Let A be a set of complex numbers. A number z is called, as in the real case, a limit point of the set A if for every (real) e > 0, there is a point a in A with |2 – a| < e but z # a. Prove the two-dimensional version of the Bolzano-Weierstrass Theorem: If A is an infinite subset of {a, b] × [c, d], then A has a limit point in [a, b] × [c, d]. Hint: First divide [a, b] × [c, d] in half by a vertical line as in Figure 7(a). Since A is infinite, at least one half contains infinitely many points of A. Divide this in half by a horizontal line, as in Figure 7(b). Continue in this way, alternately dividing by vertical and horizontal lines. (The two-dimensional bisection argument outlined in this hint is so standard that the title "Bolzano-Weierstrass" often serves to describe the method of proof, in addition to the theorem itself. See, for example, H. Petard, “A Contribution to the Mathematical Theory of Big Game Hunting," Amer. Math. Monthly, 45 (1938), 446-447.) Prove that a continuous (complex-valued) function on [a, b] × [c, d] is bounded on [a, b] × [c, d]. (Imitate Problem 21-31.)
Let A be a set of complex numbers. A number z is called, as in the real case, a limit point of the set A if for every (real) e > 0, there is a point a in A with |2 – a| < e but z # a. Prove the two-dimensional version of the Bolzano-Weierstrass Theorem: If A is an infinite subset of {a, b] × [c, d], then A has a limit point in [a, b] × [c, d]. Hint: First divide [a, b] × [c, d] in half by a vertical line as in Figure 7(a). Since A is infinite, at least one half contains infinitely many points of A. Divide this in half by a horizontal line, as in Figure 7(b). Continue in this way, alternately dividing by vertical and horizontal lines. (The two-dimensional bisection argument outlined in this hint is so standard that the title "Bolzano-Weierstrass" often serves to describe the method of proof, in addition to the theorem itself. See, for example, H. Petard, “A Contribution to the Mathematical Theory of Big Game Hunting," Amer. Math. Monthly, 45 (1938), 446-447.) Prove that a continuous (complex-valued) function on [a, b] × [c, d] is bounded on [a, b] × [c, d]. (Imitate Problem 21-31.)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
Please solve question no 5 completely
![LTE
7:29 PM 1 D Ø
VOLTE
46%
ther
Tun-
can be written as a product of linear factors z + a and quadratic
factors z? + az + b all of whose coefficients are real. (Use Problem
24-7.)
4. In this problem we will consider only polynomials with real coefficients.
Such a polynomial is called a sum of squares if it can be written as
hi? + ··· + hn² for polynomials h; with real coefficients.
(a) Prove that if f is a sum of squares, then f(x) > 0 for all x.
(b) Prove that if f and g are sums of squares, then so is f g.
(c) Suppose that f(x) > 0 for all x. Show that f is a sum of squares. Hint:
First write f(x) = x*g(x), where g(x) + 0 for all x. Then k must be
even (why?), and g(x) > 0 for all x. Now use Problem 3(b).
5. (a) Let A be a set of complex numbers. A number z is called, as in the real
case, a limit point of the set A if for every (real) & > 0, there is a
point a in A with |2 – al < e but z # a. Prove the two-dimensional
version of the Bolzano-Weierstrass Theorem: If A is an infinite subset
of {a, b] × [c, d], then A has a limit point in [a, b] × [c, d]. Hint:
First divide [a, b] × [c, d] in half by a vertical line as in Figure 7(a).
Since A is infinite, at least one half contains infinitely many points of
A. Divide this in half by a horizontal line, as in Figure 7(b). Continue
in this way, alternately dividing by vertical and horizontal lines.
(The two-dimensional bisection argument outlined in this hint is so
standard that the title "Bolzano-Weierstrass" often serves to describe
the method of proof, in addition to the theorem itself. See, for example,
H. Petard, “A Contribution to the Mathematical Theory of Big Game
Hunting," Amer. Math. Monthly, 45 (1938), 446-447.)
(b) Prove that a continuous (complex-valued) function on [a, b] X [c, d]
is bounded on [a, b] × [c, d]. (Imitate Problem 21-31.)
(c) Prove that if f is a real-valued continuous function on [a, b] × [c, d],
then f takes on a maximum and minimum value on [a, b] × [c, d].
(You can use the same trick that works for Theorem 7-3.)
eries
*6.
The proof of Theorem 2 cannot be considered to be completely ele-
mentary because the possibility of choosing y with y" = -a/ß depends
on Theorem 24-2, and thus on the trigonometric functions. It is therefore
of some interest to provide an elementary proof that there is a solution for
the equation z" – c = 0.
(a) Make an explicit computation to show that solutions of 2? – c = 0
can be found for any complex number c.
(b) Explain why the solution of z" – c = 0 can be reduced to the case
where n is odd.
(c) Let zo be the point where the function f(z) = z" – chas its minimum
absolute value. If zo z 0, show that the integer m in the proof of
Theorem 2 is equal to 1; since we can certainly find y with y! =
-a/B, the remainder of the proof works for f. It therefore suffices to
show that the minimum absolute value of f does not occur at 0.
(d) Suppose instead that f has its minimum absolute value at 0. Since z is](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff4654d3a-a61f-43c1-b797-b818a0cf5d64%2F183b99af-d666-4593-a15d-ad13735a7f13%2Fszid2gi_processed.png&w=3840&q=75)
Transcribed Image Text:LTE
7:29 PM 1 D Ø
VOLTE
46%
ther
Tun-
can be written as a product of linear factors z + a and quadratic
factors z? + az + b all of whose coefficients are real. (Use Problem
24-7.)
4. In this problem we will consider only polynomials with real coefficients.
Such a polynomial is called a sum of squares if it can be written as
hi? + ··· + hn² for polynomials h; with real coefficients.
(a) Prove that if f is a sum of squares, then f(x) > 0 for all x.
(b) Prove that if f and g are sums of squares, then so is f g.
(c) Suppose that f(x) > 0 for all x. Show that f is a sum of squares. Hint:
First write f(x) = x*g(x), where g(x) + 0 for all x. Then k must be
even (why?), and g(x) > 0 for all x. Now use Problem 3(b).
5. (a) Let A be a set of complex numbers. A number z is called, as in the real
case, a limit point of the set A if for every (real) & > 0, there is a
point a in A with |2 – al < e but z # a. Prove the two-dimensional
version of the Bolzano-Weierstrass Theorem: If A is an infinite subset
of {a, b] × [c, d], then A has a limit point in [a, b] × [c, d]. Hint:
First divide [a, b] × [c, d] in half by a vertical line as in Figure 7(a).
Since A is infinite, at least one half contains infinitely many points of
A. Divide this in half by a horizontal line, as in Figure 7(b). Continue
in this way, alternately dividing by vertical and horizontal lines.
(The two-dimensional bisection argument outlined in this hint is so
standard that the title "Bolzano-Weierstrass" often serves to describe
the method of proof, in addition to the theorem itself. See, for example,
H. Petard, “A Contribution to the Mathematical Theory of Big Game
Hunting," Amer. Math. Monthly, 45 (1938), 446-447.)
(b) Prove that a continuous (complex-valued) function on [a, b] X [c, d]
is bounded on [a, b] × [c, d]. (Imitate Problem 21-31.)
(c) Prove that if f is a real-valued continuous function on [a, b] × [c, d],
then f takes on a maximum and minimum value on [a, b] × [c, d].
(You can use the same trick that works for Theorem 7-3.)
eries
*6.
The proof of Theorem 2 cannot be considered to be completely ele-
mentary because the possibility of choosing y with y" = -a/ß depends
on Theorem 24-2, and thus on the trigonometric functions. It is therefore
of some interest to provide an elementary proof that there is a solution for
the equation z" – c = 0.
(a) Make an explicit computation to show that solutions of 2? – c = 0
can be found for any complex number c.
(b) Explain why the solution of z" – c = 0 can be reduced to the case
where n is odd.
(c) Let zo be the point where the function f(z) = z" – chas its minimum
absolute value. If zo z 0, show that the integer m in the proof of
Theorem 2 is equal to 1; since we can certainly find y with y! =
-a/B, the remainder of the proof works for f. It therefore suffices to
show that the minimum absolute value of f does not occur at 0.
(d) Suppose instead that f has its minimum absolute value at 0. Since z is
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