Prev: bpv4c problem
Next: surface
From: rc3002 C on 22 Dec 2009 11:57
I tried the following commands in Matlab:
syms x;
taylor(1/(1+x^2),5,1);

which should show me the fifth order Taylor series approximation to f=1/(1+x^2) about the point 1. Obviously, f(1)=1/2. But Matlab gives me:

(x - 1)^2/4 - x/2 - (x - 1)^4/8 + 1

It looks like it expand the series about 0 in the 'zero' term.
From: Murad Abu-Khalaf on 22 Dec 2009 14:07
The expressions generated by MATLAB is such that f(1) = 1/2. You can verify
by noting that:

(x - 1)^2/4 - x/2 - (x - 1)^4/8 + 1

is equivalent to

(x - 1)^2/4 - (x - 1)^4/8 + 1/2 + 1/2 - x/2

and is equivalent to

(x - 1)^2/4 - (x - 1)^4/8 - (x-1)*1/2 + 1/2


HTH,
Murad




"rc3002 C" <rifko(a)walla.com> wrote in message
news:hgqtou$gi8$1(a)fred.mathworks.com...
>I tried the following commands in Matlab:
> syms x;
> taylor(1/(1+x^2),5,1);
>
> which should show me the fifth order Taylor series approximation to
> f=1/(1+x^2) about the point 1. Obviously, f(1)=1/2. But Matlab gives me:
>
> (x - 1)^2/4 - x/2 - (x - 1)^4/8 + 1
>
> It looks like it expand the series about 0 in the 'zero' term.
>


 | 
Pages: 1
Prev: bpv4c problem
Next: surface