Talk:Ruffini's rule
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Archived content
As this page was 36KB long, I moved the old contents (actually, most of this page) to Talk:Ruffini's rule/Archive
- Habbit 00:01, 27 Dec 2004 (UTC)
Current content
My professor showed me this algorithm for evaluating polynomials (more) quickly, and it looks very much like synthetic division:
Suppose <math> f(x) = 3x^6 + 5x^3 + 2x^2 + 1 <math> (I just made it up), and I want to know what <math> f(2) <math> is. I setup the problem thusly,
2 | 3 0 0 5 2 0 1
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| 6 12 24 58 120 240
|-------------------------------------------
3 6 12 29 60 120 241
The 2 at the very left is the parameter, and the rest of the first row is the coefficients of the polynomial. The first coefficient drops down directly. Then I multiply it by the parameter to get 6, which goes to 2nd row, 2nd column. I add the columns (0 + 6) and drop down the sum. Multiply it by the parameter again yields 12, which goes to 2nd row, 3rd column, ... Repeat until the last column, and the last sum is <math> f(2) <math>
This algorithm needed 7 multiplies and 6 adds. Computing <math> f(x) = 3(2)^6 + 5(2)^3 + 2(2)^2 + 1 <math> directly would need 14 multiplies and 4 adds.
Is this a variant of synthetic division, thus belongs to the aritcle? If not, where else is more appropriate? IMHO, it's pretty neat and deserves mention somewhere.
madoka 00:57, Oct 6, 2004 (UTC)
- That's a modified version of the remainder theorem and yes, it has something to do with Ruffini's rule because the method used is the same. The remainder theorem is usually used to get the remainder (hence the name) of diving a polynomial P(x) by a monic linear binomial (x-a): to get such a result, you must compute P(a), which will equal 0 if a is a root.
- An advice: avoid using f(x) to represent polynomials because f(x) usually represents functions, and functions can be polynomial, but can also be radical, logarithmic, etc. Try using P(x), Q(x), R(x) and so instead.
- Habbit 01:14, 24 Dec 2004 (UTC)
- This is the Horner scheme. Daniel
Possible minor errors in "factoring over the complexes" sub-section of the "Polynomial Factoring" section.
I'm not a mathematician, so I would like to request that someone with more experience review the "factoring over the complexes" sub-section of the "Polynomial Factoring" section of Ruffini's rule. I think there might be some minor errors.
The section describes using the quadratic equation to find roots for the polynomial: 2x² - x + 4 = 0
The quadratic equation shown in the article indicates that -b = -1, but I think perhaps it is supposed to show -b = -(-1) which is 1.
The quadratic equation shows the quantity (b² - 4ac) as (1² - 4*2*4), but I think perhaps it is supposed to show ((-1)² - 4*2*4).
The quadratic equation shows the quantity (b² - 4ac) working out to (-33), but I think perhaps it is supposed to show (-31).
Thanks for looking into this, and thanks for producing this great encyclopedia!!
Newbie-vts
- You're right. Fixed now. Thanks. Daniel