylor polynomial

Pn(x)   = f (4) + f’ (4) (x – a) + f’’ (4) ((x – a)2/ 2!) + f’’’ (4) ((x – a)3/ 3!)

= 2 + - (1) (x - 4) + 2((x – 4)2 / 2!) + (-6) (x – 4)3/ 6

= 2 - x + 4 + (x2 – 8x + 16) - (x3 – 3(x2(4)) + 3(x (16)) + (43)

= 2 - x + 4 + x2 – 8x + 16 - x3 + 12x2 - 48x - 64

Pn(x) = - x3 + 13x2 - 57x – 46.