From Math 2220 Class 23 Why Cont. Fcns are Integrable Dr. Allen - - PowerPoint PPT Presentation

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

From Math 2220 Class 23

• Dr. Allen Back
• Oct. 20, 2014

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Inverse Function Theorem

The identity function Id is the function defined by f(x)=x. If we want to indicate a domain (e.g. x ∈ Rn) we might write IdRn.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Inverse Function Theorem

The inverse function theorem says that for a C 1 function f : U ⊂ Rn → Rn with f (p0) = q0, then one can locally uniquely solve (for p in terms of q with p near p0 and q near q0) the equation q = f (p) if the derivative Df (p0) is invertible.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Inverse Function Theorem

Under these circumstances, the solution g(q) = p will be a (differentiable) local inverse to f ; i.e. there are neighborhoods U0 of p0 and V0 of q0 so that

1 f : U0 ⊂ Rn → Rn has range f (U0) = V0. 2 g : V0 ⊂ Rn → Rn has range g(V0) = U0. 3 f (g(q)) = q for all q ∈ V0. (i.e f ◦ g = IdV0.) 4 g(f (p)) = p for all p ∈ U0. (i.e g ◦ f = IdU0.)

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Inverse Function Theorem

So f and g are giving a one-to-one correspondence between points of U0 and V0; to each point q of V0 there is a unique point p of U0 with f (p) = q and g(q) = p. And vice versa; to each point p of U0, . . .

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Inverse Function Theorem

Moreover the derivative of g is the inverse of the derivative of f , as the chain rule applied to f ◦ g = Id shows must be the case if a differentiable inverse exists. Dg(f (p)) = [Df (p)]−1 .

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Inverse Function Theorem

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Inverse Function Theorem

Moreover the derivative of g is the inverse of the derivative of f , as the chain rule applied to f ◦ g = Id shows must be the case if a differentiable inverse exists. Dg(f (p)) = [Df (p)]−1 . The example of inverse functions f (x) = x3 g(x) =

3

√x shows that even when Df fails to be invertible (f ′(0) = 0 here), a continuous inverse CAN still possibly exist; it just can’t be differentiable. (g′(0) does not exist.)

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Inverse Function Theorem

A Basic Inverse Function Theorem Problem

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Inverse Function Theorem

Implicit Function Theorem (1 Constraint): Suppose f : U ⊂ R3 → R is a C 1 function and p0 = (x0, y0, z0) is a point on the level set f (x, y, z) = C where ∂f ∂z

• (x0,y0,z0)

= 0. Then we can locally solve for z in terms of x and y near p0. Namely there exists a C 1 function g(x, y) with domain a neighborhood of (x0, y0) so that g(x0, y0) = z0 and f (x, y, g(x, y)) = C.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Inverse Function Theorem

One way to remember this condition is to realize that the “worst thing that can can happen” in trying to solve f (x, y, z) = C for z is for z to not really appear in the equation; e.g. f (x, y, z) = x2 + y2 = 1 CANNOT be solved for z. Notice here fz = 0; i.e. the hypothesis of fz = 0 is ruling out this extreme obstruction to being able to solve for z. The conclusion of the implicit function theorem is saying the nonvanishing of this partial derivative together with a point to get started with is enough to guarantee a locally defined implicit function.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Inverse Function Theorem

A Basic Implicit Function Theorem Problem

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Inverse Function Theorem

In cases where the implicit function theorem applies, one can use implicit differentiation to compute the partials of the implicitly defined function g. There is a formula one can memorize, but I suggest you just think of implicit diff as the result of applying e.g. ∂ ∂x to the equation f (x, y, z(x, y)) = C.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Inverse Function Theorem

For example if we think of z as a function of x in f (x, y, z) = x2 + y2 + z2 + z9 = 4 near (x, y, z) = (1, 1, 1) (so we are really thinking x2 + y2 + [z(x, y)]2 + [z(x, y)]9 = 4), we obtain 2x + 2zzx + 9z8zx = 0 and zx = − 2x 2z + 9z8 .

• r zx(1, 1) = 2

11.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Inverse Function Theorem

Note that the partial derivative assumed not to vanish in the implicit function theorem hypothesis is exactly the one you divide by in implicit differentiation.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Inverse Function Theorem

In the case of a system of m equations (where we can hope to solve for m variables z1, . . . , zm) F1(x1, . . . , xn, z1, . . . , zm) = C1 . . . = . . . Fm(x1, . . . , xn, z1, . . . , zm) = Cm the general implicit function theorem says this is locally possible near a point satisfying the system if the matrix of partials  

∂F1 ∂z1

. . .

∂F1 ∂zm

. . . . . . . . .

∂Fm ∂z1

. . .

∂Fm ∂zm

  is invertible. (One possibility is to check its determinant is nonzero.)

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Inverse Function Theorem

Two Constraint Implicit Function Theorem Problem

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Inverse Function Theorem

Another Two Constraint Implicit Function Theorem Problem

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Inverse Function Theorem

y2 + z2 = 1 x2 + z2 = 1

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Double Integrals

A regular partition P of the interval [a, b] × [c, d].

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Double Integrals

A Riemann Sum for a bounded function f : [a, b] × [c, d] → R. SP =

n

• i=1

m

• j=1

f (cij)∆xi∆yj where cij ∈ [xi−1, xi] × [yj−1, yj]. (Regular partitions have all the xi (resp. yj) equally spaced.)

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Double Integrals

A Riemann Sum Picture

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Double Integrals

The Riemann Sum is sum of these (signed) volumes.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Double Integrals

A finer partition.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Double Integrals

We say a function is integrable if the Riemann sums settle down to a limit L as the mesh diameter goes to zero. If so we write

• R

f dA = L where R denotes the rectangle [a, b] × [c, d].

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Double Integrals

Density of Foxes in England

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Double Integrals

Density of Foxes in England

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Double Integrals

The example f (x, y) = 1 if x is rational

• therwise

shows that not all functions are integrable.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Double Integrals

Theorem: If f : R → R is continuous, then f is integrable; i.e.

• R f dA exists.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Double Integrals

We actually compute double integrals by a sequence of 1 dimensional integrals using Fubini’s Theorem: Theorem: If f is continuous then

• R

f dA = b

a

d

c

f (x, y) dy

• dx

(or b

a

d

c

f (x, y) dy

• dx.)

At the Riemann sum level this comes down to organizing your sums over all rectangles by first adding up contributions from each row of rectangles and then adding up the row totals. But there are subtleties to this!

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Double Integrals

The example f (x, y) = 1 if x is rational 2y

• therwise

satisfies 1 1 f (x, y) dy

• dx = 1

even though the double integral

• R

f (x, y) dA does not exist for R the rectangle [0, 1] × [0, 1].

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Double Integrals

Defining double integrals for bounded regions D more general than a rectangle: We use a trick to relate this case to the rectangle one. Namely, find rectangle R containing D and define f ∗(p) = f (p) if p ∈ D

• therwise

The downside of this is that such an f ∗ is discontinuous. But for reasonable sets D, the discontinuities will be unions of a small number of “graphs of continuous functions.” (i.e. sets {(x, g(x)) : x ∈ [α, β]} or {(h(y), y) : y ∈ [γ, δ]}).

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Double Integrals

Particularly since continuous functions with compact domains are uniformly continuous, one can readily prove: Theorem: If a bounded function f : R ⊂ R2 → R is continuous at all points of a rectangle R except possibly for points lying a finite union of graphs of continuous functions (of one variable), then f is integrable.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Double Integrals

And Fubini works here too: Theorem: Suppose a bounded function f : R ⊂ R2 → R is continuous at all points of a rectangle R except possibly for points lying in a finite union of graphs of continuous functions (of one variable.) Suppose also that g(x) = d

c

f (x, y) dy exists for each x ∈ [a, b]. Then

• R

f dA = b

a

g(x) dx = b

a

d

c

f (x, y) dy

• dx.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Double Integrals

Another example showing Fubini is not so obviously true: Consider the (infinite to the left and down) matrix of values       . . . − 1

8

− 1

4

− 1

2

1 . . . − 1

4

− 1

2

1 . . . − 1

2

1 . . . 1 . . .       where the sum of each row is 0 while the sum of the rightmost column is 1 the next rightmost 1

2, then 1 4, . . . .

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Double Integrals

Another example showing Fubini is not so obviously true: Consider the (infinite to the left and down) matrix of values       . . . − 1

8

− 1

4

− 1

2

1 . . . − 1

4

− 1

2

1 . . . − 1

2

1 . . . 1 . . .       where the sum of each row is 0 while the sum of the rightmost column is 1 the next rightmost 1

2, then 1 4, . . . .

So the sum of the row sums is 0 while the sum of the column sums is 2.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Double Integrals

If one uses the infinite partition . . . 1 2n < . . . < 1 8 < 1 4 < 1 2 < 1

• f [0, 1] in both the x and y directions on the rectangle

[0, 1] × [0, 1], it is not hard to construct a step function f whose integrals over each sub-rectangle match the infinite matrix above.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Why Cont. Fcns are Integrable

Two Definitions Continuity at each Point of a set S: ∀x ∈ S and ∀ǫ > 0 ∃δ > 0 so that |x − y| < δ ⇒ |f (x) − f (y)| < ǫ. Uniform Continuity on a Set S: ∀ǫ > 0 ∃δ > 0 so that ∀x, y ∈ S, |x − y| < δ ⇒ |f (x) − f (y)| < ǫ.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Why Cont. Fcns are Integrable

An Important Theorem Bolzano Weierstrass: Every bounded sequence in Rn has a convergent subsequence. This theorem is also at the heart of the proof that continuous functions with closed and bounded domains are automatically bounded and attain their extrema.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Why Cont. Fcns are Integrable

The idea of why Bolzano Weirrstrass holds:

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Why Cont. Fcns are Integrable

Based on Bolzano Weierstrass, one can show that every continuous function on a closed and bounded set is uniformly continuous there.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Why Cont. Fcns are Integrable

Suppose one is considering the integrability of a continuous function f over a rectangle of area A.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Why Cont. Fcns are Integrable

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Why Cont. Fcns are Integrable

Suppose one is considering the integrability of a continuous function f over a rectangle of area A. Consider any regular partition each of whose constituent rectangles is smaller in diameter than the δ given in the definition of uniform continuity.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Why Cont. Fcns are Integrable

Suppose one is considering the integrability of a continuous function f over a rectangle of area A. Consider any regular partition each of whose constituent rectangles is smaller in diameter than the δ given in the definition of uniform continuity. Then any two Riemann sums for this partition will differ by at most ǫA. This allows one to show that all Riemann sums settle down to a single limit as the partitions becomes sufficiently fine.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Double Integral Problems

Elementary Regions: y-simple leading to b

a

h(x)

g(x)

f (x, y) dy dx. and similarly x-simple.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Double Integral Problems

Elementary Regions: y-simple leading to b

a

h(x)

g(x)

f (x, y) dy dx. and similarly x-simple. Draw a picture when setting up those limits; the inner limits encode the intersection of lines x = c with the region while the

• uter limits describe which vertical lines x = c actually meet

the region.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Double Integral Problems

Elementary Regions: y-simple leading to b

a

h(x)

g(x)

f (x, y) dy dx. and similarly x-simple. Draw a picture when setting up those limits; the inner limits encode the intersection of lines x = c with the region while the

• uter limits describe which vertical lines x = c actually meet

the region. You may have to break a region into pieces to write it as a union of y-simple and x-simple pieces.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Double Integral Problems

Calculate

• [0,1]×[1,3]

x2 + y2 dA.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Double Integral Problems

Setup the double integral for the volume bounded by the paraboloid z = 4 − x2 − y2 and the xy-plane.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Double Integral Problems

Setup the integral of f (x, y) = ex over the region bounded by the curves y = 2x and y = x3 + x. (or just the area . . . .)

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Double Integral Problems

Interchange the order of integration in 1 2x x2 dy dx. Be sure and sketch the region in a problem like this.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Double Integral Problems

Calculate 1 1

y

sin (x2) dx dy. Sometimes reversing the order of integration can help computationally.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Double Integral Problems

Interchange the order of integration in t x f (y) dy dx. and obtain an alternate formula for the double antiderivative this way.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Double Integral Problems

Calculate 1 √

1−y2

• 1 − x2 dx dy.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Double Integral Problems

Use a double integral to find the volume bounded by the four planes z = x = 1 y = 2 x + 2y + 3z = 6.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Bounding the Values of integrals

If m ≤ f (x, y) ≤ M for all (x, y) ∈ D then m · Area(D) ≤

• D

f (x, y) dA ≤ M · Area(D).

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Bounding the Values of integrals

Or more generally if f ≤ g then

• D

f (x, y) dA ≤

• D

g(x, y) dA.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Bounding the Values of integrals

Give reasonable upper and lower bounds for

• D

ex2+y2 sin x x2y2 + 1 dy dx for the disk of radius 3 centered at the origin.

From Math 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Bounding the Values of integrals

There is also a mean value theorem for integrals: Theorem: If f is continuous and D is connected, then ∃(x0, y0) ∈ D so that 1 Area(D)

• D

f (x, y) dA = f (x0, y0).