5 ms·

That can't be good for all x. You mean x=π/4 or to 1dp.

by ogogmad 4y ago

Joker_vD 4y ago
Eh, it's close enough. After all, in the time of war the value of sine can go up to 4.
- nroets 4y ago
  Only 4 ? sin(1.5708) = 1 sin(1.5708 - 1.3170 i) = 2 sin(1.5708 - 2.0634 i) = 4 ...
  - dustingetz 4y ago
    wait what? ELI grad student, I know you can plug i into the sine taylor series expansion but what is happening here
    - some_furry 4y ago
      This is radians, which can go from 0 to tau (2 * pi, about 6.28).
    - deleted 4y ago
      [deleted]
    - scythe 4y ago
      sin(x + pi/2) = cos(x) = cosh(ix) Therefore, sin (pi/2 + i arcosh(2)) = cosh(-arcosh(2)) = 2 where arcosh(y) is the familiar inverse of the hyperbolic cosine, i.e. log(y + sqrt(y^2 - 1)), so we find sin(pi/2 + i log(2 + sqrt(3))) = 2.
    - hansvm 4y ago
      Sin/cos are tightly connected to the complex exponential and exhibit a lot of such behavior. In particular, sin(x+it) = sin(x)cosh(y) + i cos(x)sinh(y). Set x something to zero out cosine, and rely on the exponential nature of cosh to find a target y value.
      - hansvm 4y ago
        Oops, autocorrect, sin(x+iy). Also x and y should be real.
    - version_five 4y ago
      The above should be intelligible to someone with high school math, no? I'd be interested in the grad student version, just out of curiosity
      - nroets 4y ago
        Most highschool students mainly use sin to calculate the lengths of sides of triangles. So it feels very counterintuitive that sin can be greater than 1. If you define sin as the solution to a particular differential equation, then it's relationship to the hyperbolic functions is much more obvious.
dodnkskd 4y ago
Well, cos(π/4) happens to be the RMS value of the function.
version_five 4y ago
It's probably a pretty good 1st order guess that a surface to air missile launcher is aimed at 45 degrees. It's not aimed straight up and it's not flat on the ground.
- Arrath 4y ago
  > It's not aimed straight up Well that certainly depends! The VLS in ships are exactly that, vertical launch. As are, I believe, most Soviet/Russian large SAM systems like the S300 and S400
paulddraper 4y ago
> That can't be good for all x. It's within 1.3.

That can't be good for all x. You mean x=π/4 or to 1dp.

Joker_vD 4y ago

Eh, it's close enough. After all, in the time of war the value of sine can go up to 4.

nroets 4y ago
Only 4 ? sin(1.5708) = 1 sin(1.5708 - 1.3170 i) = 2 sin(1.5708 - 2.0634 i) = 4 ...
- dustingetz 4y ago
  wait what? ELI grad student, I know you can plug i into the sine taylor series expansion but what is happening here
  - some_furry 4y ago
    This is radians, which can go from 0 to tau (2 * pi, about 6.28).
  - deleted 4y ago
    [deleted]
  - scythe 4y ago
    sin(x + pi/2) = cos(x) = cosh(ix) Therefore, sin (pi/2 + i arcosh(2)) = cosh(-arcosh(2)) = 2 where arcosh(y) is the familiar inverse of the hyperbolic cosine, i.e. log(y + sqrt(y^2 - 1)), so we find sin(pi/2 + i log(2 + sqrt(3))) = 2.
  - hansvm 4y ago
    Sin/cos are tightly connected to the complex exponential and exhibit a lot of such behavior. In particular, sin(x+it) = sin(x)cosh(y) + i cos(x)sinh(y). Set x something to zero out cosine, and rely on the exponential nature of cosh to find a target y value.
    - hansvm 4y ago
      Oops, autocorrect, sin(x+iy). Also x and y should be real.
  - version_five 4y ago
    The above should be intelligible to someone with high school math, no? I'd be interested in the grad student version, just out of curiosity
    - nroets 4y ago
      Most highschool students mainly use sin to calculate the lengths of sides of triangles. So it feels very counterintuitive that sin can be greater than 1. If you define sin as the solution to a particular differential equation, then it's relationship to the hyperbolic functions is much more obvious.

dodnkskd 4y ago

Well, cos(π/4) happens to be the RMS value of the function.

version_five 4y ago

It's probably a pretty good 1st order guess that a surface to air missile launcher is aimed at 45 degrees. It's not aimed straight up and it's not flat on the ground.

Arrath 4y ago
> It's not aimed straight up Well that certainly depends! The VLS in ships are exactly that, vertical launch. As are, I believe, most Soviet/Russian large SAM systems like the S300 and S400

paulddraper 4y ago

> That can't be good for all x. It's within 1.3.