Obviously the definition of 0^x is not 0 because there is an exception. If you use a power series to solve anything it is only an approximation, so I don't know why you would mention it you just wanted to simplify the expression.
Obviously you do not understand what I am getting at with the power series example at all. It has nothing to do with approximation. Let me try to make it even simpler by removing the calculus aspect. Consider the following function which I call f.
f(x) = 1 + 2x + 3x^2 + 4x^3
This is just a very simple function, but it's fairly long, we we might want to make it simpler by writing it as
f(x) = sum of ( (n + 1) * x^n ) from n = 0 to 3
This doesn't look simpler in text, but it is simpler with the actual notation.
Anyway, notice how with my new version of f(x), you can't evaluate it at 0. Let's write down what f(0) gives you for my new definition of f.
f(0) = 1 * 0^0 + 2 * 0^1 + 3 * 0 ^ 2 + 4 * 0 ^ 3
Notice how this contains a 0^0 in the first term. So, in order to use my new version of f(x), we have to define 0^0 to be 1. This is why sometimes mathematicians will define 0^0 to be 1, but except in particular contexts like this, it is generally left undefined.
Math is good in that it is in everything and is always true; people cannot write down "properties they want exponentiation to have", because exponentiation is nothing contrived by people. The properties of those functions are rules with an exception (at x=0). If those were both true, it would then be defined twice, and not undefined.
Exponentiation is very much something constructed by people. Where do you think it comes from? What a^n means for natural number n is obvious:
a^n = a * a * a * ... * a where there are n copies of a
However, to raise a to something other than a natural number, we need to get more creative. For a rational number, we define a^(1/m) so that a^(1/m) = the principal m-th root of a. But what about an irrational number? A complex number? These are all things we construct, and how do we know what their values should be? The way we decide it is by writing down the properties of exponentiation that we like, and then making a more general function that also has those properties.
Two of the properties we like are the ones I listed above. They do not apply when a = 0, because if you extend them to a = 0, you get a contradiction. As a result, either we must leave 0^0 undefined, or we must explicitly give it a value. The choice is ARBITRARY. There is no "right" choice. It is a question of definition and one you make for the particular work you are working on. As I said, most mathematicians prefer to leave it undefined except when working with series. (Again, this has nothing to do with calculus or approximation per se.)
Moving along..
x^0 = x^(1-1) = (x^1)/(x^1)
From here, if enter 0, the answer is 1. If you cancel the top and bottom, the answer is 1.
The problem here is that by writing down x^0 on the left, you have already slapped a restriction on x that it is not equal to 0, since 0^0 is undefined. You could only carry out the manipulation you go on to do if you have already defined 0^0 to be 1, so you are not proving anything.
I believe that addresses all of your questions.
I was wondering how would you change x^2 + y^2 = z^2 into spherical and cylindrical coordinates?
This is just a matter of knowing the definition of spherical and cylindrical coordinates.
For spherical coordinates, replace x by r cos A sin B, replace y by r sin A cos B, and replace z by r cos B. Then just simplify the expression.
For cylindrical coordinates, replace x by p cos A, replace y by p sin A, and z is unchanged. Then simplify.
So really, this is purely a matter of knowing the definition. If you forget it, just check Wikipedia.
![[oni]LoKo](/data/avatars/l/69/69958.jpg?1360760938){kind=avatar}
