This is a cute way of explaining limits, but there's a better explanation for 00, which comes from the definition of an exponent. He defines an as 'a * a * a * ... n times'. But we can do better,...
This is a cute way of explaining limits, but there's a better explanation for 00, which comes from the definition of an exponent.
He defines an as 'a * a * a * ... n times'. But we can do better, by defining exponentiation recursively*.
a0 = 1
an = a * an-1, when n is natural and n>0
am = 1/(a-m)
ap/q = q√(ap) (that is, the q-th root of ap)
x0 = 1 is actually a rule. 0x = 0 is not a rule, it just happens to be true for all non-negative values of x except for 0.
(Side-note: @deimos, any chance of getting latex support?)
*: This definition does not work for irrational numbers. Given a proper definition of real numbers, it can be trivially extended, but that's mostly beside the point and takes up a lot of space.
Proof sketch of 0x = 0 for any x>0: 01 = 0 * 00 = 0 * 1 = 0 For any n, if 0n = 0, then 0n+1 = 0 * 0n = 0 * 0 = 0 Therefore, 0n = 0, for all natural n>0 For any natural p>0, q>0, 0p/q = q√(0p) =...
Proof sketch of 0x = 0 for any x>0:
01 = 0 * 00 = 0 * 1 = 0
For any n, if 0n = 0, then 0n+1 = 0 * 0n = 0 * 0 = 0
You can extend it by noting that fa(x)=ax is continuous. Do do that, just note that fa(x+\epsilon) = fa(x)fa(\epsilon). Them, as \epsilon->0 fa(\epsilon)->0 taking \epsilon to be in the rationals....
You can extend it by noting that fa(x)=ax is continuous. Do do that, just note that fa(x+\epsilon) = fa(x)fa(\epsilon). Them, as \epsilon->0 fa(\epsilon)->0 taking \epsilon to be in the rationals. This implies the extension to the reals is continuous, which implies that it is 0 throughout.
If a^n = a*a^{n-1} for any n > 0, then let n be 1 so that we have a = a*a^0, which implies that it must be so that a^0 = 1 and that it’s not “just a rule”.
If a^n = a*a^{n-1} for any n > 0, then let n be 1 so that we have a = a*a^0, which implies that it must be so that a^0 = 1 and that it’s not “just a rule”.
I think it’s way more important that we make sure that a^1 = a then it is that a^0 = 1, and so in my opinion a^0 = 1 should be viewed as a consequence of a^1 = a and not the other way around.
I think it’s way more important that we make sure that a^1 = a then it is that a^0 = 1, and so in my opinion a^0 = 1 should be viewed as a consequence of a^1 = a and not the other way around.
I didn't think that was an assumption. I thought that fell out of the definition of exponents. If an is a multiplied by itself n times, then a multiplied by itself one time is a.
That relies on another assumption: that a1=a.
I didn't think that was an assumption. I thought that fell out of the definition of exponents. If an is a multiplied by itself n times, then a multiplied by itself one time is a.
That only works for integer exponentiation. When you start to evaluate, say, e^3.83, that definition no longer well defined. Usually exponentiation is defined as the infinite sum of x^k / k!
That only works for integer exponentiation. When you start to evaluate, say, e^3.83, that definition no longer well defined.
Usually exponentiation is defined as the infinite sum of x^k / k!
To clarify, the idea is that we want to define an entirely new exponentiation function on the reals, whose values happen to coincide with the integer definition.
To clarify, the idea is that we want to define an entirely new exponentiation function on the reals, whose values happen to coincide with the integer definition.
The definition of exponents isn't that an is a multiplied by itself n times. That was where we started, but then we replaced it with the more precise recursive definition. That an = a multiplied...
The definition of exponents isn't that an is a multiplied by itself n times. That was where we started, but then we replaced it with the more precise recursive definition. That an = a multiplied by itself n times is a desirable property for a definition of exponentiation to have. But it's not clear what that means in a mathematical sense; that's just the intuitive understanding.
Well, it really depends on which exponent you're talking about. Typically people talk about real/complex exponentiation, which is indeed no longer repeated multiplication, but taking an exponent...
Well, it really depends on which exponent you're talking about. Typically people talk about real/complex exponentiation, which is indeed no longer repeated multiplication, but taking an exponent is defined for everything monoid+. In groups, the exponent laws work as expected as well.
There's an issue for it on the tracker, but I posted a comment about why it's not really a simple thing to add. I'm not sure whether it'll be added, it's very niche.
(Side-note: @deimos, any chance of getting latex support?)
Looking at KaTeX's website, it looks like they support server-side rendering to html. So you could do the latex rendering at comment submission time, but since it's all html it would still be...
Looking at KaTeX's website, it looks like they support server-side rendering to html. So you could do the latex rendering at comment submission time, but since it's all html it would still be fast, respect themes, look ok on mobile, etc.
That just leaves the problem of previews. I think that having latex previews disabled by default, with a userpref to enable them would be a good solution.
Yes, the server-side rendering is the option that requires the large CSS/fonts (and possibly still needs the JS too, I'm not sure). Previews are already rendered server-side, so it shouldn't be an...
Yes, the server-side rendering is the option that requires the large CSS/fonts (and possibly still needs the JS too, I'm not sure). Previews are already rendered server-side, so it shouldn't be an issue overall to include latex in that.
It's definitely not something I'm opposed to adding, if we can find a good way to do it. It's also very low-priority though, since comments that would use it are extremely rare. It's probably a good candidate for an outside contributor adding as a merge request, if someone feels strongly about adding support and wants to work on it.
This is a cute way of explaining limits, but there's a better explanation for 00, which comes from the definition of an exponent.
He defines an as 'a * a * a * ... n times'. But we can do better, by defining exponentiation recursively*.
x0 = 1 is actually a rule. 0x = 0 is not a rule, it just happens to be true for all non-negative values of x except for 0.
(Side-note: @deimos, any chance of getting latex support?)
*: This definition does not work for irrational numbers. Given a proper definition of real numbers, it can be trivially extended, but that's mostly beside the point and takes up a lot of space.
Proof sketch of 0x = 0 for any x>0:
01 = 0 * 00 = 0 * 1 = 0
For any n, if 0n = 0, then 0n+1 = 0 * 0n = 0 * 0 = 0
Therefore, 0n = 0, for all natural n>0
For any natural p>0, q>0, 0p/q = q√(0p) = q√0
let x = q√0. xq = 0. Therefore, x=0
Therefore, 0p/q = 0
QED
That only proves it for rational x
Yes; as I mention in the GP, proving anything for reals requires a formal definition of reals, which is nontrivial.
You can extend it by noting that fa(x)=ax is continuous. Do do that, just note that fa(x+\epsilon) = fa(x)fa(\epsilon). Them, as \epsilon->0 fa(\epsilon)->0 taking \epsilon to be in the rationals. This implies the extension to the reals is continuous, which implies that it is 0 throughout.
If a^n = a*a^{n-1} for any n > 0, then let n be 1 so that we have a = a*a^0, which implies that it must be so that a^0 = 1 and that it’s not “just a rule”.
That relies on another assumption: that a1=a. You can use either one (a0=1, or a1=a) as your 'bottom' rule, but I find a0=1 more elegant.
I think it’s way more important that we make sure that a^1 = a then it is that a^0 = 1, and so in my opinion a^0 = 1 should be viewed as a consequence of a^1 = a and not the other way around.
I didn't think that was an assumption. I thought that fell out of the definition of exponents. If an is a multiplied by itself n times, then a multiplied by itself one time is a.
That only works for integer exponentiation. When you start to evaluate, say, e^3.83, that definition no longer well defined.
Usually exponentiation is defined as the infinite sum of x^k / k!
To clarify, the idea is that we want to define an entirely new exponentiation function on the reals, whose values happen to coincide with the integer definition.
The definition of exponents isn't that an is a multiplied by itself n times. That was where we started, but then we replaced it with the more precise recursive definition. That an = a multiplied by itself n times is a desirable property for a definition of exponentiation to have. But it's not clear what that means in a mathematical sense; that's just the intuitive understanding.
Well, it really depends on which exponent you're talking about. Typically people talk about real/complex exponentiation, which is indeed no longer repeated multiplication, but taking an exponent is defined for everything monoid+. In groups, the exponent laws work as expected as well.
There's an issue for it on the tracker, but I posted a comment about why it's not really a simple thing to add. I'm not sure whether it'll be added, it's very niche.
Looking at KaTeX's website, it looks like they support server-side rendering to html. So you could do the latex rendering at comment submission time, but since it's all html it would still be fast, respect themes, look ok on mobile, etc.
That just leaves the problem of previews. I think that having latex previews disabled by default, with a userpref to enable them would be a good solution.
Thoughts?
Yes, the server-side rendering is the option that requires the large CSS/fonts (and possibly still needs the JS too, I'm not sure). Previews are already rendered server-side, so it shouldn't be an issue overall to include latex in that.
It's definitely not something I'm opposed to adding, if we can find a good way to do it. It's also very low-priority though, since comments that would use it are extremely rare. It's probably a good candidate for an outside contributor adding as a merge request, if someone feels strongly about adding support and wants to work on it.
Edit: wirelyre did some testing, and it looks like the required assets might not be too bad.
Everything you've said is fair. I'll just add that LaTeX support on Tildes would easily make it my no. 1 site for math/science discussions!
I like this page on ProofWiki.
This is a really good teacher, I wish my math teacher explained as well as this guy.