Mathematics - 2020-04-05 (page 2 of 2)

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Knight

22:00

The only thing we know about $f(x)$ is that it assumes the value of zero at least once

Ted Shifrin

Fuzzy: Can you do it when there is a single Jordan block?

Thorgott

does that mean you're looking for an example of such an $f$?

FuzzyPixelz

I still don't understand why we can always find Jordan Chains, but yes

Ted Shifrin

@Knight. I hate this notation. You mean $$\int_c^x f(t)dt = f(x)$$ for all $x>c$?

FuzzyPixelz

I know the steps at least

Ted Shifrin

22:04

So you start with the eigenvector and work backwards. You should understand why it works.

Knight

@TedShifrin Yes, all I want is to prove the existence

FuzzyPixelz

I want a rigorous proof

Most proofs I've seen use weird arguments for existence and then handwave their way through examples

Ted Shifrin

Fund thm of calculus?

nitsua60

either'll work =)

Thorgott

why not take $f=0$

nitsua60

22:07

(though maybe I should say "both"...)

Knight

@TedShifrin Didn’t get you

@Thorgott I need it for a General case

The only thing we know about $f(x)$ is that it assumes the value of zero at least once

Ted Shifrin

In the case we're discussing, it's basic linear algebra. Start with the case of $2\times 2$.

nitsua60

(hrmm... maybe "either." But I've got dinner on, so I can't sketch more than this margin holds)

Ted Shifrin

Why don't you get me? What is the FTC?

Thorgott

I'm not sure what you're asking for. It's not true for all such $f$ and if you want an example, $f=0$ does the job.

Ted Shifrin

22:09

BTW, it's automatic that $f$ takes on the vslue $0$. Why?

He wants a derivation of all possible solutions, @Thorgott.

Knight

@TedShifrin $$\int_{c}^{x} f(x) dx = g(x) \\ g’(x) = f(x) $$

This is the FTC

Ted Shifrin

OK, good. So if $g=f$, now what?

Knight

$$f’(x) = f(x) $$

Ted Shifrin

Go on.

What are the solutions of that?

Knight

$e^x$

I mean $ce^x$

Thorgott

22:13

Ted Shifrin

Different letter would be good, but yes.

Now is there a point where you know the value of $f$?

Knight

Yes

$f(\alpha) =0$

It’s alpha

Ted Shifrin

Is that what you had been calling $c$?

Now finish.

Knight

Hmm.. thanks but I wanted to prove that “there exists at least one $d$ \ge $c$ such that $$\int_{c}^{d} f(x) dx = f(d)$$

(why \gt is not getting parsed?)

Ted Shifrin

Well, I asked specifically what the question was. You said it was what I wrote. Now you changed the question. Very annoying.

Knight

22:19

I apologise, am I clear now?

@skullpatrol Hello Pal

skullpatrol

Hi pal

and re-hi prof Ted

:-)

Thorgott

should be able to use the IVT for that

Ted Shifrin

So both $c$ and $d$ are to be found.

Knight

Not to be found just to prove that there exists some $c$ and $d$

Ted Shifrin

And you can take $c=d$?

Knight

22:22

@Thorgott IVT : Intermediate Value Theorem, that is function will assume a value at all points

Ted Shifrin

That was a garbage sentence.

Knight

Sir are you in a bad mood?

Ted Shifrin

You think it means something?

Knight

What?

I think the foremost question is : Can we prove it?

Is it possible to prove it?

skullpatrol

Pal, don't you think you should first try to clearly word what you're trying to prove?

7 mins ago, by Knight

@Thorgott IVT : Intermediate Value Theorem, that is function will assume a value at all points

Knight

22:30

Pal that is not a question, that’s what I understood by IVT

skullpatrol

ok, clearly word what you understand :-)

it will help later

nitsua60

@skullpatrol (would have helped earlier, too)

Knight

Intermediate Value Theorem states that a function $f(x)$ will assume a value at all the intermediate points between $(a,b)$

skullpatrol

::clap clap::

much better pal

Knight

@nitsua60 I don’t know why but your words are feeling like bullying

nitsua60

22:34

Okay, that's fair. I'll step out. Sorry, I was piling on a bit.

skullpatrol

He's talking to me pal

Thorgott

that's not the IVT

nitsua60

@skullpatrol No, @Knight's perfectly right there to call me on it. The peanut gallery (this guy) wasn't helping anything.

Thorgott

and $f(x)$ is not a function

nitsua60

(I just kept coming back to look because the problem itself caught my fancy.)

Knight

22:35

@nitsua60 No, please don’t. You’re my friend’s friend

@Thorgott What is IVT?

nitsua60

@Knight I should be finishing the job of feeding my kids, anyway.

27 mins ago, by nitsua60

(hrmm... maybe "either." But I've got dinner on, so I can't sketch more than this margin holds)

Knight

@Thorgott What? $f(x) is not a function?

skullpatrol

cya

Knight

@nitsua60 How old are they :) ?

Thorgott

Intermediate value theorem

In mathematical analysis, the intermediate value theorem states that if f is a continuous function whose domain contains the interval [a, b], then it takes on any given value between f(a) and f(b) at some point within the interval. This has two important corollaries: If a continuous function has values of opposite sign inside an interval, then it has a root in that interval (Bolzano's theorem). The image of a continuous function over an interval is itself an interval. == Motivation == This captures an intuitive property of continuous functions: given f continuous on [1, 2] with the known values...

Knight

22:38

Okay

Why $f(x)$ is not a function?

Thorgott

$f$ is a function, $f(x)$ is a specific value the function takes

Knight

Okay

nitsua60

@Knight 11, 9, 7. Home-schooling (for the time being) one and all =\

Knight

@nitsua60 Wow! All boys?

Thorgott

now try to clearly phrase the question you want to ask, because it may actually be trivial

Knight

22:48

@Thorgott There is a function $f$ Which is continuous on (a,b) and assumes the value of zero at least once in the interval. Is it possible that $$\int_{c}^{d} f(x) dx = f(d)$$ ?

Please ask me if I’m still unclear

Thorgott

This is clear and answered positively by the function $f=0$

Knight

:-)

Yes, I too think that