User:MathMan64/CasusIrreducibilis
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[edit] Casus
Some cubic equations with real solutions have solutions that can not be written exactly, without the use of imaginary expressions.
I plan to find out how to tell which a cubic equations have solutions that are totally real expressions, and to relate this topic to the exact value of trigonometric functions of angles whose degree measure is not a multiple of three.
Criteria for irreducibile polynomials are given by Eisenstein's criterion
[edit] Living with Logs
From “Death by Black Hole and other cosmic quandaries” by Neil De Grasse Tyson, Norton, 2007, page 26.
We register the world’s stimuli in logarithmic rather than linear increments. For example, if you increase the energy of a sounds volume by a factor of ten, your ears will judge this change to be rather small, (rather than tenfold.) Increase it by a factor of two and you will barely take notice. The same holds for our capacity to measure light. If you have ever viewed a total solar eclipse you may have noticed that the sun’s disk must be at least ninety percent covered by the moon before anybody comments that the sky has darkened. The stellar magnitude scale of brightness, the well-known acoustic decibel scale, and the seismic scale for earthquake severity are each logarithmic, in part because of our biological propensity to see, hear, and feel the world that way.
(Consider in addition, the pH measurements for acid and base, the musical scale, and the way we discuss the size of the universe in powers of ten, from the galaxies near the large end and the quarks near the small end.)
[edit] Piano keys
To find the piano key that corresponds to a given frequency, use the following formula.
where p is the pitch or frequency and n is the number of the key on the keyboard starting from the left and counting both white and black keys.
![b = \sqrt[12] 2 \ = 0.943874312682](../../../../math/8/2/7/8278441022d3fe789853363c4a691c94.png)
[edit] Golfing groups
[edit] Introduction
When choosing 4 people from 11; the first choice is one out of 11, the second one out of the remaining 10, third from 9 and fourth from 9. So the number of ways to do this is
But since a group consisting of ABCD is the same group as BDAC, and the number of ways to pick 4 people from a group of 4 is
So we must divide the previous number by the latter, to give us the number of ways to choose the members of the first group of 4.
The second group of four will be chosen from the remaining 7 people:
The third group has only 3 people chosen from 3 people:
[edit] Two wrong solutions
Multiplying these three products gives us:
.
(I don’t know where I got the extra factor of ten on Sunday?)
But this answer is not correct!
Let’s consider choosing 3 groups of 4 from 12 people:
.
which is also incorrect.
[edit] Correcting the errors
In the three groups of four people each, each group is identical, so choosing group ABCD first and group EFGH second, yields the same result as choosing EFGH first and ABCD second. So we must divide 34650 by 3*2*1 giving the correct answer of 5775.
Since our original problem has only 2 groups of 4, we only divide 11550 by 2*1 which amazingly enough gives the correct answer to this problem of 5775.
[edit] Similarity?
So why are two answers for these two different problems the same? I offer two reasons: First of all, the twelfth person has can go nowhere except into the group of three which needs her. Secondly, bringing in the twelfth person mathematically, we introduce a 12 into the numerator; but also into the denominator we introduce a 4 for the new group of 4, and a 3 for the third identical group.
[edit] A complication
But wait! What about Christine! She only wants to do 9 holes, and we want to put her in a group with three other people. So we choose one group of 4, and a group of 3 for Christine and a group of three without her.
.
