Hey there! Sign in to join this conversationNew here? Join for free
    Offline

    18
    ReputationRep:
    Ach, this thread is overloaded with calculus!

    Here is an interesting problem I encountered not too long ago: it is tough in my opinion, but the solution is very short.

    Problem 454**/***

    Consider \vartheta:\;\mathbb{R}\to \wp(\mathbb{R}) where \vartheta (x) finite and x\not\in \vartheta(x) for all x. The question is whether there exists S\subset\mathbb{R} s.t.:

    S\cap \vartheta(S)= \varnothing and |S| =\mathfrak{c}
    Offline

    0
    ReputationRep:
    (Original post by Lord of the Flies)
    Ach, this thread is overloaded with calculus!

    Here is an interesting problem I encountered not too long ago: it is tough in my opinion, but the solution is very short.

    Problem 454**/***

    Consider \vartheta:\;\mathbb{R}\to \wp(\mathbb{R}) where \vartheta (x) finite and x\not\in \vartheta(x) for all x. The question is whether there exists S\subset\mathbb{R} s.t.:

    S\cap \vartheta(S)= \varnothing and |S| =\mathfrak{c}
    Well, if you don't need \vartheta to be injective,
    \displaystyle \vartheta(x)=\begin{cases} \{0\} & \text{ if } x \in \mathbb{R} \setminus \{0\} \\ 

\{1 \} & \text{ if } x= 0

\end{cases}

    and S= \mathbb{R} \setminus \{0\}\subset \mathbb{R}. Obviously, \vartheta(S)= \{0\}, so S \cap \vartheta(S)= ( \mathbb{R} \setminus \{0\} ) \cap \{0\}= \emptyset, and  | \mathbb{R} \setminus \{0\} | = \aleph_1 . Thus there is some \vartheta for which this property holds.

    I guess I've made a mistake or misinterpreted the question.
    Offline

    18
    ReputationRep:
    (Original post by henpen)
    ...
    Indeed you have misunderstood the question: it asks whether for any \vartheta there exists such an S, not whether you can find both S and \vartheta satisfying the conditions (as an aside: \aleph_1 is the successor cardinal of \aleph_0 and nothing more. It seems you believe \aleph_1=\mathfrak{c}; this is an undecidable assertion).
    Offline

    17
    ReputationRep:
    Since this thread loves integration:

    Problem 455 **/***

    Let J_0(x) = \dfrac{2}{\pi} \displaystyle\int_0^{\frac{\pi}{  2}} \cos(x\cos\theta) \ d\theta. Show that

    \displaystyle\int_0^{\infty} J_0(x)e^{-ax} \ dx = \dfrac{1}{\sqrt{1+a^2}} , for a > 0


    Problem 456 **/***

    For b > a > 1 find the value of \displaystyle\int_1^{\infty} \dfrac{x^{-a}-x^{-b}}{\log(x)} \ dx


    To make it easier, you may assume both the Tonelli and Fubini theorem hold for both.
    Offline

    15
    ReputationRep:
    (Original post by Lord of the Flies)
    Ach, this thread is overloaded with calculus!

    Here is an interesting problem I encountered not too long ago: it is tough in my opinion, but the solution is very short.

    Problem 454**/***

    Consider \vartheta:\;\mathbb{R}\to \wp(\mathbb{R}) where \vartheta (x) finite and x\not\in \vartheta(x) for all x. The question is whether there exists S\subset\mathbb{R} s.t.:

    S\cap \vartheta(S)= \varnothing and |S| =\mathfrak{c}

    (Original post by henpen)
    Well, if you don't need \vartheta to be injective,
    \displaystyle \vartheta(x)=\begin{cases} \{0\} & \text{ if } x \in \mathbb{R} \setminus \{0\} \\ 

\{1 \} & \text{ if } x= 0

\end{cases}

    and S= \mathbb{R} \setminus \{0\}\subset \mathbb{R}. Obviously, \vartheta(S)= \{0\}, so S \cap \vartheta(S)= ( \mathbb{R} \setminus \{0\} ) \cap \{0\}= \emptyset, and  | \mathbb{R} \setminus \{0\} | = \aleph_1 . Thus there is some \vartheta for which this property holds.

    I guess I've made a mistake or misinterpreted the question.
    Solution 454

    As stated in the original problem, the solution is very short.

    A very short solution does not allow time to consider different cases.

    Threfore the answer is either yes or no for all such functions.

    Since we have found a function for which the answer is yes, the answer must be yes for all such functions.
    Offline

    15
    ReputationRep:
    Problem 457**

    Find all automorphisms of the real numbers

    i.e. find all functions f:\mathbb{R} \rightarrow \mathbb{R}

    such that

    f(x+y)=f(x)+f(y)

    and

    f(xy)=f(x)f(y)

    Now do the same of the complex numbers***
    • PS Helper
    • Study Helper
    Offline

    13
    ReputationRep:
    PS Helper
    Study Helper
    (Original post by james22)
    Problem 457**

    Find all automorphisms of the real numbers

    i.e. find all functions f:\mathbb{R} \rightarrow \mathbb{R}

    such that

    f(x+y)=f(x)+f(y)

    and

    f(xy)=f(x)f(y)

    Now do the same of the complex numbers***
    Done now - thanks, LotF!

    Spoiler:
    Show

    There are two real automorphisms: the constant zero one, and the identity. There is a third complex one: the conjugate.

    If any f(x) = 0, then f(xy) = 0 for all y, so the constant function f(x) = 0 works; otherwise, the only zero-value must be f(0)=0. From now on, assume we're not the constant zero function, and hence if f(x) = 0 then x=0.
    f(2x) = f(2) f(x) = 2 f(x) by the two identities, for all x, so f(2)=2. Similarly f(n) = n for all integer n > 0.

    Now we want f(-1). Have f(-1) = f(2-3) = f(2)+f(-3) = f(2)+f(-1) f(3) = 2+3 f(-1). Solving this for f(-1) gives f(-1) = -1. Hence f(n) = n for all integer n.

    We can now hit all the algebraics: f(a_n x^n + \dots + a_1 x + a_0) = a_n f(x)^n + \dots + a_0 = 0, so f(x) = x for every algebraic x. Actually, hmm - this doesn't take the multiple-roots aspect into account. Better just stick with the rationals at the moment, for which it does hold. We do have that f(\text{irrational}) is irrational, because f(z) = q = f(q) means f(z-q)=0 so z=q.

    We prove (thanks, LotF!) that f(x) \geq 0 when x \geq 0. Indeed, f(x) = f(\sqrt{x}^2) = f(\sqrt{x})^2 > 0.

    Since f(x) \geq 0 when x \geq 0, we have that f is increasing. Indeed, consider x < y. Have y-x > 0 so f(y-x) > 0, so f(y) > f(x). Hence if we let x be an arbitrary real, and x_n a sequence of rational approximations increasing to the limit x, then we have f(x_n) increasing. Moreover, if we let y_n be a sequence of rational approximations tending down to x, then f(y_n) is decreasing. We have y_n = f(y_n) > f(x) > f(x_n) = x_n, and hence f(x) = x.

    The complexes bit: f(i) = \pm i because x^2 + 1 = 0 has solutions x=\pm i, so f(x)^2 + 1 = 0 also does. This gives us a single extra possible automorphism from moving into the complexes. This is consistent because of the fact that the sum/product of conjugates is the conjugate of the product/sum.

    EDIT: As james22 points out, this is only true if we assume that a restriction to the reals of a complex automorphism is an automorphism.
    Offline

    18
    ReputationRep:
    (Original post by Smaug123)
    ...
    Hint: f(x)\geq 0 for x\geq 0. Two possible routes now: prove that a solution to Cauchy's eq. satisfying either 1. f increasing or 2. there exists a real interval on which f is bounded below [the former case is easier to deal with in my opinion], must be f=ax
    Offline

    2
    ReputationRep:
    Spoiler:
    Show
    (Original post by jjpneed1)
    if you assume the power-mean inequality I will murder you
    (Original post by Tarquin Digby)
     \displaystyle 2^n^-^1(x_1^n + x_2^n) \geq (x_1 +x_2)^n  \iff\frac{x_1^n+x_2^n}{2}\geq \left(\frac{x_1+x_2}{2}\right)^n

    By the power-mean inequality...

    (Original post by Flauta)
    Problem 439**

    Find \displaystyle\int^{\pi}_0 \frac{d \theta}{\sin \theta + \csc \theta}

    Nothing particularly ground breaking, just thought the graph looked pretty cool
    Solution 439**

    I haven't seen the graph yet, but Wolfram agrees with my final result

    Spoiler:
    Show

    \displaystyle \begin{aligned} I = \int_{0}^{\pi} \dfrac{\text{d}\theta}{\sin \theta + \csc \theta} = \int_{0}^{\pi} \dfrac{\sin \theta \text{ d}\theta}{\sin^2 \theta + 1} & \ \overset{t=\tan \frac{\theta}{2}}= \ \int_{0}^{+\infty} \dfrac{4t}{\left( t^2 + 1 \right)^2 + 4t^2} \text{ d}t \\ & \ \overset{u=2t^2}= 4 \ \int_{0}^{+\infty} \dfrac{\text{d}u}{\left( u+6 \right)^2 - \left( 4\sqrt{2} \right)^2} \\ & \  \  \  = \left \dfrac{4}{2(4\sqrt{2})} \ln \left| \dfrac{u+6-4\sqrt{2}}{u+6+4\sqrt{2}} \right| \right|_{0}^{\infty} \\ & \  \  \  = \dfrac{1}{2\sqrt{2}} \ln \left| 17 + 12\sqrt{2} \right| \  \approx \ 1.24645 \end{aligned}
    Spoiler:
    Show
    I was careful to check for any discontinuities when making the Weierstrass substitution!

    (Original post by Lord of the Flies)
    ...
    Loving your new avatar
    Offline

    15
    ReputationRep:
    (Original post by Smaug123)
    Done now - thanks, LotF!

    Spoiler:
    Show

    There are two real automorphisms: the constant zero one, and the identity. There is a third complex one: the conjugate.

    If any f(x) = 0, then f(xy) = 0 for all y, so the constant function f(x) = 0 works; otherwise, the only zero-value must be f(0)=0. From now on, assume we're not the constant zero function, and hence if f(x) = 0 then x=0.
    f(2x) = f(2) f(x) = 2 f(x) by the two identities, for all x, so f(2)=2. Similarly f(n) = n for all integer n > 0.

    Now we want f(-1). Have f(-1) = f(2-3) = f(2)+f(-3) = f(2)+f(-1) f(3) = 2+3 f(-1). Solving this for f(-1) gives f(-1) = -1. Hence f(n) = n for all integer n.

    We can now hit all the algebraics: f(a_n x^n + \dots + a_1 x + a_0) = a_n f(x)^n + \dots + a_0 = 0, so f(x) = x for every algebraic x. Actually, hmm - this doesn't take the multiple-roots aspect into account. Better just stick with the rationals at the moment, for which it does hold. We do have that f(\text{irrational}) is irrational, because f(z) = q = f(q) means f(z-q)=0 so z=q.

    We prove (thanks, LotF!) that f(x) \geq 0 when x \geq 0. Indeed, f(x) = f(\sqrt{x}^2) = f(\sqrt{x})^2 > 0.

    Since f(x) \geq 0 when x \geq 0, we have that f is increasing. Indeed, consider x < y. Have y-x > 0 so f(y-x) > 0, so f(y) > f(x). Hence if we let x be an arbitrary real, and x_n a sequence of rational approximations increasing to the limit x, then we have f(x_n) increasing. Moreover, if we let y_n be a sequence of rational approximations tending down to x, then f(y_n) is decreasing. We have y_n = f(y_n) > f(x) > f(x_n) = x_n, and hence f(x) = x.

    The complexes bit: f(i) = \pm i because x^2 + 1 = 0 has solutions x=\pm i, so f(x)^2 + 1 = 0 also does. This gives us a single extra possible automorphism from moving into the complexes. This is consistent because of the fact that the sum/product of conjugates is the conjugate of the product/sum.

    Nice solution. Although you are wrong about the automorphisms of the complex numbers. It was a bit of a trick question because you have found all the contructablle automorphisms, but the set of automorphisms of \mathbb{C} is the same size as the power set of the reals.

    EDIT: If you assume that the automorphisms must be continuous, you have them all.

    EDIT 2: Your mistake with the complex numbers bit is you assume that the restriction of an automorphism of \mathbb{C} to \mathbb{R} is an automorphism of \mathbb{R}, this is not true. In your proof for \mathbb{R} you assumed that y^2>0 for all y, but this may not be true in \mathbb{C}.
    • PS Helper
    • Study Helper
    Offline

    13
    ReputationRep:
    PS Helper
    Study Helper
    (Original post by james22)
    Nice solution. Although you are wrong about the automorphisms of the complex numbers. It was a bit of a trick question because you have found all the contructablle automorphisms, but the set of automorphisms of \mathbb{C} is the same size as the power set of the reals.

    EDIT: If you assume that the automorphisms must be continuous, you have them all.

    EDIT 2: Your mistake with the complex numbers bit is you assume that the restriction of an automorphism of \mathbb{C} to \mathbb{R} is an automorphism of \mathbb{R}, this is not true. In your proof for \mathbb{R} you assumed that y^2>0 for all y, but this may not be true in \mathbb{C}.
    Ah, that did twinge in my mind at the time, but I dismissed it because I was thinking about increasing-ness I loved the problem, though!
    Offline

    15
    ReputationRep:
    (Original post by Smaug123)
    Ah, that did twinge in my mind at the time, but I dismissed it because I was thinking about increasing-ness I loved the problem, though!
    Try and find one that is not continuous (hint: you need the axiom of hoice, in particular zorns lemma).
    • Political Ambassador
    Offline

    3
    ReputationRep:
    Political Ambassador
    Problem 458***

    For u, v >0 show that

    \displaystyle \int^{\infty}_{-\infty} \dfrac{u\cos vx - x\sin vx}{(u^2+x^2)^2} dx = \dfrac{\pi}{2u^2e^{uv}}.
    Offline

    19
    ReputationRep:
    (Original post by Lord of the Flies)
    Well done I'll type it out for you (credit to Izzy in the OP):

    \displaystyle\begin{aligned}\int  _1^n  \frac{n}{1+x^n}\,dx &=\int_{1}^n \frac{n}{x^n}\left(\frac{1 }{1+(1/x)^n}\right)dx\\&=\int_1^n\frac{  n}{x^n}\left(1-\frac{1}{x^n}+\frac{1}{x^{2n}}-\cdots\right)dx\\&=n\int_1^n \frac{1}{x^n}-\frac{1}{x^{2n}}+\frac{1}{x^{3n}  }-\cdots\,dx\\&=n\left[\frac{1}{n-1}-\frac{1}{2n-1}+\cdots\right]-\underbrace{n\left[\frac{n^{1-n}}{n-1}-\frac{n^{1-2n}}{2n-1}+\cdots\right]}_{\to\,0}\end{aligned}

    Hence the limit is \ln 2

    Replacing n with \pi obviously makes no difference.
    how does one evaluate the last line to get ln2?
    Offline

    19
    ReputationRep:
    (Original post by Khallil)
    Spoiler:
    Show







    Solution 439**

    I haven't seen the graph yet, but Wolfram agrees with my final result

    Spoiler:
    Show

    \displaystyle \begin{aligned} I = \int_{0}^{\pi} \dfrac{\text{d}\theta}{\sin \theta + \csc \theta} = \int_{0}^{\pi} \dfrac{\sin \theta \text{ d}\theta}{\sin^2 \theta + 1} & \ \overset{t=\tan \frac{\theta}{2}}= \ \int_{0}^{+\infty} \dfrac{4t}{\left( t^2 + 1 \right)^2 + 4t^2} \text{ d}t \\ & \ \overset{u=2t^2}= 4 \ \int_{0}^{+\infty} \dfrac{\text{d}u}{\left( u+6 \right)^2 - \left( 4\sqrt{2} \right)^2} \\ & \  \  \  = \left \dfrac{4}{2(4\sqrt{2})} \ln \left| \dfrac{u+6-4\sqrt{2}}{u+6+4\sqrt{2}} \right| \right|_{0}^{\infty} \\ & \  \  \  = \dfrac{1}{2\sqrt{2}} \ln \left| 17 + 12\sqrt{2} \right| \  \approx \ 1.24645 \end{aligned}
    Spoiler:
    Show
    I was careful to check for any discontinuities when making the Weierstrass substitution!



    Loving your new avatar
    nice solution though i think a more basic one would be to change the sin^2 in the denominator into 1-cos^2 and let u=cosu, then a little partial fractions and were done.
    Offline

    2
    ReputationRep:
    (Original post by newblood)
    nice solution though i think a more basic one would be to change the sin^2 in the denominator into 1-cos^2 and let u=cosu, then a little partial fractions and were done.
    I noticed that way but I really wanted to bash some of the algebra out in the integrand with the Weierstrass sub
    Offline

    19
    ReputationRep:
    Anyone like stats? Here are a couple of questions i particularly liked that id done recently [not very hard, especially for someone whos done any uni stats or S3/4 even]:

    Problem 459**
    A fair coin is tossed n times. Let un be the probability that the sequence of tosses never has two consecutive heads. Show that u_n =0.5u_(n-1)+0.25u_(n-2) .Find also u_n

    Problem 460*
    The radius of a circle has the exponential distribution with parameter lambda. What is the probability density function of the area of the circle.
    Offline

    18
    ReputationRep:
    (Original post by newblood)
    How does one evaluate the last line to get ln 2?
    \displaystyle\frac{n}{kn-1}\to \frac{1}{k}\Rightarrow \big[\cdots \big] \to \sum_{k>0} \frac{(-1)^{k+1}}{k}
    Offline

    15
    ReputationRep:
    (Original post by Lord of the Flies)
    \displaystyle\frac{n}{kn-1}\to \frac{1}{k}\Rightarrow \big[\cdots \big] \to \sum_{k>0} \frac{(-1)^{k+1}}{k}
    Why are you justified in swapping the order of limits here?
    Offline

    19
    ReputationRep:
    (Original post by james22)
    Why are you justified in swapping the order of limits here?
    if youre referring to the original integral, i think he also changed the signs as well to change the limits, just for ease of manipulation id imagine
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Would you like to hibernate through the winter months?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.