Mean average of exponential functions
I'm new to math stack exchange. Hope this goes well!
So, I have a set of exponential functions each defined in the form\begin{aligned} f_i(x)=a_ie^{-b_ix}+c_i \end{aligned}where $(a_i, b_i, c_i) \in \Bbb R^+$. (EDIT: Changed from $\Bbb R$ to $\Bbb R^+$)
Is it correct to say that:\begin{aligned} \frac{1}{n}\sum_{i=1}^{n}f_i(x) &= (\frac{1}{n}\sum_{i=1}^{n}a_i)e^{-(\frac{1}{n}\sum_{i=1}^{n}b_i)x}+\frac{1}{n}\sum_{i=1}^{n}c_i \end{aligned}
for $n \in \Bbb Z^+$?
If so, how can we prove it? If not, why not?
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$\begingroup$You wrote :
\begin{aligned} f_i(x)=a_ie^{-b_ix}+c_i \end{aligned} where $(a_i, b_i, c_i) \in \Bbb R^+$. Is it correct to say that: \begin{aligned} \frac{1}{n}\sum_{i=0}^{n}f_i(x) &= (\frac{1}{n}\sum_{i=0}^{n}a_i)e^{-(\frac{1}{n}\sum_{i=0}^{n}b_i)x}+\frac{1}{n}\sum_{i=0}^{n}c_i \end{aligned}
First there is a mistake : A sum from $i=0$ to $i=n$ counts $n+1$ terms, not $n$.
So, for average, you should write $\frac{1}{n+1}\sum_{i=0}^{n}$ or $\frac{1}{n}\sum_{i=1}^{n}$ ,but not $\frac{1}{n}\sum_{i=0}^{n}$ .
I suppose that it is a typo and that the proposed relationship is $$\frac{1}{n}\sum_{i=1}^{n}(a_ie^{-b_ix}+c_i) = (\frac{1}{n}\sum_{i=1}^{n}a_i)e^{-(\frac{1}{n}\sum_{i=1}^{n}b_i)x}+\frac{1}{n}\sum_{i=1}^{n}c_i$$
It is easy to find a lot of counter-examples .
For example, check your formula with
$$n=2\:;\:a_1=a_2=1\:;\:b_1=1\:;\:b_2=3\:;\:c _1=c_2=0$$$$\begin{cases} \frac{1}{n}\sum_{i=1}^{n}(a_ie^{-b_ix}+c_i) =\frac12 (e^{-x}+e^{-3x})\\ (\frac{1}{n}\sum_{i=1}^{n}a_i)e^{-(\frac{1}{n}\sum_{i=1}^{n}b_i)x}+\frac{1}{n}\sum_{i=1}^{n}c_i =e^{-2x} \end{cases}$$$$\frac12 (e^{-x}+e^{-3x})\neq e^{-2x}$$Thus $\quad \boxed{\frac{1}{n}\sum_{i=1}^{n}f_i(x) = (\frac{1}{n}\sum_{i=1}^{n}a_i)e^{-(\frac{1}{n}\sum_{i=1}^{n}b_i)x}+\frac{1}{n}\sum_{i=1}^{n}c_i\quad \text{is false.}}$
Another obvious counter-example, with $a_k=1\:;\:b_k=k\:;\: c_k=0$ :
$$\quad \frac{1}{n}\sum_{k=0}^{n}e^{-kx} \neq e^{-\frac{n+1}{2}x}$$because $\frac{1}{n}\sum_{k=0}^{n}e^{-kx}=\frac{e^{-x}-e^{-(n+1)x}}{n(1-e^{-x})}\neq e^{-\frac{n+1}{2}x}$ .
Note that the proposed relationship is true in the trivial case $b_1=b_2=…=b_n$ .
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