Mathematics - 2022-03-11 (page 1 of 2)

user777

00:07

Hello, I have a question, if we have this function f(x)=4x-2, then the root of this function is at x=1/2. Now, If we use bisection method, let's say a=0 and b=3, we will see that f(0)= negative-sign and f(3)=positive-sign, then this shows that the solution are between 0 and 3. The only requirement to apply bisection method is to have a continuous function. Now, If I have this function f(x) = x^2-16x.

The roots are 0 and -16. Now If I apply bisection method, given two points they always going to give me a positive-sign, even though this function is continuous. Does this mean that bisection method doesn't work for some polynomial that always outputs a positive-sign?

here is some information about bisection method.
https://en.wikipedia.org/wiki/Bisection_method

copper.hat

@PM2Ring Thanks, I will take a look!

Xander Henderson

@user777 Yes, that is the correct interpretation. In order to use the method of bisection, you need a continuous function which is positive sometimes, and negative at other times.

user777

@XanderHenderson Thank you Xander, now it makes sense!

rb3652

00:30

OK, I have a really simple question.

If the force of gravity is proportional to an objects mass, how can all masses accelerate downwards at the same rate?

user4539917

F=km

rb3652

Can you clarify?

user4539917

If the force of gravity is proportional to an objects mass

rb3652

OK ...?

user4539917

then F = km

rb3652

00:35

Do you mean F = ma = mg?

user4539917

yes

rb3652

So since $F \propto m$, wouldn't $F$ be greater for a greater $m$, resulting in greater $a$?

Nevermind, this resolved all doubts: physics.stackexchange.com/questions/468349/…

user4539917

yup, a=g=k

sorry for being so terse :-)

Koro

01:12

What is the total number of reducible polynomials over $Z_p$ of the form $x^2+ax+b$ ?

For reducibility, it suffices to consider the total number of distinct factors $(x-a)(x-b)$. This can be done in $p^2$ ways (p ways to choose a and the same holds for b).

(x-a)(x-b) is same as (x-b)(x-a) son$p^2$ must be divided by 2 but that makes no sense if p is a prime >2.

Why is this paradox?

$p+(p-1)+(p-2)+…+1= p(p+1)/2$

Ted Shifrin

Huh? You're overcounting.

There is no paradox. You have to account intelligently for the case $a=b$.

Koro

That is, choose a=0 then b has p choices. Choose a=1, then b has p-1 choices (b=0 is counted earlier), If a=2 then b has p-2 choices…These are all mutually exclusive events so the total is $ p+(p-1)+(p-2)+….+1$.

But I still don’t understand why $p^2$ ( over 2) is wrong.

Ted Shifrin

That's what my comment was for.

Division by $2$ is wrong when $a=b$.

This is precisely why you look at the dimension of the space of symmetric matrices.

Koro

@TedShifrin Thanks a lot :-).

Ted Shifrin

Yup.

robjohn

02:47

$\overbrace{\frac{p(p-1)}2}^{a\lt b}+\overbrace{\quad p\vphantom{\frac12}\quad}^{a=b}=\frac{p(p+1)}2$

Ted Shifrin

Counting the dimension of symmetric matrices :)

@robjohn Re this: I got that in my head, but I don’t immediately see how that leads to a Taylor expansion.

robjohn

@TedShifrin it won't be a Taylor expansion because there are mixed $x$ and $\log(x)$ terms.

Ted Shifrin

I know.

robjohn

$$\sqrt{\log(x)}\left(\sqrt{1+\frac{\log(1+1/x)}{\log(x)}}-1\right)$$

This can be worked into an expansion

Ted Shifrin

Oh, I did it by multiplying by the conjugate.

But they want an expansion at infinity.

robjohn

02:54

Yes. That is why I wrote it that way. I will get a few terms...

Ted Shifrin

What a horrid exercise.

robjohn

$\frac1{2x\sqrt{\log(x)}}-\frac1{4x^2\sqrt{\log(x)}}-\frac1{8x^2\log(x)^{3/2}}+O\!\left(\frac1{x^3\sqrt{\log{x}}}\right)$

it is not pretty

Ted Shifrin

But that won’t be a power series in $1/x$?

robjohn

How is that a power series in $\frac1x$ when there are the $\log(x)$s in the denominators? maybe we are working on different ideas of a power series

Ted Shifrin

Aren’t we agreeing?

robjohn

03:03

Maybe we are and I am being stupid :-)

Ted Shifrin

The original question was a Taylor expansion at $\infty$. It seemed wrong to me.

robjohn

Yes, there is no Taylor series at $\infty$

Ted Shifrin

We concur. Stooopid exercise or bad representation.

Given the asker, one never knowz.

robjohn

There are occasions like that

Ted Shifrin

Verily.

PM 2Ring

05:28

@user777 No, the roots of $f(x)=x^2-16x$ are 0 and 16. And $f(x)$ is negative between those roots. If $f(x)$ is positive for all real $x$ then it has no real roots. Here's a graph of your function:

leslie townes

06:36

seen one parabola, you seen 'em all. that's my view.

some are smiley, some are frowny. it's all the same to me.

copper.hat

depends on your focus

PM 2Ring

07:08

True, they're all the same shape. But some cross the X axis & some don't, and that's kind of important for user777's question. I'm tempted to post the graph of a cubic with repeated roots, showing that a poly can touch the X axis without crossing it, since you have to deal with that when searching for roots via bisection. And repeated roots are painful with Newton-Raphson, too.

copper.hat

The numerical crux, when is something zero...

Ted Shifrin

@copper.hat You are a man with no directrix.

copper.hat

I am equidistant from many things.

PM 2Ring

07:24

We had to do lots of curve sketching when I was at school. I guess kids these days don't do much of that, since they have computers & graphing calculators.

robjohn

08:02

@copper.hat you are central

copper.hat

08:35

:-). We had limited access to graph paper, which made it valuable. I loved drawing on graph paper.

Good night folks!

Vrouvrou

09:04

@robjohn how to get this

robjohn

6 hours ago, by robjohn

$$\sqrt{\log(x)}\left(\sqrt{1+\frac{\log(1+1/x)}{\log(x)}}-1\right)$$

6 hours ago, by robjohn

$\frac1{2x\sqrt{\log(x)}}-\frac1{4x^2\sqrt{\log(x)}}-\frac1{8x^2\log(x)^{3/2}}+O\!\left(\frac1{x^3\sqrt{\log{x}}}\right)$

Vrouvrou

oh thank you

robjohn

Using $\log(1+x)=x-\frac{x^2}2+\frac{x^3}3$ and $\sqrt{1+x}-1=\frac{x}2-\frac{x^2}8+\frac{x^3}{16}$

Vrouvrou