2 ms·

This potential identity looks interesting: phi = e^((i(pi))/5) + e^((-i(pi))/5) (From user A. Rex on the referenced Math StackExchange web page...) I don't h

This potential identity looks interesting: phi = e^((i(pi))/5) + e^((-i(pi))/5) (From user A. Rex on the referenced Math StackExchange web page...) I don't have a calculator with enough precision, or a mathematical proof to know if this potential identity is exactly correct however... any comments? Also, are there any simpler identities of phi, in terms of e only, but without i or pi? Or maybe an identity of e, in terms of phi... I'm particularly interested in the relationship between phi and e, or vice-versa...

qubex 6y ago
If it’s not algebricaly proven don’t call it an ‘identity’... even if a calculator were to confirm it to some arbitrary number of digits, that’d just be a more-or-less accurate approximation. ‘Identity’ denotes a far more profound relationship and the term should be reserved for use in such cases.
- peter_d_sherman 6y ago
  That's what I'm trying to ascertain... is this an identity or not? (Note, I changed my comment to 'potential identity' rather than 'identity'). That, and are there any simple(r) identities for phi in terms of e, and e in terms of phi...?
  - deleted 6y ago
    [deleted]
  - qubex 6y ago
    As somebody in that thread points out, using division you can trivially construct an identity by taking the definition of φ and replacing any integer with expressions such as π÷π for 1 and (e+e+e+e+e)/e for 5.
exmadscientist 6y ago
Punching that in to WolframAlpha, after rewriting it in terms of hyperbolic cosine, shows that it is indeed an identity: https://www.wolframalpha.com/input/?i=2+Cosh%5BI+%5C%5BPi%5D%2F5%5D https://www.wolframalpha.com/input/?i=2+Cosh%5BI+%5C%5BPi%5D... This occurs via the identity cosh(i x) = cos x, so 2 cosh(i pi/5) = 2 cos (pi/5) and then you go into the usual stuff to evaluate that one.

2 ms·

This potential identity looks interesting: phi = e^((i(pi))/5) + e^((-i(pi))/5) (From user A. Rex on the referenced Math StackExchange web page...) I don't h

by peter_d_sherman 6y ago

qubex 6y ago
If it’s not algebricaly proven don’t call it an ‘identity’... even if a calculator were to confirm it to some arbitrary number of digits, that’d just be a more-or-less accurate approximation. ‘Identity’ denotes a far more profound relationship and the term should be reserved for use in such cases.
- peter_d_sherman 6y ago
  That's what I'm trying to ascertain... is this an identity or not? (Note, I changed my comment to 'potential identity' rather than 'identity'). That, and are there any simple(r) identities for phi in terms of e, and e in terms of phi...?
  - deleted 6y ago
    [deleted]
  - qubex 6y ago
    As somebody in that thread points out, using division you can trivially construct an identity by taking the definition of φ and replacing any integer with expressions such as π÷π for 1 and (e+e+e+e+e)/e for 5.
exmadscientist 6y ago
Punching that in to WolframAlpha, after rewriting it in terms of hyperbolic cosine, shows that it is indeed an identity: https://www.wolframalpha.com/input/?i=2+Cosh%5BI+%5C%5BPi%5D%2F5%5D https://www.wolframalpha.com/input/?i=2+Cosh%5BI+%5C%5BPi%5D... This occurs via the identity cosh(i x) = cos x, so 2 cosh(i pi/5) = 2 cos (pi/5) and then you go into the usual stuff to evaluate that one.