Sergei Yakovenko's blog: on Math and Teaching

Monday, February 1, 2016

Finally, exam!

Filed under: lecture,Rothschild course "Analysis for high school teachers" — Sergei Yakovenko @ 3:41
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Exam

The exam is posted online on Feb 1, 2016, and must be submitted on the last day of the exams’ period, February 26. Its goals are, besides testing your acquired skills in the Analysis, to teach you a few extra things and see your ability for logical reasoning, not your proficiency in performing long computations. If you find yourself involved in heavy computations, better double check whether you understand the formulation of the problem correctly. Remember, small details sometimes matter!

Please provide argumentation, better in the form of logical formulas, not forgetting explicit or implicit quantifiers \forall and \exists. They really may change the meaning of what you write!

Problems are often subdivided into items. The order of these items is not accidental, try to solve them from the first till the last, and not in a random order (solution of one item may be a building block for the next one).

To get the maximal grade, it is not necessary to solve all problems, but it is imperative not to write stupid things. Please don’t try to shoot in the air.

The English version is the authoritative source, but if somebody translates it into Hebrew (for the sake of the rest of you) and send me the translation, I will post it for your convenience, but responsibility will be largely with the translator.

If you believe you found an error or crucial omission in the formulation of a problem, please write me. If this will be indeed the case (errare humanum est), the problem will be either edited (in case of minor omissions) or cancelled (on my account).

That’s all, folks!© Good luck to everybody!

Yes, and feel free to leave your questions/talkbacks here, whether addressed to Michal/Boaz/me or to yourself, if you feel you want to ask a relevant question.


Corrections


Correction 1

The formulation of Problem 1 was indeed incorrect. The set A' was intended to be the set of accumulation points for a set A\subseteq [0,1]. The formal definition is as follows.

Definition. A point p\in [0,1] belongs to to the set of limit points A' if and only if \forall\varepsilon>0 the intersection (p-\varepsilon,p+\varepsilon)\cap A is infinite. The point p itself may be or may be not in A.

Isolated points of A are never in A', but A' may contain points p\notin A.

Apologies for the hasty formulation.


Correction 2: Problem 3(b) cancelled!

The statement requested to prove in Problem 3(b) is wrong, and I am impressed how fast did you discover that. Actually, the problem was taken from the textbook by Zorich, vol. 1, where it appears on p. 169, sec. 4.2.3, as Problem 4.

The assertion about existence of the common fixed point of two commuting continuous functions f,g\colon [0,1]\to[0,1] becomes true if we require these functions to be continuously differentiable on [0,1] (in particular, for polynomials), but the proof of this fact is too difficult to be suggested as a problem for the exam.

Thus Problem 3(b) is cancelled.

35 Comments »

  1. maybe there is a mistake in q. no. 1a ? maybe A’ should be not equal to phi?

    Comment by ahuva gutmann — Thursday, February 4, 2016 @ 6:38 | Reply

    • The symbol of the empty set is the crossed circle. If A is a finite set, then by definition it has no accumulation points, so A’ is empty (prove!). The task is to prove the opposite: if A’ is empty, then A must be finite.

      Comment by Sergei Yakovenko — Thursday, February 4, 2016 @ 8:15 | Reply

  2. so maybe A’=A\(A dot) ?

    Comment by ahuva gutmann — Friday, February 5, 2016 @ 2:35 | Reply

    • No, A' is the set of accumulation points of A\subseteq[0,1], which are not necessary in A. E.g., for the set A=\{1/2^n: n=1,2,...\} the isolated points \dot A =(0,1] and A'=\{0\}.

      Comment by Sergei Yakovenko — Friday, February 5, 2016 @ 3:08 | Reply

  3. Thank you. It is very helpful.

    Comment by ahuva gutmann — Friday, February 5, 2016 @ 10:52 | Reply

  4. Q. About A dot:

    from the test page “Denote by A(dot) the set of isolated points of A”.
    so: if A={0.25,0.5,0.75} ,then all its points are isolated, by that definition: A(DOT)=A and A’=[0,1]\{0.25,05,0.75}.

    please advice me what I didnt understand .

    Comment by Ilan — Saturday, February 6, 2016 @ 3:01 | Reply

  5. what is the derivative of f(f(x))? is it f'(x) or (f'(x))^2 or else?

    Comment by Ilan — Sunday, February 7, 2016 @ 4:17 | Reply

    • Knowing the answer to this question is a prerequisite for the exam 😉

      Comment by Sergei Yakovenko — Monday, February 8, 2016 @ 9:38 | Reply

  6. Maybe A(dot) is also different from what is written in qu. no. 1 in the exam?
    Please look at the example you gave me in reply no. 2 anove, I think that according to the exam A(dot) is different because 1/3, for example, can not be in it.

    What am I doing wrong ?

    Comment by ahuva gutmann — Monday, February 8, 2016 @ 10:14 | Reply

    • The definition for \dot A was provisional (temporary), and this set does not appear in the questions. The really interesting object is the set A' of accumulation points of A. Now, in the correction, the accurate definition of A' is given, and it is this definition that should be used for answering the questions 1(a)-1(e).

      Comment by Sergei Yakovenko — Monday, February 8, 2016 @ 10:53 | Reply

  7. Exercise 2b:
    In William M.Boyce article: “COMMUTING WITH NO COMMON FIXED POINT”
    http://www.jstor.org/stable/1994788?seq=1#page_scan_tab_contents

    He said: that f,g [0,1]–>[0,1] like in our ex. (continuous on [0,1], commute under functional composition for all x on the [0,1]), Not always have a common fixes point.
    Is he right?

    Comment by Ilan — Wednesday, February 10, 2016 @ 12:14 | Reply

    • I am not sure in the paper you cite “the unit interval” means “closed unit interval” (for the open interval the assertions of the Problem 3 fail). The rest of the paper deals with other things so that I could not reconstruct what Boyce had in mind.

      Comment by Sergei Yakovenko — Wednesday, February 10, 2016 @ 9:54 | Reply

    • Ilan, I looked at the paper you mentioned. You are right, the statement is wrong (see the correction). Thank you! The problem is retracted.

      Comment by Sergei Yakovenko — Monday, February 15, 2016 @ 5:06 | Reply

  8. ex 5:
    what is the meaning of: smooth enough?
    (iii)- what do you want us to do exactly? and does previously mean:(i), (ii) or both?

    Comment by Ilan — Thursday, February 11, 2016 @ 9:11 | Reply

    • The preamble was aimed to explain the general nature of the claim. “Smooth enough” means “n times differentiable” for some n\ge 1. Writing up the accurate formulation of the theorem is part of the problem.

      Comment by Sergei Yakovenko — Friday, February 12, 2016 @ 10:13 | Reply

  9. In ex:9, is it true to write that: ||integral [a,b] (f(x))||<=integral [a,b] (||f(x)|| ?

    Comment by Anonymous — Thursday, February 11, 2016 @ 9:32 | Reply

    • The norm \| f\| is the notion defined for functions. For numbers we use the absolute value notation |f(x)| (depends on x) or \left|\int_a^b f(x)\,\mathrm dx\right| (for the absolute value of the integral). Then the inequality you write becomes an “infinite” version of the triangle inequality.

      Comment by Sergei Yakovenko — Friday, February 12, 2016 @ 10:17 | Reply

  10. 5 (iii) – regarding “apply the previously proven estimates”:

    1.what do you want us to do? that each segment has a length less than LAMDA_n[xn,xn+1]?
    2. previously mean:(i), (ii) or both?

    Comment by Ilan — Friday, February 12, 2016 @ 1:20 | Reply

    • Hint: try first to answer the question 12 (e) 😉

      Comment by Sergei Yakovenko — Friday, February 12, 2016 @ 4:18 | Reply

  11. בשאלה 5 סעיפים 3 ו 4 האם עדיין קיימת ההנחה שהנגזרת השנייה היא חסומה

    כי אז אנחנו מוכיחים שקבוצת הערכיים הקריטים היא מאורך אפס רק עבור סוג מסויים של פונקציות

    Comment by Anonymous — Saturday, February 13, 2016 @ 6:33 | Reply

    • Yes, the assumptions of the problem apply to all steps. The condition on differentiability can be relaxed (suffices that the first derivative is continuous), but cannot be dropped completely.

      Comment by Sergei Yakovenko — Monday, February 15, 2016 @ 5:11 | Reply

  12. האם בשאלה 5 אנחנו מתיחסים בכל הסעיפים שהנגזרת השנייה חסומה עם חסם נתון ? כי אז מה שאנחנו מוכיחים עבור אורך קבוצת הערכים הקריטים הוא מיוחד לפונקציות כאלה
    ואם הנגזרת השנייה תמיד חסומה האם זה לא אומר שהאורך של התמונה שלה בכלל הוא מאורך אפס ? ולא רק קבוצת הערכים הקריטים

    Comment by Anonymous — Saturday, February 13, 2016 @ 6:40 | Reply

  13. EX 2 (c):
    “Prove that a set A [0; 1] such that A” = 0…”,
    is 0 means: {0} or empty set?

    Comment by Ilan — Sunday, February 14, 2016 @ 5:22 | Reply

  14. FOR ALL THE EFFORTs ON 3b WE DESERVE SOME EXTRA POINTS !!

    Comment by Anonymous — Monday, February 15, 2016 @ 5:48 | Reply

    • Most certainly. The will be awarded to those who successfully solved other parts of Problem 3.

      Comment by Sergei Yakovenko — Monday, February 15, 2016 @ 6:03 | Reply

  15. Good evening,
    in ex 6

    what does it mean , if the taylor series we find (around 0?) has a convergence radius of 0 and the differential equation is Not defined on zero?

    Is in that case the taylor series is a solution at all to the differential equation?

    regardless the domain of the equation, f(x)=y(x) only for x=0?

    Comment by Anonymous — Tuesday, February 16, 2016 @ 9:48 | Reply

    • Please note that asking a question sometimes reveals your level of (mis)understanding of the material: in such cases I prefer to leave the question without answer 😦

      Comment by Sergei Yakovenko — Wednesday, February 17, 2016 @ 9:40 | Reply

      • Hello sergai
        ex3 b:
        When I prove it, I found every time a new of fixed point. (If it doesn’t the previously I found) I received an infinity series of fixed points, contained in section [0,1] (I will call to this set xn) According to Bolzano-Weierstrass –
        לכל סדרה אינסופית חסומה קיימת תת סדרה מתכנסת
        (I will call to this set xnk)
        ?מתכנסת xn האם הסדרה
        ? אם לא, איך ניתן להשלים את ההוכחה

        Comment by ‫עינב קזימירסקי‬‎ — Monday, February 22, 2016 @ 10:57

      • Shalom Einav, note that this problem was cancelled.

        Comment by Sergei Yakovenko — Tuesday, February 23, 2016 @ 10:53

  16. ex6 a.
    Hello sergei
    what is the initial condition for DE?
    What you mean when you ask about – series unique?
    did you mean existence and uniqueness of DE?
    thank you

    Comment by ‫עינב קזימירסקי‬‎ — Tuesday, February 23, 2016 @ 12:40 | Reply

  17. Shalom sergei
    ex 7
    Is the idea to the solution is to derive the equation several times?
    does lamda non-zero?
    thank you

    Comment by ‫עינב קזימירסקי‬‎ — Tuesday, February 23, 2016 @ 1:14 | Reply

  18. Shalom,
    1. ex 7: can we find a solution h(x) only on (-epsilon,+epsilon) or on the entire R?
    2. ex 6: if my taylor series converge only in x=0 (where y’ is Not defined) , is it still a solution to the differential equation?

    Comment by Anonymous — Tuesday, February 23, 2016 @ 1:30 | Reply


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