Are there any hard-to-use or unsolvable math equations that have far reaching implications?

Question

We often talk about mathematical beauty and how they end up describing incredible patterns or phenomena. But are there any mathematical equations that are extremely non-trivial and virtually unsolvable which have strong real-world applications?

Edit: I realized that to obtain appropriate answers, I must define the word "ugly." An equation is said to be ugly if it satisfies on or more conditions

1) Large Number of Variables.

2) Describes a trivial phenomena in a non-trivial manner.

3) Is derived using a brute force approach.

Answer 1

The following equation is an expression of the fact that matrix multiplication is associative: \begin{eqnarray*} \sum_{j=1}^{m}\left[a_{ij}\left(\sum_{k=1}^{n}b_{jk}c_{kl}\right)\right] & = & \sum_{k=1}^{n}\left[\left(\sum_{j=1}^{m}a_{ij}b_{jk}\right)c_{kl}\right] \end{eqnarray*}

• It is certainly far reaching; matrix computations are carried out constantly.
• As far as I know, it is usually 'derived' simply by writing out the summation on the left and rewriting it as the summation on the right ('brute force').
• Do the entries in the matrices count as variables? If so, then it involves lots of variables.
• It can be re-expressed as simply this: $(f\circ g)\circ h = f\circ(g\circ h)$, where $f,g,h$ are linear transformations between suitable vector spaces.

I would say (and indeed, your criteria seem to agree) that the above equation is "ugly". However, the fourth point above is important: usually, ugly things can be rewritten in more elegant (though, in some situations, less useful) ways.

Another example that illustrates this last point especially well is the definition of the determinant of a matrix. The usual definition that students are introduced to is as some horrible summation involving cofactors etc., rather than the more elegant view as simply the volume scale factor of the unit cube (with a few terms and conditions).

Answer 2

The utterly disgusting 8 line equation found here is what happens when you build a Lagrangian for the Standard Model of Physics.

Answer 3

I'm not sure if this qualifies as "far-reaching," but the solutions to the general quartic (degree $4$) equation is quite nasty in my opinion.

This page discusses the solution, but does not actually state an explicit formula.

I would consider this a far reaching problem, as it demostrates the power and limitations of early algebra and leads to the study of group theory, specifically Abel and Galois' work in the insolubility of the quintic.

Answer 4

I think the generic solution for cubic polynomial roots certainly counts.

[ported from Wikipedia]

$$x_1 = -\frac1{3a}\left(b+\sqrt[3]{\frac{2b^3 - 9abc + 27a^2d + \sqrt{(2b^3 - 9abc + 27a^2d)^2 - 4(b^2-3ac)^3}}{2}}+\frac{b^2-3ac}{\sqrt[3]{\frac{2b^3 - 9abc + 27a^2d + \sqrt{(2b^3 - 9abc + 27a^2d)^2 - 4(b^2-3ac)^3}}{2}}}\right)$$

$$x_2 = -\frac1{3a}\left(b+\frac{-1+i\sqrt{3}}{2}\sqrt[3]{\frac{2b^3 - 9abc + 27a^2d + \sqrt{(2b^3 - 9abc + 27a^2d)^2 - 4(b^2-3ac)^3}}{2}}+\frac{b^2-3ac}{\frac{-1+i\sqrt{3}}{2}\sqrt[3]{\frac{2b^3 - 9abc + 27a^2d + \sqrt{(2b^3 - 9abc + 27a^2d)^2 - 4(b^2-3ac)^3}}{2}}}\right)$$

$$x_3 = -\frac1{3a}\left(b+\frac{-1-i\sqrt{3}}{2}\sqrt[3]{\frac{2b^3 - 9abc + 27a^2d + \sqrt{(2b^3 - 9abc + 27a^2d)^2 - 4(b^2-3ac)^3}}{2}}+\frac{b^2-3ac}{\frac{-1-i\sqrt{3}}{2}\sqrt[3]{\frac{2b^3 - 9abc + 27a^2d + \sqrt{(2b^3 - 9abc + 27a^2d)^2 - 4(b^2-3ac)^3}}{2}}}\right)$$

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Other potential candidates:

• the PDF of the multivariate normal distribution (certainly ugly enough that we avoid presenting it in intro courses, and I would guess the percentage of people who know what the "bell curve" is intuitively but haven't the slightest of this formula is, is quite large). Not hideous, but it is undoubtedly ubiquitous:

$$\varphi(x; \mu, \Sigma, k) = \frac1{\sqrt[k]{2\pi}}|\Sigma|^{-\frac12}e^{-\frac12(x-\mu)^T\Sigma^{-1}(x-\mu)}$$

• Rademacher's partition counting approximation (relevant for all sorts of computer scientific stuff like graph theory):

$$p(n) = \frac1{\pi\sqrt{2}}\sum_{k=1}^{\infty}\sqrt{k}A_k(n)\frac{d}{dn}\left(\frac1{\sqrt{n-\frac1{24}}}\sinh\left[\frac{\pi}{k}\sqrt{\frac23\left(n-\frac1{24}\right)}\right]\right)$$

$$A_k(n) = \sum_{0\leq m<k; (m, k) = 1}e^{\pi i\left[s(m, k) - \frac{2nm}{k}\right]}$$