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Some Challenging problems

  1. Jun 8, 2007 #1
    Ladies and gentlemen,

    I'll present my choice of challenging problems here.
    Most of the problems are from math analysis so I put thread here.
    I've chosen them from various sources like college competitons,journals
    or textbooks.Some of the problems are well known,and some are
    new and original.Level of difficulty vary among them,but no problem
    is trivial.I expect that really really good math students could solve about
    50% of the problems.And if you are math analysis begginer you will hardly
    solve more than 2 problems.I showed the problems to one assistant professor
    and he told me that even a college math professors,with the help of
    reference materials ,and knowledge, would be hardly able to solve them all.
    If you're mathematician or professor ,I'm looking forward to see your feedback: how do you rate each of these problems (from easiest to the hardest)?
    Finally ,if you want, feel free to post your solution (if you think you have an elegant one).

    Let the function of real parameter u be defined by the improper integral:


    a)Calculate [itex]\lim_{u\to\infty}f(u)[/itex]
    b)Given is the series:

    Find radius of convergence and sum of the series when it converges.

    Calculate the sum:

    Prove the identity:
    [tex]\int_{0}^{1}x^x dx=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^n}[/tex]

    Let B be a Borel-measurable subset of {z[itex]\in\mathbb{C}[/itex]:0<|z|<1}.
    Prove that
    [tex]-1\leq\int_{B}\frac{z}{|z|}dz\leq 1[/tex]

    Given is the function of complex variable:
    Show that it can't be analyticaly extended to area outside the unit circle |z|=1.
    Given are two real polynomials:

    Which one has the larger coefficient [itex]a_{200}[/itex] in term [itex]a_{200}x^{200}[/itex],in canonical representation?Explain answer.


    Find minimum and maximum of real function:
    [tex]f(x)=Asin^2 x + Bsin(x)cos(x) + Ccos^2 x[/tex]

    Prove that real function f(x) has at least 40 zeroes in interval [0,1000]

    Evaluate integrals:
    [tex]\int_{D}\int \frac{arctg(x+y)}{(x^2+y^2)^2} dxdy[/tex]

    [tex]D[/tex] is the area of integration defined by {x>0.y>0,x+y>1}

    Given the condition [tex]z(x,y)=f(x)f(y)[/tex] solve PDE:

    [tex]\frac{\partial^2 z}{\partial x^2}=\frac{\partial z}{\partial y}[/tex]

    What do you think?:smile:
    Last edited: Jun 8, 2007
  2. jcsd
  3. Jun 10, 2007 #2
    Very interesting and way beyond me. :D (Give me a few more years-- I'm an undergraduate math student, so I don't know too much yet ;)
    However, X looks a little too easy-- or maybe I'm looking at it incorrectly?
    Isn't that just a basic seperation of variables problem? Hmm...
  4. Jun 10, 2007 #3

    Gib Z

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    I don't have much spare time right now so I'm just going to tell people how to do them, not show full working.

    I) a) 2 ways I can see to do it - Harder method is to evaluate the indefinite integral and substitute into the anti derivative [itex]\frac{-1}{u} e^{-ux}[/itex], easier way is just to realise e^(-x) is 1 at x=0 and rapidly decreases, so as u > infinity the values after x=0 will go to zero, either way the limit is Zero.

    b) Doesn't look that hard but I can't be bothered, I'll leave some help for someone who can - [tex] f(1/n) = -n \lim_{b\to \infty} ( e^{-b/n} - 1)[/tex]

    III) Calculate the power series for [itex]e^{x \log_e x}[/itex] and knowing that it is valid for the bounds of the integral, integrate term by term to get that series.

    VI) Try letting x=1 and see what happens.

    VII) Find the derivative, which is [itex] (A-C) \sin 2x + B \cos 2x[/itex], let 2x=u and solve setting equal to zero. We can see it has an infinite number of solutions, but they are just the same ones repeating. EDIT: Incase you are having troubles solving, we can simplify that to [tex]\sqrt{(A-C)^2 + B^2} \sin ( x + \arctan \frac{B}{A-C}) [/tex] which isn't hard to solve.

    VIII) Brute force it :D

    I can't really do any of the others, but really I would be surprised if I couldn't do them in a year, max, let alone a Professor.
  5. Jun 11, 2007 #4

    Problem X. is well known PDE from mathematical physics,and I'm sure many here will recognize it.Yes,you're right that solution is searched by separation of variables.No boundary conditions were given however.Expected is general solution,in a finite form, if z ,f ,g are real derivabile functions.
    (Correction to be made in X.:The condition given should be [tex]z(x,y)=f(x)g(y)[/tex],and not like given in my first post.My apologies.)
    This question is more test of the knowledge than of solving skills.Not the hardest,but not the easiest in my set of the problems here.No problem requires knowledge beyond scope of second year of a college (what year of collegeare you?).In spite of it ,nobody solved more than 50% of the problems (included one professional mathematician!).That's a good indicator of overall difficulty I guess.I welcome more comments from experts about difficulty and how these problems compare withsay Putnam problems.

    Well,first and in a very start that is incorrect reasoning (and result).
    This integral can't be evaluated exactly ,if you meant anything like that-antiderivative is not an elementar function there.
    Last edited: Jun 11, 2007
  6. Jun 11, 2007 #5

    Gib Z

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    Your Notation is ambiguous..

    We want [tex]\lim_{u\to\infty} \int^{\infty}_0 e^{(-x^u)}[/tex]. For 0<x<1, we can see the integrand is 1, for x>1 it is zero. Hence, the integral is 1.
    Last edited: Jun 11, 2007
  7. Jun 11, 2007 #6
    They don't look too bad-- just a real pain.
    I don't think sigma-algebras and measure are second year things. (Atleast, in my college it's more commonly third year stuff)
    I'm a first year math major-- so it isn't too surprising I can't do much of these. ;)
    On second look though, a few of them (e.g. the first one, which involves a uniformly convergent sequence of functions on 0<x<1 and 1<x<oo) look doable.
  8. Jun 12, 2007 #7
    I don't see how pluggin x=1 helps.
    Are you sure you understood the problem correctly?
    Canonical repesentation (form) of polyinomials means :
    [tex]p(x)=1+a_{1}x^1 +a_{2}x^2 + ...+a_{6000}x^{6000}[/tex]
    [tex]q(x)=1+a'_{1}x^1 + a'_{2}x^2+...+a'_{6000}x^{6000}[/tex]

    What is asked is which of two coefficients [itex]a_{200}[/itex]or[itex]a'_{200}[/itex] is larger and why.

    That one is correct (probably that is very easy problem).
    What do you mean ?
    Imagine you don't have computer,how would you solve it?
    This is one of the hardest problems (by my humble judgement) in this collection.
  9. Jun 12, 2007 #8
    write out the terms in the series for every number from 0 to one thousand.
  10. Jun 13, 2007 #9


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    I'm just a freshman math undergrad, so most of these problems are out of my league...
    I actually went about I different from GibZ, instead of looking at the behavior of the integrand, I explicitly solved the integral (it ends up as gamma((1/u)/u), and used the expression for ln(gamma(z))=-ln(z)+a function that limits to 0 as z approaches 0 to find that the integral approaches 1 as u approaches infinity.
    Problem II, the sophomore's dream (I'm one year too young to be dreaming of this lol) can be done this way I think:
    [tex] x^x = e^{x \log x} = \sum_{n=0}^\infty \frac{x^n(\log x)^n}{n!}[/tex]
    [tex]\int_0^1 x^x = \sum_{n=0}^\infty \int_0^1 \frac{x^n(\log x)^n}{n!}[/tex]
    [tex]\int x^n (\log x)^n\; dx = \frac{x^{n+1}(\log x)^n}{n+1} - \frac{n}{n+1}\int x^n (\log x)^{n-1} dx[/tex]
    [tex]=\frac{x^{m+1}}{m+1} \sum_{k=0}^n (-1)^{k}\frac{n!}{(n-k)!(m+1)^{k}} (log x)^{n-k}[/tex]
    [tex]\frac{1}{n!}\int_0^1 x^n (\log x)^n\; dx =\frac{(-1)^{n}n!}{n!(n+1)^{n+1}}[/tex]
    [tex]\int_0^1 x^x = \sum_{n=0}^\infty\frac{(-1)^{n}}{(n+1)^{n+1}}[/tex]
    [tex]= -\sum_{n=1}^\infty(\frac{-1}{n})^{n} [/tex]
    The big simplifications from the large sigma expression comes from the fact that [tex]x^{m+1}(logx)^{n-k}[/tex] approach 0 as x approaches 0, and the fact that substituting x=1 kills every term except the term where k=n, as this is the only term without a log in it. Its a rather nice looking expression imo.

    For problem VIII, the way I thought about solving i was to use euler's e^ix=cis(x) formula to sum the series to get a closed form expression for it, after which its behavior can be much more easily seen. It can easily be done if it was an n instead of an n^1.5, the problem though is that i do not know how to sum 1^1.5+2^1.5+...For problem VI, don't you just apply the multinomial theorem? I think that the coefficient of x^200 for both polynomials end up to be the same, though I am probably wrong. Is II evaluated by relating the sum to a riemann sum, and hence, an integral? IX(a) looks rather interesting, but I haven't made much headway into it, substitutions and differentiation under the integral have both made the integral a lot uglier...
    Last edited: Jun 14, 2007
  11. Jun 14, 2007 #10
    Thank you for the response.
    You solved Ia correctly (your solution is mathematically rigorous and identical to the "offical" one).
    I think III is basically OK too (althought I have to cross check it in more details).You have 2 correct and since you're sophmore you fulfiled the norm :) .Not bad for a starter!
    Coefficients in VI. aren't equal.
    Problem VIII is difficult ,as well as integrals in IX.
  12. Jun 14, 2007 #11


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    Here is my attempt at VI:
    Applying the multinomial theorem to p(x):
    General term of p(x):
    [tex] \frac{2000!}{a!b!c!}(1)^{a}(x^{2})^{b}(-x^{3})^{c}[/tex]
    where a, b and c are non-negative integers.
    To find the coefficient of [tex] x^{200} [/tex], we solve the system
    [tex] 2b+3c=200[/tex]

    [tex] a+b+c=2000[/tex]

    [tex] 2b=200-3c[/tex]

    [tex] 2a+200-3c+2c=4000[/tex]

    [tex] c=2a-3800[/tex]

    c must be even, so all coefficients of [tex] x^{200}[/tex] must be positive
    General term of q(x):
    [tex] \frac{2000!}{a!b!c!}(1)^{a}(-x^{2})^{b}(x^{3})^{c}[/tex]
    where a, b and c are non-negative integers
    The system that must be solved is the same as p(x), and c must be even. Since 2b+3c=200, if c is even, 3c is even, and 2b must also be even, since 200 is even. However, b can be odd, as 2b is always even. Therefore, there will be some negative coefficients of [tex] x^{200}[/tex] in q(x). All terms in p(x) and q(x) are identical, with the exception of the signs of their coefficients, and hence the coefficient of [tex] x^{200}[/tex] is greater in p(x) than q(x).
    I'm not sure about my answer though, maple refuses to expand the polynomials lol
  13. Jun 15, 2007 #12
    Yes,the coefficient in p(x) is greater.
    There are more ways of looking at this problem,and you found one.
    That would make 3 correct for you.
    But be aware I don't want this thread be turned in competiton (or quiz).
    My goal is to get feedback from students and mathematicians about
    how many of the problems you feel you can solve.
    After taking your time and really give a try.
    You don't have to post your solutions,if you think you really solved it
    (like yip) you probably did.
    If you don't feel comfortable publicly,letting me know by PM is ok.
    Just be honest,that's all I ask.
    I doubt I would be able of solving more than 3 or 4 myself were the problems given to me for solving by sombody else.
    Please ,in your response include:

    a) What's your education/what year of college you are

    b)Specify Problems you solved

    c) If you ever participated in math competitions ,what competition it was and what were your results.

    Up to this point,I have sample of 10 people:1 assistant professor (who solved 7) and rest are students.Distributed scores vary from 1 to 7.

    Thank you in advance,
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