Taylor's series is not too difficult to express - but is less easy to explain. Provided you are not immediately concerned about exactly how this formula is evaluated, or used, this brief description may be of interest.
Essentially, a well behaved function ( f ) can be approximated about a point x, and moment μ, by expanding it into this infinitely-long series of terms,
f(x-μ) = [ f(x)/0! ]μ^{ 0} + [ f'(x)/1! ]μ^{ 1} + [ f''(x)/2! ]μ^{ 2} + ... [ f^{ ∞}(x)/∞! ]μ^{ ∞}
If we replace the expressions in square brackets by 'constants' ( c_{1} to c_{i} ), and collect all the bits to the right of the ith-order term into a remainder ( R_{i} ), this rather frightening series simplifies to (for example)
f(x-μ) ≅ c_{0}μ^{ 0} + c_{1}μ^{ 1} + c_{2}μ^{ 2} + R_{2}
Which is recognisably a 2-order polynomial ( i = 2 ), plus a remainder term.
Now for the twiddly bits.
- μ^{ 0} = 1, μ^{ 1} = μ, μ^{ 2} = μ×μ, so μ^{ ∞} = ∞, 1, or 0
- f'(x) is, effectively a probability density, or the rate of change (or slope) of f(x) when plotted against x. For example the normal density function ( Z[x] or φ[x] ) is the rate of change (or slope) of is the slope of the cumulative normal distribution function ( Φ[x] ) at point x - so Z[x] could also be written as Φ'[x] ).
- Similarly, f''(x) is the rate of change of slope at point x. These rates of change may be evaluated using differential Calculus.
- 0! = 1, 1! = 1×1, 2! = 2×1
So anything divided by ∞! is, almost certainly, zero.
- In other words, unless you are dealing with a sincerely weird function, the higher the order of each term, the smaller it is. So, as a rule, if R_{2} is fairly small, R_{5} would be much smaller.
Confusingly, although the above series can be written more concisely as f(x-μ) = Σ( [f^{i}(x)/i!]μ^{ i} ), it may be rearranged so that f(x) = Σ( [f^{i}(μ)/i!][x-μ]^{ i} ).