2110.12 – Giant Distribution

Tags:Problem Set 4GeometrySymbol-Pushing

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?

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.