Power Series from Formulas
The
GenerateUnivariatePowerSeries package enables you to create power series
from explicit formulas for their nth coefficients. In what follows, we
construct series expansions for certain transcendental functions by giving
forumulas for their coefficients. You can also compute such series
expansions directly by simply specifying the function and the point about
which the series is to be expanded. See
Converting to Power Series for more information.
Consider the Taylor expansion of %e^x about x=0:
%e^x = 1 + x + x^2/2 + x^3/6 + ...
= sum from n=0 to n=%infinity of x^n/n!
The nth Taylor coefficient is 1/n!. This is how to create this series in
Axiom.
The first argument specifies the formula for the nth coefficient by giving
a function that maps n to 1/n!. The second argument specifies that the
series is to be expanded in powers of (x-0), that is, in powers of x. Since
we did not specify an initial degress, the first term in the series was the
term of degree 0 (the constant term). Note that the formula was given as
an anonymous function. These are discussed in
Anonymous Functions
Consider the Taylor expansion of log x about x=1:
log x = (x-1) - (x-1)^2/2 + (x-1)^3/3 - ...
= sum from n=1 to n=%infinity of (-1_^(n-1) (x-1)^n/n
If you were to evaluate the expression series(n+->(-1)^(n-1)/n,x=1) you
would get an error message because Axiom would try to calculate a term of
degree n=1,... are to be computed.
Next consider the Taylor expansion of an odd function, say, sin(x):
sin x = x = x^2/3! + x^5/5! - ...
Here every other coefficient is zero and we would like to give an explicit
formula onloy for the odd Taylor coefficients. This is one way to do it.
The third argument, 1.., specifies that the first term to be computed is
the term of degree 1. The fourth argument, 2, specifies that we increment
by 2 to find the degrees of subsequent terms, that is, the next term is of
degree 1+2, the next of degree 1+2+2, etc.
The initial degree and the increment do not have to be integers. For
example, this expression produces a series expansion of sin(x^(1/3)).
While the increment must be positive, the initial degree may be negative.
This yields the Laurent expansion of csc(x) at x=0.
Of course, the reciprocal of this power series is the Taylor expansion of
sin(x).
As a final example, here is the Taylor expansion of asin(x) about x=0.
When we compute the sine of this series, we get x (in the sense that all
higher terms computed so far are zero).
As we discussed in
Converting to Power Series, you can also use
the operations
taylor,
laurent, and
puiseux, instead of
series if you know ahead of time what
kind of exponents a series has. You can't go wrong with
series though.