On Mon, 26 Jul 2010, Julien Martin wrote:
< Hello Raymond,
<
< I want to do the following:
<
< Expand
< (1-%theta/%mu)^-1
< as a Binomial series knowing %mu is a constant.
<
< Here is what I tried in Maxima:
< (1-%theta/%mu)^-1;
< taylor(%,%theta,0,10);
<
< You see, I don't understand about which point to expand given what I want to do. What do you suggest I replace the third argument (i.e. 0) with?
<
< Julien.
<
<
<
< 2010/7/26 Raymond Toy <toy.raymond at gmail.com>
< ?On 7/26/10 10:18 AM, Julien Martin wrote:
< > If from a mathematical point of view, the binomial series yields the
< > same result as the taylor expansion around zero, then taylor suits me
< > fine. Can someone please confirm this is the one and same thing?
< Well, to me binomial series is (1+x)^n, expanded out. ?A general Taylor
< expansion about zero is much more than that, so maybe the answer is no,
< they're not the same.
<
< But certainly a taylor expansion of (1+x)^n would give the binomial
< series, since power series are unique and both the taylor series and the
< binomial series converge to the same function over a suitable domain.
The binomial series, as commonly defined in calculus texts, is the Taylor series
of (1+x)^a expanded about x=0. See
http://en.wikipedia.org/wiki/Binomial_series
In Maxima you can use either taylor or powerseries to compute it:
(%i1) f : 1/(1-x)$
(%i2) fs : taylor(f,x,0,10);
(%o2) 1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^10
(%i3) g : (1-x)^a$
(%i4) gs : taylor(g,x,0,5);
(%o4)
1-a*x+(a^2-a)*x^2/2-(a^3-3*a^2+2*a)*x^3/6+(a^4-6*a^3+11*a^2-6*a)*x^4/24
-(a^5-10*a^4+35*a^3-50*a^2+24*a)*x^5/120
(%i5) powerseries(g,x,0);
(%o5) ('sum((-1)^i1*x^i1/beta(-i1+a+1,i1+1),i1,0,inf))/(a+1)
HTH,
Leo
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