Easier way to calculate Taylor remainder in 2nd order series

Question

At the moment I have implemented the code for a Taylor 2nd order series for the function in three variables:

$x_3^3+\frac{x_1-x_2}{x_1+x_2}$

The code builds on following expression:

ClearAll["Global`*"]
Remove["Global`*"]

thetacalc[xtem_, utem_, nn_] := {

n = nn;
xx = Table[Subscript[x, i], {i, 1, n}];
uu = Table[Subscript[u, i], {i, 1, n}];

f1[x_] := ((x[[1]] - x[[2]])/(x[[1]] + x[[2]])) + (x[[3]])^3;
hh = Simplify[D[f1[xx + θ*uu], {xx, 2}]] // MatrixForm;

f2[f_, x_, u_] := 
f + Sum[uu[[j]]*D[f, {xx[[j]]}], {j, 1, n}] + 
1/2*Sum[Sum[uu[[i]]*uu[[j]]*hh[[1, i, j]], {j, 1, n}], {i, 1, n}];

sol2 = FullSimplify[f2[f1[xx], xx, uu]];
uu = utem;
xx = xtem;
subst = 
Flatten[{Table[Subscript[x, i] -> xx[[i]], {i, 1, n}], 
Table[Subscript[u, i] -> uu[[i]], {i, 1, n}]}];
subst2 = Flatten[{Table[Subscript[x, i] -> xx[[i]], {i, 1, n}]}];
diff = f1[xx + uu] - sol2 /. subst;
{sei = NSolve[diff == 0, θ, Reals], diff, FullSimplify[hh] /. subst2, sein = FullSimplify[hh] /. Flatten[{subst, sei}], Eigenvalues[sein[[1]]]}

}

thetacalc[{10, 5, 1}, {1, 0, 0}, 3] 

The output (theta, difference-function, H-Matrix [unevaluated], [evaluated], Eigenvalues) for x and u as given in thetacalc:

$\tiny \begin{array}{ccccc} \{\{\theta \to 0.326189\}\} & \frac{10}{(\theta +15)^3}-\frac{1}{360} & \left( \begin{array}{ccc} -\frac{4 \left(\theta u_2+5\right)}{\left(\theta \left(u_1+u_2\right)+15\right){}^3} & \frac{2 \left(\theta u_1-\theta u_2+5\right)}{\left(\theta \left(u_1+u_2\right)+15\right){}^3} & 0 \\ \frac{2 \left(\theta u_1-\theta u_2+5\right)}{\left(\theta \left(u_1+u_2\right)+15\right){}^3} & \frac{4 \left(\theta u_1+10\right)}{\left(\theta \left(u_1+u_2\right)+15\right){}^3} & 0 \\ 0 & 0 & 6 \left(\theta u_3+1\right) \\ \end{array} \right) & \left( \begin{array}{ccc} -0.00555556 & 0.00295899 & 0 \\ 0.00295899 & 0.0114735 & 0 \\ 0 & 0 & 6. \\ \end{array} \right) & \{6.,0.011973,-0.00605506\} \\ \end{array}$

The remainder will be adjusted with a theta between (0,1) from the mean value theorem to equal the original function (at point x with direction vector). A theta is given if the direction vector u is unequal to null-vector.

I am wondering if one could do this simpler as to avoid manual term calculation like the Hesse-matrix (in my code: hh) or even multiple sums.

Thank you for your time.

Answer 1

Probably:

p = {x, y, z};
ff[x_, y_, z_] := (x - y)/(x + y) + z^3
hhh = D[ff[x, y, z], {p, 2}] // Simplify
h1 = hhh /. MapThread[Rule, {p, p + t { u1, u2, u3}}] /. 
               Thread[p :> {10, 5, 1}]

You can get the same result you got with your code with the following (please note that most of the code is just formatting,not sure why you may want that on a function):

calc[ff_, p0_, u0_] := Module[{p = {x, y, z}, u = {u1, u2, u3}, c1, c2, t, diff, tval, mt},
  mt = Flatten@MapThread[Rule, {{p, u}, {p0, u0}}, 2];
  c1 = D[ff@p, {p, 1}] // Simplify;
  c2 = D[ff[p + t u], {p, 2}] // Simplify;
  diff = (ff[p + u] - ff[p] - u.c1 - 1/2 u.c2.u) /. mt;
  tval = First@NSolve[diff == 0, Reals];

  (*Output Formatting follows*)
  {tval /. t -> \[FormalT],
   diff /. t -> \[FormalT],
   c2 /. Thread[p :> p0] /. t -> \[FormalT] // MatrixForm,
   c2 /. mt /. tval // MatrixForm,
   Eigenvalues[c2] /. mt /. tval}]

Usage

f[{x_, y_, z_}] := (x - y)/(x + y) + z^3
calc[f, {10, 5, 1}, {1, 0, 0}]