Power series expansion
HardySpicer
d(x)= 1+d1x+d2x^2 +...+ dnx^n
and I wish to express
p0+p1x+p2x^2+...+ pmx^m. = 1/d(x)
If I multiply out and compare coefficients I can come up with a
formula but I am unsure whether the limit on the summation is
different when m>n and m<n?
H.
Ray Vickson
The reciprocal of a polynomial is an *infinite* series, so m =
infinity. For example, if |x| < 1 then 1/(1-x) = 1 + x + x^2 + x^3
+ .... with no end to the terms.
R.G. Vickson
Rob Johnson
Indeed. Say for example that we have two polynomials that were
reciprocals,
d(x) = 1 + d_1 x + d_2 x^2 + ... + d_n x^n
p(x) = 1 + p_1 x + p_2 x^2 + ... + p_m x^m
where d_n != 0 and p_m != 0. Then we have that the coefficient of
the x^{n+m} term of the product would be d_n p_m != 0. However, the
coefficient of any term, except the x^0 term, must be 0. Thus, we
reach a contradiction unless n+m = 0.
Thus, the only polynomial d(x) which can have a polynomial inverse
is d(x) = 1 where p(x) = 1.
Rob Johnson <r...@trash.whim.org>
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