expressing $\frac{a_0 + a_1 + \cdots}{b_0 + b_1 +\cdots}$ in terms of $\frac{a_0}{b_0}, \frac{a_1}{b_1},\ldots$

Question

Is it possible to rewrite fractions of the form $\frac{a_0 + a_1 + \dots + a_N}{b_0 + b_1 + \cdots + b_N}$ in terms of $\frac{a_0}{b_0}, \frac{a_1}{b_1},\ldots, \frac{a_N}{b_N}$?

under which circumstances is it possible to do so, and what portion of the expression cannot be reduced in such terms?

Answer 1

Great question. Suppose

$$\frac{a_0+a_1+\cdots+a_N}{b_0+b_1+\cdots+b_N}=n_0\frac{a_0}{b_0}+n_1 \frac{a_1}{b_1}+\cdots+n_N \frac{a_N}{b_N} \tag 1$$ For $n_0, n_1,\ldots,n_N$ in the reals.

Then certainly:

$$\frac{a_0+a_1+\cdots+a_N}{b_0+b_1+\cdots+b_N - Nx} = n_0\frac{a_0}{b_0-x} + n_1\frac{a_1}{b_1-x}+\cdots+n_N \frac{a_N}{b_N-x} \tag 2$$

For all $x \notin \left \{ b_0, b_1,\ldots,b_N \right \}$.

To find $n_i$ for given $i \in \left \{ 1, 2,\ldots,N \right \}$, multiply by $b_i-x$ and evaluate at $x=b_i$. If, for any choice of $i$, $x=b_{i}$ is not a root of the denominator of the L.H.S. in (2), we would have: $0=n_i a_i \Rightarrow n_i=0$. This would be impossible, since the L.H.S. in (1) would then contain $a_i$ where the R.H.S. would not, and the equality (1) would clearly not hold for all choices of $a_i$.

Therefore $b_i-x$ must be a factor of $b_0+b_1+\cdots+b_N-Nx$ for all $i \in \left \{ 1, 2,\ldots,N \right \}$. Since $b_0+b_1+\cdots+b_N-Nx$ is first order, it has only one root. It follows $b_i-x=b_j-x$ for all choices of $i$ and $j$. This forces $b_0=b_1=\cdots=b_N$, in which case $n_0=n_1=\cdots=n_N=\frac{1}{N}$.

So it's not possible, except in the trivial case that $b_i=b_0$ and $a_i=\frac{1}{N}$ for all $i$.

EDIT: I see I've answered the wrong question. So instead, suppose I start with a function $$\frac{a_{1}+a_{2}+...+a_{n}}{b_{1}+b_{2}+...+b_{n}}= f\left (\frac{a_{1}}{b_{1}},...,\frac{a_{n}}{b_{n}} \right )$$

Then, by letting $(a_{i}, b_{i}) \rightarrow (\frac{1}{a_{i}}, \frac{1}{b_{i}})$ for all $i$, we get:

$$\frac{\frac{1}{a_{1}}+\frac{1}{a_{2}}+...+\frac{1}{a_{n}}}{\frac{1}{b_{1}}+\frac{1}{b_{2}}+...+\frac{1}{b_{n}}} = \frac{b_{1}+b_{2}+...+b_{n}}{a_{1}+a_{2}+...+a_{n}}$$

Clearly not true for all choices of $a_{i}$ and $b_{i}$. One sufficient condition to prevent this is for $\frac{a_{i}}{b_{i}}$ to be constant, as already mentioned.

Answer 2

The only case in which the fraction $\sum a_i/\sum b_i$ can be computed in terms of $a_i/b_i$ is when $a_i/b_i=\text{const}$ in which case $\sum a_i/\sum b_i$ equals that constant. To prove this, it is sufficient to look at the case $n=2$. If $(1+a)/(1+b)$ was computable in terms of $a/b$, then it should give the same value for the pair $(2,1)$ and $(4,2)$, but it doesn't.

Answer 3

Assume that it is possible.

Then set $$\frac{a_0}{b_0}=k_0,\frac{a_1}{b_1}=k_1,\cdots,\frac {a_n}{ b_n}=k_n$$ for arbitrary constants $k_0,k_1,\cdots, k_n$.

If your request is possible, then there exist a function $f$ such that $$\frac{\sum b_m k_m}{\sum b_m}=f(k_0,k_1,\cdots,k_n)$$

For any integer $j$ with $0\le j\le n$, if we take logarithm on both sides of the above equation, and partially differentiate both sides with respect to $b_j$: $$\frac{\partial}{\partial b_j}\ln\frac{\sum b_m k_m}{\sum b_m}= \frac{\partial}{\partial b_j}\ln f(k_0,k_1,\cdots,k_n)$$ $$\implies \frac{k_j}{\sum b_m k_m}-\frac1{\sum b_m}=0$$

Take any integer $0\le j_1, j_2\le n$. Trivially, we have $$\frac{k_{j_1}}{\sum b_m k_m}-\frac1{\sum b_m}=0$$ $$\frac{k_{j_2}}{\sum b_m k_m}-\frac1{\sum b_m}=0$$

Subtract the two equations to obtain $$k_{j_1}=k_{j_2}$$

This immediately implies $$\frac {a_0}{b_0}=\cdots=\frac {a_n}{b_n}=C$$ where $C$ is some non-zero constant.

This is the sufficient condition for your request to be possible.