@TedShifrin G(x,x').

Come on, man. What do these symbols mean?

You're going to make me guess? Calculus of variations with functions of $t$, and $x’$ means $x’(t)$?

@Marco Yes, I'm very familiar with Green's functions. So $\mathbf x$ and $\mathbf x'$ are independent variables, each in $\Bbb R^n$. Nothing to do with derivatives anywhere. If you write $\partial G/\partial{\mathbf x}$, I assume you mean the gradient vector (or its transpose, properly) of $G$ with respect to the first $n$ variables.

@TedShifrin Sorry, I ended up getting confused now.

Well, I think you were confused to start with :)

So $\textbf{r}'$ have nothing to do with derivative?

Correct.

These are two independent copies of the position vector.

So what means $\textbf{r}'$?

I told you. It's an independent position vector. $G$ is a function on $\Bbb R^n\times\Bbb R^n$ (missing the diagonal).

Green's functions are standard in partial differential equations and physics.

Ok, I will study this more. It's very confuse to me, because I thought that apostrophe were a symbol to denote derivative of a variable.

It can mean lots of things. If you don't have a function of $t$, for example, what would $x'$ mean?

That's why I assumed you were right and thought you were doing calculus of variations. But you're not.

Sorry.

Context is very important in mathematics. Anyhow, happy learning!

Ok. Thanks.

@robjohn (a+b)^2 ---> ${2+2-1}\choose{2-1}$ gives three terms, not 4, so it combines like terms

If $X\sim Beta(a,b)$ is there a closed formula for $P(X>x)$?

there's a formula for when $x > 1$

Hi @robjohn May I know who was your complex analysis teacher at UCLA?

@user91500 Steven Krantz

@robjohn Thanks.

Hi, a question here: I know how to prove $a^{p-1}\equiv1\pmod p,$ given $(a,p)=1$, but how to prove $a^p\equiv a\pmod p$? (the second one without $(a,p)=1$). $p$ is prime btw.

@FtyRain either $p \mid a$ or $(a,p) = 1$

Can someone please give me a hint on a problem? Question- Suppose you have an n-digit calculator (the total number of digits before and after the decimal point add up to n). You input an arbitrary number and keep on pressing the sqrt button. What is the max number of times (in terms of n only) you have to press the sqrt button to reach 1?

@tatan: You're never going to reach $1$ exactly. So the question is: How close to $1$ do you have to be to say you're done?

@TedShifrin Right. If we have 1.000 (n-1 zeroes) xyz... we are done

that's actually exactly what I used to do when I hadn't learnt limits

I would notice that eventually I get to 1

Can I ask to ask? But I just asked to ask to ask.

no, but you can ask to ask to ask to ask

@LeakyNun ask(ask(ask(x))) is permitted you say?

Let $ f: \mathbb{R} \to (0,1)$ be a continuous function. Then, which of the following function(s) has(have) the value zero at some point in the interval (0,1) ?

1. $e^x - \int_{0}^{x} f(t) \sin t dt$

It’s a strange question. I don’t think we can much in this problem.

None of them has to be zero.

Okay, but the book says (3) and (4) have zero as their values at some point.

The book is wrong

I want the book to be wrong, they have made such a bad question.

For the option (d) they write :

If $f\le 0$ on the interval, then (d) will always be positive.

$$ Let~ g(x) = x^9 - f(x) \\ g(0) = - f(0) \lt 0 \\ g(1) = 1- f(1) \gt 0$$

Oh, I totally missed the fact that $f$ has values in $(0,1)$.

Bad Ted.

Yeah, they're right.

Now it's a good question, if one actually reads it.

What I did was that I approximated $f(x)$ with either $\sin^2 x$ or $\cos^2 x $ and noting that they cannot have value 0 and 1.

So, I thought $x^9$ increases fastly than $\sin^2 x$ or $\cos^2 x$ and hence they may not have a same value at any point.

$x^9$ is not increasing fast at all. We're on the interval $(0,1)$.

So (1) clearly needn't have a zero if $f$ is small everywhere on the interval.

(2) the expression is obviously positive.

Now what about (3)?

Possible

Well, make the argument.

You know that $f\ge\epsilon>0$ on $(0,1)$. Why?

Yes, because f cannot be zero

And it is continuous

So we consider $f$ on the compact, closed interval $[0,1]$ and get a positive minimum.

And using IVT we can have your inequality

You need more than IVT for this, though, as I just said.

Okay

But, then you know that $\int_0^{\pi/2} f(t)\cos t\,dt \ge \epsilon$. So, yeah.

Can you please please explain a little more how we eradicated the option (1)

?

Say $0<f\le \epsilon$ on $[0,1]$. What do you know about the integral?

Calculus I is done.

Basic Real Analysis is done.

Huh?

(I was answering “What do you know about integrals?”)

Which one?

Use your head, man. You were asking about (1).

The integral is $$\int_{0}^{x} f(t) \sin t dt$$

$e^x$ will always be greater than 1

Yup.

and the integral will always be less than 1

Because $x$ is between 0 and 1

No.

What if $f$ is huge?

$f$ cannot be greater than 1. And sine function is also less than 1.

Oh, right, I kept forgetting that. But, I asked what happens if $f\le \epsilon$. Not needed.

(I’m a little confused about what is not needed)

My question.

So, the integral will always be less than 1, right?

Yes.

Thanks Ted.

Sure thing.