The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

A Summer of Maths (ASoM) 2016

Scroll to see replies

Original post by Gregorius
It's the less than. When you have ∣z−a∣=b|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|z-a| = b you want the points
∣z−a∣=b′|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⇒∣∣z−3∣−5∣<1|(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+5|z-3| < 1 + \sqrt{5} and ∣z−3∣>5−1|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.
Original post by physicsmaths
Do the normal z=x+iy and look at the distances.


Posted from TSR Mobile


Yeah, I see now thanks.
Original post by Insight314
Interesting. That is what I thought exactly, but then ∣(z−3)−(−2+i)∣<1⇒∣∣z−3∣−5∣<1|(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+5|z-3| < 1 + \sqrt{5} and ∣z−3∣>5−1|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.


Posted from TSR Mobile
Original post by physicsmaths
Looks like triangle inequality to me.


Posted from TSR Mobile


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)∣≤∣∣z−3∣−5∣|(z-3) - (-2 + i)| \leq \Big| |z-3| - \sqrt{5} \Big|
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!
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)∣≤∣∣z−3∣−5∣|(z-3) - (-2 + i)| \leq \Big| |z-3| - \sqrt{5} \Big|


The triangle inequality with a difference is:
∣(z−3)−(−2+i)∣≥∣∣z−3∣−5∣|(z-3) - (-2 + i)| \geq \Big| |z-3| - \sqrt{5} \Big| , so
1>∣(z−3)−(−2+i)∣≥∣∣z−3∣−5∣1 > |(z-3) - (-2 + i)| \geq \Big| |z-3| - \sqrt{5} \Big|
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?
Original post by JackHKeynes
The triangle inequality with a difference is:
∣(z−3)−(−2+i)∣≥∣∣z−3∣−5∣|(z-3) - (-2 + i)| \geq \Big| |z-3| - \sqrt{5} \Big| , so
1>∣(z−3)−(−2+i)∣≥∣∣z−3∣−5∣1 > |(z-3) - (-2 + i)| \geq \Big| |z-3| - \sqrt{5} \Big|


Oh, whoops. That's my mistake haha. :biggrin:

Thanks m8!
V&M Example Sheet 1 - Question 5

How do I show that

sinz=∑n=0∞(−1)n(2n+1)!cos⁡(2n+1)θ+i∑n=0∞(−1)n(2n+1)!sin⁡(2n+1)θsin 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 sinz=2sin z = 2 ?

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

Tagging @Gregorius
(edited 7 years ago)
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.
Original post by Insight314
V&M Example Sheet 1 - Question 5

How do I show that

sinz=∑n=0∞(−1)n(2n+1)!cos⁡(2n+1)θ+i∑n=0∞(−1)n(2n+1)!sin⁡(2n+1)θsin 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 sinz=2sin z = 2 ?

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

Tagging @Gregorius


How about showing that

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

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

sin⁡(x+iy)=sin⁡xcosh⁡y+icos⁡xsinh⁡y \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?
Reply 152
Avatar for Zacken
Zacken
OP
22
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.
Reply 153
Avatar for Zacken
Zacken
OP
22
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. :tongue:
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.
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. :tongue:


I know you are talking about me. No problem, I will stop tagging people. :biggrin:
Reply 156
Avatar for M14B
M14B
18
Original post by physicsmaths
Ah ok. Very nyc.
Shudda thought this was undergrad type ****.



Posted from TSR Mobile


Absolutely a terrible method for this integral. Your method is the best
Original post by Gregorius
How about showing that

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

and working from there?


Sorry if I'm wrong but shouldn't it be sin(x)cosh(y) — icos(x)sinh(y)
Original post by Insight314
I know you are talking about me. No problem, I will stop tagging people. :biggrin:


Course he talking about you. U tag more people then there are in Cambridge. Lol


Posted from TSR Mobile
Original post by physicsmaths
Course he talking about you. U tag more people then there are in Cambridge. Lol


Posted from TSR Mobile


Lol.I actually only tagged one person there (Gregorius) since he knows his stuff. I can't see the problem of tagging one person, so not sure why he wanted to spank me for tagging only one person; I would understand if he told me off earlier when I tagged tons of people.

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you