2110.12 – Giant Distribution

Simplify:  x(x(x(2−x)+1)+1)+1+x(x(x(x−2)+1)+1)+1.x (x (x (2 - x) + 1) + 1 ) + 1 + x (x (x (x - 2) + 1) + 1) + 1.

This is supposed to come out to be zero. Does it? If not, can you fix it up so that it does?

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Solution

Oogh. Some students actually LIKE working these out, bless 'em. (And we need 'em!)

For convenience, let

E=x(x(x(2−x)+1)+1)+1+x(x(x(x−2)+1)+1)+1. E = x (x (x (2-x) + 1) + 1) + 1 + x (x (x (x -2) + 1) + 1) + 1.

Then here goes:

E=x(x(x(2−x)+1)+1)+1+x(x(x(x−2)+1)+1)+1=x(x(2x−x2+1)+1)+1+x(x(x2−2x+1)+1)+1=x(2x2−x3+x+1)+1+x(x3−2x2+x+1)+1=2x3−x4+x2+x+1+x4−2x3+x2+x+1=2x2 +2x +2\begin{aligned}E &= x (x (x (2-x) + 1) + 1) + 1 + x (x (x (x -2) + 1) + 1) + 1 \\&= x (x (2x- x^2 + 1) + 1) + 1 + x (x (x^2 -2x + 1) + 1) + 1 \\&= x (2x^2- x^3 + x + 1) + 1 + x (x^3 -2x^2 + x + 1) + 1 \\&= 2x^3- x^4 + x^2 + x + 1 + x^4 -2x^3 + x^2 + x + 1 \\&= 2x^2 + 2x + 2\end{aligned}

which is not zero, at least not identically.

Change the second half to:

x(x(x(x−2)−1)−1)−1=x(x(x2−2x−1)−1)−1=x(x3−2x2−x−1)−1=x4−2x3−x2−x−1.\begin{aligned}x (x (x (x -2) - 1) - 1) - 1 &= x (x( x^2 - 2x - 1) - 1) - 1 \\&= x(x^3 - 2x^2 -x - 1) -1 \\&= x^4 - 2x^3 - x^2 - x - 1.\end{aligned}

Now the sum of the two halves will equal zero.