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Your book defines the first and second order Taylor polynomials on page 196. Using more compact
notation, we can write the Taylor Forumla for the function f about the point x0 as
p2 (x) = f (x0 ) + Df (x0 )(x − x0 ) + (x − x0 )t Hf (x0 )(x − x0 )
1. Write out Taylor’s Forumla explicitly for scalar valued functions of one, two, and three variables.
In dimension 1 this is the same Taylor polynomial that we worked with in single variable calculus.
We will let our point a = a.
p1 (x) = f (a) + f 0 (a)(x − a)
p2 (x) = f (a) + f 0 (a)(x − a) +
f 00 (a)
(x − a)2
2
Now for n = 2 the two dimensional case, rather than using x1 and x2 as variables we will use x and
y, and we will write a = (a, b)
p1 (x, y) = f (a) + fx (a)(x − a) + fy (a)(y − b)
p2 (x, y) = f (a) + fx (a)(x − a) + fy (a)(x − a) +
fxx (a)
fyy (a)
(x − a)2 +
(y − b)2 + fxy (a)(x − a)(y − b)
2
2
Finally if n = 3, the 3 dimensional case, we will use x, y, z for our variables, and let the point
a = (a, b, c) we have
p1 (x, y, z) = f (a) + fx (a)(x − a) + fy (a)(y − b) + fz (a)(z − c)
p2 (x, y, z) = f (a) + fx (a)(x − a) + fy (a)(y − b) + fz (a)(z − c) +
2. Consider the function f (x, y) = sin(xy) and
• Compute the Hessian matrix of f at the point a = (0, 0).
he Hessian matrix in this situation will be given as
H=
∂2f
∂x2
∂2f ∂2f
∂y∂x ∂y 2
∂2f
∂x∂y
!
To begin we will compute
∂f
= fx (x, y) = y cos(xy)
∂x
∂f
= fy (x, y) = x cos(xy)
∂y
fxx (a)
fyy (a)
fyy (a)
(x − a)2 +
(y − b)2 +
(z − c)2 + f
2
2
2
Building off of this result, we can easily compute the whole Hessian
Which we can easily compute explicitly
H=
−y 2 sin(xy)
cos(xy) − xy sin(xy)
cos(xy) − xy sin(xy)
−x2 sin(xy)
• Compute the degree 2 Taylor Polynomial of f (x, y) at the point a = (0, 0)
Using the general form we compute above, and the Hessian matrix (which we also computed)
p2 (x, y) = sin(0 · 0) + 0 · cos(0 · 0)(x − 0) + 0 · cos(0 · 0)(y − 0) − 02 sin(0 · 0) − 02 sin(0 · 0) + cos(0 · 0) − 0 · 0 sin(x
3. Find the second order Taylor expansion about the point (0, 0) of the function
f (x, y) = exy
We begin by computing the matrix of partial derivatives of f .
Df (x, y) = (exy y, exy x)
From this we compute the Hessian matrix
Hf (x, y) =
exy y 2
xy
e + exy xy
exy + exy xy
exy x2
Then we evaluate at the point (0, 0) and find
Df (0, 0) = (0, 0)
0 1
Dg(0, 0) =
1 0
Now we put these together to compute the degree 2 Taylor polynomial
1
x
T2 (x, y) = f (0, 0) + Df (0, 0) · (x, y) + (x, y)Hf (0, 0)
y
2
Expanding out the linear algebra we obtain
T2 (x, y) = 1 + xy
Which is the degree 2 Taylor polynomial about (0, 0) of the function f (x, y) = exy
4. Find and classify all critical points of the function
f (x, y) = x3 + x2 y + y 2 + xy + x + 1
As in the last problem we will begin by computing both the matrix of partial derivatives and the
Hessian matrix.
Df (x, y) = (3x2 + 2xy + y, x2 + 2y + x)
6x + 2y 2x + 1
Hf (x, y) =
2x + 1
2
Now to find the critical values we compute where Df (x, y) = (0, 0)
We easily find the following points (0, 0), (1, −1), and (1/2, −3/8). Now using our criteria on each
of these points we find that the points (0, 0) and (1, −1) are both saddle points, while (1/2, −3/8)
is a local minimum.
5. Let F(x, y, z) = (xy 2 , yx2 , −z 2 ), use the divergence theorem to calcualte the total flux out of a
canister W , which can be expressed by the equations x2 + y 2 ≤ 9 and 1 ≤ z ≤ 3.
We will solve this problem using the divergence theorem. As the problem is stated we want compute
a flux (surface) integral over the boundary of W . That is we want to compute
ZZ
FdS
∂W
Using the divergence theorem this becomes
ZZZ
÷W dA
W
It is not difficult to compute the div of F
∂
∂
∂
xy 2 +
yx2 +
−z 2 =
∂x
∂y
∂z
2
2
= y + x − 2z
÷F =
So we setup the integral according to the divergence theorem, and writing the region in cylindrical
coordinates
ZZZ
W
= y 2 + z 2 − 2z =
Z3 Z
1
0
2π
Z
0
3
(r2 − 2z)rdrdθdz = 9π
6. Use the Divergence Theorem to compute the value of the flux integral
ZZ
FdS
S
Where F(x, y, z) = (y 3 + 3x, xz + y, z + x4 cos(x2 y)) and S is the boundary of the region bounded
by x2 + y 2 = 1, x ≥ 0, y ≥ 0 and 0 ≤ z ≤ 1
Using the divergence theorem we will compute this rather difficult integral into a (hopefully) simpler
triple integral over the divergence of F. To begin let us compute ÷F
÷(F) =
∂ ∂ ∂
,
,
∂x ∂y ∂z
· y 3 + 3x, xz + y, z + x4 cos(x2 y) = 5
We then can setup the triple integral as
ZZZ
5dV
W
Looking at the region we want to compute in cylindrical coordinates as
π
Z1 Z2 Z1
5rdrdθdz =
0
0
5π
4
0
Which is the value of the surface integral, by the Divergence theorem.