A Summer of Maths (ASoM) 2016

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    (Original post by Insight314)
    Gregorius,

    This might be a very stupid question, and it probably is, but Edexcel FP2 doesn't really teach this properly so I couldn't get the intuition behind it.

    If S is the interior of the circle |z - (1 + i)| = 1 and if z \in S then is |z-(1+i)| < 1? Or is it > sign? How do I get intuition behind this, it makes more sense for it to be less than 1 since it is the interior of the circle, but then I don't get the necessary inequalities.

    Again, I am pretty embarrassed of this, so don't judge.
    It's the less than. When you have |z-a| = b you are looking at the set of points z that are a distance b from point a. So if you want the interior of |z-a| = b you want the points
    |z-a| = b' where b' is less than b.

    OT: what has happened to the TSR editor?!
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    (Original post by Gregorius)
    It's the less than. When you have |z-a| = b you are looking at the set of points z that are a distance b from point a. So if you want the interior of |z-a| = b you want the points
    |z-a| = b' where b' is less than b.

    OT: what has happened to the TSR editor?!
    Interesting. That is what I thought exactly, but then |(z-3) - (-2+i)| < 1 \Rightarrow \Big| |z-3| - \sqrt{5} \Big| < 1 is an incorrect implication, and this is how you get to the two inequalities |z-3| < 1 + \sqrt{5} and |z-3| > \sqrt{5} - 1 . (Question 1 of Example sheet 1 for V&M)

    Yeah, they made some changes to TSR; pretty annoying if you ask me.
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    (Original post by physicsmaths)
    Do the normal z=x+iy and look at the distances.


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    Yeah, I see now thanks.
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    (Original post by Insight314)
    Interesting. That is what I thought exactly, but then |(z-3) - (-2+i)| < 1 \Rightarrow \Big| |z-3| - \sqrt{5} \Big| < 1 is an incorrect implication, and this is how you get to the two inequalities |z-3| < 1 + \sqrt{5} and |z-3| > \sqrt{5} - 1 . (Question 1 of Example sheet 1 for V&M)

    Yeah, they made some changes to TSR; pretty annoying if you ask me.
    Looks like triangle inequality to me.


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    (Original post by physicsmaths)
    Looks like triangle inequality to me.


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    Of course it does. That is how I got to that implication, but then it requres the > sign since you get |(z-3) - (-2 + i)| \leq \Big| |z-3| - \sqrt{5} \Big|
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    (Original post by Gregorius)
    OT: what has happened to the TSR editor?!
    Assuming you have the same problem I do (weird text boxes and stuff), you can switch between WYSIWYG and BBcode in a button in the "Advanced" set of options in the editor
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    (Original post by Mathemagicien)
    Assuming you have the same problem I do (weird text boxes and stuff), you can switch between WYSIWYG and BBcode in a button in the "Advanced" set of options in the editor
    PRSOM That's worked!
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    (Original post by Insight314)
    Of course it does. That is how I got to that implication, but then it requres the > sign since you get |(z-3) - (-2 + i)| \leq \Big| |z-3| - \sqrt{5} \Big|
    The triangle inequality with a difference is:
    |(z-3) - (-2 + i)| \geq \Big| |z-3| - \sqrt{5} \Big| , so
    1 > |(z-3) - (-2 + i)| \geq \Big| |z-3| - \sqrt{5} \Big|
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    (Original post by Mathemagicien)
    Assuming you have the same problem I do (weird text boxes and stuff), you can switch between WYSIWYG and BBcode in a button in the "Advanced" set of options in the editor
    How is IA bro?
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    (Original post by JackHKeynes)
    The triangle inequality with a difference is:
    |(z-3) - (-2 + i)| \geq \Big| |z-3| - \sqrt{5} \Big| , so
    1 > |(z-3) - (-2 + i)| \geq \Big| |z-3| - \sqrt{5} \Big|
    Oh, whoops. That's my mistake haha.

    Thanks m8!
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    (Original post by Gregorius)
    PRSOM That's worked!
    Glad to have helped

    (Original post by Insight314)
    How is IA bro?
    Haven't started it quite yet (bro)... just got a few (quite urgent) things to sort out after my exams (including a slight cold), so I'll probably start on it tomorrow

    How did I ever get an offer with this incredibly lazy mentality, I wonder?
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    V&M Example Sheet 1 - Question 5

    How do I show that

    sin z = \displaystyle\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} \cos(2n+1)\theta + i \displaystyle\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} \sin(2n+1)\theta

    has infinitely many solutions for sin z = 2 ?

    Or is my initial assumption that  z \in \mathbb{C} incorrect? Or is my whole method of working through this problem wrong?

    Tagging Gregorius
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    Is anyone learning Analysis I here? I have begun reading the lecture notes but am in two minds between learning Analysis I or Vectors and Matrices. Essentially I want to do the one which I will most enjoy, and hopefully attempt some of the Example Sheets. Does anyone have any recommendations or any advice having done Analysis I?

    Thanks.
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    (Original post by Insight314)
    V&M Example Sheet 1 - Question 5

    How do I show that

    sin z = \displaystyle\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} \cos(2n+1)\theta + i \displaystyle\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} \sin(2n+1)\theta

    has infinitely many solutions for sin z = 2 ?

    Or is my initial assumption that  z \in \mathbb{C} incorrect? Or is my whole method of working through this problem wrong?

    Tagging Gregorius
    How about showing that

     \displaystyle \sin(x + iy) = \sin{x} \cosh{y} + i \cos{x}\sinh{y}

    and working from there?
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    (Original post by Gregorius)
    How about showing that

     \displaystyle \sin(x + iy) = \sin{x} \cosh{y} + i \cos{x}\sinh{y}

    and working from there?
    All right, gonna try that, after I finish eating though. However, is it possible to show that it has infinite solutions using my method?
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    (Original post by tridianprime)
    Is anyone learning Analysis I here? I have begun reading the lecture notes but am in two minds between learning Analysis I or Vectors and Matrices. Essentially I want to do the one which I will most enjoy, and hopefully attempt some of the Example Sheets. Does anyone have any recommendations or any advice having done Analysis I?

    Thanks.
    I wouldn't start with Analysis. Do some other course that helps build up your proving skills first before getting into that. So I'd recommend Numbers and Sets, but I suppose you could go for V&M.
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    On a seperate note, do users have to continually tag others in every post? Posting your question here is enough, somebody will notice and respond in due course. Not sure if I'm the only one that feels that way. Feel free to disagree.
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    (Original post by Zacken)
    I wouldn't start with Analysis. Do some other course that helps build up your proving skills first before getting into that. So I'd recommend Numbers and Sets, but I suppose you could go for V&M.
    I'll give N&S a look.
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    (Original post by Zacken)
    On a seperate note, do users have to continually tag others in every post? Posting your question here is enough, somebody will notice and respond in due course. Not sure if I'm the only one that feels that way. Feel free to disagree.
    I know you are talking about me. No problem, I will stop tagging people.
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    (Original post by physicsmaths)
    Ah ok. Very nyc.
    Shudda thought this was undergrad type ****.



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    Absolutely a terrible method for this integral. Your method is the best
 
 
 
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