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liberstina [14]
3 years ago
14

You can “unfold” a cube to make six squares, shaped like a cross. You can unfold a cylinder to make a rectangle and two circles.

Is it possible to unfold a sphere? What would an unfolded sphere look like?
Mathematics
1 answer:
den301095 [7]3 years ago
8 0

As John Steele mentioned, the reason that we can’t ‘unfold’ the surface of a sphere (that is, as I take your meaning, place it onto a flat surface) has to do with the Gaussian curvature of the surface. Bending or even tearing the surface into pieces won’t change this curvature. The surface of a sphere and that of a flat surface have fundamentally different Gaussian curvatures which cause this to be impossible.

Consider extending some radius outwards from a point and drawing a circle at this radius. On a flat surface (ie euclidean) we measure this to be of length 2πr. On the surface of a sphere however we will always measure a smaller length than this. As an extreme example to illustrate this is true we can imagine starting at the north pole and having a radius that extends all the way down to the south pole - at this fixed radius the circle would have zero length - quite different than the same radius in euclidean space.

So if we want to press some section of a spherical surface onto a flat surface it becomes apparent that we will need to tear it at some point because we have a smaller circumference on our section than the flat surface has. What Gaussian curvature gets at however is that this phenomena exists in the space itself - down to an infinitesimal limit. We wouldn’t just have to tear our surface along one point to make it flatten - it ends up that every point in that space would have to be torn. I’m sure that you could imagine how this is problematic to our ideal ‘unfolding’ of the surface - its not much of a transformation if we need to tear it up into infinitesimal pieces.

https://www.quora.com/Why-cant-you-unfold-a-sphere-Isnt-the-surface-of-sphere-two-dimensional

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y=∑n=0∞anxn⟹y′=∑n=1∞nanxn−1⟹y′′=∑n=2∞n(n−1)anxn−2

Next, plug into differential equation:

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Move constants inside of summations:

∑n=2∞x⋅n(n−1)anxn−2+∑n=2∞2⋅n(n−1)anxn−2+∑n=1∞x⋅nanxn−1−∑n=0∞anxn=0

∑n=2∞n(n−1)anxn−1+∑n=2∞2n(n−1)anxn−2+∑n=1∞nanxn−∑n=0∞anxn=0

Change limits so that the exponents for  x  are the same in each summation:

∑n=1∞(n+1)nan+1xn+∑n=0∞2(n+2)(n+1)an+2xn+∑n=1∞nanxn−∑n=0∞anxn=0

Pull out any terms from sums, so that each sum starts at same lower limit  (n=1)

∑n=1∞(n+1)nan+1xn+4a2+∑n=1∞2(n+2)(n+1)an+2xn+∑n=1∞nanxn−a0−∑n=1∞anxn=0

Combine all sums into a single sum:

4a2−a0+∑n=1∞(2(n+2)(n+1)an+2+(n+1)nan+1+(n−1)an)xn=0

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We would usually let  a0  and  a1  be arbitrary constants. Then all other constants can be expressed in terms of these two constants, giving us two linearly independent solutions. However, since  a0=4a2 , I’ll choose  a1  and  a2  as the two arbitrary constants. We can still express all other constants in terms of  a1  and/or  a2 .

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We see a pattern emerging here:

an=(−1)(n+1)n−4n!a2

This can be proven by mathematical induction. In fact, this is true for all  n≥0 , except for  n=1 , since  a1  is an arbitrary constant independent of  a0  (and therefore independent of  a2 ).

Plugging back into original power series for  y , we get:

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y=(c1−3c2)x+c2(4+x2−13!x3+04!x4+15!x5−⋯)

y=c1x+c2(4−3x+x2−13!x3+04!x4+15!x5−⋯)

y=c1x+c2(−0−40!+0−31!x−2−42!x2+3−43!x3−4−44!x4+5−45!x5−⋯)

y=c1x+c2∑n=0∞(−1)n+1n−4n!xn

Learn more about constants here:

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