Chapter 4: Partial Differentiation
Section 4.7: Approximations
Example 4.7.10
At x,y=4,1, construct the second-degree Taylor polynomial for fx,y=3 x2+y3.
Solution
Mathematical Solution
The expression for the second-degree Taylor polynomial, namely,
f4,1+fx⋅x−4+fy⋅y−1+12fxx⋅x−42+2 fxy⋅x−4⋅y−1+fyy⋅y−12
where all derivatives are evaluated at 4,1, becomes
−114+127⁢x+314⁢y+3686⁢x−42−18343⁢x−4⁢y−1+5792744⁢y−12
or
−12744+594343⁢x+31372⁢y+3686⁢x2−18343⁢x y+5792744⁢y2
upon simplifying.
Maple Solution - Interactive
Initialize
Tools≻Load Package: Student Multivariate Calculus
Loading Student:-MultivariateCalculus
Context Panel: Assign Function
fx,y=3 x2+y3→assign as functionf
Obtain the second-degree Taylor polynomial
Use the tutor to obtain the first-degree Taylor polynomial. See Figure 4.7.10(a).
Alternatively, launch the tutor by typing fx,y and pressing the Enter key.
Context Panel: Student Multivariate Calculus≻Tutors≻Taylor Approximation Set x,y=4,1 and set the Degree to 2
Figure 4.7.10(a) The Taylor Approximation tutor
Obtain the second-degree Taylor polynomial from first principles
Set all subscripted instances of the symbol f as Atomic Identifiers.
Calculus palette: Partial derivative operator (as appropriate)
f__xx,y=∂∂ x fx,y→assign as functionf__x
f__yx,y=∂∂ y fx,y→assign as functionf__y
f__xxx,y=∂2∂x2 fx,y→assign as functionf__xx
f__xyx,y=∂2∂ y⁢∂ x fx,y→assign as functionf__xy
f__yyx,y=∂2∂y2 fx,y→assign as functionf__yy
Setting each subscripted f as an Atomic Identifier, write the appropriate formula and press the Enter key.
Context Panel: Simplify≻Simplify
f4,1+f__x4,1⋅x−4+f__y4,1⋅y−1+12f__xx4,1⋅x−42+2 f__xy4,1⋅x−4⋅y−1+f__yy4,1⋅y−12
49+1249⁢49⁢x−4+398⁢49⁢y−1+34802⁢49⁢x−42−182401⁢49⁢x−4⁢y−1+57919208⁢49⁢y−12
= simplify
594343⁢x+31372⁢y−12744+3686⁢x2−18343⁢y⁢x+5792744⁢y2
Maple Solution - Coded
Define f.
f≔x,y→3 x2+y3:
Apply the simplify command to the result from the TaylorApproximation command in the Student MultivariateCalculus package.
p≔Student:-MultivariateCalculus:-TaylorApproximationfx,y,x,y=4,1,2
127⁢x+314⁢y−114+3686⁢x−42−18343⁢y−1⁢x−4+5792744⁢y−12
simplifyp
Apply the simplify command to the result of setting the formula with the differential operator D.
q≔f4,1+D1f4,1⋅x−4+D2f4,1⋅y−1+12D1,1f4,1⋅x−42+2 D1,2f4,1⋅x−4⋅y−1+D2,2f4,1⋅y−12
7+1249⁢49⁢x−4+398⁢49⁢y−1+34802⁢49⁢x−42−182401⁢49⁢x−4⁢y−1+57919208⁢49⁢y−12
simplifyq
In constructing the expression for the quadratic term, note that the equality of mixed partials was assumed. The terms containing fxy and fyx were combined into 2 fxy4,1⋅x−4⋅y−1.
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