[math-fun] nice little sin(a+b+c+...) formula
n n ==== /===\ \ k | | %pi k > (- 1) | | 2 sin(a - -----) n / | | j n + 1 ==== ==== j = 0 \ k = 0 sin( > a ) = ------------------------------------, / k 2 (n + 1) ==== k = 1 where %pi n a := -----. 0 2 There are n^2 sins on the right, vs n2^(n-1) for the usual addition formula. But all n^2 sins are different, vs only 2n species for the usual formula. E.g., for n=3, %pi %pi sin(a + a + a ) = - sqrt(2) sin(a + ---) sin(a + ---) 3 2 1 1 4 2 4 %pi %pi %pi %pi sin(a + ---) - sqrt(2) sin(a - ---) sin(a - ---) sin(a - ---) 3 4 1 4 2 4 3 4 + 2 sin(a ) sin(a ) sin(a ) 1 2 3 = - sin(a ) sin(a ) sin(a ) + cos(a ) cos(a ) sin(a ) 1 2 3 1 2 3 + cos(a ) sin(a ) cos(a ) + sin(a ) cos(a ) cos(a ) 1 2 3 1 2 3 We can trade # of terms vs # of species by applying the new(?) formula recursively, e.g., a_1 = b_11+b_12+b_13, a_2= b_21+b_22+... . --rwg __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com
Dear Bill: I assume that some of the same types of formulae work for hyperbolic sines? I'm still looking for good formulae for the following cases: sinh(x*2^k) = ?? sinh(x+y) = ?? asinh(x*2^k) = ?? asinh(x+y) = ?? Thx. Henry Baker hbaker1@pipeline.com Santa Barbara, CA At 02:32 AM 10/24/2007, you wrote:
n n ==== /===\ \ k | | %pi k > (- 1) | | 2 sin(a - -----) n / | | j n + 1 ==== ==== j = 0 \ k = 0 sin( > a ) = ------------------------------------, / k 2 (n + 1) ==== k = 1 where %pi n a := -----. 0 2
There are n^2 sins on the right, vs n2^(n-1) for the usual addition formula. But all n^2 sins are different, vs only 2n species for the usual formula. E.g., for n=3,
%pi %pi sin(a + a + a ) = - sqrt(2) sin(a + ---) sin(a + ---) 3 2 1 1 4 2 4 %pi %pi %pi %pi sin(a + ---) - sqrt(2) sin(a - ---) sin(a - ---) sin(a - ---) 3 4 1 4 2 4 3 4 + 2 sin(a ) sin(a ) sin(a ) 1 2 3 = - sin(a ) sin(a ) sin(a ) + cos(a ) cos(a ) sin(a ) 1 2 3 1 2 3 + cos(a ) sin(a ) cos(a ) + sin(a ) cos(a ) cos(a ) 1 2 3 1 2 3
We can trade # of terms vs # of species by applying the new(?) formula recursively, e.g., a_1 = b_11+b_12+b_13, a_2= b_21+b_22+... . --rwg
On 10/24/07, Henry Baker <hbaker1@pipeline.com> wrote:
sinh(x*2^k) = ?? sinh(x+y) =
These two are exactly the same as in the regular sine case, e.g. sinh(2x) = 2 sinh(x) cosh(x) sinh(x+y) = sinh(x) cosh(y) + sinh(y) cosh(x) and so on.
asinh(x*2^k) = ?? asinh(x+y) = ??
I don't know. Either use the ln formulation of these or just use the fact that asinh(x) = i * asin(-ix) to change your favorite asin formulas into asinh formulas. I don't have any favorite asin formulas to use to start with, though. --Joshua Zucker
But note the very cheap recurrence %pi k 2 sin(a - -----) = y , n + 1 k which can be computed ultrastably from the coupled recurrences x = x - y , y = e x + y , with k + 1 k k k + 1 k + 1 k 2 %pi e := 4 sin -------, and 2 n + 2 %pi cos(------- + a) 2 n + 2 x = - ----------------, y = 2 sin(a). 0 %pi 0 sin ------- 2 n + 2 This gives you all n+1 sins for the cost of two sins and a cos. Note that if you algebraically eliminate x[k] in favor of a second order recurrence for y[k], you lose the reversibility and hence the stability. --rwg Bill Gosper <rwmgosper@yahoo.com> wrote: n n ==== /===\ \ k | | %pi k > (- 1) | | 2 sin(a - -----) n / | | j n + 1 ==== ==== j = 0 \ k = 0 sin( > a ) = ------------------------------------, / k 2 (n + 1) ==== k = 1 where %pi n a := -----. 0 2 There are n^2 sins on the right, vs n2^(n-1) for the usual addition formula. __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com
"Why should Johnny learn the quadratic equation? He'll never use it." Did the spectators at those Renaissance polynomial-solving contests grasp the motivation of the problem? What is the most "mundane" (whatever that means) need for solving a cubic? SteveG's irrational(!) remark could be phrased: 1) Two blips on the radar are exactly a mile away and 40 deg apart. How far apart are they? Another one: 2) A cupcake paper is made from a 6" disk. What conical frustum maximizes its batter capacity? The terser "What's the largest capacity pie tin you can make from a square foot of sheet metal?" is only biquadratic. Unless I've erred, it's 12.06" across, 2.89" deep, with only a 5.46" bottom, and holds a human blood supply(!?) (6.02qt). According to the prism(at)oid volume formula (Simpson's rule, which is exact in this case), frustum volume = pi depth (RR+Rr+rr)/3. Anyway, there must be lots of "nice" cubics out there that textbooks and puzzlists routinely shun. Suggestions? --rwg __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com
=Bill Gosper 1) Two blips on the radar are exactly a mile away and 40 deg apart. How far apart are they?
Or, "The sun is roughly 100 million miles away and looks half a degree wide. About how big is it?" But I'm not getting how these are cubic?
...the largest capacity pie tin you can make ...holds a human blood supply(!?) (6.02qt)
I've heard of blood pudding, but PIE?! Dude, I am so NOT trick 'r treating at your house!
On Saturday 27 October 2007, Marc LeBrun wrote:
=Bill Gosper 1) Two blips on the radar are exactly a mile away and 40 deg apart. How far apart are they?
Or, "The sun is roughly 100 million miles away and looks half a degree wide. About how big is it?"
But I'm not getting how these are cubic?
The answer is 2 miles * sin 20 degrees, and sin 20 degrees satisfies a cubic equation over Z because sin (3*20 degrees) = 1/2. So if knowing what equation the answer satisfies is more important to you than knowing it to (say) five significant figures, then this is a problem involving cubic equations. I'll hazard a guess that anyone who feels that way *already* sees value in understanding cubic equations. -- g
I claim to have at last slightly outsimplified Derive's monomial sqrt(-4*sqrt(sqrt(38*sqrt(17)+170)+3*sqrt(17)+17) -2*sqrt(34-2*sqrt(17))+2*sqrt(17)+30)/8 (nesting depth =4) with (-2^(3/4)*17^(1/8)*sqrt(sqrt(2)*17^(1/4)-sqrt(sqrt(17)-1))*sqrt(sqrt(17)+3) +sqrt(17)+sqrt(34-2*sqrt(17))-1)/16, a quadrinomial with depth 3. (Although Derive's cons_size is less.) --rwg __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com
rwg>The terser "What's the largest capacity pie tin you can make from a square foot of sheet metal?" is only biquadratic. Unless I've erred, it's 12.06" across, 2.89" deep, with only a 5.46" bottom, and holds a human blood supply(!?) (6.02qt). I erred. The hemispherical ideal, radius = depth = 4.79", holds only 3.98 qt. I got the base diamter right (5.46196"), but the top is 9.95648", and the depth is a more satisfyingly hemispherical 4.44215". Volume is a more plausible 3.692178 qt. The dimensions are still biquadratics/sqrt pi, but they factored out of much higher degree polynomials from clearing radicals and eliminating. --rwg __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com
On 10/26/07, Bill Gosper <rwmgosper@yahoo.com> wrote:
"Why should Johnny learn the quadratic equation? He'll never use it." Did the spectators at those Renaissance polynomial-solving contests grasp the motivation of the problem? What is the most "mundane" (whatever that means) need for solving a cubic? ... Anyway, there must be lots of "nice" cubics out there that textbooks and puzzlists routinely shun. Suggestions? --rwg
Build an emergency bivouac tent from a 2-metre square mylar sheet and 2 adjustable ski poles: locate them under the sheet so that the side edges sweep down from the central ridge to ground level; and the end edges drape over the pole tops, mid-points secured at the feet, and stretching tautly to the corners. (Or if you prefer, construct an 8-hedral teabag using a pair of similar shells of more modest dimensions.) The pole height e maximising the enclosed volume is the root in the unit interval of the cubic 3 - 4*e - 6*e^2 + 6*e^3 = 0 which alas has 3 real roots, so the Cardan (Tartaglia?) formula is not terribly informative. [In unit teabag terms, the depth equals 0.54676 and the volume 0.1454632612.] WFL
participants (6)
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Bill Gosper -
Fred lunnon -
Gareth McCaughan -
Henry Baker -
Joshua Zucker -
Marc LeBrun