MATH301 Real Analysis Tutorial Note #3

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1 MATH301 Real Analysis Tutorial Note #3 More Differentiation in Vector-valued function: Last time, we learn how to check the differentiability of a given vector-valued function. Recall a function F: is differentiable at a iff there is a linear transformation T: such that 0 We denote the transformation T DFa and the corresponding matrix to be F a, according to differentiation theorem, this matrix is given by f f F x x a f f x x which contains all the slopes of tangent planes of f, f,, f (More explanations will be given in the appendix.) The corresponding linear transformation DFax, y is given by f 1 f 1 x x 1 x n DFax, y. 1 f m f. m. x n x 1 x n In fact, if a function is differentiable, then there is ONLY ONE transformation satisfy the definition of derivative of Fx. It means the derivative of F is uniquely determined. The reason of this is given by the following example: Example 1 (Practice Exercise #3) Let : be differentiable at, explain why there cannot be two different linear transformations (say T and T ) both satisfy the definition of derivative of ) As is differentiable, then by differentiation theorem, the matrix of the transformation satisfying the definition is given by Since each entry is uniquely determined, then the matrix (and the transformation is uniquely determined, hence there is only one such transformation satisfy the definition.

2 Furthermore, we may verify a given vector-valued function is differentiable by C theorem. Recall Definition: C (or continuously differentiable) function We say F: is C at iff all exist in near and are continuous at Theorem: (C theorem) If Fx is C at, then Fx is differentiable at a. Example 2 Let F: be defined as Fx, y e y, ln1 x y Show that the function is differentiable at 1,1 and compute F a and DF a (Step 1: Check differentiability) Let f x, y e cos y and f x, y ln 1 x y, a direct computation yields f x e y, f y e, f x 2x 1 x y, f y 2y 1 x y Clearly all partial derivatives exist and continuous everywhere, hence the function is clearly C at 1,1 and therefore differentiable by C theorem (Step 2: Find F a and DFa) f F x 1,1 1,1 f x 1,1 and f y 1,1 f y 1,1 2 e 3 DF, x, y F 1,1 x e y 2 3 e 2 3 e 2 x ex ey y x 2 3 y Let us come into some complicated examples Example 3 (Practice Exercise #10) a) Show that Fx, y xy is not at 0,0 b) Show that Fx, y xy is differentiable at 0,0 c) Furthermore, compute F 0,0 and DF0,0 (Note: This example says that even a function is not C, the function can still be differentiable. That s why theorem can t be used to check a function is not differentiable.)

3 Solution a) (To see a function is NOT, we first compute the partial derivative around 0,0) We first compute, for any,, we first fix 0, 0 Then for 0, consider the case 0, near,,,, Similarly, for the case 0, we can get,, For 0, 0,,, and The it only exists when 0, for 0, 0, does not exist. So, does not near 0,0 and hence the function is not b) (Verify, is differentiable at, 0,0) Here we check by definition (Step 1: Compute the partial derivative at 0,0 0,0,, (Step 2: Check differentiability) Construct the matrix to be,, 0 and 0,0,, 0 0, 0,0 0 0, 0 0,0, 0,0 0,0,, 0 Hence the function is differentiable at, 0,0 c) From b), F 0,0 0,0, 0,0 0,0 and DF0,0x, y F 0,0 x y 0 0 Example 4 (Practice Exercise #6) Show that,, 0,0 is not C at 0,0 0, 0,0 and the function is not differentiable at 0,0

4 (The function is not C ) Once again, we shall compute a, b and a, b We focus on a, b, (The case about a, b is similar) For a, b 0,0, near a, b, fx, y, For a, b 0,0, 0,0,, 0 However, a, b is not continuous at 0,0 because,, f a, b x,, sin cos sin (the it does not exist!!) Hence the function is NOT C at x, y 0,0 a, b,, b a b sin cos sin a b (The function is not differentiable at x, y 0,0) (Step 1: Compute all partial derivatives) f, 0 0,0 0,0 0, x 0 (Step 2: Check!!) Construct the matrix to be,, cos sin f y 0, 0,0 0 0, 0 0, 0,0 0, ,0, 0,0 0,0,, sin We see the it does not exist (as it depends on θ). So f is not differentiable at 0,0 Chain Rule: Differentiation of composite function Recall in one variable calculus, when you have a function y gfx where f and g are differentiable functions, then the derivative of y is given by dy dx g faf a In general, considering 2 differentiable vector-valued function F and G, what will be the derivative of Hx G Fx GFx??

5 Theorem: Chain Rule Suppose F: R R is differentiable at x a and G: R R is differentiable at y Fa (NOT a!!). Then Hx GFx is differentiable at a and H a G FaF a (Martix multiplication) DHax DGFa DFa x (Linear Transformation Composition) Let us first have a simple example to see how the chain rule be applied Example 5 Let G: be differentiable on and Gu, v e uv, 3uv. Verify that Hx, y, z G2x 3y z, x 3y z is differentiable at 1,1,1 and compute H 1,1 and DH1,1x, y Define F: to be Fx, y, z 2x 3y z, x 3y z, then H can be written as Hx, y, z GFx Next, we shall check whether F is differentiable at 1,1,1. Note that f x 2, f y 3, f z 1, f x 2x, f y 3, f 1 z All partial derivative of f exist and continuous at 1,1, hence F is C and therefore differentiable by C theorem and f F x 1,1,1 1,1,1 f x 1,1,1 f y 1,1,1 f y 1,1,1 f z 1,1, f z 1,1, Next we shall verify Gu, v is differentiable at F1,1,1 0,3 It is easy to get e v, u, 3v, 9uv Clearly, they exist near 0,3 and continuous at 0,3 So G is C and therefore differentiable at 0,3 With G 0,3 Therefore by chain rule,, H 1,1,1 G F1,1,1F 1,1,1 G 0,3F 1,1, Finally, DH1,1, x x 12y 4z y 162x 243y 81z z *Tips: When using chain rule, it will be easier for you to do matrix multiplication than doing composite function.

6 Example 6 (Product Rule) Let f, g: R R be differentiable function at a, define hx fxgx, compute Dha and h a,,,, Define a map F: and G: be Fx fx, gx and Gu, v uv. Then hx GFx We can check by C -theorem that F is differentiable at a and G is differentiable everywhere (in particular u, v fa, ga). The corresponding matrixes are given by F a and G c, d c, d c, d d, c Hence by chain rule, h GFa is differentiable at a and h a is given by h a G FaF a G fa, gaf a f f f x ga, fa x x g g g x x x ga f g,.., f g x x x x,.,,., gaf a fag a In terms of linear transformation, Dha gadfa fadga

7 In the following, there are some suggested exercises, you should try to do them in order to understand the material. If you have any questions about them, you are welcome to find me during office hours. You are also welcome to submit your work (complete or incomplete) to me and I can give some comments to your work. Exercise 0 Determine whether the following statements are true or not. If it is true, give a short explanation, if it is false, give a counter-example. a) Let : R R be a vector value function. Then x is continuous at point a a,.., a if and only if f x is continuous at x a for i 1,2,3, (Here Fx f x, f x,.., f x) b) Given a vector-valued function :, suppose all the function is differentiable at x a exist at a R, then c) Suppose : is not C at, then is NOT differentiable at x a. d) Let : R R be a vector value function and if the martix f 1 f 1 x 1 x n J exists, then F is differentiable at x a f m f m x 1 x n Exercise 1 (Practice Exercise #5) Let F: be defined as Fx, y, z xy cosy z, ze. Show that i) The function is differentiable at any a, b, c and ii) Find F a, b, c and DFa, b, c Exercise 2 (Example 2a in Tutorial Note #1) Let a function F: be defined as Fx, y sin x y a) Show Fx, y is not C at 0,0 b) Determine whether Fx, y is differentiable at 0,0 (Hint: You may use the fact that 1) Exercise 3 Let a function F: defined as Fx, y, a) Show Fx, y is not C at 0,0 b) Determine whether Fx, y is differentiable at 0,0 Exercise 4 Let the function G: defined by

8 x y z Gx, y, z x y if x, y, z 0,0,0 z 0 if x, y, z 0,0,0 Show that the function is differentiable at x, y, z 0,0,0. Compute all G 0,0,0 and DG0,0,0. (Hint: You may use either theorem or check directly by definition) Exercise 5 Let F: R R be defined as Fx, y e cosy, e siny and G: R R be defined as Gu, v u uv, v. Verify H: R R defined as Hx, y GFx, y is differentiable at 0,0 and compute H 0,0 Exercise 6 Define a function H: R R be defined as Hx, y, z x y zsinx. Show that it is differentiable at x, y, z 0,0,0. If it is differentiable, compute H 0,0,0 and DH0,0,0 (Hint: It is rather difficult for you to check the differentiability directly, you may define F: R R as Fx, y, z x y z, sinx and G: as Gu, v uv, then H GFx.) (Note: This exercise shows a good application of chain rule) Exercise 7 a) Let f, g: be differentiable at, Compute f g a and Df ga b) Let f, g: be differentiable at, suppose ga 0. Compute a and D a. Appendix: In the following, we will give a proof to triangle inequality and also a it theorem concerning the sequence. Practice Exercise #1 For any,,, show that Let,,..,,,,,,,,..,, then the inequality becomes Here we try to work backward, we first square both side yields 2 Here we make a substitution (which make the proof easier)

9 Let and Then the expression become 2 After we expend the expression and cancel all the terms, the inequality become It is just a Cauchy-Schwarz Inequality, hence the triangle inequality is true Practice Exercise #2 Let,,,. For 1,2,3, define a sequence of vectors,,, to be,,,. We say if and only if 0. Prove that ( part) 1,2,, Suppose, we need to show (We first look at the condition and find something useful) 0 0 (Next we will show 0 Consider 0 (We add more terms) Taking, R.H.S. tends to 0 also and by sandwich theorem, we can see 0 1,2,.., ( part) Suppose, we shall show or 0 Now 0 (We have used the useful inequality in last inequality, namely, one may verify it by squaring both side and compare) Hence as, R.H.S. goes to 0 (as 0) So by sandwich theorem, 0

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