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A Summer of Maths

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  • View Poll Results: Have you studied any Group Theory already?
    Yes, I did some Group Theory during/at my A-level.
    36.99%
    No, but I plan to study some before Uni.
    21.92%
    No, I haven't.
    41.10%

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    I haven't done any maths for like 2 weeks and I already feel rusty :/
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    What is the derivative of y=x^x?

    What is the derivatie and inverse of y=x^x^x^x^... (an infinite string of x's)

    Here x^x^x=x^(x^x) not (x^x)^x

    Also for what values of x does y exist?
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    (Original post by Brit_Miller)
    I have a question but not an answer as I don't know how to do it. Hopefully someone can show how.

    Let f: \mathbb{R} \mathrm{x} \mathbb{R} \rightarrow \mathbb{R} \mathrm{x} \mathbb{R} \mathrm{be\ given\ by} f(x,y) = (x + y^3, x^5)

    \mathrm{Prove \ that}  f \mathrm{ is \ bijective \ and \ find \ its \ inverse.}
    I have never done anything like this so the following is possibly wrong. Even if it is correct I'm sure there's a much prettier way of doing it:

    Spoiler:
    Show

    (for the bijection proof)

    Suppose there exists (a,b)\in\mathbb{R},\;(a,b)\neq0 such that f(x,y)=f(x+a,y+b)

    We would have:

    (x+y^3,x^5)=(x+a+(y+b)^3,(x+a)^5  ) 

\iff\left\{\begin{array}{l l}

    x+y^3=x+a+(y+b)^3\\

    x^5=(x+a)^5\\

  \end{array} \right.

    Clearly \forall (x,a)\;  x^5\neq(x+a)^5

    Furthermore: x+y^3=x+a+(y+b)^3\Leftrightarrow a+3y^2b+3yb^2+b^3=0 \Leftrightarrow a=-(3y^2b+3yb^2+b^3)

    Now consider the functions h(a)=a and k(b)=-(3y^2b+3yb^2+b^3)

    The above equality is satisfied when h(a)=k(b)

    Obviously h is strictly increasing. Differentiating k with respect to b gives k'(b)=-3(y+b)^2\leq 0 so k decreases. Therefore h(a)=k(b) has only one solution, namely a=b=0. But this is defies our initial conditions.

    Therefore f(x,y)\neq f(x+a,y+b)\;\;\forall(a,b)\neq 0\Rightarrow f is bijective.

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    (Original post by Brit_Miller)
    I have a question but not an answer as I don't know how to do it. Hopefully someone can show how.

    Let f: \mathbb{R} \mathrm{x} \mathbb{R} \rightarrow \mathbb{R} \mathrm{x} \mathbb{R} \mathrm{be\ given\ by} f(x,y) = (x + y^3, x^5)

    \mathrm{Prove \ that}  f \mathrm{ is \ bijective \ and \ find \ its \ inverse.}
    {*} Additional:

    Spoiler:
    Show


    Lemma 1. The cube of a negative number is the negative of its magnitude cubed.
    Spoiler:
    Show

    Suppose that, for some n \in \mathbb{N}, we have

    a^{2n - 1} \equiv -(-a)^{2n - 1}

    where a \in \mathbb{R} is non-negative.


    Now, if we take for granted that (-a)^2 \equiv a^2, we continue by

    \iff\ a^{2n - 1} \cdot a^2 \equiv -(-a)^{2n-1} \cdot (-a)^2

    \iff\ a^{2(n + 1) - 1} \equiv -(-a)^{2(n+1) - 1}

    Therefore, since the result is trivially true for n=1, we get

    -(a^{2n - 1}) \equiv (-a)^{2n-1}, \ \ \forall n \in \mathbb{N}

    as desired.



    Lemma 2 (Corollary of Existence of n-th roots). Two cubes are equal if and only if the are the same.
    Spoiler:
    Show

    Existence of n-th roots:

    If x \in \mathbb{R} is positive, and n \in \mathbb{N}, then there exists exactly one positive real number y such that y^n = x.


    By Lemma 1, we can always write a^{2n - 1} = b^{2n - 1} as

    |a|^{2n - 1} = |b|^{2n - 1}

    but then the above theorem says that they must be equal.



    {**} A solution:

    Spoiler:
    Show

    We want to show that f(x,y) = (x + y^3, x^5) is injective and surjective which would imply that it is bijective.


    [1] Suppose that f(a_1, b_1) = f(a_2, b_2) with a_i, b_i \in \mathbb{R} and at least one of a_1 \not= a_2 and b_1 \not= b_2 holds.

    Then, we obtain the following two equations.

    (1)\ \ \ a_1^5 = a_2^5

    (2)\ \ \ a_1 + b_1^3 = a_2 + b_2^3


    Using Lemma 2, we see from (1) that a_1 = a_2, and by substituting into (2) we get

    a_1 + b_1^3 = a_1 + b_2^3 \iff b_1^3 = b_2^3 \iff b_1 = b_2

    which contradicts our assumption. Therefore, f is injective.


    [2] We show surjectivity by existence.

    Take any (a, b) \in \mathbb{R}^2. Then, it suffices to show that \exists x,y \in \mathbb{R} such that x + y^3 = a and x^5 = b.

    We find that x = \sqrt[5]{b} and y = \sqrt[3]{a - \sqrt[5]{b}} where x,y \in \mathbb{R} by applying Lemmas 1 & 2.

    Therefore, f is bijective.



    Now, its inverse is given by

    f^{-1}(x, y) = (\sqrt[5]{y}, \sqrt[3]{x - \sqrt[5]{y}})

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    (Original post by jack.hadamard)
    {*} Additional:

    Spoiler:
    Show


    Lemma 1. The cube of a negative number is the negative of its magnitude cubed.
    Spoiler:
    Show

    Suppose that, for some n \in \mathbb{N}, we have

    a^{2n - 1} \equiv -(-a)^{2n - 1}

    where a \in \mathbb{R} is non-negative.


    Now, if we take for granted that (-a)^2 \equiv a^2, we continue by

    \iff\ a^{2n - 1} \cdot a^2 \equiv -(-a)^{2n-1} \cdot (-a)^2

    \iff\ a^{2(n + 1) - 1} \equiv -(-a)^{2(n+1) - 1}

    Therefore, since the result is trivially true for n=1, we get

    -(a^{2n - 1}) \equiv (-a)^{2n-1}, \ \ \forall n \in \mathbb{N}

    as desired.



    Lemma 2 (Corollary of Existence of n-th roots). Two cubes are equal if and only if the are the same.
    Spoiler:
    Show

    Existence of n-th roots:

    If x \in \mathbb{R} is positive, and n \in \mathbb{N}, then there exists exactly one positive real number y such that y^n = x.


    By Lemma 1, we can always write a^{2n - 1} = b^{2n - 1} as

    |a|^{2n - 1} = |b|^{2n - 1}

    but then the above theorem says that they must be equal.



    {**} A solution:

    Spoiler:
    Show

    We want to show that f(x,y) = (x + y^3, x^5) is injective and surjective which would imply that it is bijective.


    [1] Suppose that f(a_1, b_1) = f(a_2, b_2) with a_i, b_i \in \mathbb{R} and at least one of a_1 \not= a_2 and b_1 \not= b_2 holds.

    Then, we obtain the following two equations.

    (1)\ \ \ a_1^5 = a_2^5

    (2)\ \ \ a_1 + b_1^3 = a_2 + b_2^3


    Using Lemma 2, we see from (1) that a_1 = a_2, and by substituting into (2) we get

    a_1 + b_1^3 = a_1 + b_2^3 \iff b_1^3 = b_2^3 \iff b_1 = b_2

    which contradicts our assumption. Therefore, f is injective.


    [2] We show surjectivity by existence.

    Take any (a, b) \in \mathbb{R}^2. Then, it suffices to show that \exists x,y \in \mathbb{R} such that x + y^3 = a and x^5 = b.

    We find that x = \sqrt[5]{b} and y = \sqrt[3]{a - \sqrt[5]{b}} where x,y \in \mathbb{R} by applying Lemmas 1 & 2.

    Therefore, f is bijective.



    Now, its inverse is given by

    f^{-1}(x, y) = (\sqrt[5]{y}, \sqrt[3]{x - \sqrt[5]{y}})

    Nice - and informative too! I guess I was completely off the map on this one. :rolleyes:
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    (Original post by Lord of the Flies)
    Nice - and informative too! I guess I was completely off the map on this one. :rolleyes:
    You can add some bits to it, and fix some, but in general I don't think you were off the map at all. (not that I am much on the map, but anyway)

    For instance, \mathbb{R} \times \mathbb{R} \equiv \mathbb{R}^2, so that (x,y) \in \mathbb{R}^2 is a point in the plane.
    The function is a mapping from points on the plane to points on the real plane; i.e. f: \mathbb{R}^2 \to \mathbb{R}^2.
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    (Original post by jack.hadamard)
    {*} Additional:

    Spoiler:
    Show


    Lemma 1. The cube of a negative number is the negative of its magnitude cubed.
    Spoiler:
    Show

    Suppose that, for some n \in \mathbb{N}, we have

    a^{2n - 1} \equiv -(-a)^{2n - 1}

    where a \in \mathbb{R} is non-negative.


    Now, if we take for granted that (-a)^2 \equiv a^2, we continue by

    \iff\ a^{2n - 1} \cdot a^2 \equiv -(-a)^{2n-1} \cdot (-a)^2

    \iff\ a^{2(n + 1) - 1} \equiv -(-a)^{2(n+1) - 1}

    Therefore, since the result is trivially true for n=1, we get

    -(a^{2n - 1}) \equiv (-a)^{2n-1}, \ \ \forall n \in \mathbb{N}

    as desired.



    Lemma 2 (Corollary of Existence of n-th roots). Two cubes are equal if and only if the are the same.
    Spoiler:
    Show

    Existence of n-th roots:

    If x \in \mathbb{R} is positive, and n \in \mathbb{N}, then there exists exactly one positive real number y such that y^n = x.


    By Lemma 1, we can always write a^{2n - 1} = b^{2n - 1} as

    |a|^{2n - 1} = |b|^{2n - 1}

    but then the above theorem says that they must be equal.



    {**} A solution:

    Spoiler:
    Show

    We want to show that f(x,y) = (x + y^3, x^5) is injective and surjective which would imply that it is bijective.


    [1] Suppose that f(a_1, b_1) = f(a_2, b_2) with a_i, b_i \in \mathbb{R} and at least one of a_1 \not= a_2 and b_1 \not= b_2 holds.

    Then, we obtain the following two equations.

    (1)\ \ \ a_1^5 = a_2^5

    (2)\ \ \ a_1 + b_1^3 = a_2 + b_2^3


    Using Lemma 2, we see from (1) that a_1 = a_2, and by substituting into (2) we get

    a_1 + b_1^3 = a_1 + b_2^3 \iff b_1^3 = b_2^3 \iff b_1 = b_2

    which contradicts our assumption. Therefore, f is injective.


    [2] We show surjectivity by existence.

    Take any (a, b) \in \mathbb{R}^2. Then, it suffices to show that \exists x,y \in \mathbb{R} such that x + y^3 = a and x^5 = b.

    We find that x = \sqrt[5]{b} and y = \sqrt[3]{a - \sqrt[5]{b}} where x,y \in \mathbb{R} by applying Lemmas 1 & 2.

    Therefore, f is bijective.



    Now, its inverse is given by

    f^{-1}(x, y) = (\sqrt[5]{y}, \sqrt[3]{x - \sqrt[5]{y}})

    Nicely done!
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    Is anyone interested in summing up

    \displaystyle \sum_{n=0}^{\infty} \frac{n!}{(n + x)!}

    where x is a positive integer? Ideas how we can do this?
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    (Original post by jack.hadamard)
    Is anyone interested in summing up

    \displaystyle \sum_{n=0}^{\infty} \frac{n!}{(n + x)!}

    where x is a positive integer? Ideas how we can do this?
    Without doing any mathematics, I think that should diverge. For large n the value of x can be ignored so n!/(n+x)! is approximately 1 so it must diverge.

    EDIT: After doing some maths, it seems that the above reasoning is flawed, the terms do indeed tend to 0 so it may converge.
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    (Original post by james22)
    Without doing any mathematics, I think that should diverge. For large n the value of x can be ignored so n!/(n+x)! is approximately 1 so it must diverge.
    Take x = 2, then n!/(n+2)! = 1/(n+1)(n+2) which converges.
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    (Original post by jack.hadamard)
    Is anyone interested in summing up

    \displaystyle \sum_{n=0}^{\infty} \frac{n!}{(n + x)!}

    where x is a positive integer? Ideas how we can do this?
    Spoiler:
    Show
    I'd start off by noting that \displaystyle \sum_{n=0}^{\infty} \dfrac{n!}{(n + x)!} \equiv \displaystyle \sum_{j=x}^{\infty} \dfrac{(j-x)!}{j!} \equiv \dfrac{1}{x!} \displaystyle \sum_{j=x}^{\infty} \dfrac{1}{^j\mathrm{C}_x}, which should converge for x\geq 2.

    I'll finish it off when I get back if someone hasn't already.
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    subscribing.....around 2 in the night I tend to get bored
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    (Original post by Lord of the Flies)
    Spoiler:
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    If n is the number of powers of x we can define:

    f_n(x)=x^{x^{x^{x^{...}}}}=\left  (e^{\ln x}\right)^{x^{x^{x^{...}}}}=e^{x  ^n\ln x}\Rightarrow f_n'(x)=x^{n-1}(n\ln x +1)x^{x^{x^{x^{...}}}}\;\;(In particular y=x^x\Rightarrow y'=(\ln x +1)y)

    I don't believe it is possible to express the inverse in the form f^{-1}(x)=blah though.

    Sorry I wasn't clear, there are meant to be infinite x's.

    As a side question to it, for what values of x is f(x) defined?
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    (Original post by james22)
    Sorry I wasn't clear, there are meant to be infinite x's.

    As a side question to it, for what values of x is f(x) defined?
    Ah. In any case my working is wrong, I stupidly misread the question!...

    ... which makes the question more difficult, but more interesting! Hm...
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    (Original post by jack.hadamard)
    {*} Additional:

    Spoiler:
    Show


    Lemma 1. The cube of a negative number is the negative of its magnitude cubed.
    Spoiler:
    Show

    Suppose that, for some n \in \mathbb{N}, we have

    a^{2n - 1} \equiv -(-a)^{2n - 1}

    where a \in \mathbb{R} is non-negative.


    Now, if we take for granted that (-a)^2 \equiv a^2, we continue by

    \iff\ a^{2n - 1} \cdot a^2 \equiv -(-a)^{2n-1} \cdot (-a)^2

    \iff\ a^{2(n + 1) - 1} \equiv -(-a)^{2(n+1) - 1}

    Therefore, since the result is trivially true for n=1, we get

    -(a^{2n - 1}) \equiv (-a)^{2n-1}, \ \ \forall n \in \mathbb{N}

    as desired.



    Lemma 2 (Corollary of Existence of n-th roots). Two cubes are equal if and only if the are the same.
    Spoiler:
    Show

    Existence of n-th roots:

    If x \in \mathbb{R} is positive, and n \in \mathbb{N}, then there exists exactly one positive real number y such that y^n = x.


    By Lemma 1, we can always write a^{2n - 1} = b^{2n - 1} as

    |a|^{2n - 1} = |b|^{2n - 1}

    but then the above theorem says that they must be equal.



    {**} A solution:

    Spoiler:
    Show

    We want to show that f(x,y) = (x + y^3, x^5) is injective and surjective which would imply that it is bijective.


    [1] Suppose that f(a_1, b_1) = f(a_2, b_2) with a_i, b_i \in \mathbb{R} and at least one of a_1 \not= a_2 and b_1 \not= b_2 holds.

    Then, we obtain the following two equations.

    (1)\ \ \ a_1^5 = a_2^5

    (2)\ \ \ a_1 + b_1^3 = a_2 + b_2^3


    Using Lemma 2, we see from (1) that a_1 = a_2, and by substituting into (2) we get

    a_1 + b_1^3 = a_1 + b_2^3 \iff b_1^3 = b_2^3 \iff b_1 = b_2

    which contradicts our assumption. Therefore, f is injective.


    [2] We show surjectivity by existence.

    Take any (a, b) \in \mathbb{R}^2. Then, it suffices to show that \exists x,y \in \mathbb{R} such that x + y^3 = a and x^5 = b.

    We find that x = \sqrt[5]{b} and y = \sqrt[3]{a - \sqrt[5]{b}} where x,y \in \mathbb{R} by applying Lemmas 1 & 2.

    Therefore, f is bijective.



    Now, its inverse is given by

    f^{-1}(x, y) = (\sqrt[5]{y}, \sqrt[3]{x - \sqrt[5]{y}})

    Nice

    The following result could of been quoted to make your answer alot shorter (though it's always good practice to do things straight from the definitions)
    Related exercise:
    Let f: X \rightarrow Y, g:Y \rightarrow X
    g is said to be a left [or right] inverse of f if (g \circ f)(x)=x, [(f \circ g)(x)] respectively


    Show that f is surjective iff it has a right inverse
    Show that f is injective iff it has a left inverse

    Hence a map is bijective iff it has an (left and right) inverse. (an g is said to be an inverse iff g is both a left and right inverse.)

    Using this, you only need to verify that your inverse is an inverse .
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    {*} Question:

    The polynomial x^2 + 1 is irreducible over \mathbb{R}.

    i) By completing the square, show that x^4 + 1 is not irreducible over the set of real numbers.

    Hence, derive the Sophie Germain algebraic identity

    x^4 + 4y^4\ \equiv\ (x^2 + 2xy + 2y^2)(x^2 - 2xy + 2y^2)

    by starting from the left-hand side.


    ii) Evaluate \displaystyle \sum_{k=1}^{n} \frac{4k}{4k^4 + 1}


    {**} Required:

    Spoiler:
    Show

    A polynomial is said to be irreducible over a set if it cannot be factored into polynomials with coefficients from the given set.

    As an example,

    x^2 - 2 \equiv (x - \sqrt{2})(x + \sqrt{2})

    is irreducible over the set of rational numbers denoted by \mathbb{Q}.

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    Okay, I have another question which I'm sure is relatively simple and someone will know (sorry for using the thread without answers, but it's a nice place to ask questions!)

    Consider this 2nd order differential equation:

    y'' + 2ty' + t^2y = 2, y(0) = 1, y'(0) = -1

    Write this as a system of 1st order equations with appropriate initial conditions.
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    (Original post by Brit_Miller)
    Okay, I have another question which I'm sure is relatively simple and someone will know (sorry for using the thread without answers, but it's a nice place to ask questions!)

    Consider this 2nd order differential equation:

    y'' + 2ty' + t^2y = 2, y(0) = 1, y'(0) = -1

    Write this as a system of 1st order equations with appropriate initial conditions.
    Let z=y'.

    Also quite easy: How many (real) solutions does (x + 101)^{16} + x^2 = 0 have? Knowledge required: GCSE & below.
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    In an office, at various times during the day, the boss gives the secretary a letter to type, each time putting the letter on top of the pile and types it. There are nine letters to be typed during the day, and the boss delivers them in the order 1,2,3,4,5,6,7,8,9. While leaving for lunch, the secretary tells a colleague that letter 8 has already been typed, but says nothing about the morning's typing. The colleague wonders which of the nine letters remain to be typed after lunch and in what order they will be typed. Based on the above information, how many such after-lunch typing orders are possible? (That there are no letters to be typed is on of the possibilities).

    No ugrad knowledge required beyond combinations/ permutations.
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    (Original post by electriic_ink)
    Let z=y'.

    Also quite easy: How many (real) solutions does (x + 101)^{16} + x^2 = 0 have? Knowledge required: GCSE & below.
    Thanks

    (and none surely?)

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Updated: August 12, 2013
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