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    (Original post by Indeterminate)
    Problem 372***

    Prove the following:

    If

    f : D \rightarrow \mathbb{C}

    and if f is holomorophic and

    \forall z \in D, f'(z)=0

    then

     f= constant
    What is D?
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    (Original post by Hodor)
    What is D?
    I assume it's any domain in the complex numbers on which f can de differentiable.
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    (Original post by james22)
    I assume it's any domain in the complex numbers on which f can de differentiable.
    Then I don't think it's true.
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    (Original post by Hodor)
    Then I don't think it's true.
    ..Really? If you have a real function  f:\mathbb{R} \mapsto \mathbb{R} which was differentiable everywhere, and f'=0 everywhere, wouldn't you believe that was constant?

    You'd know this was constant, since the tangent to the curve of f is always parallel to the x axis - which only happens when f is constant. The above proposition says the same is true for complex functions; and by the definition of the complex derivative, it should appeal to intuition that it is 'obviously true', as any straight line in the complex plane behaves like the real line in terms of differentiation (i.e. \mathbb{R} is not really a special line for differentiability; complex derivatives must exist and be equal in all possible paths of approach).
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    (Original post by FireGarden)
    ..Really? If you have a real function  f:\mathbb{R} \mapsto \mathbb{R} which was differentiable everywhere, and f'=0 everywhere, wouldn't you believe that was constant?

    You'd know this was constant, since the tangent to the curve of f is always parallel to the x axis - which only happens when f is constant. The above proposition says the same is true for complex functions; and by the definition of the complex derivative, it should appeal to intuition that it is 'obviously true', as any straight line in the complex plane behaves like the real line in terms of differentiation (i.e. \mathbb{R} is not really a special line for differentiability; complex derivatives must exist and be equal in all possible paths of approach).
    In the real case, let A = (-1,0) u (1,2). Then I can think of a non-constant function f: A -> R with f' = 0...
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    (Original post by Hodor)
    In the real case, let A = (-1,0) u (1,2). Then I can think of a non-constant function f: A -> R with f' = 0...
    True, I think D should also be connected.
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    Most probably D is the open 2-disk (and, if so, the proof is immediate).
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    (Original post by Mladenov)
    Most probably D is the open 2-disk (and, if so, the proof is immediate).
    What if it is something more unusual? I guess the proof for convex domains is just the same as in the real case, but what about concave domains? Thinking about it geometrically I think it is true but the proof is not as obvious.
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    (Original post by Hodor)
    In the real case, let A = (-1,0) u (1,2). Then I can think of a non-constant function f: A -> R with f' = 0...
    Oh, sure! .. please excuse the evident jadedness from always being told D is a domain!


    Anyway, some more 'different' maths. Group theory!

    Problem 373***

    Let G be a group, which acts on the set X.

    Suppose x,y \in X have the same orbit. Prove that their stabilisers are conjugates (i.e. \exists g\in G : G_y = gG_xg^{-1})


    Problem 374***

    How many ways can you colour the vertices of a square using at most three colours, which are distinct under the action of D_8? (i.e. colourings which cannot be obtained from another by rotations or reflections)
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    (Original post by FireGarden)
    Anyway, some more 'different' maths. Group theory!

    Problem 373***

    Let G be a group, which acts on the set X.

    Suppose x,y \in X have the same orbit. Prove that their stabilisers are conjugates (i.e. \exists g\in G : G_y = gG_xg^{-1})
    Finally, some Algebra.

    Solution 373

    Simple.

    From the definition of an orbit, if x, y lie in the same orbit, then, there exists a g \in G that acts on  x such that, gx=y \Rightarrow x= g^{-1}y .

    Now, suppose,

     a \in G_x, \Rightarrow ax= x 



\Rightarrow gax = gx = y \Rightarrow ga(g^{-1}y)= y 



\Rightarrow gag^{-1}y=y \Rightarrow gag^{-1} \in G_y .

    Therefore, we have,  g G_x g^{-1} \subset G_y .

    However, repeating the steps above using  g^{-1}y=x , we have,

     b \in G_y \Rightarrow g^{-1}bg \in G_x 



\Rightarrow g^{-1}G_yg \subset G_x



\Rightarrow G_y \subset g G_x g^{-1} .


    And so, by symmetry, we can assert that,  G_y = g G_x g^{-1} .
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    (Original post by Hodor)
    X
    (Original post by james22)
    X
    (Original post by Mladenov)
    X

    Oops, as I had posted it from memory (came across it a long time ago), I made some mistakes in stating it. I'll consult the source and re-post it
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    (Original post by FireGarden)
    Problem 374***

    How many ways can you colour the vertices of a square using at most three colours, which are distinct under the action of D_8? (i.e. colourings which cannot be obtained from another by rotations or reflections)
    Solution 376

    Let S be the set of colorings of the vertices.
    Firstly, observe that two colorings are essentially distinct (i.e. distinct under the action of the dihedral 8 group) if they belong to different orbits of D_{8} in S.
    Hence, we shall be done if we manage to find the number of orbits of D_{8} in the set of colorings.
    The following result is easily proved (counting in two ways, Lang's Algebra, ect.)

    Proposition:

    Let G be a finite group acting on a finite set X, and \alpha(g) is the number of elements x \in X such that \pi_{g}(x) = x. Then the number of orbits of G in X is \displaystyle \frac{1}{card (G)}\sum_{g \in G} \alpha(g).

    Since the representative does not matter, we have \displaystyle \sum_{t \in Gs} \frac{1}{card(Gt)} =1 and hence the number of orbits of G in X is \displaystyle \sum_{x \in X} \frac{1}{card(Gx)}.
    Fix x \in X. Notice that there is a bijection G/G_{x} \sim Gx, call this (1). Also, the number of g \in G such that \pi_{g}(x)=x is the order of G_{x}.

    We count the number of solutions (x,g) to the equation \pi_{g}(x)=x in two ways.

    One the one hand, there are \alpha(g) elements x \in X such that \pi_{g}(x)=x. On the other hand, each g \in G is counted \alpha(g) times by the sum \displaystyle \sum_{x \in X} card(G_{x}) since there are \alpha(g) x's in X such that \pi_{g}(x)=x. Thus \displaystyle \sum_{x \in X} card(G_{x}) = \sum_{g \in G} \alpha(g); call this (2).
    Hence, by (1) and (2), we obtain the result.

    We can interpret the elements of D_{8} as products of cycles.

    Second Proposition (too general):

    Let \pi be a product of m cycles (counting cycles of length 1). Then the number of colorings fixed by \pi is n^{m}, where n is the number of available colors.

    Although, we do not need it, I write it in case someone is interested.

    For our problem, it is sufficient to note that the vertices in a given cycle have the same color.
    Thus, if n colors are available, there are \frac{1}{8}(n^{4}+n+n^{2}+n^{2}+  n+n^{3}+n^{3}+n^{2}) essentially distinct colorings, since the elements are e, [1,2,3,4], [1,2][3,4], [1,3][2,4], [1,4,3,2], [1][3][2,4], [1,3][2][4], [1,4][2,3].
    Spoiler:
    Show
    It is left as an exercise to calculate the answer with n=3.


    As a side note, we have a problem with the numeration of the problems. It starts with Indeterminate's problem 372.

    Basic commutative algebra.

    Problem 377***

    Prove that if M is an n \times n matrix over a local ring, with coefficients in the maximal ideal J, then J+M is invertible.

    Problem 378***

    If f : A \to A is a surjective endomorphism of a Noetherian ring A, then f is an isomorphism.
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    (Original post by Lord of the Flies)
    Solution 10

    \displaystyle x\to \frac{\pi}{2}-x\Rightarrow \int_0^{\frac{\pi}{2}} \frac{\sin^{17} x\,dx}{\sin^{17} x+\cos^{17} x}= \int_0^{\frac{\pi}{2}} \frac{\cos^{17} x\,dx}{\sin^{17} x+\cos^{17} x}

    Hence \displaystyle I=\frac{1}{2}\int_0^{\frac{\pi}{  2}} dx=\frac{\pi}{4}
    Sorry, probably my maths skills have gotten rusty - I can't understand the part after "hence...". I is \dfrac{\cos^{17} x\,dx}{\sin^{17} x+\cos^{17} x}, but how does that become \frac{1}{2}\int dx?
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    (Original post by Zishi)
    Sorry, probably my maths skills have gotten rusty - I can't understand the part after "hence...". I is \dfrac{\cos^{17} x\,dx}{\sin^{17} x+\cos^{17} x}, but how does that become \frac{1}{2}\int dx?
    You have 2 experssions for I, one has sine terms on the top and the other cas cos terms. By adding them you get (sin^17+cos^17)/(sin^17+cos^17)=1 so 2I=1
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    (Original post by james22)
    You have 2 experssions for I, one has sine terms on the top and the other cas cos terms. By adding them you get (sin^17+cos^17)/(sin^17+cos^17)=1 so 2I=1
    Ahh, great! :grin: Many thanks.
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    Problem 379 **

    How many subsets of \{1,2,...,50\} are there such that the sum of the elements exceeds 637?


    Is this possible to generalise?
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    (Original post by henpen)
    Problem 379 **

    How many subsets of \{1,2,...,50\} are there such that the sum of the elements exceeds 637?


    Is this possible to generalise?
    As for each set, there is a reverse set. As 637 is slightly less than half of the sum of numbers from 1 to 50(637.5), for each pair of sets, one of them is going to satisfy the condition. So it is half of all sets, namely 2^49, 562949953421312.
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    Problem 380*

    \displaystyle\int sinxcos2x dx


    Would this be considered as trivial?
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    (Original post by henpen)
    Problem 379 **

    How many subsets of \{1,2,...,50\} are there such that the sum of the elements exceeds 637?


    Is this possible to generalise?
    It can be generalized. We have to consider two cases - \displaystyle \frac{n(n+1)}{2} \pmod 2.

    (Original post by Arieisit)
    Problem 380*

    \displaystyle\int sinxcos2x dx


    Would this be considered as trivial?
    What do you think? Using \displaystyle \sin x \cos 2x = \frac{1}{2}(\sin 3x - \sin x) and \displaystyle \int \sin nx dx = -\frac{1}{n} \cos nx + C, we are done.
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    (Original post by Mladenov)
    It can be generalized. We have to consider two cases - \displaystyle \frac{n(n+1)}{2} \pmod 2.



    What do you think? Using \displaystyle \sin x \cos 2x = \frac{1}{2}(\sin 3x - \sin x) and \displaystyle \int \sin nx dx = -\frac{1}{n} \cos nx + C, we are done.
    I only just noticed the problem and thought "oh, I can finally solve a problem on here". Too late though
    Spoiler:
    Show
    \displaystyle \int sinxcos2x=\dfrac{1}{2} \int sin 3x -sinx
    =\dfrac{1}{6}(3cosx-cos3x)+C
 
 
 
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