Proof for divisibility of polynomials.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
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Proof for divisibility of polynomials.
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
$begingroup$
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
New contributor
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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
New contributor
$endgroup$
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
New contributor
$endgroup$
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 $."
polynomials divisibility
polynomials divisibility
New contributor
New contributor
New contributor
asked 5 hours ago
HeetGorakhiyaHeetGorakhiya
203
203
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2 Answers
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$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 $$P(x)-xmid P(P(P(x)))-x$$
$endgroup$
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
5 hours 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 $,P(x)$
$endgroup$
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$begingroup$
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 $$P(x)-xmid P(P(P(x)))-x$$
$endgroup$
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
5 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 $$P(x)-xmid P(P(P(x)))-x$$
$endgroup$
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
5 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 $$P(x)-xmid P(P(P(x)))-x$$
$endgroup$
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 $$P(x)-xmid P(P(P(x)))-x$$
edited 5 hours ago
answered 5 hours ago
Maria MazurMaria Mazur
50k1361124
50k1361124
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
5 hours 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
5 hours 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
5 hours 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
5 hours 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 $,P(x)$
$endgroup$
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 $,P(x)$
$endgroup$
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 $,P(x)$
$endgroup$
$bmod P(x)!-!x!:, color#c00P(x)equiv x,Rightarrow, P(P(color#c00P(x)))equiv P(P(color#c00x)))equiv P(x)equiv x$
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 $,P(x)$
edited 5 hours ago
answered 5 hours ago
Bill DubuqueBill Dubuque
214k29196654
214k29196654
add a comment |
add a comment |
HeetGorakhiya is a new contributor. Be nice, and check out our Code of Conduct.
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