Taylor and Maclaurin Series

  • Introduction : Polynomial Functions

A polynomial function {y= f(x)} involves addition of non-negative integer powers of {x}. For example, {y = x^3 - 5x^2 + 6}. The non-polynomial functions, such as {sin (x), tan (x), e^x, ln \ x} can be expressed as polynomial functions using Taylor series and Maclaurin series. These series generally contain infinitely many terms of the form {x^n} with each term having a coefficient. The coefficients are determined by a methodology.

Note that the non-polynomial functions are not strictly polynomial functions (as the name suggests); but are expressed using infinite series.

  • Taylor’s Theorem and Series

The series was formulated by James Gregor, a British astronomer and has been named after Brook Taylor, who introduced it formally. Gregor had obtained the Taylor series of several trigonometric functions.

Let {y=f(x)} be a function of {x}, which is infinitely differentiable at {x=a}. Then, {f(x)} is given by

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

The {r}th term of the series is given by

{\frac {f^{(r-1)}}{(r-1)!} (x-a)^{(r-1)}}

The proof of this formulation is known as Taylor’s theorem.

We say that, {f(x)} is expanded about {x=a}.

Note that we have used both successive differentiation and infinite series in defining Taylor series.

  • Maclaurin’s Series

When the value of {a} is {0}, then Taylor’s series becomes Maclaurin’s series. So, the Maclaurin’s representation will be

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

  • Maclaurin’s Series for Standard Functions

I) Polynomial Functions : The function itself is its Maclaurin representation

II) Exponential Function :

{e^x = 1 + x + \frac {x^2}{2!} + \frac {x^3}{3!} + \frac {x^4}{4!} + \cdots = \sum \limits_{n=0}^{\infty} \frac {x^n}{n!}}

III) Trigonometric and Hyperbolic Functions :

The {sine} and {cosine} series, when obtained, tell us why {e^{ix}} is equal to {cos (x) + i \ sin (x)}. (Check!)

{sin (x) = x - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac {x^7}{7!} + \frac {x^9}{9!} - \cdots = \sum \limits_{n=0}^{\infty} \frac {(-1)^{n} x^{(2n-1)}}{(2n+1)!}}

{cos (x) = 1 - \frac {x^2}{2!} + \frac {x^4}{4!} - \frac {x^6}{6!} + \frac {x^8}{8!} - \cdots = \sum \limits_{n=0}^{\infty} \frac {(-1)^n x^{2n}}{(2n)!}}

{tan (x)= x + \frac {x^3}{3} + \frac 2 {15} x^5 + \frac {17}{315} x^7 + \cdots}

{sin^{-1}(x)= x + \frac {1}{6}x^3 + \frac {3}{40}x^5 + \cdots}

{cos^{-1}(x) = \frac {\pi}{2} - sin^{-1}(x) = \frac {\pi}{2} - (x + \frac {1}{6}x^3 + \frac {3}{40}x^5 + \cdots)}

{tan^{-1}(x) = x- \frac {1}{3}x^3 + \frac {1}{5}x^5 - \frac 1 7 x^7 + \cdots}

The series for {sinh (x), cosh (x)} and {tanh (x)} can be obtained easily by using their definitions. Given below are {sinh (x), sinh^{-1}(x)}.

{sinh(x) = x + \frac 1 6 x^3 + \frac 1 {120}x^5 + \cdots}

{sinh^{-1}(x)= x - \frac {1}{6}x^3 + \frac {3}{40}x^5 - \cdots}

IV) General Functions :

{\frac {1}{1-x} = 1+x+x^2+x^3+x^4+ \cdots , for \ -1 < x < 1}

Above expansion is made up of a sequence, which is a geometric progression.

{ln \ (1-x) = - \Big(x+ \frac {x^2}{2}+ \frac {x^3}{3}+ \frac {x^4}{4} + \frac {x^5}{5} + \cdots \Big)}

V) The Binomial Theorem :

The binomial theorem, which was proved earlier using mathematical induction, can also be proved (rather, gets proved) using Maclaurin series.

{(1+x)^n = 1 + nx + \frac {n(n-1)}{2!}x^2 + \frac {n(n-1)(n-2)}{3!} + \cdots}

  • Additional Useful Points

Successive differentiation is to be extensively used while solving problems. This is evident from the statement of the theorem itself.

I) Logarithmic Differentiation – A technique of differentiation, involving powers of {x}.

{y = x^x, ln (y) = x \cdot ln (x), \frac 1 y \frac {dy}{dx} = ln (x) + 1}

II) If the function is getting complicated after differentiation, it is to be modified to get {n}th derivative. This modification will be either a substitution or trigonometric substitution or actual division etc.

{y = \sqrt {1 + sin (x)}, use \ sin(x)= 2sin \ \frac x 2 \cdot cos \ \frac x 2, \ sin^2 \frac x 2 + cos^2 \frac x 2 = 1}

Then, {\ y = sin \ \frac x 2 + cos \ \frac x 2}

This can easily be expanded using series of {sine} and {cosine}.

IV) Using Taylor’s theorem, we can well-approximate a function using its Taylor series (as we did using Fourier series). If the sequence is convergent, then numerical values of functions can be computed up to any degree of accuracy.

Posted in M I

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