Problem Solving Strategies

Yashas

15:00

f```(x) = f`(x) * d[dg(f(x))/d(f(x))]/dx + dg(f(x))/d(f(x)) * f``(x)

f``` is zero becaz f`(x) and f``(x) is zero

if you observe carefully, you will notice that nth derivate will be made up of derivatives which come before the nth derivate

Madhuchhanda Mandal

Why does this not hold ? :f(x+dx)=f(x) so f'(x+dx)=f'(x)=0. Hence the samething loops : f(x+2dx)=f(x+dx) and f'(x+2dx)=f'(x+dx)=f'(x)=0...

Yashas

as all the previous derivatives before n are zero, the nth derivative is zero

therefore, the function's slope never ever changes

it has to be a straight line

Anonymous

@Yashas I don't see why this should be true...

Yashas

@MadhuchhandaMandal I still don't know what you did in the second part.

Anonymous

It could be an inflection point

Yashas

15:02

yup

Madhuchhanda Mandal

@Yashas so your derivative approach concludes the conclusion given that they are all differentiable

Anonymous

So it has an inflection point at a

Anonymous

Says nothing about other points

Yashas

it is an inflection point which never changes

Anonymous

@Yashas What?

Yashas

15:03

for an inflection point, nth derivate becomes non zero

in this case, it never becomes non-zero

Madhuchhanda Mandal

f'(x+dx)=f'(x)=0. Agree upto this?

Yashas

yea

Madhuchhanda Mandal

Had f(x+dx) been our initial point, f'(x+dx+dx)=g(f(x+dx+dx)). Agree?

Yashas

yes

Anonymous

@Yashas That is wrong. Take $f(x)=e^{-1/x^2}$. It has all derivatives 0 at 0. Yet it is not constant.

Yashas

15:07

use h instead of dx

@blue 0?

Madhuchhanda Mandal

Now f'(x+dx)=0. So f(x+dx)=f(x+dx+dx). And from first f(x+dx)=f(x) as f'(x)=0. So we see f(x+dx+dx)=f(x) So f'(x+dx+dx)=0. Agree?

Yashas

derivative does not exist at 0 I think

because you get x down

Madhuchhanda Mandal

@Yashas I exactly was going to use h :P...

Anonymous

@Yashas It does exist.

Anonymous

Check

Yashas

15:10

x is going to come downstairs

you take the derivative of the -1/x^2

you get 2/x^3

* e^{-1\x^2}

LOL

0 blows up the exponent itself

Anonymous

Wait a sec. I didn't define the full function

Yashas

I never had to take the derivative

Madhuchhanda Mandal

Now f(x+dx+dx) can again be our initial point. Hence the same thing loops

Anonymous

$f(0)=0$ and $f(x)=e^{-1/x^2}$ when x is not 0

Yashas

not differentiable

Anonymous

15:12

Oh right

Madhuchhanda Mandal

@Yashas Understood?

Yashas

I have got stuck in three contexts

Madhuchhanda Mandal

What are they?

Yashas

your proof, blue's argument and my world

Madhuchhanda Mandal

*my argument

I am not sure if it is proof of any kind

Yashas

15:17

let f(x) be a continous and differentiable function

Madhuchhanda Mandal

(Whose only first derivative exists. Lets suppose.)

Yashas

f(x) = f(a) + f`(a)(x - a) + f``(a)(x-a)^2/2! + f```(a)(x-a)^3/3! + ........

as all the derivatives of f(x) at a is zero, f(x) = f(a)

this is the best proof

Taylor expansion

one step direct

Madhuchhanda Mandal

Ok.. Are you and Blue arguing that if all nth derivative (1 to infinity) of f(x) is 0 imply f(x) is constant or not?

Yashas

yes

f(a) = 0

*f`(a)

and so forth

you can directly substitute in the Taylor expansion

it clearly gives f(x) = f(a)

Madhuchhanda Mandal

It needs to be constant

Anonymous

15:21

@Yashas Wait, that function does seem to be differentiable as rhd=lhd

Anonymous

the rhd comes out to be 0

Yashas

?

Anonymous

And the lhd comes out to be 0 too

Yashas

that function is not differentiable at all points

0 is a point where it isn't differentiable

it blows up

Anonymous

At 0 I defined it as f(x)=0

Anonymous

15:22

And at the other places I defined it as f(x)=e^(-1/x^2)

Yashas

@blue You are cheating there.

Anonymous

@Yashas How?

Yashas

becaz you haven't defined a value for f(0)

Anonymous

It is continuous and differentiable throughout

Yashas

and you define f`(0)?

Anonymous

15:23

@Yashas f(0)=0

Anonymous

@Yashas f'(0)=0

Anonymous

f'(0) is the derivative at 0 which depends on equality of lhd and rhd

Yashas

problem

f`(x) is not continous in your case

Anonymous

@Yashas Why do we need that?

Yashas

becaz I am taking infinite derivatives

Madhuchhanda Mandal

15:25

If for two functions, f'(x)=f''(x)=f'''(x)=...=f'n(x) at a point x=x0, do you agree the functions are identical?

Yashas

@MadhuchhandaMandal yes

f(x) = f(x0)

@blue look at the this

f(x) can be written as:

Madhuchhanda Mandal

@yashas Then the work is done

Yashas

f(x) = f(a) + f`(a)(x - a) + f``(a)(x - a)^2/2! + f```(a)(x-a)^3/3! + ...

Madhuchhanda Mandal

Your argument easily holds true

Yashas

If all the derivatives are zero, f(x) = f(a)

@MadhuchhandaMandal wait, you said identical

f(x) = f(x0)

Anonymous

15:28

In your proof you didn't show that f'(x) is differentiable throughout

Yashas

but f(x0) need not be the same

@blue that is included in the assumption

Anonymous

@Yashas Which assumption?

Yashas

I did not state it properly :P

it must be infinitely differentiable function

Anonymous

@Yashas Say that.

Anonymous

Makes a HUGE difference

Yashas

15:29

but see

if all the derivatives are zero for an inflection point, then the function is constant

if one of them isn't zero, then it isn't constant

Anonymous

@Yashas That is true.

Yashas

:|

you were arguing 10 mins ago

Anonymous

True, given your assumption

Madhuchhanda Mandal

Then why the hell were you arguing?

Ok.. Let's get back to the question

Yashas

@blue here

23 mins ago, by blue

@Yashas That is wrong. Take $f(x)=e^{-1/x^2}$. It has all derivatives 0 at 0. Yet it is not constant.

Anonymous

15:32

Yes. You didn't tell me that you made that assumption before.

Madhuchhanda Mandal

Let's get back to the original question. Should we?

Yashas

@blue :|

I told all the derivatives are zero

remains zero at infinite'th derivative

@MadhuchhandaMandal I am lost again.

@blue you were telling that even if infinite derivatives are zero for an inflection point, the function needn't be constant

Madhuchhanda Mandal

If f'(x) =g(f(x)) and f'(x0)=0. Will the graph of f(x) be constant from x=x0 to x=infinity??

Yashas

if f(x) is infinitely differentiable then, it is a constant

Madhuchhanda Mandal

But if is not, then what's the problem?

I mean why my looping argument don't hold?

Yashas

15:37

I still don't understand ur argument lol

I lost track again

if it is infinitely differentiable, then your claim holds; if it is not infinitely differentiable, then it does not hold until you impose more conditions

Madhuchhanda Mandal

f'(x+dx)=f'(x)=0. Agree upto this?

Yashas

how?

Madhuchhanda Mandal

f(x+dx)=f(x)

As f'(x)=0

Yashas

It logically makes sense but I would want to see a rigorous proof.

from the first principle

it should be easy

let me try

Madhuchhanda Mandal

f(x+dx)-f(x) / dx = f'(x)=0. So f(x+dx)=f(x)

Yashas

15:42

:O

I am shocked to see that

Madhuchhanda Mandal

Umm... Why?

Yashas

my intuition doesn't want to accept it

Madhuchhanda Mandal

Hmm.. Think then

Yashas

becaz it does not seem correct

idk why my instincts tell that it isn't correct

Madhuchhanda Mandal

Why?

Yashas

15:44

take a maxima of a parabola

f`(x) = 0

just before and after, f(x) changes

Madhuchhanda Mandal

f(x+dx) is indeed = f(x) at the vertex

Yashas

f(a + dx) = f(a) is true though

for dx being very small

a can be any constant

Madhuchhanda Mandal

That's why the tangent is parallel to y axis

Yashas

so f(x + h) = f(x)

now my brain is double screwd

let me ask in the h bar

Madhuchhanda Mandal

Why bro?

Let's first listen to our logic

Why won't it be right?

Yashas

15:47

becaz you dx is infinitely small

Madhuchhanda Mandal

So?

Yashas

I can argue that f(x + n*h) = f(x)

for all functions

becaz h is essentially zero

only at n = infinity, it is indeterminate

Madhuchhanda Mandal

No.. n*h is not 0

Its trends to 0

Yashas

it is zero

unless n is infinite

Madhuchhanda Mandal

*tends

Nope... Its tends to 0

Yashas

15:48

wait

it tends to zero

Madhuchhanda Mandal

Otherwise calculus will fall like anything

Yashas

something is wrong with my brain

Madhuchhanda Mandal

Ok.. Rethink

Yashas

continue to the next step

we'll see what happens next

then come back to this

Madhuchhanda Mandal

Had f(x+dx) been our initial point, f'(x+dx+dx)=g(f(x+dx+dx)). Agree

?

Yashas

15:53

This idea of using dx is not conviencing at all

Madhuchhanda Mandal

Use h instead

As you suggested

Yashas

I don't like infinitely small stuff

Madhuchhanda Mandal

f(x+dx+h)=g(f(x+dx+h))

Agree?

Again f(x+dx+h)-f(x+dx) / dx = f'(x+dx) = 0. So f(x+dx+h)=f(x+dx). Agree?

Anonymous

6

Q: All derivatives zero at a point $\implies$ constant function?

Suppose $f: \mathbb{R} \rightarrow \mathbb{R} $ is a continuous function, and there exists some $a \in \mathbb{R}$ where all derivatives of $f$ exist and are identically $0$, i.e. $f'(a) = 0, f''(a) = 0, \ldots$ Must $f$ be a constant function? or if not, are there examples of non-constant $f$ th...

calculus

Anonymous

6

A: All derivatives zero at a point $\implies$ constant function?

Cauchy's function $f(x)=e^{-1/x^2}$ for $x\ne0$ and $f(0)=0$ has all derivatives at $0$ equal to $0$, but the function is not constant on any interval, thus answering your first question. For your second question, of course if a function has first derivative equal to $0$ on an interval then the...

Anonymous

16:00

Cauchy's function is continuous at all points and so are its derivatives.

Anonymous

I'm still missing something

Yashas

WTH

that function is perfectly flat

Anonymous

@Yashas Which one?

Yashas

or the computer is approximating

Cauchy's function

Anonymous

It is approximating

Madhuchhanda Mandal

16:02

Is Cauchy's function derivable at x=0 ?

Anonymous

Yes

Madhuchhanda Mandal

It seems not using def. Of differentiation

Oh sorry

Anonymous

See Ittay's answer

Madhuchhanda Mandal

It is.

Anonymous

Oh. I misread some $a \in \mathbb{R}$ for all $a \in \mathbb{R}$. Kindly ignore my hint. — Gautam Shenoy Dec 9 '15 at 6:15

Anonymous

16:04

This is the catch ^

Yashas

what?

Madhuchhanda Mandal

0/h I was considering undefined Lol !!

Yashas

I read that comment and I couldn't make sense

Madhuchhanda Mandal

Or wait... How?

0/0 is undefined I think?

Yashas

the derivatives are also flat

this function is crazy

Madhuchhanda Mandal

16:06

Can you prove this function to be differentiable?

At x=0?

Anonymous

It is differentiable at x=0

Anonymous

Do your calc properly

Anonymous

All the answers state the same

Yashas

I can't understand

Anonymous

@Yashas What?

Yashas

16:07

those answers

Anonymous

Why not?

Yashas

I still don't understand what is wrong with the Taylor series

Anonymous

You did the same mistake which Gautam did

Yashas

I don't understand what Gautam said

He said a belongs to R and a belongs to R

what the hell?

Madhuchhanda Mandal

Is e^x != 1+x+x2/2! + x3/3!.....

?

Anonymous

16:09

@Yashas "some" and "all". Read it again.

Yashas

@MadhuchhandaMandal yes

Madhuchhanda Mandal

They aren't equal?

Yashas

equal

oh

you put !

I did not see the !

Anonymous

e^x factorial? phew.

Madhuchhanda Mandal

Okay. (By the way, Cauchy's function needs to be diff. Cause f(x+h)=f(x) at x=0

Anonymous

16:11

@MadhuchhandaMandal It is differentiable.

Madhuchhanda Mandal

Yeah.. It must be

Yashas

@blue explain Gautam's comment

Madhuchhanda Mandal

But can't prove mathematically

Anonymous

@Yashas You showed f(x)=f(a) at only one x=a

Anonymous

@Yashas Got it now?

Madhuchhanda Mandal

16:16

Can anyone prove using first principle?

Yashas

@blue no

that isn't the problem

what I did is correct

the mistake is the function is not analytic

Madhuchhanda Mandal

That it's differentiable?

Yashas

e^{-1\x^2} is not analytic function

my proof works for analytic functions only

I made such a nut move

Madhuchhanda Mandal

?

Koolman

What the hell too big discussion

Yashas

16:24

@MadhuchhandaMandal If the function is not analytic, then the Taylor series does not hold.

I learned something important today

Madhuchhanda Mandal

But can you prove existence of derivative of Cauchy's function from first principle at x=0 ?

(Or rather, if Taylor's series don't converge with f(x) , then it's non-analytic)

Anonymous

16:40

@Yashas I get it now. Phew!

Koolman

@Yashas taylor series is not in jee syllabus

Yashas

@Koolman That does not mean that you must not learn it

That helps a lot.

I was also taught hyperbolic geometry.

becaz it is so useful for integration

you can substitute x as "sin" for 1 - x^2

for 1 + x^2, it is easier if you use "sinh"

I was also taught more crazy math stuff

solving 2d coordinate geometry using 3D

Anonymous

@Yashas Use tan(y) =x

Yashas

where z = 1

@blue the process is longer

Koolman

where you have studied all this

Anonymous

16:45

@Yashas Agreed

Yashas

@Koolman Coaching

I have teachers who are from Indian Institute of Science

Koolman

which

ohh

Yashas

There is one professor who teaches all crazy things in physics and math

Madhuchhanda Mandal

Wow!!!

Yashas

The other one teaches physics toooo good

Madhuchhanda Mandal

16:46

Which state do you belong?

Yashas

@MadhuchhandaMandal Karnataka

Koolman

@MadhuchhandaMandal are you from fiitjee

Madhuchhanda Mandal

And here... Most teachers don't understand and can't teach JEE syllabus also....

@Koolman yeah

Koolman

which state

Madhuchhanda Mandal

WB

Koolman

16:47

@Yashas wbt chem

Yashas

My organic chem teacher comes from 90kms away everyday to give classes

Has patents in UK and stuff like that

His notes are better than textbooks lol

Madhuchhanda Mandal

oo

Koolman

and my organic comes from 9 km

lol

Yashas

lol

Madhuchhanda Mandal

Feeling jealous

Yashas

16:49

Some students submit research papers at 10th grade lol

last year, someone presented a paper nationally

Madhuchhanda Mandal

I have made one on Data Compression

Yashas

10th or 9th grade

@MadhuchhandaMandal :o

Madhuchhanda Mandal

I mean, working on it

Ok... Let's get back to the original question

Yashas

github.com/YashasSamaga

SA:MP Call Of Duty Global Warfare Trailer

►

I do part of the coding there

apart from doing other regular work

Madhuchhanda Mandal

I see

Yashas

16:53

since 2015

Madhuchhanda Mandal

How did u get there?

Yashas

and my code is powering hundreds of game servers

counter strike, SAMP and many others

@MadhuchhandaMandal Hmm?

I knew the nuts and bolts of C at the age of 10

at 11, I tried to make my own OS

at 12, I made a tutorial website to help others make OS

kernelx.weebly.com

Madhuchhanda Mandal

You told that that day

I mean how did you get yourself involved with Counter Strike?

*Call of Duty

Yashas

Call of duty is not the original

this is GTA San Andreas Multiplayer

with COD theme

before that, till 2014, I had my own GTA SA game server

Madhuchhanda Mandal

I see... I don't play games much :P

Yashas

16:58

Status update:

As I have addressed last month, the development is slowed down and works gradually due to low activity of Yashas who has gone for another two months due to exams, and due to me who doesn't have too much will to script as much as in previous years since SA-MP for me as a whole became too static and boring with the lack of its updates and such. Either way, I still script and will script, just less regularly, let's say an update every other week instead of every week.

The maps are still being worked on, Mircea has had exams and has therefore had to slow down with the progress,…

Koolman

@Yashas how was your test

Yashas

That was the previous announcement for SAMP COD GW

@Koolman I got a stats question wrong

Koolman

ah I mean reso mock test

Yashas

I paused it

Koolman

why

Yashas

16:59

I have 5 more mins

will check tmrow

Madhuchhanda Mandal

Ok let's get back to the question. Why looping argument is not holding?

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