Proof for divisibility of polynomials. [closed]
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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 $."
polynomials divisibility
asked Apr 7 at 18:12
HeetGorakhiyaHeetGorakhiya
263
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closed as off-topic by José Carlos Santos, Sil, John Omielan, TheSimpliFire, RRL Apr 8 at 1:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Sil, John Omielan, TheSimpliFire, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
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$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 Apr 7 at 18:12
HeetGorakhiyaHeetGorakhiya
263
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.
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closed as off-topic by José Carlos Santos, Sil, John Omielan, TheSimpliFire, RRL Apr 8 at 1:39
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
If this question can be reworded to fit the rules in the help center, please edit the question.
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 Apr 7 at 18:12
HeetGorakhiyaHeetGorakhiya
263
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 Apr 7 at 18:12
HeetGorakhiyaHeetGorakhiya
263
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 Apr 7 at 18:12
HeetGorakhiyaHeetGorakhiya
263
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 Apr 7 at 18:12
HeetGorakhiyaHeetGorakhiya
263
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 Apr 7 at 18:12
HeetGorakhiyaHeetGorakhiya
263
asked Apr 7 at 18:12
HeetGorakhiyaHeetGorakhiya
263
263
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.
closed as off-topic by José Carlos Santos, Sil, John Omielan, TheSimpliFire, RRL Apr 8 at 1:39
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
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by José Carlos Santos, Sil, John Omielan, TheSimpliFire, RRL Apr 8 at 1:39
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
If this question can be reworded to fit the rules in the help center, please edit the question.
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2 Answers
2
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oldest
votes
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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 Apr 7 at 18:16
Maria MazurMaria Mazur
49.9k1361125
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1
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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.
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– Bill Dubuque
Apr 7 at 18:36
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$bmod P(x)!-!x!:, color{#c00}{P(x)equiv x},Rightarrow, P(P(color{#c00}{P(x)}))equiv P(P(color{#c00}{x)}))equiv P(x)equiv x$
Remark $ $ The proof is a special case of: fixed points stay fixed on iteration by induction,
namely: $, $ if $ color{#c00}{f(x) = x} $ then $, {f^{large n}(x) = x},Rightarrow, f^{large n+1}(x) = f^n(color{#c00}{f(x)})=f^n(color{#c00}x)=x$
Corollary $ P(x)!-!x,$ divides $, P^n(x)!-!x,$ for all $,ninBbb N,,$ and all polynomials $,P(x)$
answered Apr 7 at 18:18
Bill DubuqueBill Dubuque
214k29197658
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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 Apr 7 at 18:16
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
Apr 7 at 18:36
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 Apr 7 at 18:16
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
Apr 7 at 18:36
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 Apr 7 at 18:16
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 Apr 7 at 18:16
Maria MazurMaria Mazur
49.9k1361125
edited Apr 7 at 18:25
answered Apr 7 at 18:16
Maria MazurMaria Mazur
49.9k1361125
answered Apr 7 at 18:16
Maria MazurMaria Mazur
49.9k1361125
answered Apr 7 at 18:16
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
Apr 7 at 18:36
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
Apr 7 at 18:36
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
Apr 7 at 18:36
$begingroup$
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
Apr 7 at 18:36
add a comment |
$begingroup$
$bmod P(x)!-!x!:, color{#c00}{P(x)equiv x},Rightarrow, P(P(color{#c00}{P(x)}))equiv P(P(color{#c00}{x)}))equiv P(x)equiv x$
Remark $ $ The proof is a special case of: fixed points stay fixed on iteration by induction,
namely: $, $ if $ color{#c00}{f(x) = x} $ then $, {f^{large n}(x) = x},Rightarrow, f^{large n+1}(x) = f^n(color{#c00}{f(x)})=f^n(color{#c00}x)=x$
Corollary $ P(x)!-!x,$ divides $, P^n(x)!-!x,$ for all $,ninBbb N,,$ and all polynomials $,P(x)$
answered Apr 7 at 18:18
Bill DubuqueBill Dubuque
214k29197658
$endgroup$
add a comment |
$begingroup$
$bmod P(x)!-!x!:, color{#c00}{P(x)equiv x},Rightarrow, P(P(color{#c00}{P(x)}))equiv P(P(color{#c00}{x)}))equiv P(x)equiv x$
Remark $ $ The proof is a special case of: fixed points stay fixed on iteration by induction,
namely: $, $ if $ color{#c00}{f(x) = x} $ then $, {f^{large n}(x) = x},Rightarrow, f^{large n+1}(x) = f^n(color{#c00}{f(x)})=f^n(color{#c00}x)=x$
Corollary $ P(x)!-!x,$ divides $, P^n(x)!-!x,$ for all $,ninBbb N,,$ and all polynomials $,P(x)$
answered Apr 7 at 18:18
Bill DubuqueBill Dubuque
214k29197658
$endgroup$
add a comment |
$begingroup$
$bmod P(x)!-!x!:, color{#c00}{P(x)equiv x},Rightarrow, P(P(color{#c00}{P(x)}))equiv P(P(color{#c00}{x)}))equiv P(x)equiv x$
Remark $ $ The proof is a special case of: fixed points stay fixed on iteration by induction,
namely: $, $ if $ color{#c00}{f(x) = x} $ then $, {f^{large n}(x) = x},Rightarrow, f^{large n+1}(x) = f^n(color{#c00}{f(x)})=f^n(color{#c00}x)=x$
Corollary $ P(x)!-!x,$ divides $, P^n(x)!-!x,$ for all $,ninBbb N,,$ and all polynomials $,P(x)$
answered Apr 7 at 18:18
Bill DubuqueBill Dubuque
214k29197658
$endgroup$
$bmod P(x)!-!x!:, color{#c00}{P(x)equiv x},Rightarrow, P(P(color{#c00}{P(x)}))equiv P(P(color{#c00}{x)}))equiv P(x)equiv x$
Remark $ $ The proof is a special case of: fixed points stay fixed on iteration by induction,
namely: $, $ if $ color{#c00}{f(x) = x} $ then $, {f^{large n}(x) = x},Rightarrow, f^{large n+1}(x) = f^n(color{#c00}{f(x)})=f^n(color{#c00}x)=x$
Corollary $ P(x)!-!x,$ divides $, P^n(x)!-!x,$ for all $,ninBbb N,,$ and all polynomials $,P(x)$
answered Apr 7 at 18:18
Bill DubuqueBill Dubuque
214k29197658
edited Apr 7 at 18:45
answered Apr 7 at 18:18
Bill DubuqueBill Dubuque
214k29197658
answered Apr 7 at 18:18
Bill DubuqueBill Dubuque
214k29197658
answered Apr 7 at 18:18
Bill DubuqueBill Dubuque
214k29197658
214k29197658
add a comment |
add a comment |
