- #1
m84uily
- 33
- 0
Hello! I want to know more ways to show the non-existence of the elementary function N(x).
Here's how N(x) is defined:
g(x) : Any elementary function.
[tex] N'(g(x)) = \frac{g(x)}{e^{N(g(x))}} [/tex]
I've only thought of a single way to show this impossibility and it doesn't really develop my understanding of the "why" behind it very well. Here's how I went about it:
If such a function N(x) were to exist then there could also be a function I(x) which could be defined as:
[tex]I(g(x)) = e^{N(g(x))} [/tex]
Which would mean:
[tex]I'(g(x)) = g(x) [/tex]
[tex]\int x^x dx = I(x^x) + C[/tex]
The integral of x^x has been proven irreducible to elementary terms by Marchisotto and Zakeri and therefore N(x) cannot exist for it allows the existence of I(x).
Would anyone else mind posting other lines of reasoning please?
Here's how N(x) is defined:
g(x) : Any elementary function.
[tex] N'(g(x)) = \frac{g(x)}{e^{N(g(x))}} [/tex]
I've only thought of a single way to show this impossibility and it doesn't really develop my understanding of the "why" behind it very well. Here's how I went about it:
If such a function N(x) were to exist then there could also be a function I(x) which could be defined as:
[tex]I(g(x)) = e^{N(g(x))} [/tex]
Which would mean:
[tex]I'(g(x)) = g(x) [/tex]
[tex]\int x^x dx = I(x^x) + C[/tex]
The integral of x^x has been proven irreducible to elementary terms by Marchisotto and Zakeri and therefore N(x) cannot exist for it allows the existence of I(x).
Would anyone else mind posting other lines of reasoning please?