The h Bar

Abcd

11:03 AM

@Blue Beyond my scope

Anonymous

At least you should be able to understand the "general proof"

Anonymous

It's just the mean value theorem

Abcd

@Blue which general proof?

Anonymous

"General form with variable limits"

Anonymous

See the Wiki page

Anonymous

11:05 AM

Anyhow, are you looking for a proof or just intuition for why that third term is necessary?

Abcd

@Blue In the third term if he has added alpha + delta alpha then in limits he should do subtraction right?

(of general form)

@Blue please see the 3rd line^

Anonymous

The limits are the limits of the variable $x$

Abcd

@Blue Why should they change when alpha is changing

Anonymous

When you change $\alpha$ by $\Delta \alpha$ they change by $\Delta a$ and $\Delta b$

Anonymous

@Abcd By definition: "where a and b are functions of α that exhibit increments Δa and Δb, respectively, when α is increased by Δα."

Anonymous

11:09 AM

It's called the "General form with variable limits"

Abcd

@Blue Okay, but we had to prove something else right?

Anonymous

No. Even here your limits are variable

Anonymous

$v(x)$ and $u(x)$

Sid

(If anyone's interested and didn't know, do look up "White Death". One of the greatest soldiers that ever lived)

Abcd

11:10 AM

@Blue But here they are depending on x and not $t$

Anonymous

Sure. Because now it's $dt$ instead of $dx$

Novalium Company

Guys, a quick and stupid question. When we say something is oxidized we mean that it has lost electron/s. For example in a wood - oxygen reaction (burning) the oxygen is the oxidizing agent and makes wood loose electrons, where wood is the reductant agent. When the wood looses electrons we say it gets oxidized and the process is oxidation?

Anonymous

Just some re-labelling is going on

Anonymous

The basic logic is still the same

Abcd

@Blue I see. let me read further

Sid

11:13 AM

@NovaliumCompany Usually, Oxidation and Reduction Take place simultaneously. So yeah, wood is getting oxidized

Novalium Company

@Sid Hmm, ok. When we see this red/orange fire over the wood, it's actually the air getting electrons from the wood? (and wood turns black because it no longer has much electrons)?

Sid

Ehhh. I am not sure about that. Charring of wood takes place because of prolonged heating

The color is usually due to emission of light by impurities in the wood

Novalium Company

Ok, I'm just trying to understand what is that fire over the wood and why does wood turn black.

"Fire is the rapid oxidation of a material in the exothermic chemical process of combustion" (Wikipedia) So that red/orange fire over the wood is basically the air getting electrons (oxidizing the wood)?

Secret

Let $\int_{v(x)}^{u(x)}f(x',t)dt = F(x',u(x))-F(x',v(x))$. Then

\begin{align}
\frac{d}{dx} (F(x',u(x))-F(x',v(x))) & = \frac{\partial F (x',u(x))}{\partial u(x)} \frac{d u(x)}{dx} - \frac{\partial F (x',v(x))}{\partial v(x)} \frac{d v(x)}{dx} + \frac{\partial F (x',u(x))}{\partial x'} - \frac{\partial F (x',v(x))}{\partial x'}\\
& = f(x',u(x)) \frac{d u(x)}{dx} - f(x',v(x)) \frac{d v(x)}{dx} + \frac{\partial }{\partial x'}F (x',t)|_{t=v(x)}^{t=u(x)}\\
& = f(x',u(x)) \frac{d u(x)}{dx} - f(x',v(x)) \frac{d v(x)}{dx} + \lim_{h\to 0} \left[\frac{F (x'+h,t)|_{t=v(x)}^{t=u(x)} - F (x',t)|_{t=v(x…

Abcd

11:28 AM

@Blue Didn't get this part^ ?

Anonymous

What is the definition of $\epsilon_1$ and $\epsilon_2$?

Anonymous

Actually what are those symbols called?

Anonymous

I forgot

Abcd

@Blue its given on the wiki page

Anonymous

I know, I was asking you ;)

Anonymous

11:31 AM

They're just using the mean value theorem

Anonymous

If $\Delta \alpha\to 0$

Anonymous

What happens?

Anonymous

$a<\epsilon<b$

Anonymous

$\Delta a\to 0$

Abcd

@Blue ya

@Blue so they just substituted 0

Anonymous

11:35 AM

So the integral $\int_a^{a+\Delta a}$ tends to 0

Anonymous

$a<\epsilon_1<a+\Delta a$

Anonymous

So it is clear that $\epsilon_1\to a$

Anonymous

It's squished between those two values

Abcd

hmm

Anonymous

After that it looks pretty simple

Abcd

11:39 AM

@Blue How did he get the last line

Anonymous

$$\lim_{\Delta \alpha\to 0} \frac{-\Delta a f(\epsilon_1,\alpha+\Delta\alpha )}{\Delta \alpha} = ?$$

Abcd

@Blue ya this part

@Blue replace epsilon by a and delta alpha by 0

Anonymous

@Abcd Right

Anonymous

And in the limit, $\Delta a/\Delta \alpha = da/d\alpha$

Abcd

@Blue Oh

Anonymous

11:43 AM

If this part seems a bit hand-wavy go through the bounded convergence theorem

Abcd

Understood the derivation @Blue. Thanks.

Anonymous

Cool!

Abcd

Now what's the intuition?

Anonymous

@Abcd To develop intuition, it's best to see the 6 solved examples on the Wiki page

Abcd

@Blue No whats the significance of those terms

Anonymous

11:45 AM

@Abcd Which terms?

Abcd

thanks for your attempt @secret

Anonymous

The one with the partial derivative?

Abcd

@Blue all terms in the formula

Anonymous

Eh, that's a strange question :P

Anonymous

It's just a formula

Secret

11:46 AM

I really need to think of a more abstract way to write $\int_{stuff}^{stuff}f(stuff,t)dt$ as a function

Anonymous

You're simply finding the derivative of an integral (which isn't a constant but rather in terms of variables)

Secret

In fact, I don't know if you can really derive that 3rd term with chain rule alone

Anonymous

For instance $\int_0^1 x dx$ is a constant and you'll get 0 when differentiating them

Anonymous

But $\int_{x}^{2x}sin(xt)dt$ isn't

Abcd

@Blue No some sort of geometrical interpretation

Secret

11:47 AM

What I really like is to be able to see what that term look like right before Dominant convergence theorem is applied to push the lim into the integral sign

Abcd

@Blue I understood almost all theorems of calculus till now using their geometrical interpretation only. That's so obvious stuff.

Shing

I am reading this, and I don't quite understand the last statement "The author declares no competing interests."

nature.com/articles/…

I don't think he uses the term "competing interests" like how the term is used in medical science

Anonymous

@Abcd That's a good and hard question. Are you looking for something like this?

Albas

Does the action angle variable and the liouville's theorem have something in common? I say this because can't liouville's theorem be stated by saying that enclosed areas remain constant in the phase space?

Abcd

@Blue ok ill see

Anonymous

11:56 AM

@Abcd Trying to find geometric intuitions behind real analysis results is obviously a good habit, but it isn't very easy and perhaps not always possible

Anonymous

I personally haven't seen any completely geometric treatment of the Feynman method aka Leibniz rule

Anonymous

You can ask on Math SE or Math Overflow if you're interested

Secret

hmmmmmm....

Abcd

Does the paper youve linked me to not cover it?
I will read that at night

Secret

Let $\int f(x,t)dt = I(f(x,t),t')$. Then $I$ has the following bizarre properties:

$\frac{\partial I(f(x,t),t')}{\partial t'} = f(x,t)$

Anonymous

12:04 PM

@Abcd For the one dimensional case it does

Anonymous

I was looking for the general version

Anonymous

But 1D will suffice for you

Secret

But... what does $\frac{\partial I(f(x,t),t')}{\partial f}$ means...?

hmm... let's wrote this formally:

For all $x$
$$\frac{\partial \cdot}{\partial x} = \lim_{h\to 0} \frac{\cdot (x+h) - \cdot x}{h}$$

Therefore:

$$\frac{\partial \int f(x,t)dt}{\partial f} = \lim_{h\to 0} \frac{\int (f+h)(x,t) dt - \int f(x,t)dt}{h}$$

Anonymous

@Secret Just use chain rule?

Anonymous

$f$ is function of $x$ and $t$

Anonymous

12:09 PM

$dI/df=dI/dx.dx/df + dI/dy.dy/df$

Anonymous

$d = \partial$

Anonymous

You could obviously substitute $f$ with a variable $z$

Anonymous

Nothing strange here

Anonymous

It's simply partial derivative of $I$ w.r.t a variable $z$

Anonymous

7

Q: Derivative of $f(x,y)$ with respect to another function of two variables $k(x,y)$

Suppose that we have a function $f(x,y)$ of two variables: $$f(x,y) = g(x) + h(y) + 5(x-y) = x^2 + y^2 + 5(x-y)$$ where $g(x) = x^2$ and $h(y) = y^2$ are also functions of $x$ and $y$, respectively. How do I take the partial derivative of $f(x,y)$ with respect to another multivariate function ...