$Amendment to "Same exponents in two nonlinear function deriving the same result?"$

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$Amendment to "Same exponents in two nonlinear function deriving the same result?"$

Sorry, I made a mistake in the first equation and forgot a bracket. In fact it has the following form, which might complicate the proof:

$f(X_n)=Y*(\sum_{n}^{N}X_n)^A$

and the 2nd one remains to be

$g(X_n)=\sum_{n}^{N}(Z_n*X_n^B)$

Could you solve this as well for a tip?

+++++++++++++

Example:

An example for $N=3$ would be

$\sum_{n}^{3}X_n=X_1+X_2+X_3=5+10+15=30$,

leading to

$f(X_n)=Y*30^A$

and

$g(X_n)=Z_1*5^B+Z_2*10^B+Z_3*15^B$.

Given that

$f(X_n)=g(X_n)$

and thus

$Y*30^A=Z_1*5^B+Z_2*10^B+Z_3*15^B$,

I'm pretty sure that it must hold that $A=B$, but I need a formal proof.

Of course, you could select $Y, Z_1, Z_2, Z_3$ for this example in a way now that the equation would hold, but my understanding is that only with $A=B$ it is possible to let the equation always be true at all levels of $X_n$ as only this leads to the needed same nonlinearity.

My gut feeling at the moment is that some sort of first derivatives should show that the exponents $A, B$ and not factors $Y, Z_1, Z_2, Z_3$ must be related for the functions to always derive the same result?

Functions

Mosi Posi

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Here $X=(X_1, X_2, \dots, X_N)$ is indeed an $N$ dimensional vector. Suppose $f(X_N)=g(X_N)$ for all such vectors. Then choosing a particular $X_N$:
\[X_N=(x, 0, \dots, 0)\]
we must have
\[f(X_N)=Y(x+0+\dots +0)^A=g(X_N)=Z_1 x^B+0+\dots+0.\]
Hence
\[Yx^A=Z_1x^B,\]
or
\[\frac{x^A}{x^B}=\frac{Z_1}{Y}=\text{constant}.\]

This implies $A=B$.

****************************************
The key point here is the assumption that $f(X_n)=g(X_n)$ for all possible vectors $X_n$. In particular, it should hold for those $X_n$ with $X_n=(x,0,0, \dots, 0)$. Of course it is possible for $f(x_n)=g(X_n)$ for some $X_n$ with all nonzero compoents. If a statement holds for some value of $X_n$ does not mean it holds for all values of $X_n$.

*******
Note that
\[x^{A-B}=\frac{x^A}{x^B}=\text{constant}\]
Hence $A-B=0$, which implies $A=B$.

Erdos

4.8K

Mosi Posi

0

I dont quite understand this solution: 1) Why does 0 show up 2 times in your vector? 2) And why can "..." be ignored in the equation after "Hence"? 3) Why does this ratio being constant imply A=B? Are you saying that A=B must hold if N=1 and hence Y*X^A=Z*X^A is the only thing being left? This is not correct: If x = 10, A = 0.5 and Y = 2, B can be 0.2 and Z_1 = 3.990525 and all equations hold without A=B.
- Erdos
  
  0
  
  I added a comment at the end of my solution.
Mosi Posi

0

Thanks, I understand the argument now. As a last question, is there an easy step to maybe show how X^A/X^B=constant requires A=B? And can I just write "=constant" as you did or is this is a somewhat informal mathematical language? Thanks again!
- Erdos
  
  0
  
  I added a comment at the end.
- Erdos
  
  0
  
  "=constant" is not informal. You can use it in a formal proof.
Mosi Posi

0

Thanks! This helped me a lot.
- Erdos
  
  0
  
  I am glad I was able to help :)

$Amendment to "Same exponents in two nonlinear function deriving the same result?"$

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