- #1

chisigma

Gold Member

MHB

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0^0

… a new chapter of the never-ending saga ‘ $0^{0}$’ has been written. Of course in the past I’m also have been involved in discussions about this specific problem and almost ever these discussions have terminated with an exchange of insults. I’m sure that MHB is no exception and that’s why I ask You to be left free to write this and the next post, where my personal opinion about the correct formulation of the problem are described. After I will be glad to answer to Your objections… if any…

Let’s start with the ‘standard definition’ of

*exponentiation*that is reported in almost all the HolyBooks…

Standard definition: exponentiation is a mathematical operation, written as $b^{n}$, involving two numbers, the base b and the exponent (or index or power) n. When n is a

*positive integer*, exponentiation corresponds to repeated multiplication. In other words, a product of n factors, each of which is equal to b (the product itself can also be called power)...

$ b^{n}= \underbrace {b \cdot b \cdot\ ... \cdot\ b}_{\text{n times}}$

Very well!... The successive step of course is to derive the rule of multiplication, division, exponentiation of powers with the same base b and doing that one 'discovers' that...

$\displaystyle \frac{b^{n}}{b^{n}}= b^{n-n}=b^{0}=1$ (1)

... so that hi can 'extend' the exponent's domain to n=0. Another property that is easily 'discovered' is that for all n>0 is $0^{n}=0$ and at this point a devil makes an entrance: '... why don't add a further 'extension' and don't find what is $0^{0}$?'... and immediately we are in panic because if we try to write the (1) with b=0 we find $\frac{0}{0}$ that is an 'indeterminate form'... and adding panic to panic we think at the well known derivation rule...

$\displaystyle \frac{d}{d x} x^{n} = n\ x^{n-1}$ (2)

... that for n=1 and x=0 gives the result $0^{0}$ but that we know is 1 for all x... so is $0^{0}=1$?... or (2) holds for all x except x=0?... a sort of 'chain reaction' with unpredictable consequences has been 'inflamed' (Wasntme)...

Now let me try do modify just an insignificant bit of the original definition of exponentiation as follows and observe the results…

Modified definition: exponentiation is a mathematical operation, written as $b^{n}$, involving two numbers, the base b and the exponent (or index or power) n. When n is a

*non negative integer,*exponentiation corresponds to repeated multiplication, in other words, a product of n+1 factors, the first equal to 1 and the others equal to b (the product itself can also be called power)...

$ b^{n}= 1 \cdot \underbrace {b \cdot b \cdot\ ... \cdot\ b}_{\text{n times}}$

What is changed?... nothing except to the fact that for all real b [including b=0...] is

*by definition*$b^{0}=1$... and the whole phobia disappears (Wasntme)...

On the basis of exponentiation's rules [in standard or modified version...] it is possible to define the

*exponential function*as...

$\displaystyle e^{x}= \lim_{n \rightarrow \infty} (1+\frac{x}{n})^{n}$ (3)

... and also its inverse that is the

*natural logarithm function*as the function $\ln x$ for which for all x is $e^{\ln x}= x$. The exponential function is then 'generalized' defining...

$\displaystyle a^{x}= e^{x\ \ln a}$ (4)

... as well as its inverse function $\log_{a} x$. On the basis of (4) the derivative of the

*generalized exponential function*is...

$\displaystyle \frac{d}{d x} a^{x} = a^{x} \ln a$ (5)

For the moment we suppose that a is a real number greater than 0…

All these definitions are a very useful preliminary to answering to the ‘critic’ question: what’s the value of $0^{0}$?... It is evident that the question itself is ambiguous because it is not specified if we are intending exponentiation or generalized exponential, so that we have two possibilities…

a) exponentiation: given a real b and a non negative integer n, for b=0 and n=0 is,

*by definition*$b^{n}=0^{0}=1$...

b) generalized exponential: given two real non negative numbers x and y, it is possible to evaluate $f(x,y)= x^{y}$ for x=y=0?...

A possible answer to the question b) will be examined in a successive post…

Kind regards

$\chi$ $\sigma$