Some questions I am not very sure about

[VietDao29]

I have some questions that I don't really know for sure, and think that it would be great if you guys can help me out.
In high school, I am taught that:
a ^ 0 = 1, \ \forall a \neq 0, and hence 0⁰ is not defined. However, I've recently read an article in Wikipedia that states a⁰ = 1, and it's also true for a = 0. So I am a little bit confused here.
Is 0⁰ defined? What's the international rule?
---------------
The second question is about the chain rule. I did see someone posted something like:
\frac{dx}{dy} \frac{dy}{dz} \frac{dz}{dx} = -1. However, I am not sure how to arrive at this equality.
I think it should be:
\frac{dx}{dy} \frac{dy}{dz} \frac{dz}{dx} = \frac{dx}{dz} \frac{dz}{dx} = \frac{dx}{dx} = 1.
I just wonder if I did make a mistake somewhere?
Thanks,

 

[quasar987]

I have some questions that I don't really know for sure, and think that it would be great if you guys can help me out.
In high school, I am taught that:
a ^ 0 = 1, \ \forall a \neq 0, and hence 0⁰ is not defined. However, I've recently read an article in Wikipedia that states a⁰ = 1, and it's also true for a = 0. So I am a little bit confused here.
Is 0⁰ defined? What's the international rule?

I've always tought 0^0 is undefined. I would be surprised if someone said otherwise. There are cases however, where we chose to regard o^0 as 1 in order to simplify the notation. Take a power serie for exemple. We write

\sum_{k=0}^{\infty}a_k x^k

But evaluate that that x=0 and your first term is a_0 0^0, which is not defined. But by convention, we chose to regard that term as just a_0 just to allow this writing. Otherwise, we'd have to always write power series as

a_0 + \sum_{k=1}^{\infty}a_k x^k

which would be tiresome.

 

[VietDao29]

Yes, I see your point of simplifying the notation. Thanks, :)
By the way, I found it here, section 1.2 Exponents one and zero.
So in that article, it's just plain wrong, right? The way I see it, the writer is defining 0⁰ to be 1, which is a bit ambiguity, and also it does not seem correct to me...
Should I correct the article?

 

[TD]

So in that article, it's just plain wrong, right? The way I see it, the writer is defining 0⁰ to be 1, which is a bit ambiguity,

I've seen it being defined as 1 before, I wouldn't correct it since there isn't an international consensus about this.

 

[shmoe]

It seems like there should be a note about how this isn't universal and it's sometimes just left as undefined. This has been discussed many times here, so search around if you want more.

 

[topsquark]

It seems like there should be a note about how this isn't universal and it's sometimes just left as undefined. This has been discussed many times here, so search around if you want more.

Warning: Some people take 0^0=1 as obvious. I once heard an NCU Physics professor chew her grad student out because he didn't "know that."

Personally, I have found good arguments to call it either 0 or 1 depending on how you want to look at the situation. So I agree: undefined. (Sorry NCU prof! )

-Dan

 

[assyrian_77]

The second question is about the chain rule. I did see someone posted something like:
\frac{dx}{dy} \frac{dy}{dz} \frac{dz}{dx} = -1. However, I am not sure how to arrive at this equality.
I think it should be:
\frac{dx}{dy} \frac{dy}{dz} \frac{dz}{dx} = \frac{dx}{dz} \frac{dz}{dx} = \frac{dx}{dx} = 1.
I just wonder if I did make a mistake somewhere?
Thanks,

Actually that looks like the cyclical relation or the cyclical rule (from what I remember). It looks like this:

\Big(\frac{dx}{dy}\Big)_z\Big(\frac{dy}{dz}\Big)_x\Big(\frac{dz}{dx}\Big)_y=-1

where the subscript means "keep fixed" (the first parenthesis means "derivate x wrt y and keep z fixed" and so on).

 

[VietDao29]

Actually that looks like the cyclical relation or the cyclical rule (from what I remember). It looks like this:

\Big(\frac{dx}{dy}\Big)_z\Big(\frac{dy}{dz}\Big)_x\Big(\frac{dz}{dx}\Big)_y=-1

where the subscript means "keep fixed" (the first parenthesis means "derivate x wrt y and keep z fixed" and so on).

Uhmm, not very sure if I understand it. Can you give me an example?
Thanks,

 

[arildno]

Sure, consider the equation for the plane,
ax+by+cz=d
and assume that a,b,c are non-zero (I'll set d=0 for simplicity.

Hence, you may if you wish solve for x in terms of y and z:
x=-\frac{b}{a}y-\frac{c}{a}z
Or if you'd rather solve for y in terms of x or z:
y=-\frac{a}{b}x-\frac{c}{b}z
Or you may solve for z in terms of x and y:
z=-\frac{a}{c}x-\frac{b}{c}y

Using the appropriate expression, we may find partial derivatives:
\frac{\partial{x}}{\partial{y}}=-\frac{b}{a},\frac{\partial{y}}{\partial{z}}=-\frac{c}{b},\frac{\partial{z}}{\partial{x}}=-\frac{a}{c}

Multiply these together and get the amusing result.

 

[assyrian_77]

Good example. The cyclical relation is quite useful in Thermodynamics / Statistical Mechanics with all the relations between the different variables. Here is an example:

\Big(\frac{\partial{S}}{\partial{T}}\Big)_{P}\Big(\frac{\partial{P}}{\partial{S}}\Big)_{T}\Big(\frac{\partial{T}}{\partial{P}}\Big)_{S}=-1

where S is the entropy, T the temperature and P the pressure.

 

[arildno]

The general case is as follows, let G be a function of n variables, and consider the equation for constant G:
G(x_{1},x_{2},...,x_{n})=K
Assume that we have a solution of this equation called \hat{x}=(\hat{x}_{1},\hat{x}_{2}...,\hat{x}_{n})[/tex].<br /> In a neighbourhood of \hat{x}, it should be possible to solve for anyone of the coordinates as some function of the other coordinates, i.e, there should exist n functions of the type:<br /> x_{i}=X_{i}(x_{1},..x_{j},..,x_{n}), j=1,..,n, j\neq{i}<br /> so that we identically have in our region:<br /> G(x_{1},..,X_{i}(x_{1},..x_{j},..,x_{n}),...,x_{n})=K<br /> <br /> By differentiating these relations, we should in particular have:<br /> \frac{\partial{X}_{i}}{\partial{x}_{i+1}}=-\frac{\frac{\partial{G}}{\partial{x}_{i+1}}}{\frac{\partial{G}}{\partial{x}_{i}}},i=1,..n, x_{n+1}\equiv{x}_{1}<br /> <br /> Multiplying all these together, you&#039;ll get:<br /> \prod_{i=1}^{n}\frac{\partial{X}_{i}}{\partial{x}_{i+1}}=(-1)^{n},x_{n+1}\equiv{x}_{1}

 

[VietDao29]

Yes, thanks for all the replies, guys. It's very clear. Thanks, :)