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

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10:00 PM

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

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

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

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

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

I know the steps at least

10:04 PM

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

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

I want a rigorous proof

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

Fund thm of calculus?

either'll work =)

why not take $f=0$

10:07 PM

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

@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

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

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

2

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

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.

10:09 PM

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

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

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

This is the FTC

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

$$f’(x) = f(x) $$

Go on.

What are the solutions of that?

$e^x$

I mean $ce^x$

10:13 PM

Ah

Different letter would be good, but yes.

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

Yes

$f(\alpha) =0$

It’s alpha

Is that what you had been calling $c$?

Now finish.

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?)

ge

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

10:19 PM

I apologise, am I clear now?

@skullpatrol Hello Pal

Hi pal

and re-hi prof Ted

:-)

should be able to use the IVT for that

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

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

And you can take $c=d$?

10:22 PM

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

That was a garbage sentence.

Sir are you in a bad mood?

You think it means something?

What?

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

Is it possible to prove it?

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

10:30 PM

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

ok, clearly word what you understand :-)

it will help later

@skullpatrol (would have helped earlier, too)

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

::clap clap::

much better pal

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

10:34 PM

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

He's talking to me pal

that's not the IVT

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

and $f(x)$ is not a function

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

10:35 PM

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

@Thorgott What is IVT?

@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)

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

cya

@nitsua60 How old are they :) ?

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...

10:38 PM

Okay

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

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

Okay

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

@nitsua60 Wow! All boys?

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

10:48 PM

@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

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

:-)

Yes, I too think that

helloo

11:04 PM

hi

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