ATAR Notes: Forum

Uni Stuff => Science => Faculties => Mathematics => Topic started by: QuantumJG on March 01, 2010, 08:03:32 pm

Title: Real Analysis
Post by: QuantumJG on March 01, 2010, 08:03:32 pm: Ok I have hit a rather embarrassing but confusing question.

What is x²?

Actually according to wikipedia,

$x^{2} = \sum _{n = 0} ^{x - 1} (2n + 1)$
Title: Re: Real Analysis
Post by: QuantumJG on March 01, 2010, 08:12:06 pm: Ok is it $x \times x$ ?
Title: Re: Real Analysis
Post by: QuantumJG on March 01, 2010, 09:04:26 pm: Another question asks me to define e^x

$e^{x} = 1 + \dfrac{x}{1!} + \dfrac{x^{2} }{2!} + \dfrac{x^{3} }{3!} + ... + \dfrac{x^{n} }{n!} , where n \in \mathbb{Z} ^{+}$

$\therefore e^{x} = \sum _{n = 0} ^{ \infty } \dfrac{x^{n} }{ n! }$

$where n! = n \times (n-1) \times (n-2) \times ... \times 2 \times 1 = 1 \times 2 \times ... \times (n-2) \times (n-1) \times n$

$\therefore n! = \prod _{k=1} ^{n} k , where k \in \mathbb{Z} \le n$

Is this definition ok for e^x?
Title: Re: Real Analysis
Post by: TrueTears on March 01, 2010, 09:10:19 pm: well the way you have defined e^x is centered at 0, which is a Maclaurin series, it could be centered around other numbers. You could also add that the radius of convergence is infinity, so the series converges for all x.

Maybe you could also use taylor's inequality and prove that e^x is indeed equal to this specific taylor expansion. Which will prove that your definition is correct.
Title: Re: Real Analysis
Post by: QuantumJG on March 01, 2010, 09:14:49 pm: We haven't looked at that stuff yet! :P
Title: Re: Real Analysis
Post by: TrueTears on March 01, 2010, 09:15:06 pm: Oh, it's fun as stuff, omg you will love it when you get to it !!

it's kinda like epsilon - delta proofs, i love it!!

but if you are not up to it, do u know WHY e^x can be expanded like that? I think that is important to know, maybe you can try prove it :P
Title: Re: Real Analysis
Post by: QuantumJG on March 01, 2010, 09:39:28 pm: Quote from: TrueTears on March 01, 2010, 09:15:06 pm
Oh, it's fun as stuff, omg you will love it when you get to it !!

it's kinda like epsilon - delta proofs, i love it!!

but if you are not up to it, do u know WHY e^x can be expanded like that? I think that is important to know, maybe you can try prove it :P

Our professor didn't discuss why, but I remember in physics our professor was saying it's from the Taylor series, is that true? Then again he didn't really explain what a Taylor series is.
Title: Re: Real Analysis
Post by: TrueTears on March 01, 2010, 09:43:07 pm: if f has a power series expansion: $f(x) = \sum_{n=0}^{\infty} \frac{f^n(a)}{n!} (x-a)^n$ then the maclaruin series is just a special case of it: namely, centered at 0. (a = 0)

but that's not the slick part, the coolest part is the relationship between the coefficients of the power series f(x) and the derivatives of f(x), thats the part i love the most!!
Title: Re: Real Analysis
Post by: humph on March 01, 2010, 11:12:21 pm: Wait until you get to Laurent series in complex analysis and its relation to residues and contour integrals. That's pretty cool ;)
Title: Re: Real Analysis
Post by: /0 on March 02, 2010, 01:19:11 am: Quote from: QuantumJG on March 01, 2010, 08:12:06 pm
Ok is it $x \times x$ ?

Isn't it just a matter of definition?
Title: Re: Real Analysis
Post by: QuantumJG on March 02, 2010, 03:01:19 pm: Quote from: TrueTears on March 01, 2010, 09:43:07 pm
if f has a power series expansion: $f(x) = \sum_{n=0}^{\infty} \frac{f^n(a)}{n!} (x-a)^{n}$ then the maclaruin series is just a special case of it: namely, centered at 0. (a = 0)

but that's not the slick part, the coolest part is the relationship between the coefficients of the power series f(x) and the derivatives of f(x), thats the part i love the most!!

So what this is impliying that if,

$f(x) = e^{x}$

$then, f(a=0) = e^{0} = 1$

So

$f(x) = \sum_{n=0}^{\infty} \frac{f^n(a)}{n!} (x-a)^{n}$ becomes $f(x) = \sum_{n=0}^{\infty} \frac{ x^{n} }{n!}$
Title: Re: Real Analysis
Post by: TrueTears on March 02, 2010, 09:33:42 pm: Quote from: QuantumJG on March 02, 2010, 03:01:19 pm
Quote from: TrueTears on March 01, 2010, 09:43:07 pm
if f has a power series expansion: $f(x) = \sum_{n=0}^{\infty} \frac{f^n(a)}{n!} (x-a)^{n}$ then the maclaruin series is just a special case of it: namely, centered at 0. (a = 0)

but that's not the slick part, the coolest part is the relationship between the coefficients of the power series f(x) and the derivatives of f(x), thats the part i love the most!!

So what this is impliying that if,

$f(x) = e^{x}$

$then, f(a=0) = e^{0} = 1$

So

$f(x) = \sum_{n=0}^{\infty} \frac{f^n(a)}{n!} (x-a)^{n}$ becomes $f(x) = \sum_{n=0}^{\infty} \frac{ x^{n} }{n!}$

don't you mean when f(x) is centered around 0? not f(a=0)?

it's not implying anything... just that the expansion is "centered" around 0 so that the power series expansion provides a good approximation around 0 xD
Title: Re: Real Analysis
Post by: QuantumJG on March 06, 2010, 09:12:53 pm: Prove that:

$tan ^{-1} x = sin ^{-1} ( \dfrac{x}{ \sqrt{ 1 + x^{2} } } )$

What would be a good way to do this proof?
Title: Re: Real Analysis
Post by: TrueTears on March 06, 2010, 09:28:11 pm: $\tan\left(\sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)\right)$

$\sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)$ Right angle triangle, opposite is $x$ , hypotenuse is $\sqrt{1+x^2}$

Thus tan of this angle is $x$ .

Thus $\tan\left(\sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)\right) = x$
Title: Re: Real Analysis
Post by: QuantumJG on March 11, 2010, 01:54:43 pm: What is the difference between $\mathbb{Q} [x]$ and $\mathbb{Q} [[x]]$ ?

Also when you are asked:

write sin(x) as an element of $\mathbb{Q} [[x]]$ is that just:

sin(x) = $x - \dfrac{x ^{3} }{3!} + \dfrac{x ^{5} }{5!} - \dfrac{x ^{7} }{7!}$

Also what is: $\mathbb{Q} (x)$ & $\mathbb{Q} ((x))$ ?
Title: Re: Real Analysis
Post by: QuantumJG on March 11, 2010, 07:22:15 pm: What is the difference between say:

Find the Taylor expansion of e^x at x = 0 & Find the Taylor series for sinx at the point a = $\dfrac{ \pi }{4}$ ?

Help my professor hasn't exactly explained what the Taylor series is.
Title: Re: Real Analysis
Post by: /0 on March 11, 2010, 07:26:35 pm: Basically it's a polynomial approximation to a function. The more terms it has the more accurate it is.

http://en.wikipedia.org/wiki/Taylor_expansion
Wikipedia has a nice gif of a taylor expansion centred at 0.

If you centre your series at x = 0 then it will be most accurate at x = 0, and as it departs from that point it will start to diverge. Likewise, if you centre at $\frac{\pi}{4}$ , it will be very accurate there, but will diverge as you go out.

If you have a taylor series with an infinite number of terms, it doesn't matter where you centre it, since it will be exactly equal to the function within a reasonable domain. However if you want the infinite term expansion then its best to centre it at x = 0 otherwise you will be expanding an infinite number of brackets of the form $(x-\frac{\pi}{4})^n$
Title: Re: Real Analysis
Post by: QuantumJG on March 12, 2010, 06:41:42 am: But what is the difference of a and x?
Title: Re: Real Analysis
Post by: TrueTears on March 12, 2010, 12:46:24 pm: a is where the function is centered at.
Title: Re: Real Analysis
Post by: QuantumJG on March 12, 2010, 08:23:23 pm: Let x $\in \mathbb{R}$ . Graph the sequence:

$a_{n} = (1 + \dfrac{x}{n} ) ^{n}$ .

How do I graph this?
Title: Re: Real Analysis
Post by: QuantumJG on March 13, 2010, 03:10:57 pm: Find the sum of the series:

$\sum ^{ \infty} _{n = 1} \dfrac{ x^{n} }{n}$

Can someone tell me what this is asking?
Title: Re: Real Analysis
Post by: TrueTears on March 13, 2010, 03:15:47 pm: Quote from: QuantumJG on March 13, 2010, 03:10:57 pm
Find the sum of the series:

$\sum ^{ \infty} _{n = 1} \dfrac{ x^{n} }{n}$

Can someone tell me what this is asking?
first of all can you prove its converging series?

if so can u find the infinite sum?

basically what i'm saying is:

if $s_n = \sum ^{ \infty} _{n = 1} \dfrac{ x^{n} }{n}$

then what value does $s_n$ converge to? [assuming the series $s_n$ does converge]
Title: Re: Real Analysis
Post by: QuantumJG on March 13, 2010, 04:30:28 pm: Ok what if I give an example like this:

$\sum ^{ \infty} _{n = 1} \dfrac{n}{ 3 ^{n-1} }$

Approaches 2.25 as n $\rightarrow \infty$ is this right?
Title: Re: Real Analysis
Post by: TrueTears on March 13, 2010, 04:40:35 pm: yeah, basically if you let $s_n = \sum ^{ \infty} _{n = 1} \dfrac{n}{ 3 ^{n-1} }$

then you are forming a new series which the elements are the partial sums of the original sequence.

if the $s_n$ sequence is converging then the value it converges to is the sum of your original sequence :)
Title: Re: Real Analysis
Post by: Pappa-Bohr on March 13, 2010, 04:49:55 pm: Quote from: QuantumJG on March 13, 2010, 04:30:28 pm
Ok what if I give an example like this:

$\sum ^{ \infty} _{n = 1} \dfrac{n}{ 3 ^{n-1} }$

Approaches 2.25 as n $\rightarrow \infty$ is this right?

I understand how you got it, but did you get 2.25 with a calculator?? I mean how are we expected to do these ugly type of q's without a calculator?
Title: Re: Real Analysis
Post by: Pappa-Bohr on March 13, 2010, 05:00:14 pm: BTW Quantum have you done question 10 from in 'Taylor series' section, of the 'expressions' sheet?
Did you use a proof by induction?
Title: Re: Real Analysis
Post by: QuantumJG on March 13, 2010, 06:23:08 pm: Quote from: Pappa-Bohr on March 13, 2010, 04:49:55 pm
Quote from: QuantumJG on March 13, 2010, 04:30:28 pm
Ok what if I give an example like this:

$\sum ^{ \infty} _{n = 1} \dfrac{n}{ 3 ^{n-1} }$

Approaches 2.25 as n $\rightarrow \infty$ is this right?

I understand how you got it, but did you get 2.25 with a calculator?? I mean how are we expected to do these ugly type of q's without a calculator?

So you are in for some fun with real analysis, at the moment I f****en hate the subject and want a professor that actually teaches you how to solve Taylor series problems or even explain what it is!

Yeah I got that value by calculator.

As for question 10 I used induction but wasn't happy with my proof, but hey meh.

This guy is not teaching us the fundementals and I hate it!
Title: Re: Real Analysis
Post by: TrueTears on March 13, 2010, 06:26:22 pm: Quote from: QuantumJG on March 13, 2010, 06:23:08 pm
Quote from: Pappa-Bohr on March 13, 2010, 04:49:55 pm
Quote from: QuantumJG on March 13, 2010, 04:30:28 pm
Ok what if I give an example like this:

$\sum ^{ \infty} _{n = 1} \dfrac{n}{ 3 ^{n-1} }$

Approaches 2.25 as n $\rightarrow \infty$ is this right?

I understand how you got it, but did you get 2.25 with a calculator?? I mean how are we expected to do these ugly type of q's without a calculator?

So you are in for some fun with real analysis, at the moment I f****en hate the subject and want a professor that actually teaches you how to solve Taylor series problems or even explain what it is!

Yeah I got that value by calculator.

As for question 10 I used induction but wasn't happy with my proof, but hey meh.

This guy is not teaching us the fundementals and I hate it!
You shouldn't be doing this if you don't know what a taylor series is lol, better read up quickly!
Title: Re: Real Analysis
Post by: /0 on March 13, 2010, 06:28:28 pm: Quantum there's a good primer in Stewart Calculus if you've got it
Title: Re: Real Analysis
Post by: TrueTears on March 13, 2010, 06:40:33 pm: yeah the chapter in stewarts on sequence and series is very well set out.
Title: Re: Real Analysis
Post by: Pappa-Bohr on March 13, 2010, 07:01:51 pm: Quote from: QuantumJG on March 13, 2010, 06:23:08 pm

So you are in for some fun with real analysis, at the moment I f****en hate the subject and want a professor that actually teaches you how to solve Taylor series problems or even explain what it is!

Yeah I got that value by calculator.

As for question 10 I used induction but wasn't happy with my proof, but hey meh.

This guy is not teaching us the fundementals and I hate it!

Yea it's a bit weird how he doesn't teach us the actual concepts...., I mean he just does random questions from the problem sheet, his website 'notes' are just a mess as-well lol..

I've read a bit of (Baby) Rudin and Spivak to help elucidate the concepts, and that's helped a little bit, but I hate self-learning mathematics from textbooks.

The other thing I find really bizarre is the length of this assignment - almost 80 questions long!, when the first assignment from last year only had 8 questions...
Title: Re: Real Analysis
Post by: Cthulhu on March 13, 2010, 07:27:25 pm: Surely you covered taylor series and series in general last year?
Title: Re: Real Analysis
Post by: QuantumJG on March 13, 2010, 07:29:40 pm: Quote from: Cthulhu on March 13, 2010, 07:27:25 pm
Surely you covered taylor series and series in general last year?

No it wasn't covered!
Title: Re: Real Analysis
Post by: QuantumJG on March 13, 2010, 08:58:09 pm: Pappa-Bohr can you back me up that we really haven't covered Taylor series.
Title: Re: Real Analysis
Post by: TrueTears on March 13, 2010, 08:59:07 pm: your lecturer is an idiot if you haven't covered taylor series amd he/she expects you to do these questions...
Title: Re: Real Analysis
Post by: QuantumJG on March 13, 2010, 09:13:39 pm: Ok I was looking through the lecture outline and apparently we go into more detail further down the track.
Title: Re: Real Analysis
Post by: Pappa-Bohr on March 13, 2010, 09:47:59 pm: Quote from: TrueTears on March 13, 2010, 08:59:07 pm
your lecturer is an idiot if you haven't covered taylor series amd he/she expects you to do these questions...

We've covered examples of the questions, so we can most them, but we haven't actually covered the theory or purpose behind them.
I mean it's like we're still waiting for the proper theory lectures to start, and up and till now we've been doing warm up questions or something..

It's a very oddly structured subject.

but hey we have a different lecturer for the next few weeks so maybe it will get better.
Title: Re: Real Analysis
Post by: QuantumJG on March 13, 2010, 10:52:42 pm: So have you been able to do q70?

He hasn't shown us how to find the sum of a series.
Title: Re: Real Analysis
Post by: Pappa-Bohr on March 13, 2010, 11:09:37 pm: i just did it on calculator and got 0.346574
Title: Re: Real Analysis
Post by: QuantumJG on March 14, 2010, 12:15:06 am: Yeah but how can we get that without a calculator?
Title: Re: Real Analysis
Post by: Pappa-Bohr on March 14, 2010, 01:22:17 pm: i dont know.
post it and see if anyone else can do it?

also did you do the q 60 in the Taylor series section?
Title: Re: Real Analysis
Post by: QuantumJG on March 14, 2010, 07:57:23 pm: Quote from: Pappa-Bohr on March 14, 2010, 01:22:17 pm
i dont know.
post it and see if anyone else can do it?

also did you do the q 60 in the Taylor series section?

I did question 60 and I'll give you a hint:

Find the series expansion of e^x in powers of 'x+3'.

As for question 70, I'm going to just leave it.
Title: Re: Real Analysis
Post by: Mao on March 15, 2010, 12:55:12 am: Did you do calculus last year? Taylor series aka power series aka power expansion (etc etc) is one of the first things they teach you in single-variable calculus at uni, along with convergence and divergence of series, it was even covered in enhancement.
Title: Re: Real Analysis
Post by: QuantumJG on March 25, 2010, 03:52:59 pm: This question is really bugging me!

Evaluate:

$\lim_{n \rightarrow \infty} (1 + \dfrac{1}{n} )^{n}$

Graphing it on my graphics calculator shows it approaches e if you don't exceed 10¹⁰, but it approaches 1 if you go past 10¹⁴.

WTF!!!!
Title: Re: Real Analysis
Post by: TrueTears on March 25, 2010, 04:20:34 pm: Isn't that another definition for e?
Title: Re: Real Analysis
Post by: Cthulhu on March 25, 2010, 04:30:32 pm: Mathematica returns e
Title: Re: Real Analysis
Post by: QuantumJG on March 25, 2010, 04:47:28 pm: Thanks guys.
Title: Re: Real Analysis
Post by: Mao on March 27, 2010, 09:58:48 pm: Quote from: QuantumJG on March 25, 2010, 03:52:59 pm
This question is really bugging me!

Evaluate:

$\lim_{n \rightarrow \infty} (1 + \dfrac{1}{n} )^{n}$

Graphing it on my graphics calculator shows it approaches e if you don't exceed 10¹⁰, but it approaches 1 if you go past 10¹⁴.

WTF!!!!

Yes, it's called truncation error. Your calculator can only store so many digits during calculations, any computations beyond that is unreliable.
Title: Re: Real Analysis
Post by: QuantumJG on March 28, 2010, 02:20:44 pm: Hey guys I'm doing sequence analysis and I just want to have some terms cleared up:

Bounded Sequence:

My understanding of this is:

A sequence is bounded if there exists a $U, L \in mathbb{R}$

such that:

$a_{n} \ge L$

$a_{n} \le U$

Increasing or Decreasing Sequence:

My understanding is that:

A sequence is increasing if:

a_n+1 > a_n

A sequence is decreasing if:

a_n+1 < a_n

But what does Cauchy, sup, inf, lim inf, lim sup and Contractive mean?
Title: Re: Real Analysis
Post by: QuantumJG on March 28, 2010, 04:10:23 pm: In a website showing a cauchy example:

this was a part of the example:

|a_m - a_n| < |a_m| + |a_n|

is that from the triangle inequality?
Title: Re: Real Analysis
Post by: /0 on March 28, 2010, 05:12:26 pm: We learnt the cauchy sequence as:

Let $(x_n)_{n=1}^\infty \subset X$ where $(X,d)$ is a metric space. Then $(x_n)$ is a Cauchy sequence if for every $\varepsilon > 0$ , there exists an integer N such that

$m, n \geq N \Rightarrow d(x_m,x_n)<\varepsilon$

i.e. $d(x_m,x_n) \to 0$ as $m,n \to \infty$

The terms of a sequence become very 'close' to each other after a certain N.

'sup' is short for 'supremum' or 'least upper bound'. In a partially ordered set $X$ ,

If there is an $\alpha$ such that $\forall y \in X,\alpha \geq y$ and any smaller $\alpha$ would not have this property, then $\alpha=\mbox{sup}X$

e.g. In a sequence $(x_n)_{n =1}^\infty= 1-\frac{1}{n}$ , the supremum would be 1.

The 'infinum' or 'greatest lower bound' is the same thing but from the bottom.

'lim inf' is the least of the limits of the subsequences of $(x_n)_{n \geq 1}$

So say you have $(\mathbf{x}_n)_{n=1}^\infty \subset \mathbb{R}^n$ and there is the limit $\mathbf{x_n} \to \mathbf{a}$ where $\mathbf{a} = (a_1,a_2,...,a_n)$ . Then the lim inf will be the least such $a_i$ .

Opposite for lim sup.

Also, $|a_m-a_n| \leq |a_m|+|a_n|$

is also from the triangle inequality: $|a_m-a_n| \leq |a_m|+|-a_n| = |a_m|+|a_n|$

Dunno what's contractive
Title: Re: Real Analysis
Post by: QuantumJG on March 28, 2010, 05:23:32 pm: Quote from: /0 on March 28, 2010, 05:12:26 pm
We learnt the cauchy sequence as:

Let $(x_n)_{n=1}^\infty \subset X$ where $(X,d)$ is a metric space. Then $(x_n)$ is a Cauchy sequence if for every $\varepsilon > 0$ , there exists an integer N such that

$m, n \geq N \Rightarrow d(x_m,x_n)<\varepsilon$

i.e. $d(x_m,x_n) \to 0$ as $m,n \to \infty$

The terms of a sequence become very 'close' to each other after a certain N.

'sup' is short for 'supremum' or 'least upper bound'. In a partially ordered set $X$ ,

If there is an $\alpha$ such that $\forall y \in X,\alpha \geq y$ and any smaller $\alpha$ would not have this property, then $\alpha=\mbox{sup}X$

e.g. In a sequence $(x_n)_{n =1}^\infty= 1-\frac{1}{n}$ , the supremum would be 1.

The 'infinum' or 'greatest lower bound' is the same thing but from the bottom.

'lim inf' is the least of the limits of the subsequences of $(x_n)_{n \geq 1}$

So say you have $(\mathbf{x}_n)_{n=1}^\infty \subset \mathbb{R}^n$ and there is the limit $\mathbf{x_n} \to \mathbf{a}$ where $\mathbf{a} = (a_1,a_2,...,a_n)$ . Then the lim inf will be the least such $a_i$ .

Opposite for lim sup.

Also, $|a_m-a_n| \leq |a_m|+|a_n|$

is also from the triangle inequality: $|a_m-a_n| \leq |a_m|+|-a_n| = |a_m|+|a_n|$

Dunno what's contractive

Cheers.

Ok so I know that Cauchy is when the distance between 2 points as n gets larger and larger, becomes smaller and smaller but how could I prove that something isn't Cauchy?
Title: Re: Real Analysis
Post by: /0 on March 28, 2010, 06:51:12 pm: Show that $|x_m-x_n|$ is not necessarily smaller than an epsilon for m,n > N.

e.g. $x_n = \sqrt{n}$

Then $|x_m-x_n| = \sqrt{m}-\sqrt{n}$ , so for any N you can choose $n= N$ and $m> N$

Then $|x_m-x_n|$ can be as large as you want, so the sequence can't be Cauchy
Title: Re: Real Analysis
Post by: QuantumJG on March 28, 2010, 07:27:44 pm: Another question,

Am I right by saying that if a sequence is non-Cauchy, then it's divergent?
Title: Re: Real Analysis
Post by: Pappa-Bohr on March 28, 2010, 07:29:28 pm: quantum how are you finding assignment 2?
Title: Re: Real Analysis
Post by: QuantumJG on March 28, 2010, 07:31:47 pm: Quote from: Pappa-Bohr on March 28, 2010, 07:29:28 pm
quantum how are you finding assignment 2?

f$&@ing hard!

How about you?
Title: Re: Real Analysis
Post by: /0 on March 28, 2010, 07:57:06 pm: Quote from: QuantumJG on March 28, 2010, 07:27:44 pm
Another question,

Am I right by saying that if a sequence is non-Cauchy, then it's divergent?

Hmm, I think so:

Theorem 8.1.3: A sequence in $\mathbb{R}^n$ converges (to a limit in $\mathbb{R}^n$ ) iff it is Cauchy.

(the proof is nearly 2 pages)

I'm also battling with my second analysis assignment,,, due tuesday :o
I spent a few days trying to understand what all the necessary terminology and theorems were... stupid topologist's sine curve...
Title: Re: Real Analysis
Post by: Pappa-Bohr on March 28, 2010, 08:08:47 pm: yea i haven't been able to make much headway (and the terrible structure of our lectures doesn't help at all)
Title: Re: Real Analysis
Post by: Pappa-Bohr on March 28, 2010, 08:21:57 pm: Quote from: /0 on March 28, 2010, 06:51:12 pm
Then $|x_m-x_n| = \sqrt{m}-\sqrt{n}$

how do you get this? shouldn't there be a 'less than or equal to' sign instead of the equality? (i.e. the triangle inequality)
Title: Re: Real Analysis
Post by: QuantumJG on March 28, 2010, 09:03:56 pm: Quote from: Pappa-Bohr on March 28, 2010, 08:08:47 pm
yea i haven't been able to make much headway (and the terrible structure of our lectures doesn't help at all)

What are you up to?

I'm just starting analysing sequences.
Title: Re: Real Analysis
Post by: /0 on March 28, 2010, 11:44:43 pm: Quote from: Pappa-Bohr on March 28, 2010, 08:21:57 pm

Quote from: /0 on March 28, 2010, 06:51:12 pm
Then $|x_m-x_n| = \sqrt{m}-\sqrt{n}$

how do you get this? shouldn't there be a 'less than or equal to' sign instead of the equality? (i.e. the triangle inequality)

I don't know if you're learning about metric spaces, but in general, a metric $d(x,y)$ is similar to the notion of 'distance' between x and y.

If we use the standard euclidean metric $d(x,y) = |\mathbf{x}-\mathbf{y}|$ . This essentially is the 'distance' between two numbers, and it is what is normally used in sequences.

Thus we wish for a cauchy that for $m,n \geq N$ , $d(x_m,x_n) =|x_m-x_n|\leq \varepsilon$ :

For $x_n = \sqrt{n}$

Then $d(x_m,x_n) = |x_m - x_n| = |\sqrt{m}-\sqrt{n}| = \sqrt{m}-\sqrt{n}$ (assuming of course $m\geq n$ ).

And since we can set $n = N$ , $m > N$ etc etc we see that it is not cauchy
Title: Re: Real Analysis
Post by: QuantumJG on March 30, 2010, 06:49:27 am: Thanks /0. In a tute yesterday I was told what those things meant.
Title: Re: Real Analysis
Post by: kamil9876 on March 30, 2010, 09:25:49 pm: Quote from: /0 on March 28, 2010, 07:57:06 pm
Quote from: QuantumJG on March 28, 2010, 07:27:44 pm
Another question,

Am I right by saying that if a sequence is non-Cauchy, then it's divergent?

Hmm, I think so:

Theorem 8.1.3: A sequence in $\mathbb{R}^n$ converges (to a limit in $\mathbb{R}^n$ ) iff it is Cauchy.

(the proof is nearly 2 pages)

The proof of that can be made simple to the point that you can recreate it yourself, if you just split it into little pieces and treat it as a sequence of little lemmas, e.g: prove for $\mathbb{R}$ first and have $\mathbb{R}^n$ as just a corollary.

I'm gonna make this conjecture now that this thread got me thinking(I've been in the mood of making conjectures lately, so far half are true): Suppose $M_1, M_2... M_n$ are complete metric spaces, then the metric space $M_1 \times M_2 \times M_3...$ with metric $d(a,b)=\sqrt{d_1(a_1,b_1)^2+d_2(a_2,b_2)^2...+d_n(a_2,b_2)^2}$ is complete.
Title: Re: Real Analysis
Post by: Pappa-Bohr on March 30, 2010, 09:27:13 pm: wat did u get for sheet 1 section 7 question 15.

i.e ''limit as n goes to infinity'' for 1+(-1)^(n+1)

(also is n assumed to be a positive integer or a real or what?)
Title: Re: Real Analysis
Post by: QuantumJG on May 16, 2010, 02:45:11 pm: What is the difference between the maxima of a set and the supremum of a set?
Title: Re: Real Analysis
Post by: /0 on May 16, 2010, 02:59:25 pm: The supremum of a set $S$ is the least possible upper bound of the set. The supremum need not be in the set.

The maximum of a set $S$ is a number $a \in S$ which is an upper bound of the set, i.e. $\forall x \in S$ , $a \geq x$ . The maximum must be an element of the set.
As a result, some kinds of sets such as open sets don't have maximums. All finite partially ordered sets have maximums.
Title: Re: Real Analysis
Post by: QuantumJG on May 30, 2010, 07:17:37 pm: What does the standard topology of a metric space mean?
Title: Re: Real Analysis
Post by: kamil9876 on May 30, 2010, 07:50:34 pm: The usual toplogy on a metric space is one where the open sets are those sets that are the unions of open balls, along with the empty set.
Title: Re: Real Analysis
Post by: dcc on August 22, 2010, 05:43:22 pm: Quote from: QuantumJG on March 13, 2010, 04:30:28 pm
$\sum ^{ \infty} _{n = 1} \dfrac{n}{ 3 ^{n-1} }$

Approaches 2.25 as n $\rightarrow \infty$ is this right?

Not sure if its still relevant, but you should be able to recognise this sum as a relative of the Geometric series - Consider the derivative of $f(z) = \sum_{n=0}^{\infty} z^n = \frac{1}{1-z}, \left|z\right| < 1$ . (Have you learnt about uniform continuity and power series?)

edit: fixed sum index.