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Re: Taylor polynomials in mathematica

  • To: mathgroup at smc.vnet.net
  • Subject: [mg108095] Re: [mg108049] Taylor polynomials in mathematica
  • From: danl at wolfram.com
  • Date: Mon, 8 Mar 2010 06:13:49 -0500 (EST)
  • References: <201003070904.EAA01180@smc.vnet.net>

> 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? =))

One important thing is to get the parentheses to balance, else everything
is a guess as to what you actually intend. Below is my variant, using
Mathematica notation.

x[0] = -c;
y[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);

First derivatives at zero (not exactly what is claimed above, but close).

In[113]:= {x'[b], y'[b]} /. b -> 0
Out[113]= {d/(3 c^2), c/(3 d^2)}

Second derivatives at zero are evaluated by substituting zero after
differentiating the first derivatives.

In[112]:= Together[{D[x'[b], b], D[y'[b], b]}] /. b -> 0
Out[112]= {-((2 (81 c^5 d^4 - 81 c d^8))/(729 c^6 d^6)), -((
  2 (-81 c^8 d + 81 c^4 d^5))/(729 c^6 d^6))}

Daniel Lichtblau
Wolfram Research





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