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Thread: Calculus & Analysis Marathon & Questions

  1. #76
    -insert title here- Paradoxica's Avatar
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    Re: First Year Uni Calculus Marathon

    Quote Originally Posted by porcupinetree View Post
    Yup. I'm guessing you guys haven't?
    shhh....

    We need to give UNSW students the impression that they're ahead....
    If I am a conic section, then my e = ∞

    Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.

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    Ancient Orator leehuan's Avatar
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    Re: First Year Uni Calculus Marathon

    Quote Originally Posted by porcupinetree View Post
    Yup. I'm guessing you guys haven't?
    Taylor series are second semester for us.

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    Executive Member KingOfActing's Avatar
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    Re: First Year Uni Calculus Marathon

    Quote Originally Posted by porcupinetree View Post


    The nth order Taylor polynomials are the just the infinite series truncated at the nth degree summand.
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    -insert title here- Paradoxica's Avatar
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    Re: First Year Uni Calculus Marathon

    Quote Originally Posted by KingOfActing View Post


    The nth order Taylor polynomials are the just the infinite series truncated at the nth degree summand.
    Don't forget to state the radius of convergence.

    Although in this case, it is fairly obvious.
    If I am a conic section, then my e = ∞

    Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.

  5. #80
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    Re: First Year Uni Calculus Marathon











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    not actually a porcupine porcupinetree's Avatar
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    Re: First Year Uni Calculus Marathon

    Quote Originally Posted by InteGrand View Post










    Nice Q's. I think I can do them all in my head but I'll have a proper go tomorrow
    Bachelor of Science (Advanced Mathematics) @ USYD

  7. #82
    Supreme Member seanieg89's Avatar
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    Re: First Year Uni Calculus Marathon

    Quote Originally Posted by seanieg89 View Post
    And some much easier ones for students who don't want to do the above questions:

    E1. Prove that if a function f: (a,b) -> R is differentiable and f'(x) is non-negative in this interval, then f is non-decreasing in this interval.

    E2. Prove that if a function f: (a,b) -> R is differentiable and f'(x) = 0 in this interval, then f is constant.
    Unsolved.

  8. #83
    not actually a porcupine porcupinetree's Avatar
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    Re: First Year Uni Calculus Marathon

    Quote Originally Posted by InteGrand View Post






















    Don't have time to attempt part (iii) right now, but I might later
    Bachelor of Science (Advanced Mathematics) @ USYD

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    Supreme Member seanieg89's Avatar
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    Re: First Year Uni Calculus Marathon

    Whilst it is fairly obvious that r^n/n! converges to zero if r is positive (in fact regardless of sign but we can wlog take it positive), I don't think it is a fact that should be glossed over by first year students.

    In any case, the methods used to bound such a term can be useful in analysing far more subtle "remainder" terms.

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    Supreme Member seanieg89's Avatar
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    Re: First Year Uni Calculus Marathon

    Quote Originally Posted by seanieg89 View Post
    (A bit tricky)

    For which positive integers k is it possible to find a continuous function f:R->R that such that

    f(x)=y

    has exactly k solutions x for every real number y?
    Also unanswered. I will get people started:

    For k=2 it is not possible.

    Proof:
    Assume f is a cts function taking every value exactly twice.
    There must exist two zeros a < b of the function f.
    Between a and b f must have fixed sign (otherwise the intermediate value theorem would provide the existence of a third zero).
    Without loss of generality, we assume f is positive in (a,b).
    Then f must attain a positive maximum m at some p in (a,b) (Extreme value theorem applied to [a,b]).

    Now this means that f(x)=2m must have its solutions OUTSIDE the interval [a,b], without loss of generality we may assume that one such solution q is greater than b. Then by the intermediate value theorem, the equation f(x)=m/2 must have:
    -at least one solution in (a,p)
    -at least one solution in (p,b)
    -at least one solution in (b,q).

    This contradiction completes the proof.


    As for k=3, it IS actually possible.
    Picture the square [0,1]x[0,1] and draw line segments between O, (1/3,1), (2/3,0), (1,1). (Looks like a zigzag.)
    Now replicate this curve in each square [k,k+1]x[k,k+1].
    The resulting curve is the graph of a continuous function with the sought property.

    What about for higher k?

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    Supreme Member seanieg89's Avatar
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    Re: First Year Uni Calculus Marathon

    One can also consider the same problem dropping the condition of surjectivity.

    (For which positive integers k is it possible to find a continuous function f defined on R that takes every value in its range exactly k times?)

  12. #87
    -insert title here- Paradoxica's Avatar
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    Re: First Year Uni Calculus Marathon

    Quote Originally Posted by seanieg89 View Post
    Also unanswered. I will get people started:

    For k=2 it is not possible.

    Proof:
    Assume f is a cts function taking every value exactly twice.
    There must exist two zeros a < b of the function f.
    Between a and b f must have fixed sign (otherwise the intermediate value theorem would provide the existence of a third zero).
    Without loss of generality, we assume f is positive in (a,b).
    Then f must attain a positive maximum m at some p in (a,b) (Extreme value theorem applied to [a,b]).

    Now this means that f(x)=2m must have its solutions OUTSIDE the interval [a,b], without loss of generality we may assume that one such solution q is greater than b. Then by the intermediate value theorem, the equation f(x)=m/2 must have:
    -at least one solution in (a,p)
    -at least one solution in (p,b)
    -at least one solution in (b,q).

    This contradiction completes the proof.


    As for k=3, it IS actually possible.
    Picture the square [0,1]x[0,1] and draw line segments between O, (1/3,1), (2/3,0), (1,1). (Looks like a zigzag.)
    Now replicate this curve in each square [k,k+1]x[k,k+1].
    The resulting curve is the graph of a continuous function with the sought property.

    What about for higher k?
    I suppose the problem splits up on the parity of k?
    If I am a conic section, then my e = ∞

    Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.

  13. #88
    -insert title here- Paradoxica's Avatar
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    Re: First Year Uni Calculus Marathon

    Use the squeeze theorem to find the derivative of ex at x=0

    Note you must differentiate from first principles.
    If I am a conic section, then my e = ∞

    Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.

  14. #89
    Supreme Member seanieg89's Avatar
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    Re: First Year Uni Calculus Marathon

    Here is a question related to a basic tool in the analysis I do.



    Last edited by seanieg89; 13 May 2016 at 11:24 AM.

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    Supreme Member seanieg89's Avatar
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    Re: First Year Uni Calculus Marathon

    Quote Originally Posted by Paradoxica View Post
    Use the squeeze theorem to find the derivative of ex at x=0

    Note you must differentiate from first principles.
    You kinda need to include the definition of e / exponentiation that you would like students to work from to make a first principles question like this well-defined.

  16. #91
    -insert title here- Paradoxica's Avatar
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    Re: First Year Uni Calculus Marathon

    Quote Originally Posted by seanieg89 View Post
    You kinda need to include the definition of e / exponentiation that you would like students to work from to make a first principles question like this well-defined.
    Alright.

    Use the limit definition of ex

    If I am a conic section, then my e = ∞

    Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.

  17. #92
    Supreme Member seanieg89's Avatar
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    Re: First Year Uni Calculus Marathon

    Related:


  18. #93
    Ancient Orator leehuan's Avatar
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    Re: First Year Uni Calculus Marathon

    I didn't even get up to the last part of my quadrature question but I feel this last part is a good marathon question so I'll drop it here... I could not get it out though.


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    Ancient Orator leehuan's Avatar
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    Re: First Year Uni Calculus Marathon

    Good exercise question from a past paper.



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    Ancient Orator leehuan's Avatar
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    Re: First Year Uni Calculus Marathon

    By the way, would you say that the fact each monotone interval on a function attains at most one zero is a corollary of Rolle's theorem?

    Quote Originally Posted by porcupinetree View Post


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    Rambling Spirit
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    Re: First Year Uni Calculus Marathon

    Quote Originally Posted by leehuan View Post
    By the way, would you say that the fact each monotone interval on a function attains at most one zero is a corollary of Rolle's theorem?
    What do you mean? Do you mean if a function is monotone on an interval, it has at most one zero on that interval? If so, that follows from the definition of monotone (if it's monotone, it can't take the same value at two different places).

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    Ancient Orator leehuan's Avatar
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    Re: First Year Uni Calculus Marathon

    Quote Originally Posted by InteGrand View Post
    What do you mean? Do you mean if a function is monotone on an interval, it has at most one zero on that interval? If so, that follows from the definition of monotone (if it's monotone, it can't take the same value at two different places).
    Oh right. Ok yep basically what I needed to see - just quote one-to-one

  23. #98
    Supreme Member seanieg89's Avatar
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    Re: First Year Uni Calculus Marathon


  24. #99
    Senior Member integral95's Avatar
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    Re: First Year Uni Calculus Marathon

    Quote Originally Posted by leehuan View Post
    Good exercise question from a past paper.











    Last edited by integral95; 6 Jun 2016 at 5:13 PM.
    “Smart people learn from their mistakes. But the real sharp ones learn from the mistakes of others.”
    ― Brandon Mull

  25. #100
    Ancient Orator leehuan's Avatar
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    Re: First Year Uni Calculus Marathon

    Quote Originally Posted by integral95 View Post








    Yeah basically, but part a) could've been smacked straight away with the first FTC and the chain rule

    Hiccup on the final answer for d) though

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