Notes for AKT-140314/0:02:29: Difference between revisions
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(Created page with "Boaz wrote: I have one comment- you claim it is weaker than Galois. At least in one important aspect I think it is much stronger. I challenge you to show with Galois theor...") |
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Boaz wrote: |
Boaz wrote: |
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:I have one comment- you claim it is weaker than Galois. At least in one important aspect I think it is much stronger. I challenge you to show with Galois theory that if we add exp(z) to our list of allowed operations, a solution cannot be found... |
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:In fact, I guess you can add to the list of allowed functions any continuous multivalued function with up to 4 values and it won't help (of course with 5 values you can solve it since the solution is such a function...). Am I right? |
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important aspect I think it is much stronger. I challenge you to show with |
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Galois theory that if we add exp(z) to our list of allowed operations, a |
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solution cannot be found... |
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In fact, I guess you can add to the list of allowed functions any continuous |
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multivalued function with up to 4 values and it won't help (of course with 5 |
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values you can solve it since the solution is such a function...). Am I right? |
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I believe Boaz is right, and he is making a very strong point - not only there is no formula for the roots of a quadratic in terms of radicals - in fact this remains true even if we are allowed to use other ''univalent'' functions. |
I believe Boaz is right, and he is making a very strong point - not only there is no formula for the roots of a quadratic in terms of radicals - in fact this remains true even if we are allowed to use other ''univalent'' functions (or even up to 4-valent). |
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Latest revision as of 13:51, 14 March 2014
Boaz wrote:
- I have one comment- you claim it is weaker than Galois. At least in one important aspect I think it is much stronger. I challenge you to show with Galois theory that if we add exp(z) to our list of allowed operations, a solution cannot be found...
- In fact, I guess you can add to the list of allowed functions any continuous multivalued function with up to 4 values and it won't help (of course with 5 values you can solve it since the solution is such a function...). Am I right?
I believe Boaz is right, and he is making a very strong point - not only there is no formula for the roots of a quadratic in terms of radicals - in fact this remains true even if we are allowed to use other univalent functions (or even up to 4-valent).
