Proof for divisibility of polynomials. [on hold] The 2019 Stack Overflow Developer Survey Results Are InShow that $a^p^n=amod p$Divisibility problemPolynomial divisibility proofPolynomials and Divisibility Rule.Induction proof, divisibilityDivisibility of a polynomial by another polynomialDoes there exist a polynomial $f(x)$ with real coefficients such that $f(x)^2$ has fewer nonzero coefficients than $f(x)$?Polynomials - Relation of DivisibilityProof using concept of polynomials.Proof of Existence of A Rational Polynomial which has Irrational Root for an EquationTricky problem of infinite harmonic sum of polynomials
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Proof for divisibility of polynomials. [on hold]
The 2019 Stack Overflow Developer Survey Results Are InShow that $a^p^n=amod p$Divisibility problemPolynomial divisibility proofPolynomials and Divisibility Rule.Induction proof, divisibilityDivisibility of a polynomial by another polynomialDoes there exist a polynomial $f(x)$ with real coefficients such that $f(x)^2$ has fewer nonzero coefficients than $f(x)$?Polynomials - Relation of DivisibilityProof using concept of polynomials.Proof of Existence of A Rational Polynomial which has Irrational Root for an EquationTricky problem of infinite harmonic sum of polynomials
$begingroup$
I have no idea how to proceed with the following question. Please help!
"Prove that for any polynomial $ P(x) $ with real coefficients, other than polynomial $x$, the polynomial $ P(P(P(x))) − x $ is divisible by $ P(x) − x $."
polynomials divisibility
asked 2 days ago
HeetGorakhiyaHeetGorakhiya
283
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put on hold as off-topic by José Carlos Santos, Sil, John Omielan, TheSimpliFire, RRL 2 days ago
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." – José Carlos Santos, Sil, John Omielan, TheSimpliFire, RRL
add a comment |
$begingroup$
I have no idea how to proceed with the following question. Please help!
"Prove that for any polynomial $ P(x) $ with real coefficients, other than polynomial $x$, the polynomial $ P(P(P(x))) − x $ is divisible by $ P(x) − x $."
polynomials divisibility
asked 2 days ago
HeetGorakhiyaHeetGorakhiya
283
New contributor
HeetGorakhiya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
put on hold as off-topic by José Carlos Santos, Sil, John Omielan, TheSimpliFire, RRL 2 days ago
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." – José Carlos Santos, Sil, John Omielan, TheSimpliFire, RRL
add a comment |
$begingroup$
I have no idea how to proceed with the following question. Please help!
"Prove that for any polynomial $ P(x) $ with real coefficients, other than polynomial $x$, the polynomial $ P(P(P(x))) − x $ is divisible by $ P(x) − x $."
polynomials divisibility
asked 2 days ago
HeetGorakhiyaHeetGorakhiya
283
New contributor
HeetGorakhiya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I have no idea how to proceed with the following question. Please help!
"Prove that for any polynomial $ P(x) $ with real coefficients, other than polynomial $x$, the polynomial $ P(P(P(x))) − x $ is divisible by $ P(x) − x $."
polynomials divisibility
polynomials divisibility
asked 2 days ago
HeetGorakhiyaHeetGorakhiya
283
New contributor
HeetGorakhiya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
HeetGorakhiyaHeetGorakhiya
283
New contributor
HeetGorakhiya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
HeetGorakhiyaHeetGorakhiya
283
New contributor
HeetGorakhiya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
HeetGorakhiyaHeetGorakhiya
283
asked 2 days ago
HeetGorakhiyaHeetGorakhiya
283
283
New contributor
HeetGorakhiya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
HeetGorakhiya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
HeetGorakhiya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by José Carlos Santos, Sil, John Omielan, TheSimpliFire, RRL 2 days ago
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." – José Carlos Santos, Sil, John Omielan, TheSimpliFire, RRL
put on hold as off-topic by José Carlos Santos, Sil, John Omielan, TheSimpliFire, RRL 2 days ago
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." – José Carlos Santos, Sil, John Omielan, TheSimpliFire, RRL
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2 Answers
2
active
oldest
votes
$begingroup$
Remember that $$a-bmid P(a)-P(b)$$
so $$P(x)-xmid P(P(x))-P(x)$$ and thus $$P(x)-xmid (P(P(x))-P(x))+ (P(x)-x)$$
so $$P(x)-xmid P(P(x))-xmid P(P(P(x)))-P(x)$$
and thus $$P(x)-xmid (P(P(P(x)))-P(x))+ (P(x)-x)$$
and finaly we have $$P(x)-xmid P(P(P(x)))-x$$
answered 2 days ago
Maria MazurMaria Mazur
49.9k1361125
$endgroup$
1
$begingroup$
Modular arithmetic was invented to clarify proofs like this where the divisibilty relation greatly obfuscates the algebraic (operational) essence of the matter - here the simple notion of a fixed point - see my answer.
$endgroup$
– Bill Dubuque
2 days ago
add a comment |
$begingroup$
$bmod P(x)!-!x!:, color#c00P(x)equiv x,Rightarrow, P(P(color#c00P(x)))equiv P(P(color#c00x)))equiv P(x)equiv x$
Remark $ $ The proof is a special case of: fixed points stay fixed on iteration by induction,
namely: $, $ if $ color#c00f(x) = x $ then $, f^large n(x) = x,Rightarrow, f^large n+1(x) = f^n(color#c00f(x))=f^n(color#c00x)=x$
Corollary $ P(x)!-!x,$ divides $, P^n(x)!-!x,$ for all $,ninBbb N,,$ and all polynomials $,P(x)$
answered 2 days ago
Bill DubuqueBill Dubuque
214k29196656
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Remember that $$a-bmid P(a)-P(b)$$
so $$P(x)-xmid P(P(x))-P(x)$$ and thus $$P(x)-xmid (P(P(x))-P(x))+ (P(x)-x)$$
so $$P(x)-xmid P(P(x))-xmid P(P(P(x)))-P(x)$$
and thus $$P(x)-xmid (P(P(P(x)))-P(x))+ (P(x)-x)$$
and finaly we have $$P(x)-xmid P(P(P(x)))-x$$
answered 2 days ago
Maria MazurMaria Mazur
49.9k1361125
$endgroup$
1
$begingroup$
Modular arithmetic was invented to clarify proofs like this where the divisibilty relation greatly obfuscates the algebraic (operational) essence of the matter - here the simple notion of a fixed point - see my answer.
$endgroup$
– Bill Dubuque
2 days ago
add a comment |
$begingroup$
Remember that $$a-bmid P(a)-P(b)$$
so $$P(x)-xmid P(P(x))-P(x)$$ and thus $$P(x)-xmid (P(P(x))-P(x))+ (P(x)-x)$$
so $$P(x)-xmid P(P(x))-xmid P(P(P(x)))-P(x)$$
and thus $$P(x)-xmid (P(P(P(x)))-P(x))+ (P(x)-x)$$
and finaly we have $$P(x)-xmid P(P(P(x)))-x$$
answered 2 days ago
Maria MazurMaria Mazur
49.9k1361125
$endgroup$
1
$begingroup$
Modular arithmetic was invented to clarify proofs like this where the divisibilty relation greatly obfuscates the algebraic (operational) essence of the matter - here the simple notion of a fixed point - see my answer.
$endgroup$
– Bill Dubuque
2 days ago
add a comment |
$begingroup$
Remember that $$a-bmid P(a)-P(b)$$
so $$P(x)-xmid P(P(x))-P(x)$$ and thus $$P(x)-xmid (P(P(x))-P(x))+ (P(x)-x)$$
so $$P(x)-xmid P(P(x))-xmid P(P(P(x)))-P(x)$$
and thus $$P(x)-xmid (P(P(P(x)))-P(x))+ (P(x)-x)$$
and finaly we have $$P(x)-xmid P(P(P(x)))-x$$
answered 2 days ago
Maria MazurMaria Mazur
49.9k1361125
$endgroup$
Remember that $$a-bmid P(a)-P(b)$$
so $$P(x)-xmid P(P(x))-P(x)$$ and thus $$P(x)-xmid (P(P(x))-P(x))+ (P(x)-x)$$
so $$P(x)-xmid P(P(x))-xmid P(P(P(x)))-P(x)$$
and thus $$P(x)-xmid (P(P(P(x)))-P(x))+ (P(x)-x)$$
and finaly we have $$P(x)-xmid P(P(P(x)))-x$$
answered 2 days ago
Maria MazurMaria Mazur
49.9k1361125
edited 2 days ago
answered 2 days ago
Maria MazurMaria Mazur
49.9k1361125
answered 2 days ago
Maria MazurMaria Mazur
49.9k1361125
answered 2 days ago
Maria MazurMaria Mazur
49.9k1361125
49.9k1361125
1
$begingroup$
Modular arithmetic was invented to clarify proofs like this where the divisibilty relation greatly obfuscates the algebraic (operational) essence of the matter - here the simple notion of a fixed point - see my answer.
$endgroup$
– Bill Dubuque
2 days ago
add a comment |
1
$begingroup$
Modular arithmetic was invented to clarify proofs like this where the divisibilty relation greatly obfuscates the algebraic (operational) essence of the matter - here the simple notion of a fixed point - see my answer.
$endgroup$
– Bill Dubuque
2 days ago
1
1
$begingroup$
Modular arithmetic was invented to clarify proofs like this where the divisibilty relation greatly obfuscates the algebraic (operational) essence of the matter - here the simple notion of a fixed point - see my answer.
$endgroup$
– Bill Dubuque
2 days ago
$begingroup$
Modular arithmetic was invented to clarify proofs like this where the divisibilty relation greatly obfuscates the algebraic (operational) essence of the matter - here the simple notion of a fixed point - see my answer.
$endgroup$
– Bill Dubuque
2 days ago
add a comment |
$begingroup$
$bmod P(x)!-!x!:, color#c00P(x)equiv x,Rightarrow, P(P(color#c00P(x)))equiv P(P(color#c00x)))equiv P(x)equiv x$
Remark $ $ The proof is a special case of: fixed points stay fixed on iteration by induction,
namely: $, $ if $ color#c00f(x) = x $ then $, f^large n(x) = x,Rightarrow, f^large n+1(x) = f^n(color#c00f(x))=f^n(color#c00x)=x$
Corollary $ P(x)!-!x,$ divides $, P^n(x)!-!x,$ for all $,ninBbb N,,$ and all polynomials $,P(x)$
answered 2 days ago
Bill DubuqueBill Dubuque
214k29196656
$endgroup$
add a comment |
$begingroup$
$bmod P(x)!-!x!:, color#c00P(x)equiv x,Rightarrow, P(P(color#c00P(x)))equiv P(P(color#c00x)))equiv P(x)equiv x$
Remark $ $ The proof is a special case of: fixed points stay fixed on iteration by induction,
namely: $, $ if $ color#c00f(x) = x $ then $, f^large n(x) = x,Rightarrow, f^large n+1(x) = f^n(color#c00f(x))=f^n(color#c00x)=x$
Corollary $ P(x)!-!x,$ divides $, P^n(x)!-!x,$ for all $,ninBbb N,,$ and all polynomials $,P(x)$
answered 2 days ago
Bill DubuqueBill Dubuque
214k29196656
$endgroup$
add a comment |
$begingroup$
$bmod P(x)!-!x!:, color#c00P(x)equiv x,Rightarrow, P(P(color#c00P(x)))equiv P(P(color#c00x)))equiv P(x)equiv x$
Remark $ $ The proof is a special case of: fixed points stay fixed on iteration by induction,
namely: $, $ if $ color#c00f(x) = x $ then $, f^large n(x) = x,Rightarrow, f^large n+1(x) = f^n(color#c00f(x))=f^n(color#c00x)=x$
Corollary $ P(x)!-!x,$ divides $, P^n(x)!-!x,$ for all $,ninBbb N,,$ and all polynomials $,P(x)$
answered 2 days ago
Bill DubuqueBill Dubuque
214k29196656
$endgroup$
$bmod P(x)!-!x!:, color#c00P(x)equiv x,Rightarrow, P(P(color#c00P(x)))equiv P(P(color#c00x)))equiv P(x)equiv x$
Remark $ $ The proof is a special case of: fixed points stay fixed on iteration by induction,
namely: $, $ if $ color#c00f(x) = x $ then $, f^large n(x) = x,Rightarrow, f^large n+1(x) = f^n(color#c00f(x))=f^n(color#c00x)=x$
Corollary $ P(x)!-!x,$ divides $, P^n(x)!-!x,$ for all $,ninBbb N,,$ and all polynomials $,P(x)$
answered 2 days ago
Bill DubuqueBill Dubuque
214k29196656
edited 2 days ago
answered 2 days ago
Bill DubuqueBill Dubuque
214k29196656
answered 2 days ago
Bill DubuqueBill Dubuque
214k29196656
answered 2 days ago
Bill DubuqueBill Dubuque
214k29196656
214k29196656
add a comment |
add a comment |
-divisibility, polynomials
