Mixing
Do the zeroes of this polynomial lie inside, outside, or on the unit circle? $P_n(z)=1^3 z + 2^3 z^2 + 3^3...
$begingroup$
For each positive integer $n$, let's define the polynomial $$P_n(z)=1^3 z + 2^3 z^2 + 3^3 z^3 + cdots + n^3 z^n$$
Do the zeroes of $P_n$ lie inside, outside, or on the unit circle $|z|=1$?
I tried to find a formula for $displaystyle sum_{k=1}^n k^3 z^k$ by repeatedly taking derivatives of $z^k$ but it was so tough. Initial investigation showed that the zeroes lie inside the unit circle but I couldn't generalize the result.
Any help would be appreciated!
Source : The Arts and Crafts of Problem Solving
polynomials power-series roots roots-of-unity
edited Jan 6 at 2:52
Blue
47.9k870152
asked Jan 5 at 9:54
Atiq RahmanAtiq Rahman
1143
$endgroup$
add a comment |
$begingroup$
For each positive integer $n$, let's define the polynomial $$P_n(z)=1^3 z + 2^3 z^2 + 3^3 z^3 + cdots + n^3 z^n$$
Do the zeroes of $P_n$ lie inside, outside, or on the unit circle $|z|=1$?
I tried to find a formula for $displaystyle sum_{k=1}^n k^3 z^k$ by repeatedly taking derivatives of $z^k$ but it was so tough. Initial investigation showed that the zeroes lie inside the unit circle but I couldn't generalize the result.
Any help would be appreciated!
Source : The Arts and Crafts of Problem Solving
polynomials power-series roots roots-of-unity
edited Jan 6 at 2:52
Blue
47.9k870152
asked Jan 5 at 9:54
Atiq RahmanAtiq Rahman
1143
$endgroup$
2
$begingroup$
hint: have you heard of Rouché's theorem? mathworld.wolfram.com/RouchesTheorem.html this might help
$endgroup$
– Pink Panther
Jan 5 at 10:35
add a comment |
$begingroup$
For each positive integer $n$, let's define the polynomial $$P_n(z)=1^3 z + 2^3 z^2 + 3^3 z^3 + cdots + n^3 z^n$$
Do the zeroes of $P_n$ lie inside, outside, or on the unit circle $|z|=1$?
I tried to find a formula for $displaystyle sum_{k=1}^n k^3 z^k$ by repeatedly taking derivatives of $z^k$ but it was so tough. Initial investigation showed that the zeroes lie inside the unit circle but I couldn't generalize the result.
Any help would be appreciated!
Source : The Arts and Crafts of Problem Solving
polynomials power-series roots roots-of-unity
edited Jan 6 at 2:52
Blue
47.9k870152
asked Jan 5 at 9:54
Atiq RahmanAtiq Rahman
1143
$endgroup$
For each positive integer $n$, let's define the polynomial $$P_n(z)=1^3 z + 2^3 z^2 + 3^3 z^3 + cdots + n^3 z^n$$
Do the zeroes of $P_n$ lie inside, outside, or on the unit circle $|z|=1$?
I tried to find a formula for $displaystyle sum_{k=1}^n k^3 z^k$ by repeatedly taking derivatives of $z^k$ but it was so tough. Initial investigation showed that the zeroes lie inside the unit circle but I couldn't generalize the result.
Any help would be appreciated!
Source : The Arts and Crafts of Problem Solving
polynomials power-series roots roots-of-unity
polynomials power-series roots roots-of-unity
edited Jan 6 at 2:52
Blue
47.9k870152
asked Jan 5 at 9:54
Atiq RahmanAtiq Rahman
1143
edited Jan 6 at 2:52
Blue
47.9k870152
asked Jan 5 at 9:54
Atiq RahmanAtiq Rahman
1143
edited Jan 6 at 2:52
Blue
47.9k870152
edited Jan 6 at 2:52
Blue
47.9k870152
edited Jan 6 at 2:52
Blue
47.9k870152
47.9k870152
asked Jan 5 at 9:54
Atiq RahmanAtiq Rahman
1143
asked Jan 5 at 9:54
Atiq RahmanAtiq Rahman
1143
asked Jan 5 at 9:54
Atiq RahmanAtiq Rahman
1143
1143
2
$begingroup$
hint: have you heard of Rouché's theorem? mathworld.wolfram.com/RouchesTheorem.html this might help
$endgroup$
– Pink Panther
Jan 5 at 10:35
add a comment |
2
$begingroup$
hint: have you heard of Rouché's theorem? mathworld.wolfram.com/RouchesTheorem.html this might help
$endgroup$
– Pink Panther
Jan 5 at 10:35
2
2
$begingroup$
hint: have you heard of Rouché's theorem? mathworld.wolfram.com/RouchesTheorem.html this might help
$endgroup$
– Pink Panther
Jan 5 at 10:35
$begingroup$
hint: have you heard of Rouché's theorem? mathworld.wolfram.com/RouchesTheorem.html this might help
$endgroup$
– Pink Panther
Jan 5 at 10:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $$g(z) = sum_{k=0}^n z^k = frac{z^{n+1}-1}{z-1}$$
It is easy to see the roots of $g(z)$ lie on the unit circle $|z|= 1 $ and all of them are simple.
By Gauss-Lucas theorem, the roots of $g'(z)$ belong to the convex hull of the roots of $g$. Since the closed unit disk $|z| le 1$ is convex, this convex hull is a subset of the closed unit disk.
Notice the convex hull is a $n$-gon which intersect the unit circle $|z| = 1$ only at the roots of $g(z)$. Since the roots of $g(z)$ are simple, none of them can be root of $g'(z)$. As a result, the roots of $g'(z)$ belongs to the open unit disk $|z| < 1$.
Since $zg'(z)$ differs from $g'(z)$ by only a root at $0$, the roots of
$$zg'(z) = sum_{k=1}^n k z^k$$
belong to the open unit disk $|z| < 1$.
Apply Gauss-Lucas theorem again and then add a root at $z = 0$, we find the roots of
$$left(zfrac{d}{dz}right)^2 g(z) = sum_{k=1}^n k^2 z^k$$
belong to the open unit disk $|z| < 1$.
Repeat this process one more time, we find all the zeros of
$$left(zfrac{d}{dz}right)^3 g(z) = sum_{k=1}^n k^3 z^k = P_n(z)$$
lie inside the unit circle.
Notes
There are other ways to arrive at same conclusion. In particular, we can use following results:
Let $displaystyle;f(z) = sum_{k=0}^m a_k z^k$ be any polynomial with real and positive coefficients.
If the coefficients $a_k$ are non-descending,
$$0 < a_0 le a_2 le cdots le a_m$$
then roots of $f(z)$ belong to the closed unit disk $|z| le 1$.
If the coefficients $a_k$ are non-ascending,
$$ a_0 ge a_1 ge cdots ge a_m > 0$$
then roots of $f(z)$ lie outside the open unit disk (i.e. $|z| ge 1$ for all the roots)
In general, the roots of $f(z)$ belong to the closed annulus
$$min_{1le k le m} frac{a_{k-1}}{a_k} le |z| le max_{1 le k le m}frac{a_{k-1}}{a_k}$$
Since it is easy to derive any one of these results from the other two,
these results are typically treated as a single theorem known as the Eneström-Kakeya Theorem.
For a proof of the first result, see answers of a related question. In particular, the answer by Ayman Hourieh which uses Rouché's theorem.
Back to the problem at hand. It is easy to see we can rewrite $P_n(z)$ as $z f(z)$ for some polynomial $f(z)$ with real and positive coefficients. Apply the third result, we immediately find aside from a root at $z = 0$, the remaining $n-1$ roots of $P_n(z)$ lies within the closed annulus $$frac18 le |z| le left(frac{n-1
}{n}right)^3 < 1$$
answered Jan 5 at 11:26
achille huiachille hui
95.7k5131258
$endgroup$
add a comment |
Your Answer
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$begingroup$
Let $$g(z) = sum_{k=0}^n z^k = frac{z^{n+1}-1}{z-1}$$
It is easy to see the roots of $g(z)$ lie on the unit circle $|z|= 1 $ and all of them are simple.
By Gauss-Lucas theorem, the roots of $g'(z)$ belong to the convex hull of the roots of $g$. Since the closed unit disk $|z| le 1$ is convex, this convex hull is a subset of the closed unit disk.
Notice the convex hull is a $n$-gon which intersect the unit circle $|z| = 1$ only at the roots of $g(z)$. Since the roots of $g(z)$ are simple, none of them can be root of $g'(z)$. As a result, the roots of $g'(z)$ belongs to the open unit disk $|z| < 1$.
Since $zg'(z)$ differs from $g'(z)$ by only a root at $0$, the roots of
$$zg'(z) = sum_{k=1}^n k z^k$$
belong to the open unit disk $|z| < 1$.
Apply Gauss-Lucas theorem again and then add a root at $z = 0$, we find the roots of
$$left(zfrac{d}{dz}right)^2 g(z) = sum_{k=1}^n k^2 z^k$$
belong to the open unit disk $|z| < 1$.
Repeat this process one more time, we find all the zeros of
$$left(zfrac{d}{dz}right)^3 g(z) = sum_{k=1}^n k^3 z^k = P_n(z)$$
lie inside the unit circle.
Notes
There are other ways to arrive at same conclusion. In particular, we can use following results:
Let $displaystyle;f(z) = sum_{k=0}^m a_k z^k$ be any polynomial with real and positive coefficients.
If the coefficients $a_k$ are non-descending,
$$0 < a_0 le a_2 le cdots le a_m$$
then roots of $f(z)$ belong to the closed unit disk $|z| le 1$.
If the coefficients $a_k$ are non-ascending,
$$ a_0 ge a_1 ge cdots ge a_m > 0$$
then roots of $f(z)$ lie outside the open unit disk (i.e. $|z| ge 1$ for all the roots)
In general, the roots of $f(z)$ belong to the closed annulus
$$min_{1le k le m} frac{a_{k-1}}{a_k} le |z| le max_{1 le k le m}frac{a_{k-1}}{a_k}$$
Since it is easy to derive any one of these results from the other two,
these results are typically treated as a single theorem known as the Eneström-Kakeya Theorem.
For a proof of the first result, see answers of a related question. In particular, the answer by Ayman Hourieh which uses Rouché's theorem.
Back to the problem at hand. It is easy to see we can rewrite $P_n(z)$ as $z f(z)$ for some polynomial $f(z)$ with real and positive coefficients. Apply the third result, we immediately find aside from a root at $z = 0$, the remaining $n-1$ roots of $P_n(z)$ lies within the closed annulus $$frac18 le |z| le left(frac{n-1
}{n}right)^3 < 1$$
answered Jan 5 at 11:26
achille huiachille hui
95.7k5131258
$endgroup$
add a comment |
$begingroup$
Let $$g(z) = sum_{k=0}^n z^k = frac{z^{n+1}-1}{z-1}$$
It is easy to see the roots of $g(z)$ lie on the unit circle $|z|= 1 $ and all of them are simple.
By Gauss-Lucas theorem, the roots of $g'(z)$ belong to the convex hull of the roots of $g$. Since the closed unit disk $|z| le 1$ is convex, this convex hull is a subset of the closed unit disk.
Notice the convex hull is a $n$-gon which intersect the unit circle $|z| = 1$ only at the roots of $g(z)$. Since the roots of $g(z)$ are simple, none of them can be root of $g'(z)$. As a result, the roots of $g'(z)$ belongs to the open unit disk $|z| < 1$.
Since $zg'(z)$ differs from $g'(z)$ by only a root at $0$, the roots of
$$zg'(z) = sum_{k=1}^n k z^k$$
belong to the open unit disk $|z| < 1$.
Apply Gauss-Lucas theorem again and then add a root at $z = 0$, we find the roots of
$$left(zfrac{d}{dz}right)^2 g(z) = sum_{k=1}^n k^2 z^k$$
belong to the open unit disk $|z| < 1$.
Repeat this process one more time, we find all the zeros of
$$left(zfrac{d}{dz}right)^3 g(z) = sum_{k=1}^n k^3 z^k = P_n(z)$$
lie inside the unit circle.
Notes
There are other ways to arrive at same conclusion. In particular, we can use following results:
Let $displaystyle;f(z) = sum_{k=0}^m a_k z^k$ be any polynomial with real and positive coefficients.
If the coefficients $a_k$ are non-descending,
$$0 < a_0 le a_2 le cdots le a_m$$
then roots of $f(z)$ belong to the closed unit disk $|z| le 1$.
If the coefficients $a_k$ are non-ascending,
$$ a_0 ge a_1 ge cdots ge a_m > 0$$
then roots of $f(z)$ lie outside the open unit disk (i.e. $|z| ge 1$ for all the roots)
In general, the roots of $f(z)$ belong to the closed annulus
$$min_{1le k le m} frac{a_{k-1}}{a_k} le |z| le max_{1 le k le m}frac{a_{k-1}}{a_k}$$
Since it is easy to derive any one of these results from the other two,
these results are typically treated as a single theorem known as the Eneström-Kakeya Theorem.
For a proof of the first result, see answers of a related question. In particular, the answer by Ayman Hourieh which uses Rouché's theorem.
Back to the problem at hand. It is easy to see we can rewrite $P_n(z)$ as $z f(z)$ for some polynomial $f(z)$ with real and positive coefficients. Apply the third result, we immediately find aside from a root at $z = 0$, the remaining $n-1$ roots of $P_n(z)$ lies within the closed annulus $$frac18 le |z| le left(frac{n-1
}{n}right)^3 < 1$$
answered Jan 5 at 11:26
achille huiachille hui
95.7k5131258
$endgroup$
add a comment |
$begingroup$
Let $$g(z) = sum_{k=0}^n z^k = frac{z^{n+1}-1}{z-1}$$
It is easy to see the roots of $g(z)$ lie on the unit circle $|z|= 1 $ and all of them are simple.
By Gauss-Lucas theorem, the roots of $g'(z)$ belong to the convex hull of the roots of $g$. Since the closed unit disk $|z| le 1$ is convex, this convex hull is a subset of the closed unit disk.
Notice the convex hull is a $n$-gon which intersect the unit circle $|z| = 1$ only at the roots of $g(z)$. Since the roots of $g(z)$ are simple, none of them can be root of $g'(z)$. As a result, the roots of $g'(z)$ belongs to the open unit disk $|z| < 1$.
Since $zg'(z)$ differs from $g'(z)$ by only a root at $0$, the roots of
$$zg'(z) = sum_{k=1}^n k z^k$$
belong to the open unit disk $|z| < 1$.
Apply Gauss-Lucas theorem again and then add a root at $z = 0$, we find the roots of
$$left(zfrac{d}{dz}right)^2 g(z) = sum_{k=1}^n k^2 z^k$$
belong to the open unit disk $|z| < 1$.
Repeat this process one more time, we find all the zeros of
$$left(zfrac{d}{dz}right)^3 g(z) = sum_{k=1}^n k^3 z^k = P_n(z)$$
lie inside the unit circle.
Notes
There are other ways to arrive at same conclusion. In particular, we can use following results:
Let $displaystyle;f(z) = sum_{k=0}^m a_k z^k$ be any polynomial with real and positive coefficients.
If the coefficients $a_k$ are non-descending,
$$0 < a_0 le a_2 le cdots le a_m$$
then roots of $f(z)$ belong to the closed unit disk $|z| le 1$.
If the coefficients $a_k$ are non-ascending,
$$ a_0 ge a_1 ge cdots ge a_m > 0$$
then roots of $f(z)$ lie outside the open unit disk (i.e. $|z| ge 1$ for all the roots)
In general, the roots of $f(z)$ belong to the closed annulus
$$min_{1le k le m} frac{a_{k-1}}{a_k} le |z| le max_{1 le k le m}frac{a_{k-1}}{a_k}$$
Since it is easy to derive any one of these results from the other two,
these results are typically treated as a single theorem known as the Eneström-Kakeya Theorem.
For a proof of the first result, see answers of a related question. In particular, the answer by Ayman Hourieh which uses Rouché's theorem.
Back to the problem at hand. It is easy to see we can rewrite $P_n(z)$ as $z f(z)$ for some polynomial $f(z)$ with real and positive coefficients. Apply the third result, we immediately find aside from a root at $z = 0$, the remaining $n-1$ roots of $P_n(z)$ lies within the closed annulus $$frac18 le |z| le left(frac{n-1
}{n}right)^3 < 1$$
answered Jan 5 at 11:26
achille huiachille hui
95.7k5131258
$endgroup$
Let $$g(z) = sum_{k=0}^n z^k = frac{z^{n+1}-1}{z-1}$$
It is easy to see the roots of $g(z)$ lie on the unit circle $|z|= 1 $ and all of them are simple.
By Gauss-Lucas theorem, the roots of $g'(z)$ belong to the convex hull of the roots of $g$. Since the closed unit disk $|z| le 1$ is convex, this convex hull is a subset of the closed unit disk.
Notice the convex hull is a $n$-gon which intersect the unit circle $|z| = 1$ only at the roots of $g(z)$. Since the roots of $g(z)$ are simple, none of them can be root of $g'(z)$. As a result, the roots of $g'(z)$ belongs to the open unit disk $|z| < 1$.
Since $zg'(z)$ differs from $g'(z)$ by only a root at $0$, the roots of
$$zg'(z) = sum_{k=1}^n k z^k$$
belong to the open unit disk $|z| < 1$.
Apply Gauss-Lucas theorem again and then add a root at $z = 0$, we find the roots of
$$left(zfrac{d}{dz}right)^2 g(z) = sum_{k=1}^n k^2 z^k$$
belong to the open unit disk $|z| < 1$.
Repeat this process one more time, we find all the zeros of
$$left(zfrac{d}{dz}right)^3 g(z) = sum_{k=1}^n k^3 z^k = P_n(z)$$
lie inside the unit circle.
Notes
There are other ways to arrive at same conclusion. In particular, we can use following results:
Let $displaystyle;f(z) = sum_{k=0}^m a_k z^k$ be any polynomial with real and positive coefficients.
If the coefficients $a_k$ are non-descending,
$$0 < a_0 le a_2 le cdots le a_m$$
then roots of $f(z)$ belong to the closed unit disk $|z| le 1$.
If the coefficients $a_k$ are non-ascending,
$$ a_0 ge a_1 ge cdots ge a_m > 0$$
then roots of $f(z)$ lie outside the open unit disk (i.e. $|z| ge 1$ for all the roots)
In general, the roots of $f(z)$ belong to the closed annulus
$$min_{1le k le m} frac{a_{k-1}}{a_k} le |z| le max_{1 le k le m}frac{a_{k-1}}{a_k}$$
Since it is easy to derive any one of these results from the other two,
these results are typically treated as a single theorem known as the Eneström-Kakeya Theorem.
For a proof of the first result, see answers of a related question. In particular, the answer by Ayman Hourieh which uses Rouché's theorem.
Back to the problem at hand. It is easy to see we can rewrite $P_n(z)$ as $z f(z)$ for some polynomial $f(z)$ with real and positive coefficients. Apply the third result, we immediately find aside from a root at $z = 0$, the remaining $n-1$ roots of $P_n(z)$ lies within the closed annulus $$frac18 le |z| le left(frac{n-1
}{n}right)^3 < 1$$
answered Jan 5 at 11:26
achille huiachille hui
95.7k5131258
edited Jan 6 at 2:49
answered Jan 5 at 11:26
achille huiachille hui
95.7k5131258
answered Jan 5 at 11:26
achille huiachille hui
95.7k5131258
answered Jan 5 at 11:26
achille huiachille hui
95.7k5131258
95.7k5131258
add a comment |
add a comment |
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$begingroup$
hint: have you heard of Rouché's theorem? mathworld.wolfram.com/RouchesTheorem.html this might help
$endgroup$
– Pink Panther
Jan 5 at 10:35