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

Question

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

I have two functions:

$f\left ( X_n \right ) =B*\sum_{n}^{N}X_n^A $

and

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

with $A$ and $B$ as the exponents of $X_n$.

It alway must hold that $f\left ( X_n \right ) = g\left ( X_n \right )$ for all possible input combinations when changing $X_n$ and keeping all other inputs constant.

I am pretty sure that for this condition to always hold, it must be that A = B. Because only those parameters drive the nonlinearity of those functions and any deviation would not allow for the condition $f\left ( X_n \right ) = g\left ( X_n \right )$, no matter what values the other variables $B$, $Z_n$ and $N$ take.

Can this intuitive result be proven mathematically in easy terms?

Thanks for your support!

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Mosi Posi

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Answer 1

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Yes. To prove it, fitst note that
$$\lim_{x_n\rightarrow \infty} \frac{x_n^A}{x_n^B}=0, \ \ \text{if} \ \ A<B$$
and
$$\lim_{x_n\rightarrow \infty} \frac{x_n^A}{x_n^B}=\infty, \ \ \text{if} \ \ A>B.$$

Thus is $f(x_n)=g(x_n)$ for all $x_n$. Then
$$1=\lim_{x_n\rightarrow \infty} \frac{f(X_n)}{g(X_n)}=0, \ \ \text{if} \ \ A<B$$
and
$$1=\lim_{x_n\rightarrow \infty} \frac{f(X_n)}{g(X_n)}=\infty, \ \ \text{if} \ \ A>B.$$

Both above cases are impossible.
Therefore we must have $A=B$.

Erdos

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Mosi Posi

0

Sorry, I made a mistake in the first equations. In fact it has the following form, which likely complicates the proof: $f(x_n)=B*(\sum_{n}^{N}X_n)^A$ Could you solve this as well for $10 tip?
Erdos

0

Could you please repost your other question? You answered my clarification question too late and the deadline passed.
- Mosi Posi
  
  0
  
  Done, I just reposted the question.

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

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