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Degree of the Zero polynomial |
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| Jul20-12, 10:47 PM | #1 |
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Degree of the Zero polynomial
I understand that mathematicians have had to define the number '0' also as a polynomial because it acts as the additive identity for the additive group of poly's.What I do not understand is why they define the degree of the zero polynomial as [ tex ]-\infty[ /tex ].
An explanation on planetMath wasn't that helpful,at the end they point-out to refer to the extended real numbers(don't they mean 'projectively extended real numbers??) |
| Jul20-12, 10:51 PM | #2 |
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Well, I guess it's similar to how one sometimes regards zero as both a real and an imaginary number because you can write 0 = 0 + i0. Similarly, you can write
0 = 0 + 0x + 0x2 + 0x3 + ... i.e., you can write '0' as an infinite degree polynomial with all coefficients zero. (There may be a more rigorous reason, but that's an intuitive one). |
| Jul20-12, 10:58 PM | #3 |
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Thanks!Could you explain what extended real numbers have got to do with this?
But,polynomials always have non-negative degrees. deg[P(x)]=+n Why would mathematicians define a polynomial with a negative degree? |
| Jul20-12, 11:00 PM | #4 |
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Degree of the Zero polynomial
Huh, I'm looking at a precalculus textbook (Larson, 8th Ed.), and it states that the zero polynomial has no degree. Is that wrong? (Note that no degree ≠ zero degree -- a polynomial that consists of a single non-zero number has a degree of zero.)
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| Jul20-12, 11:12 PM | #5 |
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The choice is pretty arbitrary. Sometimes it defined as having no degree, sometimes it's -1, sometimes it's [itex]-\infty[/itex].
A handy formula for polynomials is [tex]deg(P)+deg(Q)=deg(PQ)[/tex] If we want this formula to hold for the zero polynomial, then we see (by taking Q=0) that [tex]deg(P)+deg(0)=deg(0)[/tex] must hold for all P. This is only satisfied with [itex]deg(0)=-\infty[/itex]. This is the reason why they defined it this way. But again, it's pretty arbitrary. |
| Jul20-12, 11:22 PM | #6 |
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| Jul20-12, 11:33 PM | #7 |
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