Xjyuk

Calculus Optimization - Point on graph closest to given point

What is the white spray-pattern residue inside these Falcon Heavy nozzles?

Is Social Media Science Fiction?

A newer friend of my brother's gave him a load of baseball cards that are supposedly extremely valuable. Is this a scam?

Can I make popcorn with any corn?

Why is an old chain unsafe?

What is the meaning of "of trouble" in the following sentence?

Why don't electron-positron collisions release infinite energy?

I see my dog run

What makes Graph invariants so useful/important?

What is the offset in a seaplane's hull?

N.B. ligature in Latex

If Manufacturer spice model and Datasheet give different values which should I use?

Finding files for which a command fails

Are tax years 2016 & 2017 back taxes deductible for tax year 2018?

Motorized valve interfering with button?

How can the DM most effectively choose 1 out of an odd number of players to be targeted by an attack or effect?

How can bays and straits be determined in a procedurally generated map?

Copycat chess is back

When blogging recipes, how can I support both readers who want the narrative/journey and ones who want the printer-friendly recipe?

Is there a familial term for apples and pears?

How old can references or sources in a thesis be?

Accidentally leaked the solution to an assignment, what to do now? (I'm the prof)

Email Account under attack (really) - anything I can do?

Proof for divisibility of polynomials. [on hold]

Show 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

0

$begingroup$

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

share|cite|improve this question

asked 11 hours ago

HeetGorakhiyaHeetGorakhiya

203

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 4 hours 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

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

0

$begingroup$

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

share|cite|improve this question

asked 11 hours ago

HeetGorakhiyaHeetGorakhiya

203

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 4 hours 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

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

share|cite|improve this question

asked 11 hours ago

HeetGorakhiyaHeetGorakhiya

203

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.

$endgroup$

share|cite|improve this question

asked 11 hours ago

HeetGorakhiyaHeetGorakhiya

203

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.

share|cite|improve this question

asked 11 hours ago

HeetGorakhiyaHeetGorakhiya

203

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 11 hours ago

HeetGorakhiyaHeetGorakhiya

203

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 11 hours ago

HeetGorakhiyaHeetGorakhiya

203

2 Answers
2

active

oldest

votes

4

$begingroup$

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 $$P(x)-xmid P(P(P(x)))-x$$

share|cite|improve this answer

edited 11 hours ago

answered 11 hours ago

Maria MazurMaria Mazur

50k1361124

$endgroup$

• 
  
  
  

  
  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
  11 hours ago
  
  

  
  
  
  

add a comment | 

3

$begingroup$

$bmod P(x)!-!x!:, color#c00P(x)equiv x,Rightarrow, P(P(color#c00P(x)))equiv P(P(color#c00x)))equiv P(x)equiv x$

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 $,P(x)$

share|cite|improve this answer

edited 11 hours ago

answered 11 hours ago

Bill DubuqueBill Dubuque

214k29196655

$endgroup$

add a comment | 

4

$begingroup$

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 $$P(x)-xmid P(P(P(x)))-x$$

share|cite|improve this answer

edited 11 hours ago

answered 11 hours ago

Maria MazurMaria Mazur

50k1361124

$endgroup$

• 
  
  
  

  
  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
  11 hours ago
  
  

  
  
  
  

add a comment | 

4

$begingroup$

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 $$P(x)-xmid P(P(P(x)))-x$$

share|cite|improve this answer

edited 11 hours ago

answered 11 hours ago

Maria MazurMaria Mazur

50k1361124

$endgroup$

• 
  
  
  

  
  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
  11 hours 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 $$P(x)-xmid P(P(P(x)))-x$$

share|cite|improve this answer

edited 11 hours ago

answered 11 hours ago

Maria MazurMaria Mazur

50k1361124

$endgroup$

share|cite|improve this answer

edited 11 hours ago

answered 11 hours ago

Maria MazurMaria Mazur

50k1361124

answered 11 hours ago

Maria MazurMaria Mazur

50k1361124

answered 11 hours ago

Maria MazurMaria Mazur

50k1361124

3

$begingroup$

$bmod P(x)!-!x!:, color#c00P(x)equiv x,Rightarrow, P(P(color#c00P(x)))equiv P(P(color#c00x)))equiv P(x)equiv x$

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 $,P(x)$

share|cite|improve this answer

edited 11 hours ago

answered 11 hours ago

Bill DubuqueBill Dubuque

214k29196655

$endgroup$

add a comment | 

3

$begingroup$

$bmod P(x)!-!x!:, color#c00P(x)equiv x,Rightarrow, P(P(color#c00P(x)))equiv P(P(color#c00x)))equiv P(x)equiv x$

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 $,P(x)$

share|cite|improve this answer

edited 11 hours ago

answered 11 hours ago

Bill DubuqueBill Dubuque

214k29196655

$endgroup$

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 $,P(x)$

share|cite|improve this answer

edited 11 hours ago

answered 11 hours ago

Bill DubuqueBill Dubuque

214k29196655

$endgroup$

share|cite|improve this answer

edited 11 hours ago

answered 11 hours ago

Bill DubuqueBill Dubuque

214k29196655

answered 11 hours ago

Bill DubuqueBill Dubuque

214k29196655

answered 11 hours ago

Bill DubuqueBill Dubuque

214k29196655