User blog:Holomanga/Binomial Hypercubes

In this exercise, I want you to pay attention to the coefficients.

Take the polynomial

$$2 + x$$

Now, multiply it by 2 + x.

$$4 + 4x + x^2$$

Now, multiply it by 2 + x.

$$8 + 12x + 6x^2 + x^3$$

Are you noticing a pattern, yet? Let's try once more.

$$16 + 32x + 24x^2 + 8x^3 + x^4$$

Okay, there's no way this is a coincidence. You can find the number of subfacets in a hypercube of a certain dimensionality by looking at the binomial expansion of $${(x+2)}^{n}$$. For example, a tesseract has 16 vertices, 32 edges, 24 faces, 8 cells, and 1 teron. These are the coefficients of the last polynomial in that list above.

Neat applications of this are that you can use the binomial formula to quickly find the hypervolumes of any hypercube.

Just for fun, you could try non-integer exponents as well, like $$\frac{1}{2}$$ or $$4 + 3i$$, then use an extension of the binomial theorem to get the coefficients. Share what happens in the comments!

Update: You can use the binomial expansion if you rewrite the equation as $$2^n (1 + \frac{x}{2})^n$$. That might come in handy, since it allows for any $$n \in \mathbb{Q}$$

Update 2: You can use the Taylor Series expansion without rewriting the equation. This lets you have the exponent be any function of X, including such fun things as (log(X) + 2) dimensional hypercubes and stuff!