# Homework Help: Some questions I am not very sure about

1. Mar 8, 2006

### 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?
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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,

2. Mar 8, 2006

### quasar987

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.

3. Mar 8, 2006

### 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?

4. Mar 8, 2006

### TD

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

5. Mar 8, 2006

### 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.

6. Mar 8, 2006

### topsquark

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

7. Mar 8, 2006

### assyrian_77

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).

8. Mar 9, 2006

### VietDao29

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

9. Mar 9, 2006

### 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.

10. Mar 9, 2006

### 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.

11. Mar 9, 2006

### 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 [itex]\hat{x}=(\hat{x}_{1},\hat{x}_{2}...,\hat{x}_{n})[/tex].
In a neighbourhood of $$\hat{x}$$, it should be possible to solve for any one of the coordinates as some function of the other coordinates, i.e, there should exist n functions of the type:
$$x_{i}=X_{i}(x_{1},..x_{j},..,x_{n}), j=1,..,n, j\neq{i}$$
so that we identically have in our region:
$$G(x_{1},..,X_{i}(x_{1},..x_{j},..,x_{n}),....,x_{n})=K$$

By differentiating these relations, we should in particular have:
$$\frac{\partial{X}_{i}}{\partial{x}_{i+1}}=-\frac{\frac{\partial{G}}{\partial{x}_{i+1}}}{\frac{\partial{G}}{\partial{x}_{i}}},i=1,..n, x_{n+1}\equiv{x}_{1}$$

Multiplying all these together, you'll get:
$$\prod_{i=1}^{n}\frac{\partial{X}_{i}}{\partial{x}_{i+1}}=(-1)^{n},x_{n+1}\equiv{x}_{1}$$

12. Mar 10, 2006

### VietDao29

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