Let's say I want to divide a number $g$ according to proportions defined by a set of numbers, e.g. $\{2,1,1\}$, let's call them $x_i$. Take $g=100$. By proportionally I mean multiplying $g$ by each $x_i$ divided by $\sum{x_i}$.

So the consecutive results (denoted by $r_k$) are 50, 25, 25 in this example. However when calculating the second result number, I can subtract the first result from $g$ and divide by 2 instead of 4, because I ignore $x_1=2$ , like this: $(100-50) * 1 / 2 = 25$, which is equivalent to $100 * 1 / 4$.

How to prove that these two ways are equivalent and give the same result?

$$r_{k} = g \dfrac{x_k}{\sum\limits_{i=1}^{n}x_i}$$

This method of calculating the result $r_k$ is equivalent to:

$$r_k = (g - \sum\limits_{i=1}^{k-1} r_i) \frac{x_k}{\sum\limits_{i=k}^{n}x_i}$$

However these sums look overly complicated to me, is there any way to simplify this and show that they are equal?

  • $\begingroup$ You can get displayed equations by enclosing them in double instead of single dollar signs. You can get proper parentheses (and other paired delimiters) that adjust to the size of their content by preceding them with \left and \right. $\endgroup$
    – joriki
    Dec 23 '19 at 22:52

Well, you saw that this is the case; I think it would be more enlightening to write a proof by trying to formalize how you saw it rather than by algebraic manipulations; but to answer the question:

I'll assume $x_k\gt0$ for all $k$. The proof is by induction. The equations are clearly equivalent for $k=1$. Assume that they are equivalent up to $k-1$. Multiply by both denominators and cancel $x_k$. Then we want to show

$$ g\sum_{i=k}^nx_i=\left(g-\sum_{i=1}^{k-1}r_i\right)\sum_{i=1}^nx_i\;. $$

By the induction hypothesis, we can use the upper equation for $r_i$ so this becomes

$$ g\sum_{i=k}^nx_i=\left(g-\sum_{i=1}^{k-1}\frac{gx_i}{\sum_{j=1}^nx_j}\right)\sum_{i=1}^nx_i\;, $$

and multiplying out the right-hand side and canceling the sum transforms this to

$$ g\sum_{i=k}^nx_i=g\sum_{i=1}^nx_i-\sum_{i=1}^{k-1}gx_i\;, $$

which is obviously correct.

  • $\begingroup$ "By the induction hypothesis, we can use the upper equation" - just to clarify, I can plug in either equation there and the proof will be valid, because they both are assumed to be equivalent for all cases up to k-1, right? $\endgroup$ Dec 23 '19 at 23:39
  • $\begingroup$ @user5539357: Correct. $\endgroup$
    – joriki
    Dec 23 '19 at 23:39

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