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Homework Portfolio

1.

- 1.71. Our proof of the Cauchy-Schwarz inequality, Theorem 1.13, used that when *U* is a unit vector, $$0 \leq ||V−(U·V)U||^2 = ||V||2 −(U·V)^2$$. Therefore if *U* is a unit vector and equality holds, then *V* = (*U* · *V*)*U*. Show that equality occurs in the Cauchy Schwarz inequality for two arbitrary vectors *V* and *W* only if one of the vectors is a multiple (perhaps zero) of the other vector.

Answer: In the first case, when *W* = 0, the *W* is a multiple of *V*; In the second case, when *W* is nonzero, then consider the unit vector $U = \frac{W}{||W||}$. Then, by the result in the question, it follows that *V* = (*U* · *V*)*U*. Therefore:

$$V = (U · V)U = (\frac{W}{||W||}· V)·\frac{W}{||W||}= (\frac{W·V}{||W||^2})·W$$ As $(\frac{W·V}{||W||^2})$ is a constant, so *V* is a multiple of *W*.

2.

-2.19. Suppose C is an n by n matrix with orthonormal columns. Use Theorem 2.2 to show that $$||CX|| \leq \sqrt{n} ||X||$$ Use the Pythagorean theorem and the result of Problem 2.17 to show that in fact $$||CX|| = ||X||$$ for such a matrix.

Answer: (1) First we calculate *C*. Let *C*_{j} denote the *j*th column of C. Since C has orthonormal columns, each *C*_{j} has norm 1. Then $$||C|| = \sqrt{\mathop{\sum_{i=1}^n\sum_{j=1}^n}C_{ij}^2}$$ $$=\sqrt{\mathop{\sum_{j=1}^n(\sum_{i=1}^n}C_{ij}^2)}$$ $$=\sqrt{\sum_{j=1}^n|| C_j||^2}$$ Now by Theorem 2.2, $$||CX|| \leq ||C||||X|| = \sqrt{n}||X||$$ as desired. (2)By Problem 2.17, *C**X* = *x*_{1}*C*_{1} + *x*_{2}*C*_{2} + … + *x*_{n}*C*_{n}. To find the norm of RHS, we need to apply the Pythagorean theorem (we need that C has orthonormal columns) to get $$||x_1 C_1 + x_2 C_2 + \dots + x_n C_n||^2 = ||x_1 C_1||^2 + ||x_2 C_2||^2 + \dots + ||x_n C_n||^2$$ Now we can put the pieces together: $$||CX||^2 = ||x_1 C_1 + x_2 C_2 + \dots + x_n C_n||^2$$ $$=||x_1 C_1||^2 + ||x_2 C_2||^2+\dots+ ||x_n C_n||^n$$ $$=x_1^2||C_1||^2 + x_2^2|| C_2||^2+\dots+ x_n ^2||C_n||^n$$ $$=x_1^2 + x_2^2+\dots+ x_n ^2$$ $$= ||X||^2$$ Since norms are nonnegative, we can conclude that ||*C**X*|| = ||*X*||.

3.

-2.44. Use the Cauchy-Schwarz inequality $$|A · B| \leq||A||||B||$$

to prove: (a) the function *f*(*X*)=*C* · *X*is uniformly continuous,

(b) the function *g*(*X*, *Y*)=*X* · *Y* is continuous.

Answer: (a) In case 1, If *C* = 0 then *f*(*X*)=0 for all *X*,so |*f*(*X*)−*f*(*Y*)| = 0 < *ε* for all *ε*, *X*, *Y*.

In case 2, where *C* = (*c*1, *c*2, ..., *c**n*)≠(0, 0, ..., 0).

By the definition of f and properties of the dot product, |*f*(*X*)−*f*(*Y*)| = |*C* · *Y* − *C* · *Y*|=|*C* · (*X* − *Y*)|.

By the Cauchy-Schwartz inequality we get $$|f(X)− f(Y)|=|C·(X−Y)|\leq||C||||X−Y||$$ Let *ε* > 0 and take $δ=\frac {ε}{||C||}$

If $||X−Y||<δ= \frac {ε}{||C||}$ Then we have |*f*(*X*)−*f*(*Y*)| ≤ ||*C*||||*X* − *Y*|| < *ε* for all *X* and *Y* in *R**n*. Therefore for any C, f is uniformly continuous.

(b) Fix *V*(*A*, *B*) in *R*^{2n}, to show that *g*(*V*)=*g*(*X*, *Y*)=*X**Y* is continuous at (*a*, *b*)

Given *ϵ* > 0, we need to find *δ* > *o* If $||U-V||=\sqrt {(X-A)^2+(Y-B)^2}<\delta$

then ||*g*(*U*)−*g*(*V*)||=|*X**Y* − *A**B*|<*ϵ*

By the triangle inequality, we know that $$|XY-AB|= |[(X-A)+A][(Y-B)+B]-AB|$$ $$=|(X-A)(Y-B)+B(X-A)+A(Y-B)|$$ $$\leq||X-A||+||B|| ||X-A||+||A|| ||Y-B||$$ If $\sqrt {(X-A)^2+(Y-B)^2}<\delta$ Then ||*X* − *A*|| + ||*B*||||*X* − *A*|| + ||*A*||||*Y* − *B*|| $$\leq (1+||A||+||B||)\sqrt {(X-A)^2+(Y-B)^2}$$ $$\leq(1+||A||+||B||)\epsilon$$ $$\leq(1+||A||+||B||) \frac {\delta}{1+||A||+||B||}$$ $$=\delta$$ So, given *ϵ* > 0, set $\delta= min(1,\frac {\delta}{1+||A||+||B||})$ we have $\sqrt {(X-A)^2+(Y-B)^2}<\delta$ Therefore, for any (*A*, *B*) in *R*^{2n}, *g*(*X*, *Y*)=*X* · *Y* is continuous.

4.

-2.45. In the triangle inequality ||*A* + *B*|| ≤ ||*A*|| + ||*B*|| put *A* = *X* − *Y* and *B* = *Y*. Deduce ||*Y*|| − ||*X*|| ≤ ||*Y* − *X*||. Show that if two points are within one unit distance of each other, then the difference of their norms is less than or equal to one.

Answer: Let *A* and *B*be in *R*_{n}. Apply the triangle inequality we have $$||A + B|| \leq ||A|| + ||B||.$$ Let*X* = *A* + *B*, *Y* = *B*,so *A* = *X* − *Y*. so we have $$||X|| \leq ||X − Y|| + ||Y||, ||X|| − ||Y|| \leq ||X − Y||.$$

Exchange the symbol *X* and *Y* we get $$ ||Y|| − ||X|| \leq ||Y − X||.$$ So when *X* and *Y* are within one unit of distance of each other which means ||*Y* − *X*|| ≤ 1, by the inequality we can conclude $$||Y|| − ||X|| \leq ||Y − X||\leq 1$$

5.

-3.9. Suppose a function *F* from *R*_{n} to *R*_{m} is differentiable at *A*. Justify the following statements that prove $$L_AH = DF(A)H,$$ that is, the linear function LA in Definition 3.4 is given by the matrix of partial derivatives*D**F*(*A*).

(a) There is a matrix*C* such that *L*_{A}(*H*)=*C**H* for all *H*.

(b) Denote by *C*_{i} the *i* − *t**h* row of *C*. The fraction $$\frac{||F(A+H)−F(A)−L_A(H)||}{||H||} = \frac{||F(A+H)−F(A)−CH||}{||H||}$$ tends to zero as ||*H*|| tends to zero if and only if each component $$\frac {f_i(A+H)− f_i(A)−C_iH}{ ||H||}$$ tends to zero as ||*H*|| tends to zero.

(c) Set *H* = *h**e*_{j} in the *i* − *t**h*component of the numerator to show that the partial derivative *f*_{i, xj}(*A*) exists and is equal to the (*i*, *j*) entry of *C*.

Answer: (a) *F* from *R*_{n} to *R*_{m} is differentiable at *A*. By the definition of differentiability there is a linear function *L*_{A}(*H*) such that $$\frac{||F(A+H)−F(A)−L_A(H)||}{||H||}$$

tends to 0 as ||*H*|| tends to zero.

By Theorem 2.1, every linear function from *R*_{n} to *R*_{m} can be written as *L**A*(*H*)=*C**H* for all *H*, where *C* is some *m* × *n* matrix.

(b) The absolute value of each component of a vector is less than or equal to the norm of the vector so for each H and i we have $$0 \leq| f_i(A + H) − f_i(A) − C_i · H| \leq ||F(A + H) − F(A) − CH||$$ where *C**i* is the *i* − *t**h* row of *C*.

Since $$\frac{||F(A+H)−F(A)=LA(H)||}{||H||}$$ tends to 0,

by the squeeze theorem, both$\frac{ | f_i(A+H)− f_i(A)−C_i·H|}{||H|| }$ and$\frac{ f_i(A+H)− f_i(A)−C_i·H}{||H|| }$

(c) Let *H* = *h**e*_{j} = (0, 0, ..., 1, 0, ..., 0), the 1 in the*j* − *t**h* place. By part (b), tend to zero as ||H|| tends to zero. $$\lim_{\||H||=|h|\to\ 0 } \frac{f_i(A+he_j)− f_i(A)−hC_i ·e_j}{|h|} =0$$

Therefore $$\lim_{h\to\ 0 } \frac{f_i(A+he_j)− f_i(A)−hC_i ·e_j}{h} =0$$

Since *C*_{i} · *e*_{j} = *c*_{ij} and by the definition of partial derivatives,we have $$\lim_{h\to\ 0 } \frac{f_i(A+he_j)− f_i(A)}{h} = \frac {\partial f_i}{\partial x_j} (A)$$

So we can conclude that *c*_{ij} = *c*(*A*)

6.

6.14. Justify the following items which prove:

If f is continuous on *R*_{2} and ∫_{R}*f**d**A* = 0 for all smoothly bounded sets *R*, then is identically zero.

(a) If *f*(*a*, *b*)=*p* > 0 then there is a disc *D* of radius *r* > 0 centered at (*a*, *b*)in which $f(x,y)> \frac {1}{2}p$

(b) If f is continuous and *f*(*x*, *y*)≥*p*_{1} > 0 on a disk *R* then ∫_{R}*f**d**A* ≥ *p*_{1}(*A**r**e**a*(*R*)). ∫_{R}*f**d**A* = 0for all smoothly bounded regions *R*, then *f* cannot be positive at any point. (d) *f* is not negative at any point either. (e) *f* = 0 at all points.

Answer: First, we can assume that f is continuous on R2 and ∫_{R}*f**d**A* = 0 for all smoothly bounded sets *R*

(a) Because f is continuous at (*a*, *b*), by the definition of continuity, there is *r* > 0 such that for all (*x*, *y*) such that||(*x*, *y*)−(*a*, *b*)||<*r*, we have |*f*(*x*, *y*)−*f*(*a*, *b*)| < *p*/2.Then we assume p > 0, so *p* − *p*/2 < *f*(*x*, *y*)<*p* + *p*/2 In particular, *f*(*x*, *y*)>*p*/2

(b) As *R* is bounded, the closure of *R* is closed and bounded. So we can apply the extreme value theorem which means *f* is bounded on the closure of *R*. In particular, *f* is bounded on *R*. *f* is also integrable on *R*; in fact ∫_{R}*f**d**A* = 0. Apply the lower bound property, ∫_{R}*f**d**A* ≥ *p*_{1}(*A**r**e**a*(*R*)) holds.

(c) Suppose *f* is positive at (*a*, *b*).

From (*a*), there is a disc *R*of nonzero radius on which *f*(*x*, *y*)>*f*(*a*, *b*)/2 > 0.

From (b), ∫_{R}*f**d**A* ≥ (*f*(*a*, *b*)/2)·*a**r**e**a*(*R*)>0 But we assumed that ∫_{R}*f**d**A* = 0 for all smoothly bounded sets *R*, it comes to a contradiction. Therefore *f* cannot be positive at any point.

(d) As we know that −*f* is continuous, and that for all smoothly bounded regions *R*, by linearity, we have −*f**d**A* = −*f**d**A* = −0 = 0 . From(c),we know that −*f* cannot be positive at any point. Thus, we conclude that f cannot be negative at any point.

(e) Therefore, for any (*a*, *b*), *f*(*a*, *b*) is defined and is neither positive nor negative, so it must be 0.

7.

-6.44. Justify the following steps to prove that if *f* is integrable on *R*_{2} and *g* is a continuous function with 0 ≤ *g* ≤ *f* then *g* is integrable on *R*_{2}.

(a) ∫_{D(n)}*g**d**A* exsits

(b) 0 ≤ ∫_{D(n)}*g**d**A* ≤ ∫_{D(n)}*f**d**A*

(c) The numbers ∫_{D(n)}*g**d**A* are an increasing sequence bounded above.

(d) lim_{n → ∞}∫_{D(n)}*g**d**A* exsits

Answer:

Check *D* : *D* = *R*^{2} unbounded, g 0, continuous, so we need to prove lim_{n → ∞}∫_{D(n)}*g**d**A* exsits.

(a) *g* ≥ 0 is continuous on *R*_{2} and *D*(*n*) is bounded for each *n* so *g* is integrable over *D*(*n*)

(b) By theorem 6.9 *L**a**r**e**a*(*D*)≤*I*(*f*, *D*) and the fact 0g, we know that 0*a**r**e**a*(*D*)≤∫_{D(n)}*g**d**A* so if 0 ≤ *f*(*x*, *y*)−*g*(*x*, *y*) then $$ 0=0 area(D)\leq \int_{D(n)} f(x,y)-g(x,y) dA \leq \int_{D(n)} f dA - \int_{D(n)} g dA $$ Therefore, 0 ≤ ∫_{D(n)}*g**d**A* ≤ ∫_{D(n)}*f**d**A*

(c) Let *C*_{n} = ∫_{D(n)}*g**d**A*. Because *g* ≥ 0, *D*(*n*)≤*D*(*n* + 1). Then *C*_{1}, *C*_{2}, *C*_{3}...*C*_{n} is an increasing sequence. Since 0 ≤ ∫_{D(n)}*g**d**A* ≤ ∫_{D(n)}*f**d**A* and $$\lim_{n\to\infty} \int_{D(n)} f dA = \int_{R ^2} f dA$$ exists, We got $$\int_{D(n)} g dA \leq \int_{R ^2} f dA$$

(e) By the Monotone Convergence Theorem for sequences, ∫_{D(n)}*g**d**A* increasing and bounded above is convergent so lim_{n → ∞}∫_{D(n)}*g**d**A* = lim_{n → ∞}*C*_{n} exists

8.

6.50. Justify steps (a)–(d) to prove that if a continuous function *f* is integrable on an unbounded set *D* then |∫_{D}*f**d**A*| ≤ ∫_{D}|*f*|*d**A*

(a)∫_{D}*f**d**A* = ∫_{D}*f*_{+}*d**A* − ∫_{D}*f*_{−}*d**A* ≤ ∫_{D}*f*_{+}*d**A* + ∫_{D}*f*_{−}*d**A* = ∫_{D}|*f*|*d**A*

(b)∫_{D}(−*f*)*d**A* ≤ ∫_{D}|*f*|*d**A*

(c)−∫_{D}*f**d**A* ≤ ∫_{D}|*f*|*d**A*

(d)|∫_{D}*f**d**A*| ≤ ∫_{D}|*f*|*d**A*

(a) By Definition 6.9, if *f* is continuous and integrable on an unbounded set *D*, then |*f*| is integrable on *D*. Rewrite *f*(*x*, *y*)=*f*_{+}(*x*, *y*)−*f*_{−}(*x*, *y*) where *f*_{+}(*x*, *y*)=*f*(*x*, *y*) if *f*(*x*, *y*)≥0 and 0 otherwise, and *f*_{−}(*x*, *y*)= − *f*(*x*, *y*)if *f*(*x*, *y*)≤0 and 0 otherwise. So, by the definition of ∫_{D}*f**d**A*, $$\int_ {D} f dA=\int_ {D} f_+ dA - \int_ {D} f_- dA $$ Since ∫_{D}*f*_{−}*d**A* is nonnegtive $$\int_ {D} f_+ dA - \int_ {D} f_- dA \leq \int_ {D} f_+ dA + \int_ {D} f_- dA$$ Since *f*_{+} ≥ 0 and *f*_{−} ≥ 0 are integrable over *D* $$\int_ {D(n)} f_+ dA + \int_ {D(n)} f_- dA =\int_ {D(n)} (f_+ + f_-) dA$$ By the properties of limits of increasing sequence D(n), we know ∫_{D(n)}(*f*_{+} + *f*_{−})*d**A* converges so $$\int_ {D} f_+ dA + \int_ {D} f_- dA =\int_ {D} (f_+ + f_-) dA$$ By the equation *f*(*x*, *y*)=*f*_{+}(*x*, *y*)−*f*_{−}(*x*, *y*), we got $$\int_ {D} f dA \leq \int_ {D} \left|f \right|dA$$

(b) In the same way, we apply (a) to the functions − f to get $$\int_ {D} -f dA \leq \int_ {D} \left|-f \right|dA= \int_ {D} \left|f \right|dA$$

(c)By the properties of limits and the equation ∫_{D(n)} − *f**d**A*=_ D(n) f dA ,*w**e**g**e**t**m**a**t**h**P**l**a**c**e**h**o**l**d**e**r*46*i**d*(*d*)*I**f*ba*a**n**d* −*b* ≤ *a* then|*b*|≤*a*. From (a), we got $$\int_ {D} f dA \leq \int_ {D} \left|f \right|dA$$, From (b) and (c), we got $$- \int_ {D} f dA \leq \int_ {D} \left|f \right|dA$$ Therefore, we can conclude that $$\left| \int_ {D} f dA\right| \leq \int_ {D} \left|f \right|dA$$

9.

-4.21. Find the point on the plane $$z = x − 2y + 3$$ that is closest to the origin, by finding where the square of the distance between (0, 0) and a point (*x*, *y*) of the plane is at a minimum. Use the matrix of second partial derivatives to show that the point is a local minimum.

Let $$ D=d^2 = f(x,y)= x^2+y^2+(x-2y+3)^2 $$, to find the local extrema we let $$\triangledown f = (4x−4y+6,−4x+10y−12)=0$$ at ( − 0.5, 1). so $$ H(-0.5,1)=
\left[
\begin{array}{ c c }
4 & -4 \\
-4 & 10
\end{array} \right]
$$ Because 4 > 0 and (4)(10) − (−4)2 = 24 > 0. So by the Theorem 4.3, it is positive definite. By theorem 4.8, If ▿*f*(*A*)=0 and the Hessian matrix [*f*_{xixj}(*A*]) is positive definite at *A*, then *f*(*A*) is a local minimum. Therefore, *f* has a local minimum at point ( − 0.5, 1)

10.

-7.32. Let *S* be the unit sphere centered at the origin in *R*^{3}. Evaluate the following items, using as little calculation as possible

(a)∫_{S}1*d**σ*

(b)∫_{S}||*X*||^{2}*d**σ*

(c) Verify that ∫_{S}*x*_{1}^{2}*d**σ* = ∫_{S}*x*_{2}^{2}*d**σ* = ∫_{S}*x*_{3}^{2}*d**σ* using either a symmetric argument or parametrizations. Can you do this without evaluating them?

(d) Use the result of parts (b) and (c) to deduce the value of ∫_{S}*x*_{1}^{2}*d**σ*

Answer:

(a) In geometry, ∫_{S}1*d**σ* means the area of the unit sphere in *R*^{3} So ∫_{S}1*d**σ* = *π* · 1^{3} = 4*π*

(b) For all X S we have ||*X*||^{2} = 1, therefore ∫_{S}||*X*||^{2}*d**σ* = ∫_{S}1*d**σ* = 4*π*

(c) Rotation by /2 about the *x*_{3}-axis corresponds to some transformation on the domain of the parametrization of *S*. We know that *x*_{1} comes to the same position as *x*_{2}, Therefore ∫_{S}*x*_{1}^{2}*d**σ* = ∫_{S}*x*_{2}^{2}*d**σ* In the same way, make a rotation by *π*/2 about the *x*_{2} , we got ∫_{S}*x*_{1}^{2}*d**σ* = ∫_{S}*x*_{3}^{2}*d**σ* Therefore, ∫_{S}*x*_{1}^{2}*d**σ* = ∫_{S}*x*_{2}^{2}*d**σ* = ∫_{S}*x*_{3}^{2}*d**σ*

(d) By the definition of norm ||*X*||, we know that ||*X*|| = *x*_{1}^{2} + *x*_{2}^{2} + *x*_{3}^{2} So, $$\int_{S} ||X||^2 d\sigma= \int_{S} x_1^2 +x_2^2 +x_3^2 d\sigma= 3\int_{S} x_1^2 d\sigma = 4\pi$$ Therefore,$$ \int_{S} x_1^2 d\sigma =\frac{1}{3}\int_{S} ||X||^2 d\sigma = \frac{4\pi}{3}$$

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