Mixing
Find all polynomials $p(x)$ such that $(p(x))^2 = 1 + x cdot p(x + 1)$ for all $xin mathbb{R}$
$begingroup$
Find all polynomials $p(x)$ such that $(p(x))^2 = 1 + x cdot p(x + 1)$ for all $xin mathbb{R}$.
I assume that $O(p(x)=n)$ (Where O(p) denotes the order of p)
let $p(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2} dots dots +a_{1}x^{1}+a_{n-1}$ also $a_n neq 0$
Then we have $a_n^2x^{2n} +dots +a_0^2=1+a_nx^{n+1}+ dots a_0x$
since $a_n neq 0$ we must have $2n=n+1$ and hence $n=1$ hence the polynomial must be linear and hence there doesn't exist any solution(I proved assuming $p(x)=ax+b$)
Is it correct?
real-analysis abstract-algebra polynomials
asked Jan 10 at 0:37
SunShineSunShine
1434
$endgroup$
add a comment |
$begingroup$
Find all polynomials $p(x)$ such that $(p(x))^2 = 1 + x cdot p(x + 1)$ for all $xin mathbb{R}$.
I assume that $O(p(x)=n)$ (Where O(p) denotes the order of p)
let $p(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2} dots dots +a_{1}x^{1}+a_{n-1}$ also $a_n neq 0$
Then we have $a_n^2x^{2n} +dots +a_0^2=1+a_nx^{n+1}+ dots a_0x$
since $a_n neq 0$ we must have $2n=n+1$ and hence $n=1$ hence the polynomial must be linear and hence there doesn't exist any solution(I proved assuming $p(x)=ax+b$)
Is it correct?
real-analysis abstract-algebra polynomials
asked Jan 10 at 0:37
SunShineSunShine
1434
$endgroup$
2
$begingroup$
What about $p(x)=x+1$?
$endgroup$
– Michael Burr
Jan 10 at 0:46
4
$begingroup$
$p$ cannot be zero. If $p$ has degree $r,$ the left hand side has degree $2r$ or $1$ and the right hand side has degree $r + 1,$ hence $r = 1.$ Now propose $p(x)=ax+b$ and solve.
$endgroup$
– Will M.
Jan 10 at 0:48
5
$begingroup$
What you wrote looks good, but you should check the $ax+b$ case more carefully.
$endgroup$
– lulu
Jan 10 at 0:51
$begingroup$
@WillM. That's an ideal hint, I think---perhaps write it up as an answer?
$endgroup$
– Travis
Jan 10 at 2:57
add a comment |
$begingroup$
Find all polynomials $p(x)$ such that $(p(x))^2 = 1 + x cdot p(x + 1)$ for all $xin mathbb{R}$.
I assume that $O(p(x)=n)$ (Where O(p) denotes the order of p)
let $p(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2} dots dots +a_{1}x^{1}+a_{n-1}$ also $a_n neq 0$
Then we have $a_n^2x^{2n} +dots +a_0^2=1+a_nx^{n+1}+ dots a_0x$
since $a_n neq 0$ we must have $2n=n+1$ and hence $n=1$ hence the polynomial must be linear and hence there doesn't exist any solution(I proved assuming $p(x)=ax+b$)
Is it correct?
real-analysis abstract-algebra polynomials
asked Jan 10 at 0:37
SunShineSunShine
1434
$endgroup$
Find all polynomials $p(x)$ such that $(p(x))^2 = 1 + x cdot p(x + 1)$ for all $xin mathbb{R}$.
I assume that $O(p(x)=n)$ (Where O(p) denotes the order of p)
let $p(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2} dots dots +a_{1}x^{1}+a_{n-1}$ also $a_n neq 0$
Then we have $a_n^2x^{2n} +dots +a_0^2=1+a_nx^{n+1}+ dots a_0x$
since $a_n neq 0$ we must have $2n=n+1$ and hence $n=1$ hence the polynomial must be linear and hence there doesn't exist any solution(I proved assuming $p(x)=ax+b$)
Is it correct?
real-analysis abstract-algebra polynomials
real-analysis abstract-algebra polynomials
asked Jan 10 at 0:37
SunShineSunShine
1434
asked Jan 10 at 0:37
SunShineSunShine
1434
asked Jan 10 at 0:37
SunShineSunShine
1434
asked Jan 10 at 0:37
SunShineSunShine
1434
asked Jan 10 at 0:37
SunShineSunShine
1434
1434
2
$begingroup$
What about $p(x)=x+1$?
$endgroup$
– Michael Burr
Jan 10 at 0:46
4
$begingroup$
$p$ cannot be zero. If $p$ has degree $r,$ the left hand side has degree $2r$ or $1$ and the right hand side has degree $r + 1,$ hence $r = 1.$ Now propose $p(x)=ax+b$ and solve.
$endgroup$
– Will M.
Jan 10 at 0:48
5
$begingroup$
What you wrote looks good, but you should check the $ax+b$ case more carefully.
$endgroup$
– lulu
Jan 10 at 0:51
$begingroup$
@WillM. That's an ideal hint, I think---perhaps write it up as an answer?
$endgroup$
– Travis
Jan 10 at 2:57
add a comment |
2
$begingroup$
What about $p(x)=x+1$?
$endgroup$
– Michael Burr
Jan 10 at 0:46
4
$begingroup$
$p$ cannot be zero. If $p$ has degree $r,$ the left hand side has degree $2r$ or $1$ and the right hand side has degree $r + 1,$ hence $r = 1.$ Now propose $p(x)=ax+b$ and solve.
$endgroup$
– Will M.
Jan 10 at 0:48
5
$begingroup$
What you wrote looks good, but you should check the $ax+b$ case more carefully.
$endgroup$
– lulu
Jan 10 at 0:51
$begingroup$
@WillM. That's an ideal hint, I think---perhaps write it up as an answer?
$endgroup$
– Travis
Jan 10 at 2:57
2
2
$begingroup$
What about $p(x)=x+1$?
$endgroup$
– Michael Burr
Jan 10 at 0:46
$begingroup$
What about $p(x)=x+1$?
$endgroup$
– Michael Burr
Jan 10 at 0:46
4
4
$begingroup$
$p$ cannot be zero. If $p$ has degree $r,$ the left hand side has degree $2r$ or $1$ and the right hand side has degree $r + 1,$ hence $r = 1.$ Now propose $p(x)=ax+b$ and solve.
$endgroup$
– Will M.
Jan 10 at 0:48
$begingroup$
$p$ cannot be zero. If $p$ has degree $r,$ the left hand side has degree $2r$ or $1$ and the right hand side has degree $r + 1,$ hence $r = 1.$ Now propose $p(x)=ax+b$ and solve.
$endgroup$
– Will M.
Jan 10 at 0:48
5
5
$begingroup$
What you wrote looks good, but you should check the $ax+b$ case more carefully.
$endgroup$
– lulu
Jan 10 at 0:51
$begingroup$
What you wrote looks good, but you should check the $ax+b$ case more carefully.
$endgroup$
– lulu
Jan 10 at 0:51
$begingroup$
@WillM. That's an ideal hint, I think---perhaps write it up as an answer?
$endgroup$
– Travis
Jan 10 at 2:57
$begingroup$
@WillM. That's an ideal hint, I think---perhaps write it up as an answer?
$endgroup$
– Travis
Jan 10 at 2:57
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
After observing that the degree is at most $1$ you can simplify the rest of the argument as follows: $p(0)^{2}=1$ from the equation, so $p(0)=pm 1$. Also $p(-1)^{2}=1-p(0)$. By considering the cases $p(0)=1$ and $p(0)=-1$ separately it is quite easy to conclude that the only solution is $p(x)=x+1$.
answered Jan 10 at 6:26
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
$endgroup$
add a comment |
$begingroup$
Our OP SunShine has correctly deduced that $deg p(x) = 1$, so that
$p(x) = ax + b; tag 1$
we compute:
$p^2(x) = a^2x^2 + 2ab x + b^2; tag 2$
$p(x + 1) = ax + a + b; tag 3$
$xp(x + 1) = ax^2 + ax + bx; tag 4$
$xp(x + 1) + 1 = ax^2 + ax + bx + 1; tag 5$
$a^2 x^2 + 2ab x + b^2 = ax^2 + ax + bx + 1; tag 6$
$(a^2 - a)x^2 + 2abx + b^2 = ax + bx + 1; tag 7$
$a^2 = a, ; b^2 = 1, ; 2ab = a + b; tag 8$
$a = 0, 1; ; b = pm 1; tag 9$
$b = 1 Longrightarrow 2a = a + 1 Longrightarrow a = 1 in {0, 1 }; tag{10}$
$b = -1 Longrightarrow -2a = a - 1 Longrightarrow a = dfrac{1}{3} notin {0, 1 }; tag{11}$
thus the only solution is $a = b = 1$, so that
$p(x) = x + 1. tag{12}$
CHECK:
$p^2(x) = (x + 1)^2 = x^2 + 2x + 1$
$= x(x + 2) + 1 = x((x + 1) + 1) + 1 = xp(x + 1) + 1; tag{13}$
it thus appears that the only solution is
$p(x) = x + 1. tag{14}$
answered Jan 10 at 2:15
Robert LewisRobert Lewis
45.8k23065
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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$begingroup$
After observing that the degree is at most $1$ you can simplify the rest of the argument as follows: $p(0)^{2}=1$ from the equation, so $p(0)=pm 1$. Also $p(-1)^{2}=1-p(0)$. By considering the cases $p(0)=1$ and $p(0)=-1$ separately it is quite easy to conclude that the only solution is $p(x)=x+1$.
answered Jan 10 at 6:26
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
$endgroup$
add a comment |
$begingroup$
After observing that the degree is at most $1$ you can simplify the rest of the argument as follows: $p(0)^{2}=1$ from the equation, so $p(0)=pm 1$. Also $p(-1)^{2}=1-p(0)$. By considering the cases $p(0)=1$ and $p(0)=-1$ separately it is quite easy to conclude that the only solution is $p(x)=x+1$.
answered Jan 10 at 6:26
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
$endgroup$
add a comment |
$begingroup$
After observing that the degree is at most $1$ you can simplify the rest of the argument as follows: $p(0)^{2}=1$ from the equation, so $p(0)=pm 1$. Also $p(-1)^{2}=1-p(0)$. By considering the cases $p(0)=1$ and $p(0)=-1$ separately it is quite easy to conclude that the only solution is $p(x)=x+1$.
answered Jan 10 at 6:26
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
$endgroup$
After observing that the degree is at most $1$ you can simplify the rest of the argument as follows: $p(0)^{2}=1$ from the equation, so $p(0)=pm 1$. Also $p(-1)^{2}=1-p(0)$. By considering the cases $p(0)=1$ and $p(0)=-1$ separately it is quite easy to conclude that the only solution is $p(x)=x+1$.
answered Jan 10 at 6:26
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
answered Jan 10 at 6:26
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
answered Jan 10 at 6:26
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
answered Jan 10 at 6:26
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
57.7k42160
add a comment |
add a comment |
$begingroup$
Our OP SunShine has correctly deduced that $deg p(x) = 1$, so that
$p(x) = ax + b; tag 1$
we compute:
$p^2(x) = a^2x^2 + 2ab x + b^2; tag 2$
$p(x + 1) = ax + a + b; tag 3$
$xp(x + 1) = ax^2 + ax + bx; tag 4$
$xp(x + 1) + 1 = ax^2 + ax + bx + 1; tag 5$
$a^2 x^2 + 2ab x + b^2 = ax^2 + ax + bx + 1; tag 6$
$(a^2 - a)x^2 + 2abx + b^2 = ax + bx + 1; tag 7$
$a^2 = a, ; b^2 = 1, ; 2ab = a + b; tag 8$
$a = 0, 1; ; b = pm 1; tag 9$
$b = 1 Longrightarrow 2a = a + 1 Longrightarrow a = 1 in {0, 1 }; tag{10}$
$b = -1 Longrightarrow -2a = a - 1 Longrightarrow a = dfrac{1}{3} notin {0, 1 }; tag{11}$
thus the only solution is $a = b = 1$, so that
$p(x) = x + 1. tag{12}$
CHECK:
$p^2(x) = (x + 1)^2 = x^2 + 2x + 1$
$= x(x + 2) + 1 = x((x + 1) + 1) + 1 = xp(x + 1) + 1; tag{13}$
it thus appears that the only solution is
$p(x) = x + 1. tag{14}$
answered Jan 10 at 2:15
Robert LewisRobert Lewis
45.8k23065
$endgroup$
add a comment |
$begingroup$
Our OP SunShine has correctly deduced that $deg p(x) = 1$, so that
$p(x) = ax + b; tag 1$
we compute:
$p^2(x) = a^2x^2 + 2ab x + b^2; tag 2$
$p(x + 1) = ax + a + b; tag 3$
$xp(x + 1) = ax^2 + ax + bx; tag 4$
$xp(x + 1) + 1 = ax^2 + ax + bx + 1; tag 5$
$a^2 x^2 + 2ab x + b^2 = ax^2 + ax + bx + 1; tag 6$
$(a^2 - a)x^2 + 2abx + b^2 = ax + bx + 1; tag 7$
$a^2 = a, ; b^2 = 1, ; 2ab = a + b; tag 8$
$a = 0, 1; ; b = pm 1; tag 9$
$b = 1 Longrightarrow 2a = a + 1 Longrightarrow a = 1 in {0, 1 }; tag{10}$
$b = -1 Longrightarrow -2a = a - 1 Longrightarrow a = dfrac{1}{3} notin {0, 1 }; tag{11}$
thus the only solution is $a = b = 1$, so that
$p(x) = x + 1. tag{12}$
CHECK:
$p^2(x) = (x + 1)^2 = x^2 + 2x + 1$
$= x(x + 2) + 1 = x((x + 1) + 1) + 1 = xp(x + 1) + 1; tag{13}$
it thus appears that the only solution is
$p(x) = x + 1. tag{14}$
answered Jan 10 at 2:15
Robert LewisRobert Lewis
45.8k23065
$endgroup$
add a comment |
$begingroup$
Our OP SunShine has correctly deduced that $deg p(x) = 1$, so that
$p(x) = ax + b; tag 1$
we compute:
$p^2(x) = a^2x^2 + 2ab x + b^2; tag 2$
$p(x + 1) = ax + a + b; tag 3$
$xp(x + 1) = ax^2 + ax + bx; tag 4$
$xp(x + 1) + 1 = ax^2 + ax + bx + 1; tag 5$
$a^2 x^2 + 2ab x + b^2 = ax^2 + ax + bx + 1; tag 6$
$(a^2 - a)x^2 + 2abx + b^2 = ax + bx + 1; tag 7$
$a^2 = a, ; b^2 = 1, ; 2ab = a + b; tag 8$
$a = 0, 1; ; b = pm 1; tag 9$
$b = 1 Longrightarrow 2a = a + 1 Longrightarrow a = 1 in {0, 1 }; tag{10}$
$b = -1 Longrightarrow -2a = a - 1 Longrightarrow a = dfrac{1}{3} notin {0, 1 }; tag{11}$
thus the only solution is $a = b = 1$, so that
$p(x) = x + 1. tag{12}$
CHECK:
$p^2(x) = (x + 1)^2 = x^2 + 2x + 1$
$= x(x + 2) + 1 = x((x + 1) + 1) + 1 = xp(x + 1) + 1; tag{13}$
it thus appears that the only solution is
$p(x) = x + 1. tag{14}$
answered Jan 10 at 2:15
Robert LewisRobert Lewis
45.8k23065
$endgroup$
Our OP SunShine has correctly deduced that $deg p(x) = 1$, so that
$p(x) = ax + b; tag 1$
we compute:
$p^2(x) = a^2x^2 + 2ab x + b^2; tag 2$
$p(x + 1) = ax + a + b; tag 3$
$xp(x + 1) = ax^2 + ax + bx; tag 4$
$xp(x + 1) + 1 = ax^2 + ax + bx + 1; tag 5$
$a^2 x^2 + 2ab x + b^2 = ax^2 + ax + bx + 1; tag 6$
$(a^2 - a)x^2 + 2abx + b^2 = ax + bx + 1; tag 7$
$a^2 = a, ; b^2 = 1, ; 2ab = a + b; tag 8$
$a = 0, 1; ; b = pm 1; tag 9$
$b = 1 Longrightarrow 2a = a + 1 Longrightarrow a = 1 in {0, 1 }; tag{10}$
$b = -1 Longrightarrow -2a = a - 1 Longrightarrow a = dfrac{1}{3} notin {0, 1 }; tag{11}$
thus the only solution is $a = b = 1$, so that
$p(x) = x + 1. tag{12}$
CHECK:
$p^2(x) = (x + 1)^2 = x^2 + 2x + 1$
$= x(x + 2) + 1 = x((x + 1) + 1) + 1 = xp(x + 1) + 1; tag{13}$
it thus appears that the only solution is
$p(x) = x + 1. tag{14}$
answered Jan 10 at 2:15
Robert LewisRobert Lewis
45.8k23065
edited Jan 10 at 2:41
answered Jan 10 at 2:15
Robert LewisRobert Lewis
45.8k23065
answered Jan 10 at 2:15
Robert LewisRobert Lewis
45.8k23065
answered Jan 10 at 2:15
Robert LewisRobert Lewis
45.8k23065
45.8k23065
add a comment |
add a comment |
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2
$begingroup$
What about $p(x)=x+1$?
$endgroup$
– Michael Burr
Jan 10 at 0:46
4
$begingroup$
$p$ cannot be zero. If $p$ has degree $r,$ the left hand side has degree $2r$ or $1$ and the right hand side has degree $r + 1,$ hence $r = 1.$ Now propose $p(x)=ax+b$ and solve.
$endgroup$
– Will M.
Jan 10 at 0:48
5
$begingroup$
What you wrote looks good, but you should check the $ax+b$ case more carefully.
$endgroup$
– lulu
Jan 10 at 0:51
$begingroup$
@WillM. That's an ideal hint, I think---perhaps write it up as an answer?
$endgroup$
– Travis
Jan 10 at 2:57