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  1. $\prod\limits_{i\in I}(1+x_i)=\sum\limits_{J\subseteq I}\prod\limits_{j\in J}x_j$– user940 Oct 1 '13 at 12:13

My child's 16. How can we intuit or prove this most easily? We prefer a Story Proof, rather than tedious algebraic manipulations or induction.

One hitch is that $|I|$ may $\neq |J|$. LHS $ = (1 + x_1) \dots (1+x_{|I|}) $.

$\begin{align} RHS & = \sum\limits_{J\subseteq I} \color{limegreen}{x_1}\color{red}{x_2}\dots x_{|J|} \\ & = \color{limegreen}{x_1} + \color{limegreen}{x_1}\color{red}{x_2}+ \color{limegreen}{x_1}\color{red}{x_2}x_3 + \dots + \color{limegreen}{x_1}\color{red}{x_2}x_3 \dots x_{|J|} \\ & = \color{limegreen}{x_1(}1 + \color{red}{x_2} + x_2x_3 + \dots + x_2x_3 \dots x_{|J|}\color{limegreen}{)} \\ & = \color{limegreen}{x_1(}1 + \color{red}{x_2[}1 + x_3 +\dots x_{|J|}\color{red}{]}\color{limegreen}{)} \end{align}$. Now what?

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    At some point you will be forced to do some algebraic manipulations because you need to manipulate the terms on the LHS to get to the RHS. In my opinion, induction is a rather easy way because you can prove your hypotheses for small sets $I$ until your child sees that there is something which is systematic and then you let him/her generalize this observation. – Jfischer Jul 21 '21 at 08:22
  • Your formula for RHS is wrong, unless every $J$ containers $1$ and $2.$ – Thomas Andrews Jul 21 '21 at 08:25
  • Yeah, it's hard to prove something if all the tools for proving it are taken away from you. Here the tools being induction and algebraic manipulation. Also I don't think a child would have a hard time understanding induction. – Jakobian Jul 21 '21 at 08:25
  • This is one of those theorem which is sort of obvious if you write out a few examples, but will require formal tools like induction to prove. The underlying reason is the distributive law. – Thomas Andrews Jul 21 '21 at 08:29
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    I don't think age is a sufficient information about background. – VIVID Jul 21 '21 at 08:52
  • @ThomasAndrews I edited my post. Is my formula for RHS still wrong? –  Jul 21 '21 at 23:16
  • @JeanMarie This isn't a duplicate, because that other answer uses math beyond the level of a 16 y.o. –  Jul 21 '21 at 23:20
  • Yes, still wrong. @codidact also, I don’t understand what the color coding means. The (original) formula can be rewritten as:$$\sum_{k=0}^n\sum_{1\leq i_1<i_2<\cdots<i_k\leq n} x_{i_1}x_{i_2}\cdots x_{i_k}$$ In particular, there is no reason $x_1$ is a term in the product. – Thomas Andrews Jul 21 '21 at 23:30
  • In your first question on this site, your child was 15 years old. Now he suddenly jumped to 16? – PatrickR Jul 22 '21 at 06:20

2 Answers2

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For an answer that a 16 year old may understand, you can consider this interpretation. Take a box with $n$ objects, and give the $i$th object the label $x_i$ (which you can think of as a letter).

Now consider picking (in no particular order) some elements in that box, and recording your choice. If you picked elements labeled $x_{i_1},\ldots,x_{i_k}$, then assign it the word $x_{i_1} \ldots x_{i_k}$ in the letters as above. This word corresponds to a set $J = \{i_1,\ldots, i_k\}$ with $k$ distinct elements contained in $\{1,\ldots,n\}$.

Takeaway 1. for each subset $J$ of the labels $\{1,\ldots,n\}$, we have created a word $x_J = x_{i_1},\ldots,x_{i_k} = \prod_{j\in J} x_j$, corresponding to a choice of objects in your box.

Now consider the expression $(1+x_1)\cdots (1+x_n)$, and what happens when you multiply it out. For each term $1+x_i$, you either have to choose to take $1$ or $x_i$. This choice corresponds to looking at the object $i$, and deciding if you are picking it from the box (take the term $x_i$) or not (take the term $1$).

Takeaway 2. Doing this in all possible ways ---that is, computing the product--- then effectively gives you the sum of all the words $x_J$ as $J$ runs through sets contained in $\{1,\ldots,n\}$.

It follows that $(1+x_1) \cdots (1+x_n) = \sum_{J\subseteq [n]} x_J$, like you wanted. You can also embellish this a bit: $$(1+tx_1) \cdots (1+tx_n) = \sum_{J\subseteq [n]} x_J t^{|J|}$$ where now the $t$ variable keeps track of the number of times you did choose to pick an object of the box. In particular, the empty set $\varnothing$ corresponds to the term $1$, obtained by always picking $1$ in the left hand side. As an example, if you have three objects $x$, $y$ and $z$, you get

$$(1+tx)(1+ty)(1+tz) = 1 + (x+y+z)t +(xy+xz+yz) t^2+ xyzt^3.$$

Pedro
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So, here is my attempt at giving a visual intuition, which of course is not a proof. Let's start with the case, where we have only two variables, $x_1$ and $x_2$. We can imagine the product $(1+x_1)(1+x_2)$ as the area of a rectangle with sidelengths $1+x_1$ and $1+x_2$, like here: You can see from here that we can divide our rectangle into $4$ different rectangles of area $1\cdot 1$, $1\cdot x_1$, $1\cdot x_2$ and $x_1\cdot x_2$ and so $$(1+x_1)(1+x_2) = 1 + x_1+x_2+x_1 x_2.$$ How about three variables, $x_1,x_2$ and $x_3$? Well, now the product $(1+x_1)(1+x_2)(1+x_3)$ describes a cuboid with sidelengths $1+x_1, 1+x_2$ and $1+x_3$ as shown here:

The $1\cdot 1\cdot 1$ cube is hidden behind the remaining cuboids. Again, counting the volume of the other cuboids, we get $$(1+x_1)(1+x_2)(1+x_3) = 1 + x_1+x_2+x_3 + x_1x_2+x_2x_3 + x_3x_1 + x_1x_2x_3.$$

For higher dimensions, things get a bit messy with the visualisation.

Pengin
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