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.
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  1. Brit_Miller's Avatar
    61 02-07-2012  20:30
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    Re: A Summer of Maths
    (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?)
  2. electriic_ink's Avatar
    62 02-07-2012  20:33
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    Re: A Summer of Maths
    (Original post by Brit_Miller)
    Thanks

    (and none surely?)
    Yeah
  3. DJMayes's Avatar
    63 02-07-2012  21:30
    • Overlord in Training
    • Posts: 2,563
    Re: A Summer of Maths
    (Original post by 4ever_drifting)
    Love these kind of problems Just a note though - could you not observe that you need the product of the roots of the quadratic equation, which is given by c/a in 0=ax^2+bx+c, so you know the answer is 26 without having to find the roots then multiply them together?
    You could in this case because the quadratic simultaneous equations you get when substituting for X and when substituting for Y are the same. I preferred not to take that shortcut on the off-chance that substituting for Y generated a different quadratic than substituting for X.

    I think you could probably solve it geometrically as well, the method I posted was just the one I prefer to use.
  4. Intriguing Alias's Avatar
    64 02-07-2012  21:54
    • TSR Idol
    • Location: Yorkshire
    Re: A Summer of Maths
    (Original post by DJMayes)
    At the Nottingham University Open Day there was a "Maths Trail" with several interesting questions on it. The questions ranged from requiring a working knowledge of arithmetic progressions and combinations to lowest common multiples and counting squares; and more emphasis was put on thinking about them than slogging through endless manipulation. I thought I'd share one with you. The question is of the kind that could be set in C1, but is an interesting one:

    A rectangle is inscribed inside a circle of radius 6 units such that each of the vertices of the rectangle lie on the circumference of the circle. Given that the perimeter of the rectangle is 28 units, what is the area?

    Required Knowledge:
    Spoiler:
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    - Pythagoras' Theorem
    - Simultaneous Quadratic Equations


    Hints:
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    What information are you given? You have both the perimeter of the rectangle, and the radius of the circle it is inscribed inside. How can you relate these in terms of the length and width of the rectangle?
    Spoiler:
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    Consider the diagonal of the rectangle



    Full Solution:
    Spoiler:
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    Let X and Y represent the length and width of the rectangle. Using this and the perimeter we can automatically deduce an equation:

    2X + 2Y = 28

    Which simplifies to X + Y = 14

    As we are told that the radius of the circle is 6, we also know that its diameter must be 12. This diameter forms the diagonal of the rectangle. From this, another equation can be deduced using Pythagoras' Theorem:

    X^2 + Y^2 = 144

    We now have a set of simultaneous equations to solve. Re-arranging the first to leave Y in terms of X and substituting in the second equation leaves you with this:

    X^2 + (14 - X)^2 = 144

    This expands to:

    X^2 + X^2 - 28X + 196 = 144

    Which then simplifies to:

    2X^2 - 28X + 52 = 0

    or:

    X^2 - 14X + 26 = 0

    You can then complete the square or use the quadratic formula to get the result:

    X = 7 + rt(23) or X = 7 - rt(23)

    Using this you can then confirm that Y has the same values (i.e. if X = 7 + rt(23) then Y = 7 + rt(23) and vice-versa)

    Now you know both side lengths, you simply multiply them together to get the area. The final answer is 26.
    Spoiler:
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    This could be made a lot neater. Starting as you did, x + y = 14, and x^2 + y^2 = 144. Now see that (x+y)^2 = x^2 + y^2 + 2xy and so area = xy = 0.5(196 - 144) = 26
    Post rating:
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  5. DJMayes's Avatar
    65 02-07-2012  21:58
    • Overlord in Training
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    Re: A Summer of Maths
    (Original post by hassi94)
    Spoiler:
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    This could be made a lot neater. Starting as you did, x + y = 14, and x^2 + y^2 = 144. Now see that (x+y)^2 = x^2 + y^2 + 2xy and so area = xy = 0.5(196 - 144) = 26
    Spoiler:
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    That's a very clever trick for doing it actually! Unfortunately I have an insistence on solving all simultaneous equations (Even linear ones) through substitution instead of subtraction or manipulation which prevents me from spotting things like this
  6. jack.hadamard's Avatar
    66 02-07-2012  22:51
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    Re: A Summer of Maths
    This one is a bit tricky, but I think it has something to teach; use hints if you get stuck.

    {*} Question:

    Let f : \mathbb{R} \to \mathbb{R}\setminus \{3\} be a function with the property that there exists \omega > 0 such that

    \displaystyle f(x + \omega) = \frac{f(x) - 5}{f(x) - 3}

    for all x \in \mathbb{R}.

    Prove that f is periodic.



    {**} Required: (A-level)

    Spoiler:
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    A function is periodic, with period p, if for every x in its domain one has f(x) \equiv f(x + n p) where n is an integer.


    {***} Hints:

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    (1)
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    What about f(x) = 2?


    (2)
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    How does f(x + 2\omega) look like?

  7. Rahul.S's Avatar
    67 02-07-2012  23:23
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    Re: A Summer of Maths
    (Original post by hassi94)
    Spoiler:
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    This could be made a lot neater. Starting as you did, x + y = 14, and x^2 + y^2 = 144. Now see that (x+y)^2 = x^2 + y^2 + 2xy and so area = xy = 0.5(196 - 144) = 26
    wasnt this in the online lecture thing?
  8. jack.hadamard's Avatar
    68 02-07-2012  23:35
    • Benevolent Member
    • Posts: 696
    Re: A Summer of Maths
    (Original post by james22)
    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?
    Some thoughts under cruel assumptions.

    Spoiler:
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    Without any justification, I claim that \displaystyle y = x^y.

    Maybe the inverse will be x = y^{y^{-1}}


    Now, note that \displaystyle \frac{dy}{dx} = y \frac{d}{dx}\left[ \ln(y) \right], and hence,

    \displaystyle \frac{dy}{dx} = x^y \left[\frac{dy}{dx}\ln(x) + \frac{y}{x} \right]

    which gives as a final result

    \displaystyle \frac{dy}{dx} = \frac{y x^{y-1}}{1 - x^{y}\ln(x)}
  9. jack.hadamard's Avatar
    69 02-07-2012  23:45
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    Re: A Summer of Maths
    I wonder how people on here find Analysis in general.

    Do you like it (assuming you have seen some), or do you think it is difficult to understand and apply?
  10. Blutooth's Avatar
    70 02-07-2012  23:57
    • Peer Of The TSR Realm
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    Re: A Summer of Maths
    (Original post by jack.hadamard)
    This one is a bit tricky, but I think it has something to teach; use hints if you get stuck.

    {*} Question:

    Let f : \mathbb{R} \to \mathbb{R}\setminus \{3\} be a function with the property that there exists \omega > 0 such that

    \displaystyle f(x + \omega) = \frac{f(x) - 5}{f(x) - 3}

    for all x \in \mathbb{R}.

    Prove that f is periodic.



    {**} Required: (A-level)

    Spoiler:
    Show

    A function is periodic, with period p, if for every x in its domain one has f(x) \equiv f(x + n p) where n is an integer.


    {***} Hints:

    Spoiler:
    Show

    (1)
    Spoiler:
    Show

    What about f(x) = 2?


    (2)
    Spoiler:
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    How does f(x + 2\omega) look like?

    Didn't seem that hard.
    Spoiler:
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    Let f(x)=F
    subbing x for x+w we obtain f(x+2w)=(2F-5)/(F-2)
    subbing x for x+w again we obtain f(x+3w)=(-3F+5)/(-F+1)
    subbing x for x+w again we obtain f(x+4w)=F=f(x)
    so clearly f(x)=f(x+4nw) for every n. Period is 4w.
    Thus we are done.
    Last edited by Blutooth; 03-07-2012 at 02:20.
  11. james22's Avatar
    71 02-07-2012  23:58
    • Peer Of The TSR Realm
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    Re: A Summer of Maths
    (Original post by jack.hadamard)
    Some thoughts under cruel assumptions.

    Spoiler:
    Show

    Without any justification, I claim that \displaystyle y = x^y.

    Maybe the inverse will be x = y^{y^{-1}}


    Now, note that \displaystyle \frac{dy}{dx} = y \frac{d}{dx}\left[ \ln(y) \right], and hence,

    \displaystyle \frac{dy}{dx} = x^y \left[\frac{dy}{dx}\ln(x) + \frac{y}{x} \right]

    which gives as a final result

    \displaystyle \frac{dy}{dx} = \frac{y x^{y-1}}{1 - x^{y}\ln(x)}
    You have the right inverse and are right in the simplification of the original equation. I cannot remember the derivative but your answer looks about right. Any luck with the values of x for which this converges?
  12. Lord of the Flies's Avatar
    72 02-07-2012  23:59
    • The foul fiend Flibbertigibbet
    • Location: Paris, France
    Re: A Summer of Maths
    (Original post by jack.hadamard)
    I wonder how people on here find Analysis in general.

    Do you like it (assuming you have seen some), or do you think it is difficult to understand and apply?
    I find analysis a bit boring to be honest. Perhaps because it is not my strong point by any means, I don't know. I prefer "even purer maths"... They seem to have more depth and involve more creativity.
  13. jack.hadamard's Avatar
    73 03-07-2012  01:44
    • Benevolent Member
    • Posts: 696
    Re: A Summer of Maths
    (Original post by Blutooth)
    ...
    Can you use a spoiler, please. Thanks.

    Spoiler:
    Show

    Can f(x) take the values 1 or 2?


    (Original post by james22)
    Any luck with the values of x for which this converges?
    I don't have a good reason to believe what I derived works, so I have only rough ideas.


    (Original post by Lord of the Flies)
    I find analysis a bit boring to be honest. I prefer "even purer maths"... They seem to have more depth and involve more creativity.
    I have studied a bit of it, and I find it interesting; some of Cauchy's stuff is amazing and I would say it is creative.
    However, I do get stuck from time to time -- I thought I understood uniform continuity two months ago, but now I have to read it again.
  14. Blutooth's Avatar
    74 03-07-2012  01:56
    • Peer Of The TSR Realm
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    Re: A Summer of Maths
    (Original post by jack.hadamard)
    Can you use a spoiler, please. Thanks.

    Spoiler:
    Show

    Can f(x) take the values 1 or 2?


    Spoiler:
    Show

    No, by considering the denominator of f(x+2w), f(x+3w) etc. orplugging f(x+w)=3 and working out the value of f(x).
    Last edited by Blutooth; 03-07-2012 at 02:11.
  15. Blutooth's Avatar
    75 03-07-2012  03:30
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    Re: A Summer of Maths
    (Original post by jack.hadamard)
    {*} 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}.

    Spoiler:
    Show

    \text{i)} x^4+1=(x^2+1)^2-2x^2= \newline ((x^2+1)-\sqrt{2}x)((x^2+1)^2+\sqrt{2}x) \text{ reduced as required.} \newline x^4+4y^4=(x^2+2y^2)^2-4(xy)^2=((x^2+2y^2)-2xy)((x^2+2y^2)+2xy) \newline \newline \text{ii) By the result above } 4k^4+1=(2k^2-2k+1)(2k^2+2k+1) \newline \text{Also note, } (2k^2+2k+1)-(2k^2-2k+1)=4k, \text{ hence} \newline \displaystyle \frac{4k}{4k^4+1}=\displaystyle \frac{1}{(2k^2-2k+1)} + \displaystyle \frac{-1}{(2k^2+2k+1)}=\displaystyle \frac{1}{2(k-\frac{1}{2})^2+\frac{1}{2}} - \displaystyle \frac{1}{2(k+\frac{1}{2})^2 + \frac{1}{2}} \newline \Rightarrow \sum_{1}^{n}\frac{4k}{4k^4+1}= \sum_{1}^{n} \left (\displaystyle\frac{1}{2(k-\frac{1}{2})^2 + \frac{1}{2}}-\displaystyle\frac{1}{2(k+\frac{ 1}{2})^2+\frac{1}{2}} \right )=\frac{1}{2(\frac{1}{2})^2 + \frac{1}{2}}-\frac{1}{2(n+\frac{1}{2})^2 + \frac{1}{2}}=1-\frac{1}{2n^2+2n+1}=\frac{2n^2+2 n}{2n^2+2n+1}
  16. Lord of the Flies's Avatar
    76 03-07-2012  07:30
    • The foul fiend Flibbertigibbet
    • Location: Paris, France
    Re: A Summer of Maths
    (Original post by james22)
    You have the right inverse and are right in the simplification of the original equation. I cannot remember the derivative but your answer looks about right. Any luck with the values of x for which this converges?
    Spoiler:
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    Domain and convergence

    \bullet\;\;x<0 clearly both f^{-1} and f are undefined.

    \bullet\;\;0\leq x\leq 1\Rightarrow 0\leq f^{-1}\leq 1 and f^{-1} strictly increases \Rightarrow f strictly increases so f is injective for 0\leq x\leq 1

    \bullet\;\;x>1 \Rightarrow 1<f^{-1}\leq \sqrt[e]{e} and f^{-1} strictly increases for 1<x<\sqrt[e]{e} and strictly decreases for x>\sqrt[e]{e}.

    Also, \displaystyle \lim_{+\infty}f^{-1}=1. Therefore f is undefined for 1<x< \sqrt[e]{e} since it would return exactly two values.

    Additionally, since the maximum of f^{-1} occurs only for x=\sqrt[e]{e}\Rightarrow f^{-1}(x)=\sqrt[e]{e} has exactly one solution so f is injective for x=\sqrt[e]{e} and f is undefined for all x>\sqrt[e]{e}

    Therefore convergence occurs when 0\leq x\leq 1\; or \;x=\sqrt[e]{e}


    If that is what you meant by convergence?
    Last edited by Lord of the Flies; 03-07-2012 at 09:04.
  17. Lord of the Flies's Avatar
    77 03-07-2012  09:35
    • The foul fiend Flibbertigibbet
    • Location: Paris, France
    Re: A Summer of Maths
    Here's an easy one!

    Question

    Evaluate \displaystyle\lim_{x\to0}\frac{ \sin x^n}{\sin^n x}\quad (n>0)

    Required

    Spoiler:
    Show

    L'Hôpital's rule
    Last edited by Lord of the Flies; 03-07-2012 at 09:42.
  18. james22's Avatar
    78 03-07-2012  10:52
    • Peer Of The TSR Realm
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    Re: A Summer of Maths
    (Original post by Lord of the Flies)
    Spoiler:
    Show

    Domain and convergence

    \bullet\;\;x<0 clearly both f^{-1} and f are undefined.

    \bullet\;\;0\leq x\leq 1\Rightarrow 0\leq f^{-1}\leq 1 and f^{-1} strictly increases \Rightarrow f strictly increases so f is injective for 0\leq x\leq 1

    \bullet\;\;x>1 \Rightarrow 1<f^{-1}\leq \sqrt[e]{e} and f^{-1} strictly increases for 1<x<\sqrt[e]{e} and strictly decreases for x>\sqrt[e]{e}.

    Also, \displaystyle \lim_{+\infty}f^{-1}=1. Therefore f is undefined for 1<x< \sqrt[e]{e} since it would return exactly two values.

    Additionally, since the maximum of f^{-1} occurs only for x=\sqrt[e]{e}\Rightarrow f^{-1}(x)=\sqrt[e]{e} has exactly one solution so f is injective for x=\sqrt[e]{e} and f is undefined for all x>\sqrt[e]{e}

    Therefore convergence occurs when 0\leq x\leq 1\; or \;x=\sqrt[e]{e}


    If that is what you meant by convergence?
    That is what I mean by convergence.

    You are almost right, but it converges for all 0<x<e^(1/e)
  19. Lord of the Flies's Avatar
    79 03-07-2012  11:06
    • The foul fiend Flibbertigibbet
    • Location: Paris, France
    Re: A Summer of Maths
    (Original post by james22)
    That is what I mean by convergence.

    You are almost right, but it converges for all 0<x<e^(1/e)
    Really? I don't see how... When 1&lt;x&lt;e^{1/e},\;f(x) returns two values, no?
  20. Brit_Miller's Avatar
    80 03-07-2012  11:19
    • Benevolent Member
    • Location: Bristol
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    Re: A Summer of Maths
    (Original post by Lord of the Flies)
    Here's an easy one!

    Question

    Evaluate \displaystyle\lim_{x\to0}\frac{ \sin x^n}{\sin^n x}\quad (n&gt;0)

    Required

    Spoiler:
    Show

    L'Hôpital's rule
    Spoiler:
    Show
    Is it undefined? I make lim x->0 f'(x)/g'(x) 0/0, so obviously g'(x)=0
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