I found the following cool-looking formulas in [1].
(n + 1 nested roots)
(n + 2 nested roots)
(n + 2 nested roots)
To prove them, we use the following form double angle formula for cosine.
Now we evaluate this for different values of 
Secondly,
so that from (1),
To find 
Using the double and triple angle formulae for cosine this becomes
From this the only valid (positive) solution is 
Hence applying (1) again,
Now we set up a hypothesis for mathematical induction. Assume that
Note that these are equivalent to our initial three formulas. In (2)-(4) we have seen that these results are true for 
Similarly,
and
In each case we have shown that if the result is true for 
Reference
[1] E. Maor, Trigonometric Delights, Princeton University Press, 1998.
are collinear or in other words 
is a real number when 
.
are four points in the plane,
. The ratio 
is a complex number with argument 
while the ratio 
has argument 
. Hence for the product of these ratios to be real means 
is a multiple of 
. In other words, the points 
are four points in the plane, 
, or
![\begin{aligned} & a^3(b-c) + b^3(c-a) + c^3(a-b) \\&= (a + b + c)(a^2(b-c) + b^2(c-a) + c^2(a-b)) - \left[ (b+c) a^2(b-c) + (c + a) b^2 (c-a) + (a+b)c^2(a-b) \right]\\ &= -(a+b+c)(a-b)(b-c)(c-a) + \left[ a^2(b^2-c^2) + b^2(c^2 - a^2) + c^2(a^2-b^2) \right] \quad \text{(from the equality in 3.)}\\ &= -(a-b)(b-c)(c-a)(a + b + c)\end{aligned}](http://s0.wp.com/latex.php?latex=%5Cbegin%7Baligned%7D+%26+a%5E3%28b-c%29+%2B+b%5E3%28c-a%29+%2B+c%5E3%28a-b%29+%5C%5C%26%3D+%28a+%2B+b+%2B+c%29%28a%5E2%28b-c%29+%2B+b%5E2%28c-a%29+%2B+c%5E2%28a-b%29%29+-+%5Cleft%5B+%28b%2Bc%29+a%5E2%28b-c%29+%2B+%28c+%2B+a%29+b%5E2+%28c-a%29+%2B+%28a%2Bb%29c%5E2%28a-b%29+%5Cright%5D%5C%5C+%26%3D+-%28a%2Bb%2Bc%29%28a-b%29%28b-c%29%28c-a%29+%2B+%5Cleft%5B+a%5E2%28b%5E2-c%5E2%29+%2B+b%5E2%28c%5E2+-+a%5E2%29+%2B+c%5E2%28a%5E2-b%5E2%29+%5Cright%5D+%5Cquad+%5Ctext%7B%28from+the+equality+in+3.%29%7D%5C%5C+%26%3D+-%28a-b%29%28b-c%29%28c-a%29%28a+%2B+b+%2B+c%29%5Cend%7Baligned%7D&bg=ffffff&fg=000000&s=0 "\begin{aligned} & a^3(b-c) + b^3(c-a) + c^3(a-b) \\&= (a + b + c)(a^2(b-c) + b^2(c-a) + c^2(a-b)) - \left[ (b+c) a^2(b-c) + (c + a) b^2 (c-a) + (a+b)c^2(a-b) \right]\\ &= -(a+b+c)(a-b)(b-c)(c-a) + \left[ a^2(b^2-c^2) + b^2(c^2 - a^2) + c^2(a^2-b^2) \right] \quad \text{(from the equality in 3.)}\\ &= -(a-b)(b-c)(c-a)(a + b + c)\end{aligned}")
is the centroid of 
(i.e. the vector sum 
is 
).
runs. Later on I checked that this was in fact the case. I thought it was cool that from a single average to two decimal places one could make a good guess of both the number of dismissals and the runs scored by the player (i.e. both the dividend and divisor), just based on the assumption that the ratio is simple. Of course if the decimal had been something like 50.25, it would have been more difficult to guess whether the he had scored 201, 402, 603, … runs for 4, 8, 12, … dismissals.
or equivalently 
. We first find a fraction that is close to m/n, in this example 1/4 is appropriate. Using this, we set 
(more generally for the approximation p/q choose 
) and find that 
or equivalently,
, 
ranges from 5 to 12, corresponding to the fractions 
(note that 
which rounds to 0.26 while 
which rounds down to 0.25).
, 
. We can continue to find all fractions satisfying the above inequality but it is apparent that the simplest will correspond to 
and 
.
we wish to find
and 
).
for which the peak value of 
is maximised is 
as illustrated below (again 
).
, resulting in a minimum value of 
for the sum. However the maximum value is only 
for the case 
. Hence we make the observation that finding the lowest peak is different from finding the minimum deviation from zero.
where 
and 
are variables and 
are constants. Hence we wish to maximise 
subject to 
. By the method of Lagrange multipliers, at the maximum the gradient vector of the objective function is parallel to the gradient of the constraint. In other words, 
. Hence there exists a constant 
so that
and 
gives 
and 
. Then the condition 
, one alternative approach is to proceed as follows. We firstly note that for 
,
or 
depending on how large 
since when 
, we can achieve a minmax value of 
by aligning the peak of 
with the trough of 
(e.g. for 
. This is at the local maximum since here,
.
(i.e. 
is kept constant). Then 
,
and
. Then
, and note that 
. We wish to show if 
there exists 
. We use the fact that near 
, 
while 
. In particular for all 
. Furthermore choose 
so that 
(i.e. 
) and 
. We then find
,
, where 
can be chosen to be positive for some 
. Hence we have 
as desired and
has been exceeded. We conclude that we must have 
and 
(e.g. when 
. For the examples plotted above, 
as we found in the third graph.
where 
, we simply write this as 
where 
and use the above result to obtain the following:
in the plane there exist points 
and choose 
so that 
and the triangles have the same orientation (i.e. vertices labelled in the same direction clockwise or anticlockwise). This can be done in one way only. We now wish to show that in this configuration we have the third condition 
. We shall use complex number geometry to show this.
represent the spiral similarity that maps 
to 
.
represent the spiral similarity that maps 
.
represent the spiral similarity that maps 
. But since 
, 
to 
. Similarly 
.
maps 
).
