Yes, the golden ratio appears a lot in the pentagon.
It's possible to get the sin/cos/tan of all integers angles (in degrees), because sin(2x) is a 2nd degree function of sin(x) and sin(3x) is a 3rd degree function of sin(x).
A table with all values:
http://intmstat.com/blog/2011/06/exact-values-sin-degrees.pdf
By the way, all the regular polygons with a mersenne fermat prime number of sides (5, 17, 257, ...) can be drawn with ruler and compass. So, something like that can be done (in a much more complicated fashion).
I can't even begin to imagine how you solve the exact value of something like sin(82°).
Btw, I can't help to notice how much sqrt(5) appears everywhere. It seems to appear much more often than other square roots. Is there something inherent in the sine of integer degrees that somehow links it to sqrt(5)?
I've seen that table before. If you look carefully, there are terms of "I" sprinkled throughout. Is this the imaginary unit or is it some other constant? If it is the imaginary unit, there must be some wonderful cancellation in here.
Is it the case, then, that the sine of any rational number (or, since we really ought to be talking in radians, a rational multiple of pi) is an algebraic number? And if so, is there an algorithm that can calculate it?
Entering the value of sin(50°) from that pdf into wolfram alpha replacing the I with the imaginary unit i gives a result with an alternate form of cos(2*pi/9) which is cos(40°) which is sin(50°), so I is in fact the imaginary unit i.
Warning: Might glitch to creditsI will finish this ACE soon as possible
(or will I?)
I is the imaginary unit in those formulas, it appears frequently in the casus irreducibilis of 3rd degree and better polynomials.
While it is possible to use trig substitutions to eliminate the imaginary unit, the point of the chart is to give a solution under radicals, and these numbers simply cannot be simplified further under this restriction. However, the imaginary parts do completely cancel in all cases.
A good minimal example of such a number* to play around with is (1 + i*sqrt(7))^1/3 + (1 - i*sqrt(7))^1/3
Which is approximately 2.6016791318831542525
This is also equal to 2 sqrt(2) cos(1/3 arctan(sqrt(7)))
(Math Challenge: Can you prove this relationship?)
* It's generating cubic is x3 - 6 x - 2 = 0
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thatguy wrote:
Is it the case, then, that the sine of any rational number (or, since we really ought to be talking in radians, a rational multiple of pi) is an algebraic number? And if so, is there an algorithm that can calculate it?
Sin of pi times a rational is always algebraic indeed, due to the formula sin(m+n) = sin(m)cos(n) + sin(n)cos(m). It allows you write sin of (pi/b) as the root of a (b-1)th degree polynomial, and then to sum it to itself a times to get a polynomial of degree b-1 with one root equal to sin(a*pi/b)
It's possible to get the sin/cos/tan of all integers angles (in degrees), because sin(2x) is a 2nd degree function of sin(x) and sin(3x) is a 3rd degree function of sin(x).
sin/cos/tan of all integer degree angles are solvable by radicals, but not because sin(2x) is a 2nd degree function of sin(x) and sin(3x) is a 3rd degree function of sin(x). (A now-deleted post by OmnipotentEntity pointed out you need sin(5x), a 5th degree function. In general, 5th degree polynomials are not solvable by radicals.) It is because all roots of unity (of angle θ which is a rational multiple of π) are solvable by radicals, and so are their real part (cos θ), imaginary part (sin θ) and the ratio of the two (tan θ).
Even though every root of unity is solvable by radicals, some of them involve things like radicals of complex numbers, as you can see from the posted table. (There is no rule against taking radicals of complex numbers, however mind-bending that is.) If you want only solvability by radicals using only real numbers, this restricts things. Correct me if I am wrong, but I think the following applies (not necessarily exhaustive):
- You are allowed to solve any 2nd degree equations.
- You are allowed to solve irreducible 3rd degree equations as long as it does not have three distinct real roots, which is the "casus irreducibilis" (but the equation for sin(3x) in terms of sin(x) is always casus irreducibilis, since you can always get three different angles from the equation).
I believe that in order for the sin/cos/tan of an nth root of unity to be solvable by radicals in real numbers, the Euler phi number φ(n), which is how many numbers between 1 and n are relatively prime to n, must be a power of 2. This occurs whenever n is a product of zero or more distinct Fermat primes times a power of 2. (Ex. from the table, sin 3° is represented in radicals using only real numbers, n=120=23*3*5. However sin 1° fails because n=360=23*32*5, and 3 occurs as a factor twice.) However, I can't prove it.
Edit: Fixed some numbers.
(A now-deleted post by OmnipotentEntity pointed out you need sin(5x), a 5th degree function. In general, 5th degree polynomials are not solvable by radicals.)
I deleted it because it seemed like there was something else going on, to wit you never see 4th or 5th roots in those equations despite needing to solve sin(5x). But nothing was coming to mind, and rather than be potentially misleading, I thought it better to retract.
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FractalFusion: you're right. Although some 5th degree polynomials are impossible to solve with radicals (x^5 - x + 1 = 0, for instance), all sin(pi/n) can be solved using the n-th root of the unity.
For me there's no point doing that though, or using i in the answer.
And if n is a product of zero or more distinct Fermat primes times a power of 2 it's even better: you can express the sin with only square roots (no cubic roots) and arithmetic operations on integers.
https://mathoverflow.net/questions/36276/when-is-sinr-pi-expressible-in-radicals-for-r-rational
Maybe someone of you can solve it:
floor(n/x) == floor(n/(x+1))
for fixed positive n, find minimum positive x. in whole integers. (both n and x)
I don't have an answer.
I'm interested only in rigorous proof.
Numerical solution is easy.
Maybe someone of you can solve it:
floor(n/x) == floor(n/(x+1))
for fixed positive n, find minimum positive x. in whole integers. (both n and x)
I don't have an answer.
I'm interested only in rigorous proof.
Numerical solution is easy.
I didn't understand this question until OmnipotentEntity's post, which included a link to the relevant OEIS sequence. Allow me to clarify the question through a few examples.
We'll start with n=1:
x = 1: ⌊1/1⌋ = 1
x = 2: ⌊1/2⌋ = 0
x = 3: ⌊1/3⌋ = 0
and so we conclude that x=2 is the minimum integer we seek, since ⌊1/2⌋ = ⌊1/3⌋.
Here's n=2:
x = 1: ⌊2/1⌋ = 2
x = 2: ⌊2/2⌋ = 1
x = 3: ⌊2/3⌋ = 0
x = 4: ⌊2/4⌋ = 0
So for n = 2, x = 3.
Now n=3:
x = 1: ⌊3/1⌋ = 3
x = 2: ⌊3/2⌋ = 1
x = 3: ⌊3/3⌋ = 1
and therefore, when n = 3, x = 2.
One more time with a larger n. I notice the sequence has an interesting jump when n=90, so let's do that:
...
x = 10: ⌊90/10⌋ = 9
x = 11: ⌊90/11⌋ = 8
x = 12: ⌊90/12⌋ = 7
x = 13: ⌊90/13⌋ = 6
x = 14: ⌊90/14⌋ = 6
so for n = 90, x = 13.
I hope that helps others bite into this problem.
Edit: Typo. When n=2, x=3, not 4.
Edit 2: By the way, for those who haven't noticed or bothered to look, OEIS produces graphs of its sequences. Here's the graph for this sequence. I'm pretty sure the graph is closely approximated by sqrt(n) (which, in just 30 seconds of thought, kind of makes "intuitive sense" to me). I'd begin my investigation by plotting x(n) - sqrt(n) versus n; i.e., I'd plot the residuals and look for patterns. At the very least, I'm confident we can put lower- and upper-bounds on x(n).
I didn't know that this is famous sequence.
Bobo the King wrote:
Edit 2: By the way, for those who haven't noticed or bothered to look, OEIS produces graphs of its sequences. Here's the graph for this sequence. I'm pretty sure the graph is closely approximated by sqrt(n) (which, in just 30 seconds of thought, kind of makes "intuitive sense" to me).
I had prove some facts, including that lower bound is floor(sqrt(x)) (easy proof).
Good eye Bobo,
Here is the sequence graphed with what seems to be both a valid lower bound of sqrt(x) and a valid upper bound of sqrt(x) + ln(x). The upper bound fails for x < 3 (which are oddball anyway.)
(Click for larger)
The upper bound of ln(x) does not hold.
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The upper bound of [sqrt(n) +ln(n)] does not hold.
Sure it does! Just for n>3. Bounds like this are constructed all the time, such as in Goldbach's conjecture: every even number greater than 2 can be written as the sum of two prime numbers.
I wouldn't put too much stock into a minor violation early on when it looks like sqrt(n) + ln(n) does a fantastic job of not only providing an upper bound, but apparently a pretty good supremum as well.
Edit: I forgot Goldbach's conjecture. I really have to remember to check Wikipedia, then post...
Edit 2: Concerning this part of your post...
OmnipotentEntity wrote:
The upper bound of ln(x) does not hold.
What's being graphed here? Is the blue area x(n) while the green line is sqrt(n) + ln(n)? If so, then I misinterpreted this graph and sqrt(n) + ln(n) is, in fact, not an upper bound.
If anyone else wants to investigate the sequence, but doesn't want to generate a bunch of numbers themself. Here're the first 1e7. (555kB gzipped, 47MB raw)
http://fixed.space/files/10tothe7.txt.gz
I've got the first billion running on my beefier computer. Should be done in about a day?
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(Reposting an updated version.)
I've got a formula for a(n).
Note that floor(x) is the greatest integer less than or equal to x, ceil(x) is the least integer greater than or equal to x.
Let k=floor(sqrt(n)). If k2 ≤ n < k(k+1), then:
a(n) = ceil( ［2k+1+sqrt((2k+1)2-4n)］/2 ) - 1. (*)
If k(k+1) ≤ n < (k+1)2, then:
a(n) = ceil( ［2k+2+sqrt((2k+2)2-4n)］/2 ) - 1. (**)
-------------
By plugging in n=k(k+1)-1, n=k2 into (*) and n=(k+1)2-1, n=k(k+1) into (**), we get the following bounds:
For (*): k ≤ a(n) ≤ ceil( ［2k+1+sqrt(4k+1)］/2 ) - 1 = ceil( k+0.5+sqrt(k+0.25) ) - 1.
For (**): k+1 ≤ a(n) ≤ ceil( ［2k+2+sqrt(4k+4)］/2 ) - 1 = ceil( k+1+sqrt(k+1) ) - 1.
Since k=floor(sqrt(n)), this confirms that the lower bound grows as sqrt(n) and the upper bound grows as sqrt(n)+sqrt(sqrt(n)). OmnipotentEntity has the right guess for the upper bound.
-------------
@r57shell: Since you wanted a rigorous proof, I will try to provide one. Can't guarantee that you will be satisfied though.
Fix integer n>0. I will use the real-valued functions fr(x)=n/x and gr(x)=x+n/x, and the integer-valued functions f(x)=floor(n/x) and g(x)=floor(x+n/x).
We want to find minimum d such that f(d)=f(d+1). I claim that this is the minimum d such that g(d)<g(d+1). Indeed, g(x)=x+f(x), and so g(d+1)-g(d)=1+f(d+1)-f(d). Note that, since fr(x)=n/x is a decreasing function on (0,∞), f(x) is a weakly decreasing (non-increasing) function ; that is, f(d+1)-f(d)≤0. So g(d+1)-g(d)≤1.
This gives: g(d)<g(d+1) ⇔ g(d+1)-g(d)=1 ⇔ f(d+1)-f(d)=0 ⇔ f(d)=f(d+1). This proves the claim.
Now let k=floor(sqrt(x)), so that k2≤x<(k+1)2. I claim that the minimum of g(x) occurs at x=k or x=k+1 (not necessarily exclusively). This is because gr(x)=x+n/x is decreasing on (0,sqrt(n)］ and increasing on ［sqrt(n),∞) (by calculus), which mean g(x) is weakly decreasing up to x=k and weakly increasing from x=k+1 onward. So the claim holds.
Let C=min(g(x)). The minimum d such that g(d)<g(d+1) occurs at ceil(α)-1, where α is the larger real solution to x+n/x=C+1.
Two cases:
Case 1: k2≤x<k(k+1).
C=g(k)=g(k+1)=2k. Solving x+n/x=2k+1 gives α=［2k+1+sqrt((2k+1)2-4n)］/2, and so the minimum d occurs at ceil(［2k+1+sqrt((2k+1)2-4n)］/2)-1, which proves (*).
Case 2: k(k+1)≤x<(k+1)2.
g(k+1)=2k+1 and g(k)≥2k+1, so C=2k+1. Solving x+n/x=2k+2 gives α=［2k+2+sqrt((2k+2)2-4n)］/2, and so the minimum d occurs at ceil(［2k+2+sqrt((2k+2)2-4n)］/2)-1, which proves (**).
QED
and the integer-valued functions f(x)=floor(n/x) and g(x)=floor(x+n/x).
Just wanna confirm that they are both Z->Z (integer -> integer),
because:
FractalFusion wrote:
Indeed, g(x)=x+f(x), and so g(d+1)-g(d)=1+f(d+1)-f(d).
Тhis works only for x from Z (integer).
FractalFusion wrote:
which mean g(x) is weakly decreasing up to x=k and weakly increasing from x=k+1 onward.
It's true, just wanna add comment: we don't know difference between g(k) and g(k+1).
FractalFusion wrote:
Let C=min(g(x)). The minimum d such that g(d)<g(d+1) occurs at ceil(α)-1, where α is the larger real solution to x+n/x=C+1.
This is out of space. This is where rigorous proof ends.
If here would be also proof, that it is indeed occurs at ceil(a) -> it will be truly rigorous.
Indeed, g(x)=x+f(x), and so g(d+1)-g(d)=1+f(d+1)-f(d).
Тhis works only for x from Z (integer).
Given g(x) = x+f(x) with x ∈ ℝ.
We have by simple substitution:
g(d) = d + f(d)
g(d+1) = d + 1 + f(d+1)
Now subtract the first line from the second:
g(d + 1) - g(d) = d + 1 + f(d+1) - (d + f(d))
Simplify:
g(d + 1) - g(d) = d + 1 + f(d+1) - d - f(d)
g(d + 1) - g(d) = d - d + 1 + f(d+1) - f(d)
g(d + 1) - g(d) = 1 + f(d+1) - f(d)
QED.
Considering that g is defined to have a domain of ℝ, nothing in these steps requires d ∈ ℤ.
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and the integer-valued functions f(x)=floor(n/x) and g(x)=floor(x+n/x).
Just wanna confirm that they are both Z->Z (integer -> integer),
Yeah, that's what I meant. Apparently I forgot that "S-valued function" does not mean "function from S to S".
r57shell wrote:
FractalFusion wrote:
Let C=min(g(x)). The minimum d such that g(d)<g(d+1) occurs at ceil(α)-1, where α is the larger real solution to x+n/x=C+1.
This is out of space. This is where rigorous proof ends.
If here would be also proof, that it is indeed occurs at ceil(a) -> it will be truly rigorous.
Proof: gr(x)=x+n/x goes to infinity as x→∞. Since C+1=min(g(x)+1)≥min(gr(x)), there exists at least one solution to gr(x)=C+1. There are two solutions for x; one in (0,sqrt(n)) and one in (sqrt(n),∞) so we α to be the one in (sqrt(n),∞) (the larger one).
Since gr(α)=C+1 and gr(x)=x+n/x is increasing on ［sqrt(n),∞), it follows that gr(ceil(α))≥C+1, so g(ceil(α))≥C+1. If ceil(α)-1≥sqrt(n), then gr(ceil(α)-1)<C+1, and so g(ceil(α)-1)=C. (If not, then ceil(α)-1=floor(sqrt(n)), and it must follow that g(ceil(α)-1)=C; otherwise min(g(x))=C+1, contradicting that the minimum is C.) Therefore the minimum d such that g(d)<g(d+1) is d=ceil(α)-1.
OmnipotentEntity: Yes, that is true (if g were to have a domain of ℝ). However, there are a couple reasons I would like g to have a domain of ℤ in the proof: (1) If n=k(k+1), min(g(x))=min(floor(x+n/x)) would not have the same value if it were defined with a domain of ℝ as it would with a domain of ℤ. (2) I don't have to say something like "min over ℤ" every time.
The upper bound of ln(x) does not hold.
