8 ms·

If you consider solutions to one of the simplest non-trival second-order differential equation, f''(x) = -f(x), then you will find that the solutions all have p

If you consider solutions to one of the simplest non-trival second-order differential equation, f''(x) = -f(x), then you will find that the solutions all have period 2π, no geometry needed.

hatsunearu 5y ago
Well, the thing about a circle is that the very core of the definition has the same sort of "information" as whatever abstract "information" f'' = -f contains.
- goldenkey 5y ago
  Yeah, a circle is basically a particular exchange between the velocity in the x and velocity in the y direction, offset by a phase equal to half of the period. With different kind of exchanges, you can get squircles and all types of other pseudo-circular shapes.
- joppy 5y ago
  I guess this must be true, but from the sidelines it seems that “the set of points 1 metre from this particular point” and “the solution to this particular second-order differential equation” are quite different things. Any competent student of mathematics can connect them for sure, but it’s still quite remarkable they’re related isn’t it?
gspr 5y ago
Oh I don't know about that. The solution involves trigonometric functions. Are they not geometric in nature? Said differently: periodic things in some sense express motions on a circle.
- mnw21cam 5y ago
  No. Trig functions can be defined independently of geometry. One way is by considering the differential equation above. Another is through the Taylor expansion of the sin(x), which is periodic as a completely emergent property. Likewise, it's hard to imagine what would happen to that very famous equation e^iπ + 1 = 0 if π were something different.
  - gspr 5y ago
    I know. But they do also reflect geometry.
- HPsquared 5y ago
  In a 1-dimensional spring-mass system there are no circles, but π emerges nonetheless. Edit: natural frequency ω = Sqrt(k/m). ω comes out in rad/s, to convert to more natural units (cycles per second) π is required.
  - gspr 5y ago
    Well, you might say that the circle emerges from the dynamics of that system. Two such spring-masses out of phase draw a circle.
  - BoiledCabbage 5y ago
    To some degree, isn't that essentially just "adding one more turtle"? Now instead of space being Euclidean, you're saying space-time is? And effectively if space-time were no longer Euclidean, then the conversion to cycles/sec would use the "alternate pi" of this non-euclidean space time wouldn't it?
- kortex 5y ago
  You can derive pi/4 from summing alternating reciprocals of odd numbers. No geometry. You do need limits (since pi is transcendental, there are no purely algebraic solutions. You can use special functions, but those have limits or other transcendentals "in them"). https://en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80 https://en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80
  - gspr 5y ago
    Indeed, good point.
- rssoconnor 5y ago
  I'd say periodic things express motions on a loop, in the topological sense of looping back to itself. But because topology is only defined up to deformations, thus you cannot really say that loops are circles. However, there is indeed a connection to circles. One way of seeing this by looking at the solutions to this differential equation over the complex "plane" and looking at the exp(i x) and exp(-i x) solutions, and seeing how these functions wrap the real number line around in a circle. The complex plane fundamentally has a Euclidean geometry to it. So much so that if you grew up in a world with a strong non-euclidean geometry, or even grew up as some sort of digital being with no real notion of space at all, so long as you are able perform moderately sophisticated mathematics you are going to wind up discovering the complex numbers, and things like their absolute value and multiplication and the exponential function and how they map values around. Perhaps you never mentally arrange the complex numbers into a "plane", but none the less, all your basic geometric concepts are embedded into the algebraic operations on these complex numbers, and you will wind up effectively doing Euclidean geometry, even if you never identify it as such.
beebmam 5y ago
I assume that you mean solutions where x is in R here
- mjburgess 5y ago
  And in my view, `R` is "fundamentally" a geometrical construct.
  - zarzavat 5y ago
    R is an analytical construct, it exists to make limits meaningful. You don't need all of R to do geometry, if you are Ancient Greek you don't need any field at all.
  - contravariant 5y ago
    It's the only complete ordered field, so fundamentally it's an algebraic structure (with some topology thrown in but basic enough that it can be phrased without referring to geometry).
    - mjburgess 5y ago
      It depends what you give "priority" to -- my view is a philosophical one. I think we ought really see `R` as an algebraic stand-in for talking about geometry.
      - contravariant 5y ago
        Eh, if R is geometric then so is algebra. The only thing geometric about R is that you can define an absolute value (i.e. it is an ordered ring) which is a property it inherits from the fractions. The fractions in turn are a basic consequence of the definition of a field (the fractions have a unique image into all fields). If you've got addition and multiplication then you can get the reals by just adding additive inverse, multiplicative inverses and completeness, none of which are inherently geometric properties. So any geometry must come from the distributive property, because there's not much else it could be.
0-_-0 5y ago
Or just rely on Euler's formula: e^(iπ)=-1
hyperpallium2 5y ago
That's really freaky, though no more so than that for DE f'(x) = f(x), the solution is e^x The self-similarity in these two DEs is why Euler's formula works. Are there any other such DEs?
- contravariant 5y ago
  In general any such linear differential equation f'(t) = M f(t) can be solved as follows f(t) = e^Mt f(0) where M is a matrix and f(0) is a vector, and the matrix exponential is defined using the Taylor series of e^x. Something like f''(t) = -f(t) can be split up as follows: f(t) = h'(t) h'(t) = -f(t) which corresponds to the matrix M = [[ 0 1] [-1 0]] which not coincidentally is a matrix representation of 'i' (it satisfies M^2 = -I), so e^Mt basically generates the same values as e^it (except as 2d vectors instead of complex numbers). Since the above is also the differential equation for constant rotation in the 2D plane this is another way of deriving the relation between the complex exponential and the trigonometric functions.

If you consider solutions to one of the simplest non-trival second-order differential equation, f''(x) = -f(x), then you will find that the solutions all have period 2π, no geometry needed.

hatsunearu 5y ago

Well, the thing about a circle is that the very core of the definition has the same sort of "information" as whatever abstract "information" f'' = -f contains.

goldenkey 5y ago
Yeah, a circle is basically a particular exchange between the velocity in the x and velocity in the y direction, offset by a phase equal to half of the period. With different kind of exchanges, you can get squircles and all types of other pseudo-circular shapes.
joppy 5y ago
I guess this must be true, but from the sidelines it seems that “the set of points 1 metre from this particular point” and “the solution to this particular second-order differential equation” are quite different things. Any competent student of mathematics can connect them for sure, but it’s still quite remarkable they’re related isn’t it?

gspr 5y ago

Oh I don't know about that. The solution involves trigonometric functions. Are they not geometric in nature? Said differently: periodic things in some sense express motions on a circle.

mnw21cam 5y ago
No. Trig functions can be defined independently of geometry. One way is by considering the differential equation above. Another is through the Taylor expansion of the sin(x), which is periodic as a completely emergent property. Likewise, it's hard to imagine what would happen to that very famous equation e^iπ + 1 = 0 if π were something different.
- gspr 5y ago
  I know. But they do also reflect geometry.
HPsquared 5y ago
In a 1-dimensional spring-mass system there are no circles, but π emerges nonetheless. Edit: natural frequency ω = Sqrt(k/m). ω comes out in rad/s, to convert to more natural units (cycles per second) π is required.
- gspr 5y ago
  Well, you might say that the circle emerges from the dynamics of that system. Two such spring-masses out of phase draw a circle.
- BoiledCabbage 5y ago
  To some degree, isn't that essentially just "adding one more turtle"? Now instead of space being Euclidean, you're saying space-time is? And effectively if space-time were no longer Euclidean, then the conversion to cycles/sec would use the "alternate pi" of this non-euclidean space time wouldn't it?
kortex 5y ago
You can derive pi/4 from summing alternating reciprocals of odd numbers. No geometry. You do need limits (since pi is transcendental, there are no purely algebraic solutions. You can use special functions, but those have limits or other transcendentals "in them"). https://en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80 https://en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80
- gspr 5y ago
  Indeed, good point.
rssoconnor 5y ago
I'd say periodic things express motions on a loop, in the topological sense of looping back to itself. But because topology is only defined up to deformations, thus you cannot really say that loops are circles. However, there is indeed a connection to circles. One way of seeing this by looking at the solutions to this differential equation over the complex "plane" and looking at the exp(i x) and exp(-i x) solutions, and seeing how these functions wrap the real number line around in a circle. The complex plane fundamentally has a Euclidean geometry to it. So much so that if you grew up in a world with a strong non-euclidean geometry, or even grew up as some sort of digital being with no real notion of space at all, so long as you are able perform moderately sophisticated mathematics you are going to wind up discovering the complex numbers, and things like their absolute value and multiplication and the exponential function and how they map values around. Perhaps you never mentally arrange the complex numbers into a "plane", but none the less, all your basic geometric concepts are embedded into the algebraic operations on these complex numbers, and you will wind up effectively doing Euclidean geometry, even if you never identify it as such.

beebmam 5y ago

I assume that you mean solutions where x is in R here

mjburgess 5y ago
And in my view, `R` is "fundamentally" a geometrical construct.
- zarzavat 5y ago
  R is an analytical construct, it exists to make limits meaningful. You don't need all of R to do geometry, if you are Ancient Greek you don't need any field at all.
- contravariant 5y ago
  It's the only complete ordered field, so fundamentally it's an algebraic structure (with some topology thrown in but basic enough that it can be phrased without referring to geometry).
  - mjburgess 5y ago
    It depends what you give "priority" to -- my view is a philosophical one. I think we ought really see `R` as an algebraic stand-in for talking about geometry.
    - contravariant 5y ago
      Eh, if R is geometric then so is algebra. The only thing geometric about R is that you can define an absolute value (i.e. it is an ordered ring) which is a property it inherits from the fractions. The fractions in turn are a basic consequence of the definition of a field (the fractions have a unique image into all fields). If you've got addition and multiplication then you can get the reals by just adding additive inverse, multiplicative inverses and completeness, none of which are inherently geometric properties. So any geometry must come from the distributive property, because there's not much else it could be.

0-_-0 5y ago

Or just rely on Euler's formula: e^(iπ)=-1

hyperpallium2 5y ago

That's really freaky, though no more so than that for DE f'(x) = f(x), the solution is e^x The self-similarity in these two DEs is why Euler's formula works. Are there any other such DEs?

contravariant 5y ago
In general any such linear differential equation f'(t) = M f(t) can be solved as follows f(t) = e^Mt f(0) where M is a matrix and f(0) is a vector, and the matrix exponential is defined using the Taylor series of e^x. Something like f''(t) = -f(t) can be split up as follows: f(t) = h'(t) h'(t) = -f(t) which corresponds to the matrix M = [[ 0 1] [-1 0]] which not coincidentally is a matrix representation of 'i' (it satisfies M^2 = -I), so e^Mt basically generates the same values as e^it (except as 2d vectors instead of complex numbers). Since the above is also the differential equation for constant rotation in the 2D plane this is another way of deriving the relation between the complex exponential and the trigonometric functions.