Re: Taylor polynomials in mathematica
- To: mathgroup at smc.vnet.net
- Subject: [mg108092] Re: Taylor polynomials in mathematica
- From: "Sjoerd C. de Vries" <sjoerd.c.devries at gmail.com>
- Date: Mon, 8 Mar 2010 06:13:16 -0500 (EST)
- References: <hmvq9c$18t$1@smc.vnet.net>
If you want to get x''(b) given x'(b) all you have t use is D:
D[x'[b],b] or, in your case, D[(b*x[b] - 3*((y[b]))^3)/(9 (x[b])^2
(y[b])^2 - b^2), b]. In the resulting expression, substitute the
values of x'[b] and y'[b] with the known expressions for these terms.
Use RelaceAll (/.) and Rule (->) for this, as in
5 x^2 /. x->(n+m)
5 (m + n)^2
Taylor series are done with the Series command. You may need to look
this up in Mathematica's help centre.
In[25]:= Series[x[b], {b, 0, 2}]
Out[25]= x[0]+(x^\[Prime])[0] b+1/2 (x^\[Prime]\[Prime])[0] b^2+O[b]^3
With x[0] known, x'[b] same, and x''[b] as above from D you're ready
for a second order Taylor series near 0.
Note that functions gave square brackets in Mathematica, not round
ones.
Cheers -- Sjoerd
On Mar 7, 11:05 am, Halla Gralla <fedd... at hotmail.com> wrote:
> What I have is two unknown functions of b, x(b) and y(b). I do know their values in b=0, their first derivative and the value of their first derivatives in b=0. >
> x(0) = -c
>
> x(0) = -d
>
> x'(b) = (b*x(b) - 3*((y(b))^3) / (9(x(b))^2(y(b))^2 - b^2)
>
> y'(b) = (b*y(b) - 3*((x(b))^3) / (9(x(b))^2(y(b))^2 - b^2)
>
> it follows that:
>
> x'(0) = c / 3d^2
>
> y'(0) = d / 3x^2
>
> Now I don't now how to find x''(b) or y''(b), which would be needed for a second ordered taylor polynomial, but mathematica knows (and perhaps does this automatically when setting up taylor polynomials?). Problem is, I don't know how to use mathematica very well. I need to make a "program" in mathematica where I can change values for n (order of taylor polynomial) and for c and d (which are constants in the functions).
>
> Any help? =))