Some questions I am not very sure about

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In summary, according to this equation, a chain rule equation can be found by solving for x in terms of y and z, y in terms of x or z, or z in terms of x and y.
  • #1
VietDao29
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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:
[tex]a ^ 0 = 1, \ \forall a \neq 0[/tex], 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:
[tex]\frac{dx}{dy} \frac{dy}{dz} \frac{dz}{dx} = -1[/tex]. However, I am not sure how to arrive at this equality.
I think it should be:
[tex]\frac{dx}{dy} \frac{dy}{dz} \frac{dz}{dx} = \frac{dx}{dz} \frac{dz}{dx} = \frac{dx}{dx} = 1[/tex].
I just wonder if I did make a mistake somewhere? :confused:
Thanks, :smile:
 
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  • #2
VietDao29 said:
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:
[tex]a ^ 0 = 1, \ \forall a \neq 0[/tex], 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

[tex]\sum_{k=0}^{\infty}a_k x^k[/tex]

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

[tex]a_0 + \sum_{k=1}^{\infty}a_k x^k[/tex]

which would be tiresome.
 
  • #3
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
VietDao29 said:
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.
 
  • #5
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
shmoe said:
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! o:) )

-Dan
 
  • #7
VietDao29 said:
The second question is about the chain rule. I did see someone posted something like:
[tex]\frac{dx}{dy} \frac{dy}{dz} \frac{dz}{dx} = -1[/tex]. However, I am not sure how to arrive at this equality.
I think it should be:
[tex]\frac{dx}{dy} \frac{dy}{dz} \frac{dz}{dx} = \frac{dx}{dz} \frac{dz}{dx} = \frac{dx}{dx} = 1[/tex].
I just wonder if I did make a mistake somewhere? :confused:
Thanks, :smile:

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

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

where the subscript means "keep fixed" (the first parenthesis means "derivate x wrt y and keep z fixed" and so on).
 
  • #8
assyrian_77 said:
Actually that looks like the cyclical relation or the cyclical rule (from what I remember). It looks like this:

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

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,
 
  • #9
Sure, consider the equation for the plane,
[tex]ax+by+cz=d[/tex]
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:
[tex]x=-\frac{b}{a}y-\frac{c}{a}z[/tex]
Or if you'd rather solve for y in terms of x or z:
[tex]y=-\frac{a}{b}x-\frac{c}{b}z[/tex]
Or you may solve for z in terms of x and y:
[tex]z=-\frac{a}{c}x-\frac{b}{c}y[/tex]

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

Multiply these together and get the amusing result.
 
  • #10
Good example. The cyclical relation is quite useful in Thermodynamics / Statistical Mechanics with all the relations between the different variables. Here is an example:

[tex]\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[/tex]

where S is the entropy, T the temperature and P the pressure.
 
  • #11
The general case is as follows, let G be a function of n variables, and consider the equation for constant G:
[tex]G(x_{1},x_{2},...,x_{n})=K[/tex]
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 [tex]\hat{x}[/tex], 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:
[tex]x_{i}=X_{i}(x_{1},..x_{j},..,x_{n}), j=1,..,n, j\neq{i}[/tex]
so that we identically have in our region:
[tex]G(x_{1},..,X_{i}(x_{1},..x_{j},..,x_{n}),...,x_{n})=K[/tex]

By differentiating these relations, we should in particular have:
[tex]\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}[/tex]

Multiplying all these together, you'll get:
[tex]\prod_{i=1}^{n}\frac{\partial{X}_{i}}{\partial{x}_{i+1}}=(-1)^{n},x_{n+1}\equiv{x}_{1}[/tex]
 
  • #12
Yes, thanks for all the replies, guys. It's very clear. Thanks, :)
 

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