| Table of contents | |
| Maclaurin series (special case a = 0) | |
| Applications of Taylor series | |
| Problems and Solutions | |
| Notes and useful remarks |
Taylor series is the representation of a function as an infinite sum of terms, each term being a constant multiplied by a power of (x - a). Each successive term has a higher power (exponent) than the previous term. The Taylor series expansion is centred at a point a. When the centre a = 0, the series is called the Maclaurin series.

The Taylor series of a function f(x) that is infinitely differentiable at a point a is the power series

In sigma notation this is written as

Here f(n)(a) denotes the nth derivative of f evaluated at x = a, and n! is the factorial of n. This representation gives a polynomial approximation of increasing degree; under suitable conditions the infinite sum converges to the value f(x).
Coefficients: The coefficient of (x - a)n is f(n)(a)/n!. Convergence: The radius of convergence of the Taylor series depends on the nearest singularity of f in the complex plane; inside that radius the series converges absolutely. If the series converges to f(x) for all x in some interval containing a, f is called analytic at a.
To measure the difference between f(x) and the n-th degree Taylor polynomial T_n(x), define the remainder R_n(x) by
f(x) = Tn(x) + Rn(x).
One common form of the remainder (Lagrange form) is
Rn(x) = f(n+1)(ξ)/(n+1)! · (x - a)n+1 for some ξ between a and x.
This form is useful to bound the error when the (n+1)th derivative is known or can be bounded on the interval between a and x.
Begin by assuming f can be expanded as a power series about x = 0 (Maclaurin form):

Differentiate term-by-term.
Evaluate the series and its derivatives at x = 0 to obtain the coefficients.
For example, f(0) = a0. f′(0) = a1. f″(0) = 2 a2 ⇒ a2 = f″(0)/2!. By induction, an = f(n)(0)/n!.
Substituting these coefficients into the power series gives the Maclaurin series:

To generalise about a point a, replace x by (x - a) and proceed similarly. Write
f(x) = b + b1(x - a) + b2(x - a)2 + b3(x - a)3 + ...
Evaluating derivatives at x = a gives bn = f(n)(a)/n!. Substituting yields the Taylor series about a:

Derivatives cycle every four steps:
f(x) = sin x
f′(x) = cos x
f″(x) = - sin x
f‴(x) = - cos x
f⁽⁴⁾(x) = sin x, and so on.
Evaluating at x = 0 and substituting into the Maclaurin formula gives only odd powers of x with alternating signs:

The Maclaurin series for sin x converges for all real x (radius of convergence ∞).
Functions of several variables also admit Taylor expansions when sufficiently differentiable. For a function f(x, y) expanded about (a, b), the second-order form is

The full multivariable expansion contains mixed partial derivatives and terms of higher total degree. Multivariate Taylor polynomials are widely used in optimisation, numerical methods, and approximations in engineering problems.
If the Taylor series is centred at 0 (a = 0) it is called the Maclaurin series. The general Maclaurin form is

Example: For |x| < />
1/(1 - x) = 1 + x + x2 + x3 + x4 + ...
This geometric series converges for |x| < 1. many standard maclaurin expansions x, sin x, cos x, ln(1 + x) for |x| < 1, (1 +α by binomial series) are used as building blocks in analysis and approximation.
Question 1: Determine the Taylor series at x=0 for f(x) = ex
Soln:
Write the function: f(x) = ex.
Differentiate repeatedly: f′(x) = ex; f″(x) = ex; f‴(x) = ex; and so on.
Evaluate derivatives at x = 0: f(0) = 1; f′(0) = 1; f″(0) = 1; f‴(0) = 1; ...
Substitute into the Maclaurin formula f(x) = Σ f(n)(0)/n! · xn to get the series.

Therefore,
ex = 1 + x + x2/2! + x3/3! + ..., for all real x.
Question 2: EValuate the Taylor Series for f ( x ) = cos ( x ) for x = 0.
Soln:
Compute derivatives and evaluate at 0:
f(x) = cos x ⇒ f(0) = 1
f′(x) = - sin x ⇒ f′(0) = 0
f″(x) = - cos x ⇒ f″(0) = -1
f‴(x) = sin x ⇒ f‴(0) = 0
f⁽⁴⁾(x) = cos x ⇒ f⁽⁴⁾(0) = 1
The derivatives repeat every four steps.
Substitute into the Maclaurin series formula.


Hence the Maclaurin series is
cos x = 1 - x2/2! + x4/4! - x6/6! + ..., valid for all real x.
Question 3: Evaluate the Taylor Series for f (x) = x3 - 10x2 + 6 at x = 3.
Soln:
Compute the function value and derivatives at x = 3.
f(x) = x3 - 10x2 + 6 ⇒ f(3) = 27 - 90 + 6 = -57
f′(x) = 3x2 - 20x ⇒ f′(3) = 27 - 60 = -33
f″(x) = 6x - 20 ⇒ f″(3) = 18 - 20 = -2
f‴(x) = 6 ⇒ f‴(3) = 6
Higher derivatives are zero: f⁽⁴⁾(x) = 0 and beyond.
Form the Taylor series about a = 3 using T_n(x) = Σ f(n)(3)/n! · (x - 3)n.

Thus the series ends at the third derivative (polynomial):
f(x) = f(3) + f′(3)(x - 3) + f″(3)/2!(x - 3)2 + f‴(3)/3!(x - 3)3.
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| 2. How is the Maclaurin series a special case of Taylor's series? | ![]() |
| 3. Can you derive the Taylor series for sin x? | ![]() |
| 4. What are some applications of Taylor series? | ![]() |
| 5. How do you compute Taylor series in several variables? | ![]() |