Can Properties Be Discovered From Maclaurin Series English Language Essay

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MacLaurin and Taylor series are polynomials which approximate a function about a specific point. These expansions are determined by the derivatives of the function at the point which it is being approximated around. MacLaurin polynomials are centered around 0 and Taylor polynomials are centered around a point x = a where a ≠ 0. The closer the value for which you are approximating is to a the better your approximation will be.

MacLaurin Series gives an approximation for a function about the point x=0. The polynomial that results from the MacLaurin expansion approximates the original function to within an error bound which is gets smaller as the degree of the polynomial increases. However, in order for an approximation to be made, the series must converge.

The formula for the MacLaurin expansion of f(x) is:

f(x)=f(0)+f'(0)x+f''(0)\frac{x^2}{2!}+f'''(0)\frac{x^3}{3!}+...+f^n(0)\frac{x^n}{n!}+...\,

MacLaurin Series Examples

Find the 4th degree MacLaurin polynomial for the function f(x)=\sin(5x)\,.

We know that the MacLaurin expansion of sin(x) is because it is a common expansion; to find the expansion of sin(5x), we simply take the expansion for sin(x) and plug 5x in wherever we see an x. The MacLaurin series that results is:

\sin(5x)=(5x)-\frac{(5x)^3}{3!}+\frac{(5x)^5}{5!}-\frac{(5x)^7}{7!}+...\,

or

However the problem asks for the fourth degree polynomial so the answer is:

\sin(5x)=5x-\frac{125x^3}{3!}\,

because the fourth degree polynomial is actually:

\sin(5x)=5x-\frac{125x^3}{3!}+\frac{0x^4}{4!}\,

According to Taylor series If we want a good approximation to the function in the region near x = a, we need to find the first, second, third (and so on) derivatives of the function and substitute the value of a. Then we need to multiply those values by corresponding powers of (x − a), giving us the Taylor Series expansion of the function f(x) about x = a:

taylors

We now take a particular case of Taylor Series, in the region near x = 0. Such a polynomial is called the Maclaurin Series.

The infinite series expansion for f(x) about x = 0 becomes:

maclaurin

f '(0) is the first derivative evaluated at x = 0, f ''(0) is the second derivative evaluated at x = 0, and so on.

Maclaurin series of a function  up to order  may be found using Series[f, x, 0, n]. The th term of a Maclaurin series of a function  can be computed in Mathematica using SeriesCoefficient[f, x, 0, n] and is given by the inverse Z-transform.

Maclaurin series are a type of series expansion in which all terms are nonnegative integer powers of the variable. Other more general types of series include the Laurent series and the Puiseux series.

Neither the function nor any of its derivatives exist at x = 0, so there is no polynomial Maclaurin expansion of the natural logarithm function ln x.

Maclaurin Series of inverse trigonometrical functions

Series for inverse trigonometrical functions can be complicated to find directly since successive differentials become unmanageable quite quickly. The idea of differentiability of functions and their series is useful in finding series of inverse trigonometrical functions.

In the maclaurin series the strict conditions about differentiability and existence at x = 0.

Common MacLaurin Polynomials

e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...+\frac{x^n}{n!}+...\,

\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...+\frac{(-1)^{n+1}x^{2n-1}}{(2n-1)!}+...\,

\frac{1}{1-x}=1+x+x^2+x^3+...+x^n+...\,

From Functions to Polynomial Series Expansions

A Polynomial Series is a mathematical expression consisting of added terms, terms which consist of a constant multiplier and one or more variables raised to integral powers. For example, 3x2 - 2x + 7 and 5y + 8x3z are closed polynomials (i.e. polynomials containing a finite number of terms).

Functions of the type (1 + x)n and their Polynomial Representation

Under certain conditions mathematical functions can be equated to polynomial series- a simple example being the quadratic function f(x) = (x +1)2, in which the brackets can be expanded giving f(x) = x2 + 2x + 1. Here x2 + 2x + 1 is a polynomial series.

When considering polynomial expansions of this type, the order is usually reversed, i.e.

(1 + x)2 = 1+2x+ x2

Similarly,

(1 + x)3 = 1+3x+3 x2+x3

(1 + x)4 = 1+4x+6 x2+4x3 +x4

Maclaurin Series expansion for f(x) = sin x.

Answer

maclaurin

Conditions of Maclaurin's series

Maclaurin's series may be used to represent any function, say f (x), as a power series provided that at x =0 the following three conditions are met:

(a) f (0) _= ∞

For example, for the function f (x)= cos x, f (0)= cos 0=1, thus cos x meets the condition.

However, if f (x)= ln x, f (0)= ln 0=−∞,

thus ln x does not meet this condition.

(b) The resultant Maclaurin's series must be convergent

In general, this means that the values of the terms, or groups of terms, must get progressively smaller and the sum of the terms must reach a limiting value.

Can properties of a function be discovered from its Maclaurin series?

Yes. If the Maclaurin expansion of a function locally converges to the function, then you know the function is smooth. In addition, if the residual of the Maclaurin expansion converges to 0, the function is analytic .And also all of the derivatives at 0 are given.

So its value at 0, slope at 0, concavity at 0 (if coefficient of x^2 is not 0).

No If f(x)=e^(-1/x^2), for x not 0, and f(0)=0. This is a well-known example of a function with a Maclaurin series, but the resulting series does NOT represent f(x)! The series has all 0 coefficients and can give no other properties than the derivatives mentioned under "Yes".

So, to find the value of a function using its Maclaurin Series to a given accuracy, one only needs to use the number of terms that give the appropriate accuracy required.