@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 ?

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

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.

@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 $N$).

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

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

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

@BalarkaSen Yup

I'm not sure how to show that

Yes

Umm, okay

Ooooo

So we can choose $n$ now

Take log and done

:P

Guess so

$(n+1)\log|x| < \log (2\epsilon)$

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

as $|x|<1$

But yeah, from here we can choose $n$

Uh, why so?

Oh, hm I see

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 $-1< -s \leq x \leq s <1$

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

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

@BalarkaSen Huh

I was referring to this:

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

@BalarkaSen Right

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

Right, gotcha

I was stuck on another similar problem too

@BalarkaSen yup!

mhm

that was indeed clever

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

There's a jump at 1

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

okay, sure

Yes, I had seen that earlier :)

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

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

But not sure about uniform convergence again

Perhaps I should try plotting the graph

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

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

So all terms are $(1/2)$

So the sum is $n/2$ for $n$ terms

It doesn't converge

Sure!

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

BTW the question only mentioned "for which positive values of $x$, the series converges"

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

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

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

Alright, now the uniform convergence thingy

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

From desmos

(it seems)

desmos.com/calculator/qddvowqbwg

@BalarkaSen Yeah, so basically that is $x>1+\lambda$ and $x<1-\lambda$

Right

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

We need to choose $n$ now

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

I mean the terms keep reducing

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

There might be some trick

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

But it looks so...

From the graphs..

Okay, go on

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

I'm a bit confused at the moment

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

@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!

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

Legendre series convergence proof

Then the Chebyshev equation and Gegenbauer function

Ok :)

@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).

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?

Here's the question for reference

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

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

Right

They explicitly mentioned that we should check end-points

I'm not sure how to deal with that

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

It's inconclusive there

Any ideas for that condition? :P

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

@BalarkaSen Yep

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

Oh, yeah

:/

We can try bounding it by something

Yes

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

We need to have sufficiently large $j$

For that to not happen

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

Then the ratio would no longer be <1

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

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

Is $(\frac{((2+1)(2+2)-1000(1000+1))}{(2+2)(2+3)})^2$ lesser than $(\frac{(2+1)(2+2)}{(2+2)(2+3)})^2$ ?

:P

Okaies. Any other ideas to bound that?

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

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

Right

It is greater than 1/3

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

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

I get that it tends to 1

Then?

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

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

That's indeterminate

$e^3$

Yeah

So okay, it works

Yes, phew

Finally

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

Something to do with Fourier Legendre series it seems

There's a chapter on Legendre polynomials in the book

Later on

Probably useful in physics

Even the Hermite polynomial stuff

I was learning the various integrals which represent the Dirac Delta function yesterday

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

@BalarkaSen School still on?

When are your boards?

@BalarkaSen ah! Lot's of time. :D

good luck for practicals...I always hated those

:P

(Especially chemistry)

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

Irodov and Cengage have some of the best problems

Even the FIITJEE papers have some good problems

For 1.2.7

I get the limit as:

$\lim_{j\to\infty}(\frac{(k+j)(k+j+2\alpha)}{(k+j+1)(k+j+2)})^{j}$ as

$e^{2\alpha-3}$

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

Oh, yes

But how did they get $\alpha <1$

Is convergent

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

So it changes some things

Just ratio test gives $1$

So that is inconclusive

@BalarkaSen any suggestion?

I think it should always diverge

But maybe I'm missing something

Nothing is mentioned about k :/

Actually let's think about this tomorrow. Feeling too sleepy now. Goodnight !