Mixing
How prove that: $left | a right |+left | b right |leq 5$ for $P(x)=x^{3}+ax^{2}+bx+1$?
$begingroup$
Let $P(x)=x^{3}+ax^{2}+bx+1$ and $left | P(x) right |leq 1$ for all x such that $left | x right |leq 1$. How prove that: $left | a right |+left | b right |leq 5$?
polynomials
asked Jul 23 '14 at 19:01
piteerpiteer
2,917619
$endgroup$
|
show 2 more comments
$begingroup$
Let $P(x)=x^{3}+ax^{2}+bx+1$ and $left | P(x) right |leq 1$ for all x such that $left | x right |leq 1$. How prove that: $left | a right |+left | b right |leq 5$?
polynomials
asked Jul 23 '14 at 19:01
piteerpiteer
2,917619
$endgroup$
$begingroup$
You have two inequalities: $x^3 + ax^2 + bx le 0 quadforall |x|le 1$ and $x^3 + ax^2 + bx ge -2 quadforall |x|le 1$
$endgroup$
– AlexR
Jul 23 '14 at 19:05
1
$begingroup$
And $a$ and $b$ are real too?
$endgroup$
– Daniel Fischer♦
Jul 23 '14 at 19:09
$begingroup$
$a+bleq 1$ and $a+b ge -3$ right?
$endgroup$
– piteer
Jul 23 '14 at 19:10
$begingroup$
We have $P(0) = 1$ and $P'(0) = b$. Since we have $|P(x)|le 1$ and $0 in [-1,1]^circ$, doesn't this imply that $b=0$?
$endgroup$
– copper.hat
Jul 23 '14 at 19:14
3
$begingroup$
First, $P(0) = 1$ and $b = P'(0)$ implies $b = 0$. $P''(0) = 2a$ implies $a leqslant 0$. $P(1) = 2+a leqslant 1$ implies $a leqslant -1$, and $P(-1) = a geqslant -1$ implies $a geqslant -1$.
$endgroup$
– Daniel Fischer♦
Jul 23 '14 at 19:14
|
show 2 more comments
$begingroup$
Let $P(x)=x^{3}+ax^{2}+bx+1$ and $left | P(x) right |leq 1$ for all x such that $left | x right |leq 1$. How prove that: $left | a right |+left | b right |leq 5$?
polynomials
asked Jul 23 '14 at 19:01
piteerpiteer
2,917619
$endgroup$
Let $P(x)=x^{3}+ax^{2}+bx+1$ and $left | P(x) right |leq 1$ for all x such that $left | x right |leq 1$. How prove that: $left | a right |+left | b right |leq 5$?
polynomials
polynomials
asked Jul 23 '14 at 19:01
piteerpiteer
2,917619
asked Jul 23 '14 at 19:01
piteerpiteer
2,917619
edited Feb 14 at 18:25
piteer
asked Jul 23 '14 at 19:01
piteerpiteer
2,917619
asked Jul 23 '14 at 19:01
piteerpiteer
2,917619
asked Jul 23 '14 at 19:01
piteerpiteer
2,917619
2,917619
$begingroup$
You have two inequalities: $x^3 + ax^2 + bx le 0 quadforall |x|le 1$ and $x^3 + ax^2 + bx ge -2 quadforall |x|le 1$
$endgroup$
– AlexR
Jul 23 '14 at 19:05
1
$begingroup$
And $a$ and $b$ are real too?
$endgroup$
– Daniel Fischer♦
Jul 23 '14 at 19:09
$begingroup$
$a+bleq 1$ and $a+b ge -3$ right?
$endgroup$
– piteer
Jul 23 '14 at 19:10
$begingroup$
We have $P(0) = 1$ and $P'(0) = b$. Since we have $|P(x)|le 1$ and $0 in [-1,1]^circ$, doesn't this imply that $b=0$?
$endgroup$
– copper.hat
Jul 23 '14 at 19:14
3
$begingroup$
First, $P(0) = 1$ and $b = P'(0)$ implies $b = 0$. $P''(0) = 2a$ implies $a leqslant 0$. $P(1) = 2+a leqslant 1$ implies $a leqslant -1$, and $P(-1) = a geqslant -1$ implies $a geqslant -1$.
$endgroup$
– Daniel Fischer♦
Jul 23 '14 at 19:14
|
show 2 more comments
$begingroup$
You have two inequalities: $x^3 + ax^2 + bx le 0 quadforall |x|le 1$ and $x^3 + ax^2 + bx ge -2 quadforall |x|le 1$
$endgroup$
– AlexR
Jul 23 '14 at 19:05
1
$begingroup$
And $a$ and $b$ are real too?
$endgroup$
– Daniel Fischer♦
Jul 23 '14 at 19:09
$begingroup$
$a+bleq 1$ and $a+b ge -3$ right?
$endgroup$
– piteer
Jul 23 '14 at 19:10
$begingroup$
We have $P(0) = 1$ and $P'(0) = b$. Since we have $|P(x)|le 1$ and $0 in [-1,1]^circ$, doesn't this imply that $b=0$?
$endgroup$
– copper.hat
Jul 23 '14 at 19:14
3
$begingroup$
First, $P(0) = 1$ and $b = P'(0)$ implies $b = 0$. $P''(0) = 2a$ implies $a leqslant 0$. $P(1) = 2+a leqslant 1$ implies $a leqslant -1$, and $P(-1) = a geqslant -1$ implies $a geqslant -1$.
$endgroup$
– Daniel Fischer♦
Jul 23 '14 at 19:14
$begingroup$
You have two inequalities: $x^3 + ax^2 + bx le 0 quadforall |x|le 1$ and $x^3 + ax^2 + bx ge -2 quadforall |x|le 1$
$endgroup$
– AlexR
Jul 23 '14 at 19:05
$begingroup$
You have two inequalities: $x^3 + ax^2 + bx le 0 quadforall |x|le 1$ and $x^3 + ax^2 + bx ge -2 quadforall |x|le 1$
$endgroup$
– AlexR
Jul 23 '14 at 19:05
1
1
$begingroup$
And $a$ and $b$ are real too?
$endgroup$
– Daniel Fischer♦
Jul 23 '14 at 19:09
$begingroup$
And $a$ and $b$ are real too?
$endgroup$
– Daniel Fischer♦
Jul 23 '14 at 19:09
$begingroup$
$a+bleq 1$ and $a+b ge -3$ right?
$endgroup$
– piteer
Jul 23 '14 at 19:10
$begingroup$
$a+bleq 1$ and $a+b ge -3$ right?
$endgroup$
– piteer
Jul 23 '14 at 19:10
$begingroup$
We have $P(0) = 1$ and $P'(0) = b$. Since we have $|P(x)|le 1$ and $0 in [-1,1]^circ$, doesn't this imply that $b=0$?
$endgroup$
– copper.hat
Jul 23 '14 at 19:14
$begingroup$
We have $P(0) = 1$ and $P'(0) = b$. Since we have $|P(x)|le 1$ and $0 in [-1,1]^circ$, doesn't this imply that $b=0$?
$endgroup$
– copper.hat
Jul 23 '14 at 19:14
3
3
$begingroup$
First, $P(0) = 1$ and $b = P'(0)$ implies $b = 0$. $P''(0) = 2a$ implies $a leqslant 0$. $P(1) = 2+a leqslant 1$ implies $a leqslant -1$, and $P(-1) = a geqslant -1$ implies $a geqslant -1$.
$endgroup$
– Daniel Fischer♦
Jul 23 '14 at 19:14
$begingroup$
First, $P(0) = 1$ and $b = P'(0)$ implies $b = 0$. $P''(0) = 2a$ implies $a leqslant 0$. $P(1) = 2+a leqslant 1$ implies $a leqslant -1$, and $P(-1) = a geqslant -1$ implies $a geqslant -1$.
$endgroup$
– Daniel Fischer♦
Jul 23 '14 at 19:14
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
We have: $|P(1)| leq 1 to |1+a+b+1| leq 1 to -1 leq a+b+2 leq 1 to -3 leq a+b leq -1$.
Also $|P(-1)| leq 1 to |-1+a-b+1| leq 1 to -1 leq a-b leq 1$.
Add two inequalities we have: $-4 leq 2a leq 0 to -2 leq a leq 0 to |a| leq 2$.
Also:
$-1leq b-a leq 1$.
Add this and the first inequality we have: $-4 leq 2b leq 0 to -2 leq b leq 0 to |b| leq 2$.
Thus: $|a| + |b| leq 2 + 2 = 4 < 5$.
edited Jul 23 '14 at 19:40
AlexR
22.7k12349
answered Jul 23 '14 at 19:32
DeepSeaDeepSea
71.3k54487
$endgroup$
$begingroup$
The comments above show that $a=-1, b=0$.
$endgroup$
– copper.hat
Jul 23 '14 at 19:33
add a comment |
$begingroup$
Using the two inequalities
$$begin{align*}
x^3 + ax^2 + bx & le 0 & forall |x|le 1\
x^3 + ax^2 + bx & ge -2 & forall |x|le 1
end{align*}$$
We can see that $x=0$ imposes no condition on $a,b$, so we may assume $xne 0$ to obtain
$$x^2 + ax + b le 0 quadforall |x|le 1$$
Can you calculate
$$M = max_{|x|le 1} x^2+ax+b$$
As a function of $a,b$? You will then find $M(a,b)le 0$ as the first restriction. Similar thoughts will give you another bound from the second inequality.
answered Jul 23 '14 at 19:09
AlexRAlexR
22.7k12349
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
We have: $|P(1)| leq 1 to |1+a+b+1| leq 1 to -1 leq a+b+2 leq 1 to -3 leq a+b leq -1$.
Also $|P(-1)| leq 1 to |-1+a-b+1| leq 1 to -1 leq a-b leq 1$.
Add two inequalities we have: $-4 leq 2a leq 0 to -2 leq a leq 0 to |a| leq 2$.
Also:
$-1leq b-a leq 1$.
Add this and the first inequality we have: $-4 leq 2b leq 0 to -2 leq b leq 0 to |b| leq 2$.
Thus: $|a| + |b| leq 2 + 2 = 4 < 5$.
edited Jul 23 '14 at 19:40
AlexR
22.7k12349
answered Jul 23 '14 at 19:32
DeepSeaDeepSea
71.3k54487
$endgroup$
$begingroup$
The comments above show that $a=-1, b=0$.
$endgroup$
– copper.hat
Jul 23 '14 at 19:33
add a comment |
$begingroup$
We have: $|P(1)| leq 1 to |1+a+b+1| leq 1 to -1 leq a+b+2 leq 1 to -3 leq a+b leq -1$.
Also $|P(-1)| leq 1 to |-1+a-b+1| leq 1 to -1 leq a-b leq 1$.
Add two inequalities we have: $-4 leq 2a leq 0 to -2 leq a leq 0 to |a| leq 2$.
Also:
$-1leq b-a leq 1$.
Add this and the first inequality we have: $-4 leq 2b leq 0 to -2 leq b leq 0 to |b| leq 2$.
Thus: $|a| + |b| leq 2 + 2 = 4 < 5$.
edited Jul 23 '14 at 19:40
AlexR
22.7k12349
answered Jul 23 '14 at 19:32
DeepSeaDeepSea
71.3k54487
$endgroup$
$begingroup$
The comments above show that $a=-1, b=0$.
$endgroup$
– copper.hat
Jul 23 '14 at 19:33
add a comment |
$begingroup$
We have: $|P(1)| leq 1 to |1+a+b+1| leq 1 to -1 leq a+b+2 leq 1 to -3 leq a+b leq -1$.
Also $|P(-1)| leq 1 to |-1+a-b+1| leq 1 to -1 leq a-b leq 1$.
Add two inequalities we have: $-4 leq 2a leq 0 to -2 leq a leq 0 to |a| leq 2$.
Also:
$-1leq b-a leq 1$.
Add this and the first inequality we have: $-4 leq 2b leq 0 to -2 leq b leq 0 to |b| leq 2$.
Thus: $|a| + |b| leq 2 + 2 = 4 < 5$.
edited Jul 23 '14 at 19:40
AlexR
22.7k12349
answered Jul 23 '14 at 19:32
DeepSeaDeepSea
71.3k54487
$endgroup$
We have: $|P(1)| leq 1 to |1+a+b+1| leq 1 to -1 leq a+b+2 leq 1 to -3 leq a+b leq -1$.
Also $|P(-1)| leq 1 to |-1+a-b+1| leq 1 to -1 leq a-b leq 1$.
Add two inequalities we have: $-4 leq 2a leq 0 to -2 leq a leq 0 to |a| leq 2$.
Also:
$-1leq b-a leq 1$.
Add this and the first inequality we have: $-4 leq 2b leq 0 to -2 leq b leq 0 to |b| leq 2$.
Thus: $|a| + |b| leq 2 + 2 = 4 < 5$.
edited Jul 23 '14 at 19:40
AlexR
22.7k12349
answered Jul 23 '14 at 19:32
DeepSeaDeepSea
71.3k54487
edited Jul 23 '14 at 19:40
AlexR
22.7k12349
edited Jul 23 '14 at 19:40
AlexR
22.7k12349
edited Jul 23 '14 at 19:40
AlexR
22.7k12349
22.7k12349
answered Jul 23 '14 at 19:32
DeepSeaDeepSea
71.3k54487
answered Jul 23 '14 at 19:32
DeepSeaDeepSea
71.3k54487
answered Jul 23 '14 at 19:32
DeepSeaDeepSea
71.3k54487
71.3k54487
$begingroup$
The comments above show that $a=-1, b=0$.
$endgroup$
– copper.hat
Jul 23 '14 at 19:33
add a comment |
$begingroup$
The comments above show that $a=-1, b=0$.
$endgroup$
– copper.hat
Jul 23 '14 at 19:33
$begingroup$
The comments above show that $a=-1, b=0$.
$endgroup$
– copper.hat
Jul 23 '14 at 19:33
$begingroup$
The comments above show that $a=-1, b=0$.
$endgroup$
– copper.hat
Jul 23 '14 at 19:33
add a comment |
$begingroup$
Using the two inequalities
$$begin{align*}
x^3 + ax^2 + bx & le 0 & forall |x|le 1\
x^3 + ax^2 + bx & ge -2 & forall |x|le 1
end{align*}$$
We can see that $x=0$ imposes no condition on $a,b$, so we may assume $xne 0$ to obtain
$$x^2 + ax + b le 0 quadforall |x|le 1$$
Can you calculate
$$M = max_{|x|le 1} x^2+ax+b$$
As a function of $a,b$? You will then find $M(a,b)le 0$ as the first restriction. Similar thoughts will give you another bound from the second inequality.
answered Jul 23 '14 at 19:09
AlexRAlexR
22.7k12349
$endgroup$
add a comment |
$begingroup$
Using the two inequalities
$$begin{align*}
x^3 + ax^2 + bx & le 0 & forall |x|le 1\
x^3 + ax^2 + bx & ge -2 & forall |x|le 1
end{align*}$$
We can see that $x=0$ imposes no condition on $a,b$, so we may assume $xne 0$ to obtain
$$x^2 + ax + b le 0 quadforall |x|le 1$$
Can you calculate
$$M = max_{|x|le 1} x^2+ax+b$$
As a function of $a,b$? You will then find $M(a,b)le 0$ as the first restriction. Similar thoughts will give you another bound from the second inequality.
answered Jul 23 '14 at 19:09
AlexRAlexR
22.7k12349
$endgroup$
add a comment |
$begingroup$
Using the two inequalities
$$begin{align*}
x^3 + ax^2 + bx & le 0 & forall |x|le 1\
x^3 + ax^2 + bx & ge -2 & forall |x|le 1
end{align*}$$
We can see that $x=0$ imposes no condition on $a,b$, so we may assume $xne 0$ to obtain
$$x^2 + ax + b le 0 quadforall |x|le 1$$
Can you calculate
$$M = max_{|x|le 1} x^2+ax+b$$
As a function of $a,b$? You will then find $M(a,b)le 0$ as the first restriction. Similar thoughts will give you another bound from the second inequality.
answered Jul 23 '14 at 19:09
AlexRAlexR
22.7k12349
$endgroup$
Using the two inequalities
$$begin{align*}
x^3 + ax^2 + bx & le 0 & forall |x|le 1\
x^3 + ax^2 + bx & ge -2 & forall |x|le 1
end{align*}$$
We can see that $x=0$ imposes no condition on $a,b$, so we may assume $xne 0$ to obtain
$$x^2 + ax + b le 0 quadforall |x|le 1$$
Can you calculate
$$M = max_{|x|le 1} x^2+ax+b$$
As a function of $a,b$? You will then find $M(a,b)le 0$ as the first restriction. Similar thoughts will give you another bound from the second inequality.
answered Jul 23 '14 at 19:09
AlexRAlexR
22.7k12349
answered Jul 23 '14 at 19:09
AlexRAlexR
22.7k12349
answered Jul 23 '14 at 19:09
AlexRAlexR
22.7k12349
answered Jul 23 '14 at 19:09
AlexRAlexR
22.7k12349
22.7k12349
add a comment |
add a comment |
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$begingroup$
You have two inequalities: $x^3 + ax^2 + bx le 0 quadforall |x|le 1$ and $x^3 + ax^2 + bx ge -2 quadforall |x|le 1$
$endgroup$
– AlexR
Jul 23 '14 at 19:05
1
$begingroup$
And $a$ and $b$ are real too?
$endgroup$
– Daniel Fischer♦
Jul 23 '14 at 19:09
$begingroup$
$a+bleq 1$ and $a+b ge -3$ right?
$endgroup$
– piteer
Jul 23 '14 at 19:10
$begingroup$
We have $P(0) = 1$ and $P'(0) = b$. Since we have $|P(x)|le 1$ and $0 in [-1,1]^circ$, doesn't this imply that $b=0$?
$endgroup$
– copper.hat
Jul 23 '14 at 19:14
3
$begingroup$
First, $P(0) = 1$ and $b = P'(0)$ implies $b = 0$. $P''(0) = 2a$ implies $a leqslant 0$. $P(1) = 2+a leqslant 1$ implies $a leqslant -1$, and $P(-1) = a geqslant -1$ implies $a geqslant -1$.
$endgroup$
– Daniel Fischer♦
Jul 23 '14 at 19:14