My Math Forum Exponential identity

 Algebra Pre-Algebra and Basic Algebra Math Forum

 January 15th, 2018, 10:32 PM #1 Senior Member   Joined: May 2015 From: Arlington, VA Posts: 464 Thanks: 29 Math Focus: Number theory Exponential identity 0 is the additive identity; i.e. 0+R=R 1 is the multiplicative identity; i.e. 1*R=R 0 is an "exponential inverse"; i.e. R^0=1 for R≠0 1 is an "exponential identity"; i.e. R^1=R 2 is a "universal identity"; i.e. 2+2=2*2=2^2
 January 15th, 2018, 10:55 PM #2 Global Moderator   Joined: Dec 2006 Posts: 21,035 Thanks: 2272 What is your question?
 January 16th, 2018, 04:42 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,691 Thanks: 2670 Math Focus: Mainly analysis and algebra 2 is not an identity. An identity leaves all elements unchanged. 2 doesn't do this.
 January 16th, 2018, 09:55 AM #4 Senior Member   Joined: May 2015 From: Arlington, VA Posts: 464 Thanks: 29 Math Focus: Number theory Are there any accepted names for the properties of each of the last three lines? Do the relations shown on the last line give the highest arithmetic symmetry for any positive integer (here 2), or zero?
January 16th, 2018, 10:05 AM   #5
Senior Member

Joined: Oct 2009

Posts: 886
Thanks: 340

Quote:
 Originally Posted by Loren 0 is an "exponential inverse"; i.e. R^0=1 for R≠0
Just accept the convention that $0^0=1$ (which many mathematicians do, but not everybody), and it'll hold everywhere.

 January 16th, 2018, 02:13 PM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,691 Thanks: 2670 Math Focus: Mainly analysis and algebra $0^0$ is an indeterminate form. It's undefined.
January 16th, 2018, 06:16 PM   #7
Senior Member

Joined: Oct 2009

Posts: 886
Thanks: 340

Quote:
 Originally Posted by v8archie $0^0$ is an indeterminate form. It's undefined.
Not everybody agrees. Look it up.
https://en.wikipedia.org/wiki/Zero_to_the_power_of_zero

Last edited by Micrm@ss; January 16th, 2018 at 06:18 PM.

 January 16th, 2018, 06:44 PM #8 Senior Member   Joined: Oct 2009 Posts: 886 Thanks: 340 If you think $0^0 = 1$ is false (which is what many mathematicians think), then don't write $$e^x = \sum_{n=0}^{+\infty} \frac{x^n}{n!}$$ anymore. You'd have to write $$e^x = 1 + \sum_{n=1}^{+\infty} \frac{x^n}{n!}$$ everytime, or specifiy $x\neq 0$ every time (which is the value around which you apply Taylor to begin with!). There are other reasons to adopt $0^0$ which are more theoretical. In set theory, numbers are defined in a very interesting way: $$0=\emptyset, 1 = \{\emptyset\},...$$ and operations on these are also defined set theoretically: - $n+m$ is the cardinality of the disjoint union of $n$ and $m$ - $n\cdot m$ is the cardinality of the product $n\times m$ - $n^m$ is the cardinality of the set of functions $m\rightarrow n$ If we follow this, we get $0^0 = |\emptyset^\emptyset| = 1$. And then there's category theory. If we look at any bicartesian closed category (and hence every topos), that $0^0 =1$, where $0$ is the initial object, $1$ the terminal object and $x^y$ the exponential object. In particular, this applies to the full subcategory of SET consisting of the natural numbers $\emptyset$, $\{\emptyset\}$, ... It is very understandable that $0^0=1$ is debatable because of limits such as $\lim_{x\rightarrow 0} 0^x$, and because the function $x^y$ has ahas a rather unpleasant behavior around $(x,y)=(0,0)$, allowing the limit to be everything. So sure, everybody agrees that if you compute a limit and find $0^0$ then you got an indeterminate form. But that doesn't mean to everybody to leave the form undefined! Take Knuth (the inventor of LaTeX and writer of the Art of Computer Programming), who writes in his excellent book Concrete Mathematics: Some textbooks leave the quantity $0^0$ undefined, because the functions $0^x$ and $x^0$ have different limiting values when $x$ decreases to $0$. But this is a mistake. We must define $x^0=1$ for all $x$ , if the binomial theorem is to be valid when $x=0$ , $y=0$ , and/or $x=-y$ . The theorem is too important to be arbitrarily restricted! By contrast, the function $0^x$ is quite unimportant. The situation is very similar to the truth table of the implication in logic! Many people have already noted a similarity between logic and arithmetic. The value $0$ is logic is False, the value $1$ is True. The exponential $x^y$ in logic would be the implication $y\Rightarrow x$ (this can be rigorized by seeing a Heyting algebra as a category and using categorical definitions!). So $T\Rightarrow T = T$ corresponds to $1^1= 1$, $T\Rightarrow F = F$ corresponds to $0^1 = 0$, $F\Rightarrow T$ corresponds to $1^0 = 1$ and finally our infamous $0^0=1$ corresponds to $F\rightarrow F = T$. And then there's lambda calculus and Church numeral, who also defined exponentiation in a very nice way, and in which $0^0=1$ is a theorem you can prove I probably changed nobody's mind with this argument. If you still feel $0^0$ is undefined, go ahead. But you'd have to admit not everybody agrees. And that's ok. Thanks from Maschke and Loren Last edited by skipjack; January 17th, 2018 at 01:43 AM.
 January 16th, 2018, 11:19 PM #9 Senior Member   Joined: Oct 2009 Posts: 886 Thanks: 340 In the end it's just a definition. There is no right or wrong in this. It's just a convention that people need to agree to. And some people don't and some people do. Personally, I think that adopting the convention $0^0=1$ far outweighs its negative issues. It's very inconvenient to write summation symbols taking the $0^0$ case into account. There is this joke about scientists applying to be the referee of a baseball match. There was a situation were it was unclear whether the player was in or out. The physicist started doing various calculation of the player's speed and the resistive medium of sand. After a long time he said the player was out. The mathematician on the other hand immediately said the player was out. They asked him how he could be so sure. He replied that the player was out because he said he was out. This is the idea behind mathematics. Mathematics needs to be internally consistent, but is not discovering something actually true. So whether $0^0= 1$ is not something that can be empirically verified at all. It just needs to be internally consistent: whatever you choose, be consistent and keep adopting the standard you choose. And of course when communicating to others, be very clear what standard you adopt. This applies of course to all of math, if you wish to work in a natural number system where $1+1=3$, you can do this if you are clear about it and if you are consistent. But that system is not going to be very useful. The $0^0=1$ rule however IS useful and has benefits. Lastly, I might want to put a different light on the issue. Maybe the $0^0=1$ thing is a type issue. We often identify the $0$ of the naturals with the $0$ of the reals, for good reasons. But pure formally they are not the same. We construct the naturals directly from sets. Then we introduce the integers as pairs with an equivalence relation. We do the same for rationals, and then we use Dedekind cuts to make a version of the reals. If we adopt this practice, then the naturals, the integers, the rationals and the reals are all very distinct sets which have typically no element in common. So the $0$ of the naturals is very different from the $0$ of the reals. (This issue can be circumvented by adopting the reals as the standard set and then finding the naturals inside it). Where am I going with this? Well, I think the value of $0^0=1$ might depend on whether the exponent is natural or real. Indeed, if we work solely with natural numbers $n$, then $x^n$ should be $1$ every time if $x=0$ and $n=0$. But if the exponent $n$ can take on every real value, then $x^n$ might be better undefined. More formally, we have a monoid-action $\mathbb{N}\times \mathbb{R}\rightarrow \mathbb{R}: (n,x)\rightarrow x^n$, and another monoid action $\mathbb{R}^+\times \mathbb{R}\rightarrow \mathbb{R}: (n,x)\rightarrow x^n$. But perhaps for the first monoid action, we want to define $x^n = 0$, while for the second we won't. So the second monoid action is not an extension of the first. This is a very crazy state of affairs, but it makes sense to me: if we take a Taylor expansion, we have $$e^x = \sum_{n=0}^{+\infty} \frac{x^n}{n!}$$ we want $0^0 = 1$ in order for this formula to hold everywhere. But this is ok since $n$ only takes on natural values, so we can use the action $\mathbb{N}\times \mathbb{R}\rightarrow \mathbb{R}: (n,x)\rightarrow x^n$. On the other hand, if we investigate the function $(x,y)\rightarrow x^y$, then the exponent should take on more than natural values, so we should leave it undefined. In fact ALL the benefits of $0^0=1$ only occur if the exponent is natural only. So this kind of typing does make sense I think. In any case, all of this is very unclear to me. It is clear how $\mathbb{N}$ is a category and how the categorical operations force $0^0=1$. My goal is to see $\mathbb{R}$ as a category too and perhaps to obtain that $0^0$ should be undefined in this category. this would be the optimal solution for me. But I'm not there yet.
 January 17th, 2018, 03:22 AM #10 Math Team     Joined: May 2013 From: The Astral plane Posts: 2,304 Thanks: 961 Math Focus: Wibbly wobbly timey-wimey stuff. Hmmmmm.... Mathematica does not like this one. See here. -Dan

 Tags exponential, identity

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post magnanimous Trigonometry 5 December 18th, 2015 04:06 PM Dacu Algebra 3 March 15th, 2014 06:45 AM kfarnan Algebra 6 October 27th, 2010 04:42 PM sharmouc Calculus 2 September 26th, 2010 12:19 AM Soha Algebra 6 December 14th, 2006 11:32 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top