From Math 2220 Class 22 Double Integrals Double Integral Dr. - PowerPoint PPT Presentation

From Math 2220 Class 22 V1cc IFT Why the IFT? From Math 2220 Class 22 Double Integrals Double Integral Dr. Allen Back Problems Oct. 17, 2014

Inverse Function Theorem From Math 2220 Class 22 V1cc IFT The identity function Id is the function defined by f(x)=x. If we Why the IFT? want to indicate a domain (e.g. x ∈ R n ) we might write Id R n . Double Integrals Double Integral Problems

Inverse Function Theorem The inverse function theorem says that for a C 1 function From Math 2220 Class 22 f : U ⊂ R n → R n with f ( p 0 ) = q 0 , then one can locally V1cc uniquely solve (for p in terms of q with p near p 0 and q near IFT q 0 ) the equation Why the IFT? q = f ( p ) Double Integrals if the derivative Df ( p 0 ) is invertible. Double Integral Problems

Inverse Function Theorem From Math 2220 Class 22 Under these circumstances, the solution g ( q ) = p will be a V1cc (differentiable) local inverse to f ; i.e. there are neighborhoods IFT U 0 of p 0 and V 0 of q 0 so that Why the IFT? 1 f : U 0 ⊂ R n → R n has range f ( U 0 ) = V 0 . Double Integrals 2 g : V 0 ⊂ R n → R n has range g ( V 0 ) = U 0 . Double Integral 3 f ( g ( q )) = q for all q ∈ V 0 . (i.e f ◦ g = Id V 0 . ) Problems 4 g ( f ( p )) = p for all p ∈ U 0 . (i.e g ◦ f = Id U 0 . )

Inverse Function Theorem From Math 2220 Class 22 V1cc IFT So f and g are giving a one-to-one correspondence between Why the IFT? points of U 0 and V 0 ; to each point q of V 0 there is a unique Double point p of U 0 with f ( p ) = q and g ( q ) = p . And vice versa; to Integrals each point p of U 0 , . . . Double Integral Problems

Inverse Function Theorem From Math 2220 Class 22 V1cc 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 Why the IFT? case if a differentiable inverse exists. Double Integrals Dg ( f ( p )) = [ Df ( p )] − 1 . Double Integral Problems

Inverse Function Theorem From Math 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Inverse Function Theorem Moreover the derivative of g is the inverse of the derivative of From Math 2220 Class 22 f , as the chain rule applied to f ◦ g = Id shows must be the V1cc case if a differentiable inverse exists. IFT Dg ( f ( p )) = [ Df ( p )] − 1 . Why the IFT? Double Integrals Double The example of inverse functions Integral Problems x 3 f ( x ) = √ x 3 g ( 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.)

Inverse Function Theorem From Math A Basic Inverse Function Theorem Problem 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Inverse Function Theorem Implicit Function Theorem (1 Constraint): Suppose From Math 2220 Class 22 f : U ⊂ R 3 → R is a C 1 function and p 0 = ( x 0 , y 0 , z 0 ) is a V1cc point on the level set f ( x , y , z ) = C where IFT � ∂ f Why the IFT? � � = 0 . � ∂ z Double � ( x 0 , y 0 , z 0 ) Integrals Double Then we can locally solve for z in terms of x and y near p 0 . Integral Namely there exists a C 1 function g ( x , y ) with domain a Problems neighborhood of ( x 0 , y 0 ) so that g ( x 0 , y 0 ) = z 0 and f ( x , y , g ( x , y )) = C .

Inverse Function Theorem From Math One way to remember this condition is to realize that the 2220 Class 22 “worst thing that can can happen” in trying to solve V1cc 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 . Why the IFT? Notice here f z = 0; i.e. the hypothesis of f z � = 0 is ruling out Double Integrals this extreme obstruction to being able to solve for z . Double The conclusion of the implicit function theorem is saying the Integral Problems nonvanishing of this partial derivative together with a point to get started with is enough to guarantee a locally defined implicit function.

Inverse Function Theorem From Math A Basic Implicit Function Theorem Problem 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Inverse Function Theorem From Math 2220 Class 22 V1cc In cases where the implicit function theorem applies, one can use implicit differentiation to compute the partials of the IFT Why the IFT? implicitly defined function g . Double There is a formula one can memorize, but I suggest you just Integrals ∂ think of implicit diff as the result of applying e.g. ∂ x to the Double Integral Problems equation f ( x , y , z ( x , y )) = C .

Inverse Function Theorem For example if we think of z as a function of x in From Math 2220 Class 22 f ( x , y , z ) = x 2 + y 2 + z 2 + z 9 = 4 V1cc 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 Why the IFT? Double 2 x + 2 zz x + 9 z 8 z x = 0 Integrals Double and Integral 2 x Problems z x = − 2 z + 9 z 8 . or z x (1 , 1) = 2 11 .

Inverse Function Theorem From Math 2220 Class 22 V1cc IFT Note that the partial derivative assumed not to vanish in the Why the IFT? implicit function theorem hypothesis is exactly the one you Double Integrals divide by in implicit differentiation. Double Integral Problems

Inverse Function Theorem In the case of a system of m equations (where we can hope to From Math 2220 Class 22 solve for m variables z 1 , . . . , z m ) V1cc F 1 ( x 1 , . . . , x n , z 1 , . . . , z m ) = C 1 IFT . . . = . . . Why the IFT? Double F m ( x 1 , . . . , x n , z 1 , . . . , z m ) = C m Integrals Double the general implicit function theorem says this is locally possible Integral Problems near a point satisfying the system if the matrix of partials ∂ F 1 ∂ F 1  . . .  ∂ z 1 ∂ z m . . . . . . . . .   ∂ 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 22 V1cc Two Constraint Implicit Function Theorem Problem IFT Why the IFT? Double Integrals Double Integral Problems

Inverse Function Theorem From Math 2220 Class 22 V1cc Another Two Constraint Implicit Function Theorem Problem IFT Why the IFT? Double Integrals Double Integral Problems

Inverse Function Theorem From Math y 2 + z 2 = 1 x 2 + z 2 = 1 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Why the IFT? From Math 2220 Class 22 V1cc IFT Existence of solutions to f ( p ) = q for p near p 0 and q near Why the IFT? q 0 = f ( p 0 ) may be established by showing that Newton’s Double Integrals method with initial choice p 0 converges to a solution. Double Integral Problems

Why the IFT? From Math 2220 Class 22 We want f ( p ) = q or f ( p ) − f ( p 0 ) = q − q 0 . V1cc IFT ( Df ( p 0 )) ( p − p 0 ) ∼ q − f ( p 0 ) Why the IFT? ( Df ( p 0 )) − 1 ( q − f ( p 0 ))+ Set p 1 = p 0 Double Integrals ( Df ( p 1 )) ( p − p 1 ) ∼ q − f ( p 1 ) Double ( Df ( p 1 )) − 1 ( q − f ( p 1 ))+ Integral Set p 2 = p 1 Problems . . . ∼ . . . Set p n = . . . + . . .

Why the IFT? From Math Uniqueness of solutions to f ( p ) = q for p near p 0 and q near 2220 Class 22 q 0 = f ( p 0 ) may be established for a C 2 function using Taylor’s V1cc formula with remainder. For example IFT Why the IFT? � f ( p ) − f ( r ) − Df ( p )( r − p ) � ≤ C � p − r � 2 Double Integrals ⇒ � f ( p ) − f ( r ) � ≥ 1 2 � Df ( p )( r − p ) � Double Integral Problems for p and r close enough implying f ( p ) � = f ( r ) when p � = r in this neighborhood.

Why the IFT? From Math 2220 Class 22 The implicit function theorem for the case f ( x , y , z ( x , y )) = C V1cc may be established by applying the inverse function theorem to IFT the system Why the IFT? u = x Double Integrals v = y Double Integral w = f ( x , y , z ) Problems and extracting z ( x , y ) from the local inverse which arises.

Double Integrals From Math A regular partition P of the interval [ a , b ] × [ c , d ] . 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Double Integrals From Math 2220 Class 22 A Riemann Sum for a bounded function f : [ a , b ] × [ c , d ] → R . V1cc IFT n m � � Why the IFT? S P = f ( c ij )∆ x i ∆ y j Double i =1 j =1 Integrals Double where c ij ∈ [ x i − 1 , x i ] × [ y j − 1 , y j ] . Integral Problems (Regular partitions have all the x i (resp. y j ) equally spaced.)

Double Integrals From Math A Riemann Sum Picture 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Double Integrals From Math The Riemann Sum is sum of these (signed) volumes. 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Double Integrals From Math A finer partition. 2220 Class 22 V1cc IFT Why the IFT? Double Integrals Double Integral Problems

Double Integrals From Math 2220 Class 22 V1cc 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 Why the IFT? �� Double f dA = L Integrals R Double Integral Problems where R denotes the rectangle [ a , b ] × [ c , d ] .