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

From Math 2220 Class 23 V1 IFT Double Integrals From Math 2220 Class 23 Why Cont. Fcns are Integrable Dr. Allen Back Double Integral Problems Bounding the Values of integrals Oct. 20, 2014

Inverse Function Theorem From Math 2220 Class 23 V1 IFT The identity function Id is the function defined by f(x)=x. If we Double Integrals want to indicate a domain (e.g. x ∈ R n ) we might write Id R n . 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 From Math 2220 Class 23 f : U ⊂ R n → R n with f ( p 0 ) = q 0 , then one can locally V1 uniquely solve (for p in terms of q with p near p 0 and q near IFT q 0 ) the equation Double q = f ( p ) Integrals Why Cont. if the derivative Df ( p 0 ) is invertible. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Inverse Function Theorem From Math 2220 Class 23 Under these circumstances, the solution g ( q ) = p will be a V1 (differentiable) local inverse to f ; i.e. there are neighborhoods IFT U 0 of p 0 and V 0 of q 0 so that Double 1 f : U 0 ⊂ R n → R n has range f ( U 0 ) = V 0 . Integrals Why Cont. 2 g : V 0 ⊂ R n → R n has range g ( V 0 ) = U 0 . Fcns are Integrable 3 f ( g ( q )) = q for all q ∈ V 0 . (i.e f ◦ g = Id V 0 . ) Double Integral 4 g ( f ( p )) = p for all p ∈ U 0 . (i.e g ◦ f = Id U 0 . ) Problems Bounding the Values of integrals

Inverse Function Theorem From Math 2220 Class 23 V1 IFT So f and g are giving a one-to-one correspondence between Double points of U 0 and V 0 ; to each point q of V 0 there is a unique Integrals point p of U 0 with f ( p ) = q and g ( q ) = p . And vice versa; to Why Cont. Fcns are each point p of U 0 , . . . Integrable Double Integral Problems Bounding the Values of integrals

Inverse Function Theorem From Math 2220 Class 23 V1 Moreover the derivative of g is the inverse of the derivative of IFT f , as the chain rule applied to f ◦ g = Id shows must be the Double case if a differentiable inverse exists. Integrals Why Cont. Dg ( f ( p )) = [ Df ( p )] − 1 . 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 From Math 2220 Class 23 f , as the chain rule applied to f ◦ g = Id shows must be the V1 case if a differentiable inverse exists. IFT Dg ( f ( p )) = [ Df ( p )] − 1 . Double Integrals Why Cont. Fcns are Integrable The example of inverse functions Double x 3 f ( x ) = Integral Problems √ x 3 g ( x ) = Bounding the Values of integrals 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.)

Inverse Function Theorem From Math A Basic Inverse Function Theorem Problem 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 From Math 2220 Class 23 f : U ⊂ R 3 → R is a C 1 function and p 0 = ( x 0 , y 0 , z 0 ) is a V1 point on the level set f ( x , y , z ) = C where IFT � ∂ f Double � � = 0 . � Integrals ∂ z � ( x 0 , y 0 , z 0 ) Why Cont. Fcns are Integrable Then we can locally solve for z in terms of x and y near p 0 . Namely there exists a C 1 function g ( x , y ) with domain a Double Integral Problems neighborhood of ( x 0 , y 0 ) so that g ( x 0 , y 0 ) = z 0 and Bounding the Values of f ( x , y , g ( x , y )) = C . integrals

Inverse Function Theorem From Math One way to remember this condition is to realize that the 2220 Class 23 “worst thing that can can happen” in trying to solve V1 f ( x , y , z ) = C for z is for z to not really appear in the equation; IFT e.g. f ( x , y , z ) = x 2 + y 2 = 1 CANNOT be solved for z . Double Integrals Notice here f z = 0; i.e. the hypothesis of f z � = 0 is ruling out Why Cont. this extreme obstruction to being able to solve for z . Fcns are Integrable The conclusion of the implicit function theorem is saying the Double nonvanishing of this partial derivative together with a point to Integral Problems get started with is enough to guarantee a locally defined Bounding the implicit function. Values of integrals

Inverse Function Theorem From Math A Basic Implicit Function Theorem Problem 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 In cases where the implicit function theorem applies, one can use implicit differentiation to compute the partials of the IFT Double implicitly defined function g . Integrals There is a formula one can memorize, but I suggest you just Why Cont. ∂ Fcns are think of implicit diff as the result of applying e.g. ∂ x to the Integrable Double equation f ( x , y , z ( x , y )) = C . Integral Problems Bounding the Values of integrals

Inverse Function Theorem For example if we think of z as a function of x in From Math 2220 Class 23 f ( x , y , z ) = x 2 + y 2 + z 2 + z 9 = 4 V1 near ( x , y , z ) = (1 , 1 , 1) (so we are really thinking IFT x 2 + y 2 + [ z ( x , y )] 2 + [ z ( x , y )] 9 = 4), we obtain Double Integrals 2 x + 2 zz x + 9 z 8 z x = 0 Why Cont. Fcns are Integrable and 2 x Double Integral z x = − 2 z + 9 z 8 . Problems Bounding the or z x (1 , 1) = 2 Values of integrals 11 .

Inverse Function Theorem From Math 2220 Class 23 V1 IFT Note that the partial derivative assumed not to vanish in the Double implicit function theorem hypothesis is exactly the one you Integrals Why Cont. divide by in implicit differentiation. 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 From Math 2220 Class 23 solve for m variables z 1 , . . . , z m ) V1 F 1 ( x 1 , . . . , x n , z 1 , . . . , z m ) = C 1 IFT . . . = . . . Double Integrals F m ( x 1 , . . . , x n , z 1 , . . . , z m ) = C m Why Cont. Fcns are Integrable the general implicit function theorem says this is locally possible Double near a point satisfying the system if the matrix of partials Integral Problems ∂ F 1 ∂ F 1  . . .  Bounding the ∂ z 1 ∂ z m Values of . . . . . . . . . integrals   ∂ F m ∂ F m . . . ∂ z 1 ∂ z m is invertible. (One possibility is to check its determinant is nonzero.)

Inverse Function Theorem From Math 2220 Class 23 V1 Two Constraint Implicit Function Theorem Problem 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 Another Two Constraint Implicit Function Theorem Problem IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Inverse Function Theorem From Math y 2 + z 2 = 1 x 2 + z 2 = 1 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Double Integrals From Math A regular partition P of the interval [ a , b ] × [ c , d ] . 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

Double Integrals From Math 2220 Class 23 A Riemann Sum for a bounded function f : [ a , b ] × [ c , d ] → R . V1 IFT n m � � Double S P = f ( c ij )∆ x i ∆ y j Integrals i =1 j =1 Why Cont. Fcns are Integrable where c ij ∈ [ x i − 1 , x i ] × [ y j − 1 , y j ] . Double Integral (Regular partitions have all the x i (resp. y j ) equally spaced.) Problems Bounding the Values of integrals

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

Double Integrals From Math The Riemann Sum is sum of these (signed) volumes. 2220 Class 23 V1 IFT Double Integrals Why Cont. Fcns are Integrable Double Integral Problems Bounding the Values of integrals

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

Double Integrals From Math 2220 Class 23 V1 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 IFT Double Integrals �� f dA = L Why Cont. Fcns are R Integrable Double where R denotes the rectangle [ a , b ] × [ c , d ] . Integral Problems Bounding the Values of integrals

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