# 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.

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

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.