What is the method of polynomial division? [closed]
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What is Horner's method? I can't understand Wikipedia's language. Is this only to be used for monomial divisors? How is synthetic division different from it?
Can anyone please elaborate?
polynomials
edited Dec 26 '18 at 12:50
Brahadeesh
6,46942363
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closed as off-topic by Math_QED, Saad, Davide Giraudo, egreg, Leucippus Dec 27 '18 at 0:08
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add a comment |
$begingroup$
What is Horner's method? I can't understand Wikipedia's language. Is this only to be used for monomial divisors? How is synthetic division different from it?
Can anyone please elaborate?
polynomials
edited Dec 26 '18 at 12:50
Brahadeesh
6,46942363
$endgroup$
closed as off-topic by Math_QED, Saad, Davide Giraudo, egreg, Leucippus Dec 27 '18 at 0:08
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Math_QED, Saad, Davide Giraudo, egreg, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
What is Horner's method? I can't understand Wikipedia's language. Is this only to be used for monomial divisors? How is synthetic division different from it?
Can anyone please elaborate?
polynomials
edited Dec 26 '18 at 12:50
Brahadeesh
6,46942363
$endgroup$
What is Horner's method? I can't understand Wikipedia's language. Is this only to be used for monomial divisors? How is synthetic division different from it?
Can anyone please elaborate?
polynomials
polynomials
edited Dec 26 '18 at 12:50
Brahadeesh
6,46942363
edited Dec 26 '18 at 12:50
Brahadeesh
6,46942363
edited Dec 26 '18 at 12:50
Brahadeesh
6,46942363
edited Dec 26 '18 at 12:50
Brahadeesh
6,46942363
edited Dec 26 '18 at 12:50
Brahadeesh
6,46942363
6,46942363
asked Dec 26 '18 at 12:48
user629353
closed as off-topic by Math_QED, Saad, Davide Giraudo, egreg, Leucippus Dec 27 '18 at 0:08
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Math_QED, Saad, Davide Giraudo, egreg, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Math_QED, Saad, Davide Giraudo, egreg, Leucippus Dec 27 '18 at 0:08
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Math_QED, Saad, Davide Giraudo, egreg, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
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votes
$begingroup$
Horner's method is not essentially different than synthetic division. It's simply an algorithmic method to compute, for a polynomial $p(x)$ and a point $a$, the value $p(a)$ and, simultaneously, to obtain the coefficients of the polynomial $q(x)$ such that $p(x)=q(x)(x-a)+p(a)$. It was shown that Horner's method is optimal both in terms of number of additions and of number of multiplications requires to perform the evaluation/division.
answered Dec 26 '18 at 13:13
Ittay WeissIttay Weiss
64k7102183
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$begingroup$
How we have reached from older algorithmic form to modern division
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– user629353
Dec 26 '18 at 16:52
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Horner's method is not essentially different than synthetic division. It's simply an algorithmic method to compute, for a polynomial $p(x)$ and a point $a$, the value $p(a)$ and, simultaneously, to obtain the coefficients of the polynomial $q(x)$ such that $p(x)=q(x)(x-a)+p(a)$. It was shown that Horner's method is optimal both in terms of number of additions and of number of multiplications requires to perform the evaluation/division.
answered Dec 26 '18 at 13:13
Ittay WeissIttay Weiss
64k7102183
$endgroup$
$begingroup$
How we have reached from older algorithmic form to modern division
$endgroup$
– user629353
Dec 26 '18 at 16:52
add a comment |
$begingroup$
Horner's method is not essentially different than synthetic division. It's simply an algorithmic method to compute, for a polynomial $p(x)$ and a point $a$, the value $p(a)$ and, simultaneously, to obtain the coefficients of the polynomial $q(x)$ such that $p(x)=q(x)(x-a)+p(a)$. It was shown that Horner's method is optimal both in terms of number of additions and of number of multiplications requires to perform the evaluation/division.
answered Dec 26 '18 at 13:13
Ittay WeissIttay Weiss
64k7102183
$endgroup$
$begingroup$
How we have reached from older algorithmic form to modern division
$endgroup$
– user629353
Dec 26 '18 at 16:52
add a comment |
$begingroup$
Horner's method is not essentially different than synthetic division. It's simply an algorithmic method to compute, for a polynomial $p(x)$ and a point $a$, the value $p(a)$ and, simultaneously, to obtain the coefficients of the polynomial $q(x)$ such that $p(x)=q(x)(x-a)+p(a)$. It was shown that Horner's method is optimal both in terms of number of additions and of number of multiplications requires to perform the evaluation/division.
answered Dec 26 '18 at 13:13
Ittay WeissIttay Weiss
64k7102183
$endgroup$
Horner's method is not essentially different than synthetic division. It's simply an algorithmic method to compute, for a polynomial $p(x)$ and a point $a$, the value $p(a)$ and, simultaneously, to obtain the coefficients of the polynomial $q(x)$ such that $p(x)=q(x)(x-a)+p(a)$. It was shown that Horner's method is optimal both in terms of number of additions and of number of multiplications requires to perform the evaluation/division.
answered Dec 26 '18 at 13:13
Ittay WeissIttay Weiss
64k7102183
answered Dec 26 '18 at 13:13
Ittay WeissIttay Weiss
64k7102183
answered Dec 26 '18 at 13:13
Ittay WeissIttay Weiss
64k7102183
answered Dec 26 '18 at 13:13
Ittay WeissIttay Weiss
64k7102183
64k7102183
$begingroup$
How we have reached from older algorithmic form to modern division
$endgroup$
– user629353
Dec 26 '18 at 16:52
add a comment |
$begingroup$
How we have reached from older algorithmic form to modern division
$endgroup$
– user629353
Dec 26 '18 at 16:52
$begingroup$
How we have reached from older algorithmic form to modern division
$endgroup$
– user629353
Dec 26 '18 at 16:52
$begingroup$
How we have reached from older algorithmic form to modern division
$endgroup$
– user629353
Dec 26 '18 at 16:52
add a comment |
