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Harry Gindi
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Lagrange interpolation is a special case of the Chinese remainder theorem. (Fixing a dead link: https://artofproblemsolving.com/community/c1157h990758_the_chinese_remainder_theorem_and_lagrange_interpolation )

The Jordan normal form can be proven extremely quickly using the Chinese remainder theorem for modules over a commutative ring. This proceeds by first proving the Jordan-Chevalley decomposition, and then the rest is a simple exercise of showing what the Jordan blocks actually look like.

The first one is very surprising to people, but if you state Lagrange interpolation correctly, it's easy to see that the idea is not only similar but identical.

Lagrange interpolation is a special case of the Chinese remainder theorem.

The Jordan normal form can be proven extremely quickly using the Chinese remainder theorem for modules over a commutative ring. This proceeds by first proving the Jordan-Chevalley decomposition, and then the rest is a simple exercise of showing what the Jordan blocks actually look like.

The first one is very surprising to people, but if you state Lagrange interpolation correctly, it's easy to see that the idea is not only similar but identical.

Lagrange interpolation is a special case of the Chinese remainder theorem. (Fixing a dead link: https://artofproblemsolving.com/community/c1157h990758_the_chinese_remainder_theorem_and_lagrange_interpolation )

The Jordan normal form can be proven extremely quickly using the Chinese remainder theorem for modules over a commutative ring. This proceeds by first proving the Jordan-Chevalley decomposition, and then the rest is a simple exercise of showing what the Jordan blocks actually look like.

The first one is very surprising to people, but if you state Lagrange interpolation correctly, it's easy to see that the idea is not only similar but identical.

Post Made Community Wiki by Anton Geraschenko
Source Link
Harry Gindi
  • 19.6k
  • 16
  • 123
  • 215

Lagrange interpolation is a special case of the Chinese remainder theorem.

The Jordan normal form can be proven extremely quickly using the Chinese remainder theorem for modules over a commutative ring. This proceeds by first proving the Jordan-Chevalley decomposition, and then the rest is a simple exercise of showing what the Jordan blocks actually look like.

The first one is very surprising to people, but if you state Lagrange interpolation correctly, it's easy to see that the idea is not only similar but identical.