There’s the quadratic formula and the one for 3 and 4 factor ones (apologies if I’m wording this wrong), but I just heard that apparently there isn’t anything like a quintic(?) formula and so on. Why is this?? Googling gives me a bunch of confusing terminology that’s difficult to parse.
In: 75
First, when they say “no formula”, they mean “no formula with this limitations”:
* only contains 4 arithmetic operations and roots (of any degree)
* only finite amount of operations (no infinite sums or such)
It is absolutely possible to create a quintic formula by violating those rules. Quintics can be, for example, solved with Newton’s method (which involves an infinitely repeating steps, that converge to the answer).
Now, let’s examine the relationship between the roots and coefficients. We have to use *complex numbers* here – unfortunately, many nice polynomial properties just doesn’t work when observed only on real numbers. Just pretend, that coefficients and roots are 2D points on the plane, instead of points on the line.
1. The relationship is **continuous** – it has no sudden jumps. Slightly “wiggling” coefficients will only slightly “wiggle” the roots, and the reverse is also true.
2. The coefficients are **ordered** – swapping two coefficients changes the roots.
3. The roots are **unordered** – swapping roots does not change the coefficients
Now, what properties should a root formula have?
1. If the formula is continuous (has no sudden jumps), then it must be **multi-valued** – it must produce all roots at once. Otherwise, it would be impossible to gradually swap the roots – continuity demands the formula to track the root as it moves, but single-valueness demands it to return to the original value.
2. If we run formula “in reverse”, it must take any possible rearrangement of roots, and “consume” it – reorder back into a fixed order of coefficients.
3. The n-th root in complex numbers has exactly “n” answers. However, it has only one “coefficient” to play with (the number under the root), so the answers of the root cannot go all possible rearrangements – they can only go **cyclic rearrangements**.
4. The 4 arithmetic operations are single-valued and cannot rearrange roots.
So all boils down to one question – can rearrangements of N elements be “consumed” by only cyclic rearrangements?
* For N=2, 3, 4 the answer is “yes”.
* However, starting with N=5 there is a problem: a cycle of 3 elements can be broken down into two other cycles of 3 elements. That means, each time we “consume” a 3-cycle, we break 2 other ones. So it is impossible to consume all 3-cycles in 5 element group – therefore a 5-degree solution formula with roots cannot exist.
Latest Answers