Anonymous
7:15 AM
@BalarkaSen BTW in which book and which chapter is the things about determinant of A, elliptic hyperbolic ODE and analysing the determinant of A to find type of equilibrium point given ?

Anonymous
Is it in Hirsch Smale? If so, which chapter ?

9:24 AM
@Blue Ch. 3, "Phase Portraits for Planar Systems"
You won't find the proof of the thing I said about linearization of hyperbolic ODEs in there though
That is very complicated to prove
It's in Katok-Hasselblatt but I do not recommend reading it
I have been writing some notes myself, but they are terse and I am not sure how much you will get out of them

3 hours later…
Anonymous
12:12 PM
Okay, got it. I'll try to read through those bits of Hirsch Smale for now. Later on someday you can explain me the linearization thing.

Anonymous
5:14 PM
@BalarkaSen Need some help with uniform convergence. The question is: "For what range of $x$, is $\sum_{n=0}^{n=\infty}x^n$ uniformly convergent?" So, basically it is easy to show that it converges if $|x|<1$. But I'm having a bit of trouble with "uniform convergence". Basically we need to show that for any $\epsilon>0$, $|S-S_n|<\epsilon$ $\forall n\geq N$. Or alternatively $|x^{N+1}+x^{N+2}+...|<\epsilon$ for any choice of $\epsilon>0$ (and for a choice of $N$).

Anonymous
I can simplify that to $|x^{N+1}||\frac{1}{1-x}|<\epsilon$ if $|x|<1$

Anonymous
But then I'm not sure how to express the choice of $N$ explicitly for a given $\epsilon$

Well, you want to show for a given $\epsilon > 0$ there is an $N$ such that $|x^{n+1}|/|1 - x| < \epsilon$ for all $n \geq N$
Where $|x| < 1$
And you want this bound to be uniform, i.e., for all $x$

Anonymous
@BalarkaSen You mean for $|x|<1$? (by all x)

Anonymous
@BalarkaSen Yup

Anonymous
5:21 PM
I'm not sure how to show that

@Blue Yes, all $x$ in the domain of convergence
So let's see
$|x^{n+1}| < \epsilon |1 - x|$

Anonymous
Yes

And $|1 - x| < 2$

Anonymous
Umm, okay

So $|x|^{n+1} < 2\epsilon$
I want to have this bound

Anonymous
5:23 PM
Ooooo

Anonymous
So we can choose $n$ now

I think so

Anonymous
Take log and done

Anonymous
:P

Anonymous
Guess so

Anonymous
5:25 PM
$(n+1)\log|x| < \log (2\epsilon)$

Anonymous
That $\log|x|$ will be negative I think

Anonymous
as $|x|<1$

Anonymous
But yeah, from here we can choose $n$

Wait
No
This is gotta be false
I do not think $\sum_{n = 0}^\infty x^n$ converges uniformly on $|x|< 1$

Anonymous
Uh, why so?

5:29 PM
@BalarkaSen Ignore this. That doesn't mean anything; I'm making the inequality weaker.

Anonymous
9 mins ago, by Balarka Sen
$|x^{n+1}| < \epsilon |1 - x|$

Anonymous
Oh, hm I see

@Blue It's a good exercise to prove this. But what's happening is that the series converges slower and slower for values of $x$ closer and closer to $\pm 1$
That completely destroys uniformness
I mean, $|S_n(x) - S(x)| = |x|^{n+1}/|1 - x|$ like you said. If $x$ is a fixed value very close to $1$, this number is very very very very large.
Of course it becomes zero as you let $n \to \infty$, but the rate of going to zero is different at different points
In particular, the rate of going to zero is larger as you move $x$ closer to $1$
That is contradictory to the spirit of uniform convergence. It says you have a fixed upper bound on the rate of convergence for all $x$
Does that make sense?

Anonymous
Yeah, it does indeed make some sense. But the problem is the numerator goes to $0$ as $n\to\infty$. Otherwise we could have confidently claimed that it doesn't uniformly converge. According to the answer key in the textbook, the series converges for $-1< -s \leq x \leq s <1$

Anonymous
I don't know why they put $s$ there though

5:38 PM
Yes, exactly, it convergence uniformly on $|x| \leq \lambda$ for any fixed $\lambda < 1$, but not on $|x| < 1$
@Blue Well, no, the numerator is not a problem.

Anonymous
@BalarkaSen Okay, makes sense. But how do we prove that rigorously?

Anonymous
@BalarkaSen Huh

Anonymous
I was referring to this:

Anonymous
5 mins ago, by Balarka Sen
Of course it becomes zero as you let $n \to \infty$, but the rate of going to zero is different at different points

@Blue It's really simple! $|S_n(x) - S(x)| = |x|^{n+1}/|1 - x|$. Now $|x| \leq \lambda$, and $|1 - x| \geq |1 - |x|| \geq |1 - \lambda| = 1 - \lambda$
So $|S_n(x) - S(x)| \leq \lambda^{n+1}/(1 - \lambda)$
So we want to find an $N$ such that for all $n \geq N$, $\lambda^n/(1 - \lambda) < \epsilon$

Anonymous
5:45 PM
@BalarkaSen Okay so far, but still expressing $n$ for a given choice of $\epsilon$ looks a bit tough. Any ideas for that?

Anonymous
@BalarkaSen Right

You can take logs now! $\lambda$ is a fixed number

Anonymous
Oh, okay, lol. I was being stupid. Got it

Choose $N$ to be any integer greater than $\log(\epsilon(1 - \lambda))/\log(\lambda)$
The next integer number, say

Anonymous
Right, gotcha

5:47 PM
The crucial step was

Anonymous
I was stuck on another similar problem too

3 mins ago, by Balarka Sen
So $|S_n(x) - S(x)| \leq \lambda^{n+1}/(1 - \lambda)$

Anonymous
@BalarkaSen yup!

Because you bounded the rate of convergence by a fixed number

Anonymous
mhm

Anonymous
5:47 PM
that was indeed clever

@Blue Have you seen pictures of non-uniform convergence?
Like, $f_n(x) = x^n$?

Anonymous
@BalarkaSen Oh, yes. For [0,1]

Anonymous
There's a jump at 1

Anonymous
The continuous function series has a discontinuous function as limit (i.e. piecewise function)

Yeah but that's not a picture
Let me give you a desmos link

Anonymous
5:50 PM
okay, sure

Anonymous
Yes, I had seen that earlier :)

Ah good
So what happens is that $f_n$ converges to $0$ very fast in the middle area
but very slow near $\pm 1$
That's why it's not uniform

Anonymous
So I was stuck on a similar problem: $\sum_{n=0}^{n=\infty}\frac{1}{1+x^n}$

Anonymous
By comparing it with an integral I think I can show it converges

Anonymous
5:53 PM
But not sure about uniform convergence again

Anonymous
Perhaps I should try plotting the graph

Anonymous
The integral I evaluated was $\int_{0}^{\infty}\frac{1}{1+n^x}dx$

Well it doesn't converge for $x = 1$ clearly
Wait, $n^x$ or $x^n$?

Anonymous
Sorry, had to go for a while. Umm so for x=1, $1/(1+1^0), 1/(1+1^1), 1/(1+1^2),...$

Anonymous
So all terms are $(1/2)$

Anonymous
6:01 PM
So the sum is $n/2$ for $n$ terms

Right, so it doesn't converge

Anonymous
It doesn't converge

I'll be back after I finish my dinner

Anonymous
Sure!

Anonymous
@BalarkaSen The integral has $n^x$, where the limits of $x$ is $0$ to $\infty$

Anonymous
6:07 PM
BTW the question only mentioned "for which positive values of $x$, the series converges"

Anonymous
$\int\frac{1}{1+n^x}dx=x-\frac{\log(n^x+1)}{\log(n)}$

Anonymous
So, clearly for $n=1$, the integral is undefined...

Anonymous
So, other than for $x=1$, it seems $\sum_{n=0}^{n=\infty}\frac{1}{1+x^n}$ converges

Ok, I'm back
@Blue Right

Anonymous
Alright, now the uniform convergence thingy

6:19 PM
And it uniformly converges on $|x - 1| > \lambda$

Anonymous
And also for $|x|\leq s<1$

Anonymous
From desmos

Anonymous
(it seems)

:41410898
That is contained in my region

Anonymous

Anonymous
6:24 PM
@BalarkaSen Yeah, so basically that is $x>1+\lambda$ and $x<1-\lambda$

Anonymous
Right

Anonymous
$|\frac{1}{1+x^{n+1}}+\frac{1}{1+x^{n+2}}+...\frac{1}{1+x^{\infty}}|<\epsilon$

Anonymous
We need to choose $n$ now

Right. Uh

Anonymous
That looks fine for $x>1+\lambda$

Anonymous
6:29 PM
I mean the terms keep reducing

Anonymous
But still it's difficult to find $n$ explicitly

Anonymous
There might be some trick

yeah...
I am not sure of a slick way to do it

Anonymous
For $1-\lambda$ the terms actually increase....

Well it's not true that it converges everywhere except $x = 1$

Anonymous
6:32 PM
But it looks so...

Anonymous
From the graphs..

Anonymous
Okay, go on

Take $0 < x < 1$
Then $1/(1 + x^n) \to 1$
as $n \to \infty$
For a sum $\sum a_n$ to converge you have to have $a_n \to 0$ as $n \to \infty$
It doesn't converge on the closed interval $[-1, 1]$

Anonymous
@BalarkaSen Is that true even for series of "functions"?

Fix $x$. You have a series of numbers.

Anonymous
6:39 PM
I'm a bit confused at the moment

Anonymous
Why does the graph show that it is uniformly convergent, then?

You are graphing 1/(1+x^n), not the sum

Anonymous
@BalarkaSen Oh, phew. Finally got it. For the series to converge $S_n$ and $S_{n-1}$ must have same limit (as $n\to \infty$). So obviously the last term must be $0$, as $n\to \infty$. I was thinking of something weird. Lol. Got it!

Cool
I didn't notice it at first
So you're fine

Anonymous
Yep, thanks! There are a few more exercises on this. I'll tell you if I get stuck :P

Anonymous
6:54 PM
Legendre series convergence proof

Aha OK

Anonymous
Then the Chebyshev equation and Gegenbauer function

Anonymous
Ok :)

Woo cool stuff

Anonymous
7:41 PM
@BalarkaSen The Legendre series $\sum_{\text{j even}} u_j(x)$ satisfies the recurrence relation $u_{j+2}(x)=\frac{(j+1)(j+2)-l(l+1)}{(j+2)(j+3)} x^2 u_j(x)$. So, basically $|\frac{u_{j+2}(x)}{u_j(x)}|=|\frac{(j+1)(j+2)-l(l+1)}{(j+2)(j+3)} x^2|$. For sufficiently large $j$, $|\frac{(j+1)(j+2)-l(l+1)}{(j+2)(j+3)} x^2|<1$ if $|x^2|<1$. So I think it will be absolutely convergent if $|x|<1$. BTW also it is stated that $j$ is even and $l$ is some constant (but NOT a non-negative odd integer).

Anonymous
I need to find the range of $x$ for which the Legendre series is convergent and test the endpoints. I showed the condition for absolute convergence but not sure if for some other range of $x$ conditional convergence is possible. Any ideas?

Let's see
So you're doing the ratio test there?

Anonymous

Anonymous
Here's the question for reference

Anonymous
@BalarkaSen Yeah, the absolute convergence test i.e.

7:55 PM
@Blue Well, absolute convergence implies convergence so I think you are fine

Anonymous
@BalarkaSen But, it can be conditionally convergent at some place even though it is not absolutely convergent there

Oh, so now you want to look at $|x| \geq 1$?

Anonymous
Right

Hmm

Anonymous
They explicitly mentioned that we should check end-points

Anonymous
7:57 PM
I'm not sure how to deal with that

$x = \pm 1$?
I am not sure what is meant by end-points

Anonymous
When we put $j\to\infty$, the ratio test just produces $1$ in case $x=\pm 1$

Anonymous
It's inconclusive there

Ah
Got it
Yes

Anonymous
Any ideas for that condition? :P

8:01 PM
So you have $u_{j+2}(1) = \frac{(j+1)(j+2) - C}{(j+2)(j+3)} u_j(1)$
$C = l(l+1)$

Anonymous
Right (BTW I don't know why they mentioned $l$ can't be a non-negative odd number)

I want to look at $u_0(1) + u_2(1) + u_4(1) + \cdots$, right?
I think you just get a geometric series

Anonymous
@BalarkaSen Yep

Let $K = \frac{(j+1)(j+2) - l(l+1)}{(j+2)(j+3)}$
Then $u_p(1) = K^p u_0(1)$
So it's the sum $u_0(1) \cdot (1 + K + K^2 + \cdots)$
Which is $\frac{u_0(1)}{1 - K}$

Anonymous
Oh, right. If $(j+1)(j+2)>l(l+1)$, $K$ will be a positive ratio. We can separate out the terms for which $(j+1)(j+2)<l(l+1)$ I guess

8:04 PM
Wait, no, I am dumb
The sum is indexed over $j$
$K$ has $j$ terms in it
It's not a constant

Anonymous
Oh, yeah

Anonymous
:/

Anonymous
We can try bounding it by something

The sum is $\displaystyle u_0(1) \sum_{j \, \text{even}} \left( \frac{(j+1)(j+2) - l(l+1)}{(j+2)(j+3)} \right )^p$
Right?

Anonymous
Yes

8:07 PM
I can bound the sum by $\displaystyle \sum_{j \, \text{even}} \left( \frac{(j+1)(j+2)}{(j+2)(j+3)} \right )^p$

Anonymous
I think we need to be a bit careful here. Come to think of the cases when $l(l+1)>>(j+1)(j+2)$

Anonymous
We need to have sufficiently large $j$

Anonymous
For that to not happen

Well, that's not the issue. (j+1)(j+2) - l(l+1) is smaller than (j+1)(j+2) whenever l > 0
But l is not > 0 here, is it?

Anonymous
@BalarkaSen The numerator could have a large negative magnitude...

Anonymous
8:10 PM
Then the ratio would no longer be <1

I thought we were not testing for absolute convergence?

Anonymous
Wait a bit. What is $p$ ? Odd or even ?

Ah, $p = j$.
Because if I go from $u_j$ to $u_0$ I get multiplied by $K$ $j$ times
so $K^j$
So negativity doesn't matter anyway

Anonymous
Right. Since you are summing over even j's, you can't write $\left( \frac{(j+1)(j+2) - l(l+1)}{(j+2)(j+3)} \right )^p<\left( \frac{(j+1)(j+2)}{(j+2)(j+3)} \right )^p$

? Why not. $(j+1)(j+2) - l(l+1) < (j+1)(j+2)$
Do you agree with that?
If $l > 0$ that is true.
Subtract any number from A that's smaller than A
Doesn't matter if it's negative or not

Anonymous
8:15 PM
Is $(\frac{((2+1)(2+2)-1000(1000+1))}{(2+2)(2+3)})^2$ lesser than $(\frac{(2+1)(2+2)}{(2+2)(2+3)})^2$ ?

ops
moving on
let me get to that box of shame

Anonymous
:P

okkay moooving onnn

Anonymous
Okaies. Any other ideas to bound that?

Anonymous
Actually the initial few terms don't matter anyway. For sufficiently large $j$, what you wrote is true

8:21 PM
true
Well

Anonymous
So well, that is $\frac{j+1}{j+3}$ after cancellation

(j+1)(j+2)/(j+2)(j+3) = (j+1)/(j+3)
which is 1/(1 + 3/(j+1))

Anonymous
Right

Anonymous
It is greater than 1/3

Anonymous
Uh, that's not an upper bound...

8:25 PM
well as j --> infty that tends to 1
this does not converge
so the convergence domain is indeed |x| < 1. at the endpoints you don't even converge conditionally

Anonymous
@BalarkaSen Wait a bit. How did you conclude that?

Anonymous
I get that it tends to 1

Anonymous
Then?

@Blue If $\sum a_n$ converges, $a_n \to 0$ as $n \to \infty$...
In this case, the terms tend to 1 as j --> infty
so

Anonymous
$\displaystyle u_0(1) \sum_{j \, \text{even}} \left( \frac{(j+1)(j+2) - l(l+1)}{(j+2)(j+3)} \right )^j$

8:27 PM
Just edit that p out and write j

Anonymous
You get something like of the form $1^{\infty}$

Anonymous
That's indeterminate

Oh, I missed that. Hm, redo. I have (1/(1+3/(j+1))^j
what's the limit as j --> infty?
1/(1 + 3/j)^j
lim j --> infty (1 + 3/j)^j
that's e^3 or something

Anonymous
$e^3$

Anonymous
Yeah

8:29 PM
So you still have nonzero limit

Anonymous
So okay, it works

Anonymous
Yes, phew

Anonymous
Finally

lol
what are these exercises man
these are like large manipulations

Anonymous
@BalarkaSen Yeah, but it requires an amalgamation of small concepts to solve it. Not bad. It's from Arfken

8:30 PM
True
It's kind of fun
I suppose those functions $u_j(x)$ are useful?
I have never encountered these before

Anonymous
Something to do with Fourier Legendre series it seems

Anonymous
There's a chapter on Legendre polynomials in the book

Anonymous
Later on

Anonymous
Probably useful in physics

Anonymous
Even the Hermite polynomial stuff

Anonymous
8:33 PM
I was learning the various integrals which represent the Dirac Delta function yesterday

Anonymous
Got stuck in some parts...will ask you tomorrow :P

Ah I see
OK

Anonymous
@BalarkaSen School still on?

Anonymous

nope; test exams went over
there's still the practicals
@Blue boards are in 27th march iirc

Anonymous
8:36 PM
@BalarkaSen ah! Lot's of time. :D

Anonymous
good luck for practicals...I always hated those

Anonymous
:P

Anonymous
(Especially chemistry)

me too
@Blue Yep!
I have been looking at some JEE level physics out of fun

Anonymous
JEE physics problems are indeed fun. Did you try Irodov?

Anonymous
8:38 PM
Irodov and Cengage have some of the best problems

Anonymous
Even the FIITJEE papers have some good problems

I didn't. I'll note that down
Thanks!

Anonymous

Anonymous
For 1.2.7

Anonymous
I get the limit as:

Anonymous
8:48 PM
$\lim_{j\to\infty}(\frac{(k+j)(k+j+2\alpha)}{(k+j+1)(k+j+2)})^{j}$ as

Anonymous
$e^{2\alpha-3}$

Anonymous
So I guess it will be convergent if $2\alpha-3<0$

Well, no, $\lim a_n$ has to be $0$
e^anything can never be zero

Anonymous
Oh, yes

Anonymous
But how did they get $\alpha <1$

Anonymous
8:52 PM
Is convergent

Anonymous
A difference to note here is that $x^2$ term is absent

Anonymous
So it changes some things

Anonymous
Just ratio test gives $1$

Anonymous
So that is inconclusive

Anonymous
@BalarkaSen any suggestion?

Anonymous
9:05 PM
I think it should always diverge

Ok, let's see

Anonymous
But maybe I'm missing something

9:19 PM
@Blue It's not clear to me what is happening either
$\sum a_j$ seems divergent
Wait, what is $k$?

Anonymous
Nothing is mentioned about k :/

Anonymous