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Taylor's Theorem for Functions in Two Variables
Let be a function in two variables,. Suppose the partial derivatives of of all orders up to exist and are continuous at all points in an open ball of positive radius centred at , then for , we have:
where:
and:
for some. The polynomial is called the ‑th Taylor Polynomial of about.
Week 12
Taylor's Theorem
Local Extrema
f(x, y) n ∈ N f
n + 1 B
(a, b) (x, y) ∈ B
f(x, y) = pn(x, y) + +Rn(x, y),
pn(x, y) =
n
k=
k
j=
∣(a,b)
(x − a)k−j(y − b)j
k!
k
j
∂kf
∂xk−j∂yj
= f(a, b) + fx(a, b)(x − a) + fy(a, b)(y − b)
+ (fxx(a, b)(x − a)^2 + 2fxy(a, b)(x − a)(y − b) + fyy(a, b)(y − b)^2 )
+ (fxxx(a, b)(x − a)^3 + 3fxxy(a, b)(x − a)^2 (y − b)
+3fxyy(a, b)(x − a)(y − b)^2 + fyyy(a, b)(y − b)^3 ) + ⋯ ,
Rn(x, y) =
n+ ∑ j=
∣(a+c(x−a),b+c(x−b))
(^1) (x − a)n+1−j(y − b)j, (n + 1)!
n + 1 j
∂n+1f ∂xn+1−j∂yj
c ∈ (0, 1) pn(x, y) n f(x, y) (a, b)
Example.
Let. Approximate the value of using the second Taylor Polynomial of about. We have:
Hence, the second Taylor Polynomial of about is:
So, is approximately equal to.
The error of the approximation is:
for some.
Computing the 3‑rd order partial derivatives of , we have:
since the sine and cosine functions have absolute values less than or equal to.
Example.
Find the 3 rd Taylor polynomial of at the point.
In general, for a function in variables, its ‑th Taylor polynomial at a point is:
f(x, y) = sin x sin y f(0.01, −0.2) f (0, 0)
fx(x, y) = cos x sin y, fy(x, y) = sin x cos y, fxx(x, y) = − sin x sin y, fxy(x, y) = cos x cos y, fyy(x, y) = − sin x sin y.
f (0, 0)
p(x, y) = f(0, 0) + fx(0, 0)x + fy(0, 0)y
- (fxx(0, 0)x^2 + 2fxy(0, 0)xy + fyy(0, 0)y^2 )
= 0 + 0 + 0 + (0 + 2 ⋅ 1 ⋅ xy + 0) = xy.
f(0.01, −0.2) p(0.01, −0.2) = (0.01)(−0.2) = −0.
|f(0.01, −0.2) − p(0.01, −0.2)| = |R 2 (0.01, −0.2)| = ∣∣∣ (fxxx(0.01c, −0.2c)(0.01)^3 + 3fxxy(0.01c, −0.2c)(0.01)^2 (−0.2)
+3fxyy(0.01c, −0.2c)(0.01)(−0.2)^2 + fyyy(0.01c, −0.2c)(−0.2)^3 )∣∣ ,
c ∈ (0, 1)
f
|R 2 (0.01, −0.2)| = ∣∣∣ (− cos(0.01c) sin(−0.2c)(0.01)^3 − 3 sin(−0.01c) cos(−0.2c)(0.01)^2 (−0.2)
−3 cos(0.01c) sin(−0.2c)(0.01)(−0.2)^2 − sin(0.01c) cos(−0.2c)(−0.2)^3 )∣∣
≤ (|0.01|^3 + 3|0.01|^2 |−0.2| + 3 |0.01| |−0.2|^2 + |−0.2|^3 ) ,
f(x, y) = ln(2x + y) (0, 1)
f : Rn^ ⟶ R n l a⃗ = (a 1 ,^ a 2 , … ,^ an)
Theorem.
(Second Derivative Test) Suppose is a critical point of , and the first and second order partial derivatives of are continuous on an open neighborhood of (in particular ). Then: If : If , then has a local minimum at. If , then has a local maximum at. If : has a saddle point at.
If , The second derivative test is inconclusive.
Example.
Let:
Classify the critical points of.
which is defined for all.
Solving:
We obtain:
Hence,
Evaluating at the critical points, we have:
This implies that:
corresponds to a saddle point,
and that corresponds to either a local maximum or minimum.
(a, b) f f (a, b) ∇f(a, b) =⃗ 0 D(a, b) > 0 fxx(a, b) > 0 f (a, b) fxx(a, b) < 0 f (a, b) D(a, b) < 0 f (a, b)
D(a, b) = 0
f(x, y) = 3y^2 − 2y^3 − 3x^2 + 6xy.
f
∇f(x, y) = ⟨−6x + 6y, 6y − 6y^2 + 6x⟩,
(x, y)
∇f(x, y) = ⟨0, 0⟩,
(x, y) = (0, 0) or (2, 2).
fxx = −6, fxy = 6, fyy = 6 − 12y.
D(x, y) = fxxfyy − f (^) xy^2 = 72(y − 1).
D(x, y) D(0, 0) = −72 < 0. D(2, 2) = 72 > 0.
Since, , we conclude that:
corresponds to a local maximum.
Idea Behind the Second Derivative Test
Let be the critical point under consideration. By Taylor's Theorem, over a small neighborhood of , is closely approximated by the polynomial:
The polynomial is of degree , and the graphs of such polynomials fall into 3 categories: Downward paraboloid. This corresponds to ,.
Upward paraboloid. This corresponds to ,.
Hyperbolic paraboloid. This corresponds to.
fxx(2, 2) = −6 < 0
(2, 2)
(a, b) (a, b) f(x, y) Q(x, y) = f(a, b) + f x(a, b) =
(x − a) + fy(a, b) =
(y − b)
- 12 (fx(a, b)(x − a)^2 + 2fxy(a, b)(x − a)(y − b) + fyy(a, b)(y − b)^2 ).
Q 2 D(a, b) > 0 fxx(a, b) < 0
D(a, b) > 0 fxx(a, b) > 0
D(a, b) < 0
Here, should be viewed as the integral of a one‑variable function in , with fixed. In other words:
where is a function in two variables such that.
Hence,
which is an integral of a one‑variable function in.
Likewise,
where , and:
which is an integral of a one‑variable function in.
Theorem.
(Fubini's Theorem) If is continuous over , then:
Example.
Compute:
∫ (^) cd f(x, y) dy f(x, y) y x
∫
y=d y=c
f(x, y) dy = F (x, y)∣∣∣
y=d y=c
= F (x, c) − F (x, d),
F (x, y) ∂F = f(x, y) ∂y
b a
d c
f(x, y) dy dx = ∫
b a
[F (x, c) − F (x, d)] dx,
x
x=b x=a
f(x, y) dx = G(x, y)∣∣∣
x=b x=a
= G(a, y) − G(b, y),
∂G = f(x, y) ∂x
d c
b a
f(x, y) dx dy = ∫
d c
[G(a, y) − G(b, y)] dy,
y
f(x, y) R = [a, b] × [c, d]
∬ R
f(x, y) dA = ∫
b a
d c
f(x, y) dy dx = ∫
d c
b a
f(x, y) dx dy
1 0
4 2
3 x^2 y dy dx
1 −
3 2
xyex^ dx dy
[0,1]×[0,2]
xy dA 2 (x^2 + y)^2
