Finding Quadratic Taylor Polynomial of f(x,y) = eˣ cos y at (0,0)

Calculus 3 · Axiom Academy

EXAMPLE Finding the Quadratic Taylor Polynomial of at (0,0) Build the second-order Taylor polynomial of a two-variable function using its gradient and Hessian Find the quadratic (second-degree) Taylor polynomial T_2(x,y) that approximates near the point (0,0) . How close is the approximation near (0,0) ? Same-level contours of the exact surface (solid) and its quadratic Taylor polynomial (dashed). Near (0,0) they nearly coincide; the gap widens as you move away — exactly what "a good approximation near the center" means. Nice work! You built the quadratic Taylor polynomial of from its gradient and Hessian. Here's what carries over to any second-order Taylor problem: Systematic approach: compute every partial derivative up to second order, then evaluate them all at the center point (a,b) . First-order (linear) terms: the x and y terms come from f_x(a,b) and f_y(a,b) — the components of the gradient. Second-order (curvature) terms: f_ xx , f_ yy , and f_ xy (the Hessian) supply the quadratic terms and are divided by 2! . Mixed partial: the f_ xy term carries a factor of 2 inside the bracket because it feeds both the xy and yx contributions — here f_ xy (0,0)=0 , so it drops out entirely. Approximation quality: T_2 hugs f near (0,0) . At (0.1,0.1) , T_2 = 1.100 while — agreement to about . The finished polynomial says it all: e^ x contributes the growing curvature while contributes the downward curvature — a saddle-shaped fit to the surface at the origin.

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