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Polynomial cannot have all roots real?
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Let $P in mathbb R[x]$ be a degree-$n$ polynomial with real coefficients such that $P(a) neq 0$, where $a$ is real. If $P'(a) = P ''(a) = 0$ then prove that $P$ cannot have all roots real.
Can someone suggest a possible solution using Rolle's Theorem.
All I could gather was that $P'(x) = 0$ has a repeated root by Rolle's Theorem. But I am stuck after this
calculus polynomials roots
edited Aug 16 at 20:08
Rodrigo de Azevedo
12.7k41751
asked Aug 16 at 7:13
Ramanujam Ganit Prashikshan Ke
713
add a comment |Â
up vote
14
down vote
favorite
Let $P in mathbb R[x]$ be a degree-$n$ polynomial with real coefficients such that $P(a) neq 0$, where $a$ is real. If $P'(a) = P ''(a) = 0$ then prove that $P$ cannot have all roots real.
Can someone suggest a possible solution using Rolle's Theorem.
All I could gather was that $P'(x) = 0$ has a repeated root by Rolle's Theorem. But I am stuck after this
calculus polynomials roots
edited Aug 16 at 20:08
Rodrigo de Azevedo
12.7k41751
asked Aug 16 at 7:13
Ramanujam Ganit Prashikshan Ke
713
If $P(x)$ can have $n$ real roots, then by Rolle's theorem you would get $P^(n)$ has some real root, which is impossible, cause $P^(n)$ is a nonzero constant.
– xbh
Aug 16 at 7:25
@Arthur Sorry about the ambiguity. If $P$ has $n$ real roots and $P'(a)= P''(a)=0$ where $P(a)neq 0$, then $P'$ would have $n$ roots [count multiplicity], then $P^(n)$ would have root. Am I correct now?
– xbh
Aug 16 at 7:31
@xbh If you can prove that $P'$ has $n$ real roots, then you don't need to go to $P^(n)$ to prove contradiction; $P'$ has degree $n-1$ and therefore cannot have $n$ roots. I think the main point of this exercise, however, is proving that $P'$ has at least $n$ roots in this case. You've just pointed it out as though it's a triviality and then spent a few lines proving the part which is (relatively) trivial.
– Arthur
Aug 16 at 7:39
@Arthur Thanks. I do not know what the OP has learned, so I wrote a few more lines. If s/he knew about the fact, then my lines are truly not necessary.
– xbh
Aug 16 at 7:44
add a comment |Â
up vote
14
down vote
favorite
up vote
14
down vote
favorite
Let $P in mathbb R[x]$ be a degree-$n$ polynomial with real coefficients such that $P(a) neq 0$, where $a$ is real. If $P'(a) = P ''(a) = 0$ then prove that $P$ cannot have all roots real.
Can someone suggest a possible solution using Rolle's Theorem.
All I could gather was that $P'(x) = 0$ has a repeated root by Rolle's Theorem. But I am stuck after this
calculus polynomials roots
edited Aug 16 at 20:08
Rodrigo de Azevedo
12.7k41751
asked Aug 16 at 7:13
Ramanujam Ganit Prashikshan Ke
713
Let $P in mathbb R[x]$ be a degree-$n$ polynomial with real coefficients such that $P(a) neq 0$, where $a$ is real. If $P'(a) = P ''(a) = 0$ then prove that $P$ cannot have all roots real.
Can someone suggest a possible solution using Rolle's Theorem.
All I could gather was that $P'(x) = 0$ has a repeated root by Rolle's Theorem. But I am stuck after this
calculus polynomials roots
edited Aug 16 at 20:08
Rodrigo de Azevedo
12.7k41751
asked Aug 16 at 7:13
Ramanujam Ganit Prashikshan Ke
713
edited Aug 16 at 20:08
Rodrigo de Azevedo
12.7k41751
edited Aug 16 at 20:08
Rodrigo de Azevedo
12.7k41751
edited Aug 16 at 20:08
Rodrigo de Azevedo
12.7k41751
12.7k41751
asked Aug 16 at 7:13
Ramanujam Ganit Prashikshan Ke
713
asked Aug 16 at 7:13
Ramanujam Ganit Prashikshan Ke
713
asked Aug 16 at 7:13
Ramanujam Ganit Prashikshan Ke
713
713
If $P(x)$ can have $n$ real roots, then by Rolle's theorem you would get $P^(n)$ has some real root, which is impossible, cause $P^(n)$ is a nonzero constant.
– xbh
Aug 16 at 7:25
@Arthur Sorry about the ambiguity. If $P$ has $n$ real roots and $P'(a)= P''(a)=0$ where $P(a)neq 0$, then $P'$ would have $n$ roots [count multiplicity], then $P^(n)$ would have root. Am I correct now?
– xbh
Aug 16 at 7:31
@xbh If you can prove that $P'$ has $n$ real roots, then you don't need to go to $P^(n)$ to prove contradiction; $P'$ has degree $n-1$ and therefore cannot have $n$ roots. I think the main point of this exercise, however, is proving that $P'$ has at least $n$ roots in this case. You've just pointed it out as though it's a triviality and then spent a few lines proving the part which is (relatively) trivial.
– Arthur
Aug 16 at 7:39
@Arthur Thanks. I do not know what the OP has learned, so I wrote a few more lines. If s/he knew about the fact, then my lines are truly not necessary.
– xbh
Aug 16 at 7:44
add a comment |Â
If $P(x)$ can have $n$ real roots, then by Rolle's theorem you would get $P^(n)$ has some real root, which is impossible, cause $P^(n)$ is a nonzero constant.
– xbh
Aug 16 at 7:25
@Arthur Sorry about the ambiguity. If $P$ has $n$ real roots and $P'(a)= P''(a)=0$ where $P(a)neq 0$, then $P'$ would have $n$ roots [count multiplicity], then $P^(n)$ would have root. Am I correct now?
– xbh
Aug 16 at 7:31
@xbh If you can prove that $P'$ has $n$ real roots, then you don't need to go to $P^(n)$ to prove contradiction; $P'$ has degree $n-1$ and therefore cannot have $n$ roots. I think the main point of this exercise, however, is proving that $P'$ has at least $n$ roots in this case. You've just pointed it out as though it's a triviality and then spent a few lines proving the part which is (relatively) trivial.
– Arthur
Aug 16 at 7:39
@Arthur Thanks. I do not know what the OP has learned, so I wrote a few more lines. If s/he knew about the fact, then my lines are truly not necessary.
– xbh
Aug 16 at 7:44
If $P(x)$ can have $n$ real roots, then by Rolle's theorem you would get $P^(n)$ has some real root, which is impossible, cause $P^(n)$ is a nonzero constant.
– xbh
Aug 16 at 7:25
If $P(x)$ can have $n$ real roots, then by Rolle's theorem you would get $P^(n)$ has some real root, which is impossible, cause $P^(n)$ is a nonzero constant.
– xbh
Aug 16 at 7:25
@Arthur Sorry about the ambiguity. If $P$ has $n$ real roots and $P'(a)= P''(a)=0$ where $P(a)neq 0$, then $P'$ would have $n$ roots [count multiplicity], then $P^(n)$ would have root. Am I correct now?
– xbh
Aug 16 at 7:31
@Arthur Sorry about the ambiguity. If $P$ has $n$ real roots and $P'(a)= P''(a)=0$ where $P(a)neq 0$, then $P'$ would have $n$ roots [count multiplicity], then $P^(n)$ would have root. Am I correct now?
– xbh
Aug 16 at 7:31
@xbh If you can prove that $P'$ has $n$ real roots, then you don't need to go to $P^(n)$ to prove contradiction; $P'$ has degree $n-1$ and therefore cannot have $n$ roots. I think the main point of this exercise, however, is proving that $P'$ has at least $n$ roots in this case. You've just pointed it out as though it's a triviality and then spent a few lines proving the part which is (relatively) trivial.
– Arthur
Aug 16 at 7:39
@xbh If you can prove that $P'$ has $n$ real roots, then you don't need to go to $P^(n)$ to prove contradiction; $P'$ has degree $n-1$ and therefore cannot have $n$ roots. I think the main point of this exercise, however, is proving that $P'$ has at least $n$ roots in this case. You've just pointed it out as though it's a triviality and then spent a few lines proving the part which is (relatively) trivial.
– Arthur
Aug 16 at 7:39
@Arthur Thanks. I do not know what the OP has learned, so I wrote a few more lines. If s/he knew about the fact, then my lines are truly not necessary.
– xbh
Aug 16 at 7:44
@Arthur Thanks. I do not know what the OP has learned, so I wrote a few more lines. If s/he knew about the fact, then my lines are truly not necessary.
– xbh
Aug 16 at 7:44
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
25
down vote
Assume that $P$ has degree $n$ and let $x_1,x_2,dots,x _n$ be all its roots (repetitions are allowed). Then $P(x)=cprod_k=1^n (x-x_k)$,
and if $x$ is not a root of $P$ we have that
$$fracP'(x)P(x)=sum_k=1^n frac1x-x_k.$$
After taking the derivative we obtain
$$fracP''(x)P(x)-(P'(x))^2(P(x))^2=-sum_k=1^n frac1(x-x_k)^2.$$
Finally by letting $x=a$ (which is not a root) we get a contradiction:
$$0=fracP''(a)P(a)-(P'(a))^2(P(a))^2=-sum_k=1^n frac1(a-x_k)^2<0$$
where the right-hand side is negative because $a, x_1,x_2,dots,x _n$ are all real.
answered Aug 16 at 7:56
Robert Z
85.2k1055123
3
Brilliant solution!
– xbh
Aug 16 at 7:58
How dis you get the first equation?
– Szeto
Aug 16 at 8:39
1
@Szeto It is the Logarithmic derivative: en.wikipedia.org/wiki/Logarithmic_derivative
– Robert Z
Aug 16 at 8:42
Beautiful solution! $Szeto you may want to take a look here: math.stackexchange.com/q/2660247/515527
– Zacky
Aug 16 at 20:12
add a comment |Â
up vote
10
down vote
Sketch of proof: Assume all roots of $P$ are real, and let $x_1leq x_2leq ldotsleq x_n$ be the $n$ roots (with repetition if $P$ has repeated roots). What does Rolle's theorem say about the roots of $P'$? How many roots does $P'$ have (counted with multiplicity)? Can $P'$ have a repeated root which is not one of the $x_i$?
answered Aug 16 at 7:23
Arthur
101k795176
add a comment |Â
up vote
3
down vote
I guess $P$ is not constant, otherwise the statement is false: all the roots of the constant $a$ polynomial are real, as everything holds for the elements of the emptyset.
Translate the polynomial by $a$, i.e., $Q(x):= P(x-a)$.
Then the conditions can be rephrased to $Q$ equivalently as follows: $Q(0)neq 0$, $Q'(0)=Q''(0)=0$.
In other words, $Q(x)= a_nx^n+ cdots +a_3x^3 + a_2x^2+a_1x+a_0$, where $a_0neq 0$ and $a_1=a_2=0$.
Write the Viete formulas for the roots $x_i$:
$prod x_i= (-1)^na_0neq 0, prod x_i cdot sum 1/x_i=0$, and $prod x_i cdot sumlimits_ineq j 1/(x_ix_j)=0$.
Put $y_i=1/x_i$ (possible, as $0$ is not a root, as the constant term is nonzero), then after simplification, you obtain $sum y_i= sumlimits_ineq j y_iy_j=0$, but then $sum y_i^2= 0$. So if these are real numbers, then all the $y_i$ are zero, a contradiction.
answered Aug 16 at 7:32
A. Pongrácz
4,322725
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
25
down vote
Assume that $P$ has degree $n$ and let $x_1,x_2,dots,x _n$ be all its roots (repetitions are allowed). Then $P(x)=cprod_k=1^n (x-x_k)$,
and if $x$ is not a root of $P$ we have that
$$fracP'(x)P(x)=sum_k=1^n frac1x-x_k.$$
After taking the derivative we obtain
$$fracP''(x)P(x)-(P'(x))^2(P(x))^2=-sum_k=1^n frac1(x-x_k)^2.$$
Finally by letting $x=a$ (which is not a root) we get a contradiction:
$$0=fracP''(a)P(a)-(P'(a))^2(P(a))^2=-sum_k=1^n frac1(a-x_k)^2<0$$
where the right-hand side is negative because $a, x_1,x_2,dots,x _n$ are all real.
answered Aug 16 at 7:56
Robert Z
85.2k1055123
3
Brilliant solution!
– xbh
Aug 16 at 7:58
How dis you get the first equation?
– Szeto
Aug 16 at 8:39
1
@Szeto It is the Logarithmic derivative: en.wikipedia.org/wiki/Logarithmic_derivative
– Robert Z
Aug 16 at 8:42
Beautiful solution! $Szeto you may want to take a look here: math.stackexchange.com/q/2660247/515527
– Zacky
Aug 16 at 20:12
add a comment |Â
up vote
25
down vote
Assume that $P$ has degree $n$ and let $x_1,x_2,dots,x _n$ be all its roots (repetitions are allowed). Then $P(x)=cprod_k=1^n (x-x_k)$,
and if $x$ is not a root of $P$ we have that
$$fracP'(x)P(x)=sum_k=1^n frac1x-x_k.$$
After taking the derivative we obtain
$$fracP''(x)P(x)-(P'(x))^2(P(x))^2=-sum_k=1^n frac1(x-x_k)^2.$$
Finally by letting $x=a$ (which is not a root) we get a contradiction:
$$0=fracP''(a)P(a)-(P'(a))^2(P(a))^2=-sum_k=1^n frac1(a-x_k)^2<0$$
where the right-hand side is negative because $a, x_1,x_2,dots,x _n$ are all real.
answered Aug 16 at 7:56
Robert Z
85.2k1055123
3
Brilliant solution!
– xbh
Aug 16 at 7:58
How dis you get the first equation?
– Szeto
Aug 16 at 8:39
1
@Szeto It is the Logarithmic derivative: en.wikipedia.org/wiki/Logarithmic_derivative
– Robert Z
Aug 16 at 8:42
Beautiful solution! $Szeto you may want to take a look here: math.stackexchange.com/q/2660247/515527
– Zacky
Aug 16 at 20:12
add a comment |Â
up vote
25
down vote
up vote
25
down vote
Assume that $P$ has degree $n$ and let $x_1,x_2,dots,x _n$ be all its roots (repetitions are allowed). Then $P(x)=cprod_k=1^n (x-x_k)$,
and if $x$ is not a root of $P$ we have that
$$fracP'(x)P(x)=sum_k=1^n frac1x-x_k.$$
After taking the derivative we obtain
$$fracP''(x)P(x)-(P'(x))^2(P(x))^2=-sum_k=1^n frac1(x-x_k)^2.$$
Finally by letting $x=a$ (which is not a root) we get a contradiction:
$$0=fracP''(a)P(a)-(P'(a))^2(P(a))^2=-sum_k=1^n frac1(a-x_k)^2<0$$
where the right-hand side is negative because $a, x_1,x_2,dots,x _n$ are all real.
answered Aug 16 at 7:56
Robert Z
85.2k1055123
Assume that $P$ has degree $n$ and let $x_1,x_2,dots,x _n$ be all its roots (repetitions are allowed). Then $P(x)=cprod_k=1^n (x-x_k)$,
and if $x$ is not a root of $P$ we have that
$$fracP'(x)P(x)=sum_k=1^n frac1x-x_k.$$
After taking the derivative we obtain
$$fracP''(x)P(x)-(P'(x))^2(P(x))^2=-sum_k=1^n frac1(x-x_k)^2.$$
Finally by letting $x=a$ (which is not a root) we get a contradiction:
$$0=fracP''(a)P(a)-(P'(a))^2(P(a))^2=-sum_k=1^n frac1(a-x_k)^2<0$$
where the right-hand side is negative because $a, x_1,x_2,dots,x _n$ are all real.
answered Aug 16 at 7:56
Robert Z
85.2k1055123
edited Aug 16 at 9:30
answered Aug 16 at 7:56
Robert Z
85.2k1055123
answered Aug 16 at 7:56
Robert Z
85.2k1055123
answered Aug 16 at 7:56
Robert Z
85.2k1055123
85.2k1055123
3
Brilliant solution!
– xbh
Aug 16 at 7:58
How dis you get the first equation?
– Szeto
Aug 16 at 8:39
1
@Szeto It is the Logarithmic derivative: en.wikipedia.org/wiki/Logarithmic_derivative
– Robert Z
Aug 16 at 8:42
Beautiful solution! $Szeto you may want to take a look here: math.stackexchange.com/q/2660247/515527
– Zacky
Aug 16 at 20:12
add a comment |Â
3
Brilliant solution!
– xbh
Aug 16 at 7:58
How dis you get the first equation?
– Szeto
Aug 16 at 8:39
1
@Szeto It is the Logarithmic derivative: en.wikipedia.org/wiki/Logarithmic_derivative
– Robert Z
Aug 16 at 8:42
Beautiful solution! $Szeto you may want to take a look here: math.stackexchange.com/q/2660247/515527
– Zacky
Aug 16 at 20:12
3
3
Brilliant solution!
– xbh
Aug 16 at 7:58
Brilliant solution!
– xbh
Aug 16 at 7:58
How dis you get the first equation?
– Szeto
Aug 16 at 8:39
How dis you get the first equation?
– Szeto
Aug 16 at 8:39
1
1
@Szeto It is the Logarithmic derivative: en.wikipedia.org/wiki/Logarithmic_derivative
– Robert Z
Aug 16 at 8:42
@Szeto It is the Logarithmic derivative: en.wikipedia.org/wiki/Logarithmic_derivative
– Robert Z
Aug 16 at 8:42
Beautiful solution! $Szeto you may want to take a look here: math.stackexchange.com/q/2660247/515527
– Zacky
Aug 16 at 20:12
Beautiful solution! $Szeto you may want to take a look here: math.stackexchange.com/q/2660247/515527
– Zacky
Aug 16 at 20:12
add a comment |Â
up vote
10
down vote
Sketch of proof: Assume all roots of $P$ are real, and let $x_1leq x_2leq ldotsleq x_n$ be the $n$ roots (with repetition if $P$ has repeated roots). What does Rolle's theorem say about the roots of $P'$? How many roots does $P'$ have (counted with multiplicity)? Can $P'$ have a repeated root which is not one of the $x_i$?
answered Aug 16 at 7:23
Arthur
101k795176
add a comment |Â
up vote
10
down vote
Sketch of proof: Assume all roots of $P$ are real, and let $x_1leq x_2leq ldotsleq x_n$ be the $n$ roots (with repetition if $P$ has repeated roots). What does Rolle's theorem say about the roots of $P'$? How many roots does $P'$ have (counted with multiplicity)? Can $P'$ have a repeated root which is not one of the $x_i$?
answered Aug 16 at 7:23
Arthur
101k795176
add a comment |Â
up vote
10
down vote
up vote
10
down vote
Sketch of proof: Assume all roots of $P$ are real, and let $x_1leq x_2leq ldotsleq x_n$ be the $n$ roots (with repetition if $P$ has repeated roots). What does Rolle's theorem say about the roots of $P'$? How many roots does $P'$ have (counted with multiplicity)? Can $P'$ have a repeated root which is not one of the $x_i$?
answered Aug 16 at 7:23
Arthur
101k795176
Sketch of proof: Assume all roots of $P$ are real, and let $x_1leq x_2leq ldotsleq x_n$ be the $n$ roots (with repetition if $P$ has repeated roots). What does Rolle's theorem say about the roots of $P'$? How many roots does $P'$ have (counted with multiplicity)? Can $P'$ have a repeated root which is not one of the $x_i$?
answered Aug 16 at 7:23
Arthur
101k795176
answered Aug 16 at 7:23
Arthur
101k795176
answered Aug 16 at 7:23
Arthur
101k795176
answered Aug 16 at 7:23
Arthur
101k795176
101k795176
add a comment |Â
add a comment |Â
up vote
3
down vote
I guess $P$ is not constant, otherwise the statement is false: all the roots of the constant $a$ polynomial are real, as everything holds for the elements of the emptyset.
Translate the polynomial by $a$, i.e., $Q(x):= P(x-a)$.
Then the conditions can be rephrased to $Q$ equivalently as follows: $Q(0)neq 0$, $Q'(0)=Q''(0)=0$.
In other words, $Q(x)= a_nx^n+ cdots +a_3x^3 + a_2x^2+a_1x+a_0$, where $a_0neq 0$ and $a_1=a_2=0$.
Write the Viete formulas for the roots $x_i$:
$prod x_i= (-1)^na_0neq 0, prod x_i cdot sum 1/x_i=0$, and $prod x_i cdot sumlimits_ineq j 1/(x_ix_j)=0$.
Put $y_i=1/x_i$ (possible, as $0$ is not a root, as the constant term is nonzero), then after simplification, you obtain $sum y_i= sumlimits_ineq j y_iy_j=0$, but then $sum y_i^2= 0$. So if these are real numbers, then all the $y_i$ are zero, a contradiction.
answered Aug 16 at 7:32
A. Pongrácz
4,322725
add a comment |Â
up vote
3
down vote
I guess $P$ is not constant, otherwise the statement is false: all the roots of the constant $a$ polynomial are real, as everything holds for the elements of the emptyset.
Translate the polynomial by $a$, i.e., $Q(x):= P(x-a)$.
Then the conditions can be rephrased to $Q$ equivalently as follows: $Q(0)neq 0$, $Q'(0)=Q''(0)=0$.
In other words, $Q(x)= a_nx^n+ cdots +a_3x^3 + a_2x^2+a_1x+a_0$, where $a_0neq 0$ and $a_1=a_2=0$.
Write the Viete formulas for the roots $x_i$:
$prod x_i= (-1)^na_0neq 0, prod x_i cdot sum 1/x_i=0$, and $prod x_i cdot sumlimits_ineq j 1/(x_ix_j)=0$.
Put $y_i=1/x_i$ (possible, as $0$ is not a root, as the constant term is nonzero), then after simplification, you obtain $sum y_i= sumlimits_ineq j y_iy_j=0$, but then $sum y_i^2= 0$. So if these are real numbers, then all the $y_i$ are zero, a contradiction.
answered Aug 16 at 7:32
A. Pongrácz
4,322725
add a comment |Â
up vote
3
down vote
up vote
3
down vote
I guess $P$ is not constant, otherwise the statement is false: all the roots of the constant $a$ polynomial are real, as everything holds for the elements of the emptyset.
Translate the polynomial by $a$, i.e., $Q(x):= P(x-a)$.
Then the conditions can be rephrased to $Q$ equivalently as follows: $Q(0)neq 0$, $Q'(0)=Q''(0)=0$.
In other words, $Q(x)= a_nx^n+ cdots +a_3x^3 + a_2x^2+a_1x+a_0$, where $a_0neq 0$ and $a_1=a_2=0$.
Write the Viete formulas for the roots $x_i$:
$prod x_i= (-1)^na_0neq 0, prod x_i cdot sum 1/x_i=0$, and $prod x_i cdot sumlimits_ineq j 1/(x_ix_j)=0$.
Put $y_i=1/x_i$ (possible, as $0$ is not a root, as the constant term is nonzero), then after simplification, you obtain $sum y_i= sumlimits_ineq j y_iy_j=0$, but then $sum y_i^2= 0$. So if these are real numbers, then all the $y_i$ are zero, a contradiction.
answered Aug 16 at 7:32
A. Pongrácz
4,322725
I guess $P$ is not constant, otherwise the statement is false: all the roots of the constant $a$ polynomial are real, as everything holds for the elements of the emptyset.
Translate the polynomial by $a$, i.e., $Q(x):= P(x-a)$.
Then the conditions can be rephrased to $Q$ equivalently as follows: $Q(0)neq 0$, $Q'(0)=Q''(0)=0$.
In other words, $Q(x)= a_nx^n+ cdots +a_3x^3 + a_2x^2+a_1x+a_0$, where $a_0neq 0$ and $a_1=a_2=0$.
Write the Viete formulas for the roots $x_i$:
$prod x_i= (-1)^na_0neq 0, prod x_i cdot sum 1/x_i=0$, and $prod x_i cdot sumlimits_ineq j 1/(x_ix_j)=0$.
Put $y_i=1/x_i$ (possible, as $0$ is not a root, as the constant term is nonzero), then after simplification, you obtain $sum y_i= sumlimits_ineq j y_iy_j=0$, but then $sum y_i^2= 0$. So if these are real numbers, then all the $y_i$ are zero, a contradiction.
answered Aug 16 at 7:32
A. Pongrácz
4,322725
edited Aug 16 at 7:38
answered Aug 16 at 7:32
A. Pongrácz
4,322725
answered Aug 16 at 7:32
A. Pongrácz
4,322725
answered Aug 16 at 7:32
A. Pongrácz
4,322725
4,322725
add a comment |Â
add a comment |Â
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If $P(x)$ can have $n$ real roots, then by Rolle's theorem you would get $P^(n)$ has some real root, which is impossible, cause $P^(n)$ is a nonzero constant.
– xbh
Aug 16 at 7:25
@Arthur Sorry about the ambiguity. If $P$ has $n$ real roots and $P'(a)= P''(a)=0$ where $P(a)neq 0$, then $P'$ would have $n$ roots [count multiplicity], then $P^(n)$ would have root. Am I correct now?
– xbh
Aug 16 at 7:31
@xbh If you can prove that $P'$ has $n$ real roots, then you don't need to go to $P^(n)$ to prove contradiction; $P'$ has degree $n-1$ and therefore cannot have $n$ roots. I think the main point of this exercise, however, is proving that $P'$ has at least $n$ roots in this case. You've just pointed it out as though it's a triviality and then spent a few lines proving the part which is (relatively) trivial.
– Arthur
Aug 16 at 7:39
@Arthur Thanks. I do not know what the OP has learned, so I wrote a few more lines. If s/he knew about the fact, then my lines are truly not necessary.
– xbh
Aug 16 at 7:44