A Summer of Maths (ASoM) 2016

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Reply 580

newblood

Original post by Gregorius

The fewer people who do stats, the higher my employment market value is...

:mwuaha:

What is your line of work, if you dont mind me asking

Reply 581

drandy76

Original post by Mathemagicien

Which is why I am aiming to specialise in statistics.

Now delete ur posts plz, before other students get this idea :mad:

Too late

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Reply 582

Gregorius

Original post by newblood

What is your line of work, if you dont mind me asking

I work in cancer research, particularly on the epidemiological side. Predictive survival models are my speciality.

Reply 583

Gregorius

Original post by shamika

Even in academia, I'm guessing even the likes of Tibshirani or Efron aren't masters of the entire field (they're the two I know today who are alive and eminent, not sure if they're considered the "most eminent" today).

They're still considered near the top of the tree!

Reply 584

shamika

Original post by Gregorius

They're still considered near the top of the tree!

That was me caveating in case there are some hotshots I don't know about since I finished my degree (wow that's made me feel old). Out of curiosity, who would you consider to be at the very top (who is still alive)?

Reply 585

Gregorius

Original post by shamika

I've been pondering this question for a while...it's really hard to answer! I could probably list out twenty or thirty ststisticians/probabilists who have done vital work in defining how things are done these days - but it's much harder to pick out, say, four or five of them as standing above the rest.

Perhaps it's a reflection of the state of evolution of the subject (which is not so old, after all); the foundations having been laid, much of the work now is bricklaying.

Reply 586

A Slice of Pi

Original post by A Slice of Pi

If anyone is still in the mood for some good old-fashioned pure mathematics then here's a question that might benefit a lot of first year students.

Given that there exists a bijection

f : X \mapsto Y

between sets

[br]X

and

Y

and that

x \in X

and

y \in Y

, prove that there must always exist a bijection

g : X \mapsto Y

such that

g(x) = y

Still no takers? I didn't think it was such an unpopular topic

Reply 587

username1162972

Original post by A Slice of Pi

I tried to find the official solution after I uploaded mine, but couldn't. One I came across used the prime factorisation method and this was quite a long way of doing it. In my method, I used two different expressions for the RHS and one showed that the expression was even, and the conclusion followed quickly from that. I haven't seen anyone else use that method.

Sorry just seen the topic now.
I have the official solution if you want it.

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Reply 588

InOrbit

Original post by A Slice of Pi

Still no takers? I didn't think it was such an unpopular topic

Maybe something like this:

Consider

x,f^{-1}(y) \in X

and

f(x),y \in Y

, which exist as

f: X \to Y

is a bijection.

Then the function

h: X \setminus \{x \cap f^{-1}(y)\} \to Y \setminus \{f(x) \cap y\}

, defined by

h: z \mapsto f(z)

,
with

z \in X \setminus \{x \cap f^{-1}(y)\}

, is a bijection.

Then the function

g: X \to Y

, defined by

Unparseable latex formula:

g: z \mapsto \begin{cases}[br]f(z) & z \in X \setminus \{x \cap f^{-1}(y)\} \\[br]f(x) & z = x\\[br]y & z = f^{-1}(y)[br]\end{cases}[br]

is a bijection.

(edited 7 years ago)

Reply 589

A Slice of Pi

Original post by Alex:

Maybe something like this:

Consider

x,f^{-1}(y) \in X

and

f(x),y \in Y

, which exist as

f: X \to Y

is a bijection.

Then the function

h: X \setminus \{x \cap f^{-1}(y)\} \to Y \setminus \{f(x) \cap y\}

, defined by

h: z \mapsto f(z)

z \in X \setminus \{x \cap f^{-1}(y)\}

,
is a bijection.

Then the function

g: X \to Y

, defined by

g: z \mapsto f(z)

z \in X \setminus \{x \cap f^{-1}(y)\}

\hphantom{g}: x \mapsto y

\hphantom{g}: f^{-1}(y) \mapsto f(x)

,
is a bijection.

That's a nice way of doing it. This was the method I used:

There are two cases.
1.) Suppose that

f(x) = y

, then

g = f

.
2.) Suppose instead that

f(t) = y

, where

t \neq x

. Then we could define a new function

g

as follows

Unparseable latex formula:

g(i) = \begin{cases}[br]f(t) \hspace{14pt} i = x \\[br]f(x) \hspace{11.5pt} i = t \\[br]f(i) \hspace{14pt} i \neq x \wedge i \neq t \\[br]\end{cases}

From here it's a fairly simple exercise to show that

g

is bijective, since

f

is bijective.

(edited 7 years ago)

Reply 590

ThatPerson

Original post by shamika

Why do people still suggest Rudin anyway? As Alex says, it's*awful.

I'm curious since I just started looking through Rudin today. What do you find so bad about it? (I'm going into 2nd year, so this isn't my first exposure to analysis).

Reply 591

InOrbit

Original post by ThatPerson

I'm curious since I just started looking through Rudin today. What do you find so bad about it? (I'm going into 2nd year, so this isn't my first exposure to analysis).

It's unnecessarily terse and dense. If you're someone who hasn't had exposure to real analysis before, then it's an incredibly time inefficient way to learn, as you'll be stuck there on the number of exercises left to the reader.

He also starts off with metric spaces for sequences and continuity, whilst hopping between metric spaces, the real line and the complex plane, and then goes back to the real numbers for differentiation. If you're trying to learn REAL analysis, this is incredibly unhelpful. Save complex numbers for complex analysis, and metric spaces for a point-set topology course dedicated to them.

His chapter on differential forms and chains rather is out of place and I honestly have no idea what it is on about, and the measure theory chapter is not great either.

Something like Pugh's real mathematical analysis is probably just as rigorous, but much more informative and gives more motivation.

Reply 592

ThatPerson

Original post by Alex:

Thanks for the explanation. I looked into Pugh, but solutions to the problems aren't easily available, so I don't think it's suitable for my purposes.

Reply 593

A Slice of Pi

A quick-ish brain teaser:

Following his latest defeat against a tortoise, Achilles has 11 weeks to prepare for his next race. He decides to train for at least one hour every day, but for no more than 12 hours per week. Show that there exists a succession of days during which Achilles will have trained for exactly 21 hours.

Reply 594

A Slice of Pi

Original post by Mathemagicien

I think I've got it...

Spoiler

Yes

Reply 595

Concav

Is tautology and contradiction taught in y1 uni or is it something you are naturally expected to know?

Reply 596

Zacken

Original post by Concav

Is tautology and contradiction taught in y1 uni or is it something you are naturally expected to know?

It's taught quickly in the first few N&S lectures

Reply 597

Enigmatically

Just as an alternative for you guys if you want to use it, I have stumbled upon another way of integrating by parts/evaluating indefinite integrals.
Might not be always as convenient as the normal means, but here goes. Good luck at Uni :wink: