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A generalized inequality for the polar derivative of a polynomial

Abstract

In this paper, we extend a result recently proved by Liman et al. (Complex Anal. Oper. Theory 6:1199-1209, 2012. doi:10.1007/s11785-010-0120-3) to the polar derivative of a polynomial and thereby obtain some more general results for polynomials with restricted zeros.

MSC:30A10, 30C10, 30C15.

1 Introduction and statement of results

Let P(z):= j = 0 n a j z j be a polynomial of degree n, and let P (z) be its derivative, then

max | z | = 1 | P ( z ) | n max | z | = 1 | P ( z ) | .
(1)

Inequality (1) is a famous result due to Bernstein and is best possible with equality holding for the polynomial P(z)=λ z n , where λ is a complex number.

If we restrict ourselves to a class of polynomials having no zeros in |z|<1, then the above inequality can be sharpened. In fact, Erdös conjectured and later Lax [1] proved that if P(z)0 in |z|<1, then

max | z | = 1 | P ( z ) | n 2 max | z | = 1 | P ( z ) | .
(2)

As a refinement of (2), Aziz and Dawood [2] proved that if P(z) is a polynomial of degree n having no zeros in |z|<1, then

max | z | = 1 | P ( z ) | n 2 { max | z | = 1 | P ( z ) | min | z | = 1 | P ( z ) | } .
(3)

As an improvement of (3), Dewan and Hans [3] proved that if P(z) is a polynomial of degree n having no zeros in |z|<1, then for any β with |β|1 and |z|=1,

(4)

Let D α P(z) denote the polar derivative of the polynomial P(z) of degree n with respect to α, then

D α P(z)=nP(z)+(αz) P (z).

The polynomial D α P(z) is of degree at most n1 and it generalizes the ordinary derivative in the sense that

lim α D α P ( z ) α = P (z).

As an extension of (1) to the polar derivative, Aziz and Shah ([4], Theorem 4 with k=1) showed that if P(z) is a polynomial of degree n, then for every complex number α with |α|>1,

| D α P ( z ) | n|α| max | z | = 1 | P ( z ) | for |z|=1.
(5)

Inequality (5) becomes equality for P(z)=a z n , a0.

If we divide the two sides of (5) by |α| and let |α|, we get inequality (1).

Aziz and Shah [5] proved that if P(z) is a polynomial of degree n that does not vanish in |z|<1, then for every complex number α with |α|1,

max | z | = 1 | D α P ( z ) | n 2 { ( | α | + 1 ) max | z | = 1 | P ( z ) | ( | α | 1 ) min | z | = 1 | P ( z ) | } .
(6)

The estimate (6) is best possible with equality for P(z)= z n +1. If we divide both sides of (6) by |α| and make |α|, we get inequality (3).

As an improvement and generalization to (6) and (4), Liman et al. [6] recently proved the following theorem.

Theorem 1 If P(z) is a polynomial of degree n that does not vanish in |z|<1, then for every complex number α, β with |α|1, |β|1 and |z|=1,

| z D α P ( z ) + n β | α | 1 2 P ( z ) | n 2 { ( | α + β | α | 1 2 | + | z + β | α | 1 2 | ) max | z | = 1 | P ( z ) | ( | α + β | α | 1 2 | | z + β | α | 1 2 | ) min | z | = 1 | P ( z ) | } .
(7)

In this paper, we prove the following more general result which is an extension as well as generalization of Theorem 1 and yields a number of known polynomial inequalities.

Theorem 2 Let P(z) be a polynomial of degree n that does not vanish in |z|<k, k1, then for all real or complex numbers α i with | α i |k, k1, i=1,2,,t, tn1 and for any real or complex number β with |β|1 and for |z|=1,

(8)

Remark Theorem 1 is a special case of Theorem 2 when we take t=k=1.

If we take t=1 in Theorem 2, we get the following corollary.

Corollary 1 If P(z) is a polynomial of degree n that does not vanish in |z|<k, k1, then for all complex numbers α, β with |α|k, k1, |β|1, and for |z|=1,

| z D α P ( z ) + n β | α | k 1 + k P ( z ) | n 2 { ( 1 k n | α + β | α | k 1 + k | + | z + β | α | k 1 + k | ) max | z | = 1 | P ( z ) | ( 1 k n | α + β | α | k 1 + k | | z + β | α | k 1 + k | ) min | z | = k | P ( z ) | } .

If we take k=1 in Theorem 2, we get the following result.

Corollary 2 Let P(z) be a polynomial of degree n that does not vanish in |z|1, then for all real or complex numbers α i with | α i |1, i=1,2,,t, tn1 and for any real or complex number β with |β|1 and for |z|=1,

(9)

For β=0 and t=1 in Theorem 2, we get the following.

Corollary 3 Let P(z) be a polynomial of degree n that does not vanish in |z|<k, k1, then for any real or complex number α with |α|k, k1,

max | z | = 1 | D α P ( z ) | n 2 { ( | α | k n + 1 ) max | z | = 1 | P ( z ) | ( | α | k n 1 ) min | z | = k | P ( z ) | } .
(10)

If we take k=1 in Corollary 3, then (10) reduces to (6).

By taking t=1 in (8), dividing both sides of (8) by |α| and letting |α|, we have the following generalization of inequality (4).

Corollary 4 Let P(z) be a polynomial of degree n that does not vanish in |z|<k, k1, then for any real or complex number β with |β|1and |z|=1,

| z P ( z ) + n β 1 + k P ( z ) | n 2 { ( 1 k n | 1 + β 1 + k | + | β 1 + k | ) max | z | = 1 | P ( z ) | ( 1 k n | 1 + β 1 + k | | β 1 + k | ) min | z | = k | P ( z ) | } .
(11)

Taking β=0 and k=1 in Corollary 4, (11) reduces to (3).

2 Lemmas

We require the following lemmas. The first lemma follows from Laguerre’s theorem [[7], p.52] (see also [8]).

Lemma 1 If all the zeros of the nth degree polynomial P(z) lie in a circular region C, and if ξ is any zero of

D α P(z)=nP(z)+(αz) P (z),

the polar derivative of P(z), then both points ξ and α may not lie outside of C.

By repeated applications of Lemma 1, we get the following result, when the circular region C is the circle |z|r.

Lemma 2 If all the zeros of the nth degree polynomial P(z) lie in |z|r and if none of the points α 1 , α 2 ,, α t lie in |z|r, then each of the polar derivatives D α t D α 2 D α 1 P(z), t=1,2,,n1, has all its zeros in |z|r.

Lemma 3 If P(z):= a 0 + j = μ n a j z j , 1μn, is a polynomial of degree n having no zeros in the disk |z|<k, k1, then

k μ | P ( z ) | | Q ( z ) | for |z|=1,

where Q(z)= z n P ( 1 z ¯ ) ¯ and μ n | a μ a 0 | k μ 1.

The above lemma is due to Chan and Malik [9].

Lemma 4 If P(z):= a n z n + j = μ n a n j z n j is a polynomial of degree n having all its zeros in the disk |z|k1, then

| Q ( z ) | k μ | P ( z ) | for |z|=1,1μn,
(12)

where Q(z)= z n P ( 1 z ¯ ) ¯ .

Proof of Lemma 4 Since all the zeros of P(z) lie in |z|k1, therefore all the zeros of Q(z)= z n P ( 1 z ¯ ) ¯ lie in |z| 1 k 1. Hence applying Lemma 3 to the polynomial Q(z):= a ¯ n + j = μ n a ¯ n j z j , we get

1 K μ | Q ( z ) | | P ( z ) | .

Hence, inequality (12) follows. □

Lemma 5 Let P(z)= a n z n + Σ j = μ n a n j z n j , 1μn, be a polynomial of degree n having all its zeros in the disk |z|k, k1, then for every real or complex number α with |α|k, k1 and for |z|=1,

| D α P ( z ) | n ( | α | k μ 1 + k μ ) | P ( z ) | .

Proof of Lemma 5 Let Q(z)= z n P ( 1 z ¯ ) ¯ , we have P(z)= z n Q ( 1 z ¯ ) ¯ . Then it can be easily verified that

| Q ( z ) | = | n P ( z ) z P ( z ) | for |z|=1.
(13)

Since P(z) has all its zeros in |z|k1, by Lemma 4, we get

| Q ( z ) | k μ | P ( z ) | for |z|=1.

This implies

| P ( z ) | + | Q ( z ) | ( 1 + k μ ) | P ( z ) | .
(14)

Also, for |z|=1, by using (13), we have

n | P ( z ) | = | n P ( z ) + z P ( z ) z P ( z ) | | n P ( z ) z P ( z ) | + | P ( z ) | = | Q ( z ) | + | P ( z ) | .

Using (14) in the above inequality, we get

n | P ( z ) | ( 1 + k μ ) | P ( z ) | ,

or

| P ( z ) | n ( 1 + k μ ) | P ( z ) | for |z|=1.
(15)

For every real or complex number α with |α|k, k1, we have

| D α P ( z ) | = | n P ( z ) + ( α z ) P ( z ) | .

Now, by using Lemma 4 and (15), we have

| D α P ( z ) | = | n P ( z ) + ( α z ) P ( z ) | = | α P ( z ) + n P ( z ) z P ( z ) | | α | | P ( z ) | | n P ( z ) z P ( z ) | = | α | | P ( z ) | | Q ( z ) | for  | z | = 1 | α | | P ( z ) | k μ | P ( z ) | = ( | α | k μ ) | P ( z ) | n ( | α | k μ 1 + k μ ) | P ( z ) | for  | z | = 1 .

This completes the proof of Lemma 5. □

Lemma 6 Let P(z)= a n z n + j = μ n a n j z n j , 1μn, be a polynomial of degree n having all its zeros in the disk |z|k, k1, then for every real or complex number α i with | α i |k, k1, i=1,2,,t, tn1 and for |z|=1,

(16)

Proof of Lemma 6 If | α i |=k for at least one i, 1it, then inequality (16) is trivial. Thus, we assume that | α i |>k, k1, for all 1it. We proceed by the principle of mathematical induction. The result is true for t=1 by Lemma 5, that is, if | α 1 |>k, then

| D α 1 P ( z ) | n ( | α 1 | k μ 1 + k μ ) | P ( z ) | .
(17)

Now, for t=2 and for | α 1 |>k, D α 1 P(z) will be a polynomial of degree at most n1. Since all the zeros of P(z) in |z|k, k1, therefore, by applying Lemma 1, all the zeros of D α 1 P(z) lie in |z|k, k1, then using Lemma 5 for the polynomial D α 1 P(z) of degree at most n1, and for | α 2 |>k, we have

| D α 2 { D α 1 P ( z ) } | (n1) ( | α 2 | k μ 1 + k μ ) | D α 1 P ( z ) | .
(18)

Combining (17) and (18), we get

| D α 2 D α 1 P ( z ) | n(n1) { ( | α 1 | k μ ) ( | α 2 | k μ ) ( 1 + k μ ) 2 } | P ( z ) | .

So, the result is true for t=2. Now, we assume that the result is true for t=υ<n; that is, for |z|=1,

(19)

We need to show that the result is true for t=υ+1.

Now corresponding to an n th degree polynomial P(z) whose all zeros lie in the disk |z|k, k1, we construct D α υ D α 2 D α 1 P(z) a polynomial of degree at most nυ for all real or complex numbers α i with | α i |k, k1, i=1,2,,υ (υ<n) whose all zeros lie in |z|k. Therefore, for | α υ + 1 |>k, by applying Lemma 5 to D α υ D α 2 D α 1 P(z), we get

(20)

Combining (19) and (20), we obtain

This implies that the result is true for t=υ+1 and this completes the proof of Lemma 6. □

Lemma 7 Let P(z)= a n z n + Σ j = μ n a n j z n j , 1μn, be a polynomial of degree n having all its zeros in the disk |z|k, k1. Then for every real or complex number α i with | α i |k, k1, i=1,2,,t, tn1, and for any real or complex number β with |β|1 and for |z|=1,

(21)

Proof of Lemma 7 The result is clear if P(z) has a zero on |z|=k, then m= min | z | = k |P(z)|=0. We now suppose that all the zeros of P(z) lie in |z|<k, then m>0, and we have m|P(z)| for |z|=k. Hence, for every λ with |λ|<1, we have |P(z)|>|mλ ( z k ) n | for |z|=k. Therefore, it follows by Rouche’s theorem that the polynomial G(z)=P(z)mλ ( z k ) n has all its zeros in |z|<k, k1. As α 1 , α 2 ,, α t are complex numbers with | α i |k, k1, i=1,2,,t, tn1, then by Lemma 2 all the zeros of

lie in |z|<k. Applying Lemma 6 to the polynomial G(z), we get

(22)

Since z t D α t D α 2 D α 1 G(z) has all its zeros in |z|<k, k1, therefore by Rouche’s theorem, it follows from inequality (22) that the polynomial

has all its zeros in |z|<1, where |β|<1.

By substituting for G(z), we conclude that the polynomial

T ( z ) = z t D α t D α 2 D α 1 P ( z ) + n ( n 1 ) ( n t + 1 ) × β ( ( | α 1 | k μ ) ( | α 2 | k μ ) ( | α t | k μ ) ( 1 + k μ ) t ) P ( z ) λ m ( z k ) n n ( n 1 ) ( n t + 1 ) × { α 1 α 2 α t + β ( ( | α 1 | k μ ) ( | α 2 | k μ ) ( | α t | k μ ) ( 1 + k μ ) t ) }
(23)

will have no zeros in |z|1. This implies that for every β with |β|<1 and |z|1,

(24)

If (24) is not true, then there exists a point z=ω with |ω|1 such that

(25)

We take

λ= ω t D α t D α 2 D α 1 P ( ω ) + n ( n 1 ) ( n t + 1 ) β { ( | α 1 | k μ ) ( | α 2 | k μ ) ( | α t | k μ ) ( 1 + k μ ) t } P ( ω ) n ( n 1 ) ( n t + 1 ) m ( ω k ) n ( α 1 α 2 α t + β { ( | α 1 | k μ ) ( | α 2 | k μ ) ( | α t | k μ ) ( 1 + k μ ) t } )

so that |λ|<1, and with this choice of λ, we have T(ω)=0 for |ω|1 from (23). But this contradicts the fact that T(z)0 for |z|1. For β with |β|=1, (24) follows by continuity. This completes the proof of Lemma 7. □

Lemma 8 Let P(z)= a n z n + Σ j = μ n a n j z n j , 1μn, be a polynomial of degree n. Then for all real or complex numbers α i with | α i |k, k1, i=1,2,,t, tn1, and for any real or complex number β with |β|1 and for |z|=1,

Proof of Lemma 8 Let M= max | z | = k |P(z)|, if |λ|<1, then |λP(z)|<|M ( z k ) n | for |z|=k. Therefore, it follows by Rouche’s theorem that the polynomial G(z)=M ( z k ) n λP(z) has all its zeros in |z|<k, k1. As α 1 , α 2 ,, α t are complex numbers with | α i |k, k1, i=1,2,,t, tn1, then by Lemma 2 all the zeros of

D α t D α 2 D α 1 G(z)= M k n { n ( n 1 ) ( n t + 1 ) α 1 α 2 α t } z n t λ D α t D α 2 D α 1 P(z)

lie in |z|<k. Applying Lemma 6 to the polynomial G(z), we have

 □

Now, if we proceed similarly as in Lemma 7, the result follows.

Lemma 9 Let P(z) be a polynomial of degree n, then for all real or complex numbers α i with | α i |k, i=1,2,,t, and for any real or complex number β with |β|1,

where

Q(z):= ( z k ) n P ( k 2 z ¯ ) ¯ .

Proof of Lemma 9 Let M= max | z | = k |P(z)|, then |P(z)|M for |z|k. If λ is any real or complex number with |λ|>1, then by Rouche’s theorem the polynomial G(z)=P(z)λM does not vanish in |z|<k. Consequently, the polynomial

H(z):= ( z k ) n G ( k 2 z ¯ ) ¯

has all zeros in |z|k and |G(z)|=|H(z)| for |z|=k. Since all the zeros of H(z) lie in |z|k, k1, therefore, for δ with |δ|>1, by Rouche’s theorem, all the zeros of G(z)+δH(z) lie in |z|k. Hence, by taking μ=1 in Lemma 6 and using it, for every real or complex number α i with | α i |k, i=1,2,,t (1t<n), k1, and |z|=1, we have

Also, by Lemma 2, all the zeros D α t D α 2 D α 1 (G(z)+δH(z)) lie in |z|<k1, where | α i |k, i=1,2,,t (1t<n). Therefore, for any β with |β|<1, Rouche’s theorem implies that all the zeros of

lie in |z|<1.

So, the polynomial

T ( z ) = z t D α t D α 2 D α 1 G ( z ) + β n ( n 1 ) ( n t + 1 ) × { ( | α 1 | k ) ( | α 2 | k ) ( | α t | k ) ( 1 + k ) t } G ( z ) + δ ( z t D α t D α 2 D α 1 H ( z ) + β n ( n 1 ) ( n t + 1 ) × { ( | α 1 | k ) ( | α 2 | k ) ( | α t | k ) ( 1 + k ) t } H ( z ) )

will have no zeros in |z|1. Now, using a similar argument as that in the proof of Lemma 7 and taking μ=1, we get for |z|1,

(26)

Now,

H(z):= ( z k ) n G ( k 2 z ¯ ) ¯ = ( z k ) n P ( k 2 z ¯ ) ¯ λ ¯ ( z k ) n M=Q(z) λ ¯ ( z k ) n M.

On substituting G(z) and H(z) in (26), we obtain the following:

This implies that

(27)

As |P(z)|=|Q(z)| for |z|=k, that is, M= max | z | = k |P(z)|= max | z | = k |Q(z)|, by taking μ=1 in Lemma 8 and using it to the polynomial Q(z), we obtain the following:

Thus, taking the argument of λ suitably, we obtain

(28)

Using (28) in (27), we get for |z|=1 and |β|<1,

That is,

Taking |λ|1, we get

(29)

Then, by applying the maximum modulus principle for the polynomial P(z) when |k|1,

max | z | = k | P ( z ) | max | z | = 1 | P ( z ) | .

This in conjunction with (29) and the argument of continuity gives the result. □

Lemma 10 Let H(z) be a polynomial of degree n having all its zeros in |z|k, k1, and let G(z) be a polynomial of degree not exceeding that of H(z). If |G(z)||H(z)| for |z|=k, k1, then for all real or complex numbers α i with | α i |k, i=1,2,,t, and for any real or complex number β with |β|1, and |z|=1, we have

Proof of Lemma 10 For any real or complex number λ with |λ|<1, we have |λG(z)|<|G(z)||H(z)| for |z|=k, k1; therefore, by Rouche’s theorem H(z)λG(z) and H(z) have the same number of zeros in |z|<k. Also, as |G(z)||H(z)| for |z|=k, k1, any zero of H(z) that lies on |z|=k is also a zero of G(z). Therefore, H(z)λG(z) has all its zeros in the closed disk |z|k. Therefore, using Lemma 6 with μ=1, we have for every real or complex number α i with | α i |k, i=1,2,,t, k1 (1t<n) and |z|=1,

Now, by a similar argument as that used in the proof of Lemma 7, for any real or complex number β with |β|<1, we get

for |z|=1, which implies

T ( z ) = z t D α t D α 2 D α 1 H ( z ) λ z t D α t D α 2 D α 1 G ( z ) + n ( n 1 ) ( n t + 1 ) β { ( | α 1 | k ) ( | α 2 | k ) ( | α t | k ) ( 1 + k ) t } × ( H ( z ) λ G ( z ) ) 0

for |z|=1.

That is,

T ( z ) = z t D α t D α 2 D α 1 H ( z ) + n ( n 1 ) ( n t + 1 ) × β { ( | α 1 | k ) ( | α 2 | k ) ( | α t | k ) ( 1 + k ) t } H ( z ) λ [ z t D α t D α 2 D α 1 G ( z ) + n ( n 1 ) ( n t + 1 ) × β { ( | α 1 | k ) ( | α 2 | k ) ( | α t | k ) ( 1 + k ) t } G ( z ) ] 0
(30)

for |z|=1.

So, we conclude that

(31)

for |z|=1.

If (31) is not true, then there exists a point z=ω with |ω|=1 such that

Take

λ= ω t D α t D α 2 D α 1 H ( ω ) + n ( n 1 ) ( n t + 1 ) β { ( | α 1 | k ) ( | α 2 | k ) ( | α t | k ) ( 1 + k ) t } H ( ω ) ω t D α t D α 2 D α 1 G ( ω ) + n ( n 1 ) ( n t + 1 ) β { ( | α 1 | k ) ( | α 2 | k ) ( | α t | k ) ( 1 + k ) t } G ( ω ) ,

then |λ|<1, and with this choice of λ, we have from (30) T(ω)=0 for |ω|=1. But this contradicts the fact that T(z)0 for |z|=1. For β with |β|=1, (31) follows by continuity. This completes the proof of Lemma 10. □

3 Proof of the theorem

Since the polynomial P(z) has no zeros in the disk |z|<k, and therefore, if m= min | z | = k |P(z)|, then m|P(z)| for |z|k. If λ is any real or complex number with |λ|<1, we have

|λm|<m | P ( z ) | for |z|=k.

Thus, by Rouche’s theorem, the polynomial G(z)=P(z)λm does not vanish in |z|<k. Consequently, the polynomial

H(z):= ( z k ) n G ( k 2 z ¯ ) ¯ =Q(z) λ ¯ m ( z k ) n

has all its zeros in |z|k, where Q(z)= ( z k ) n P ( k 2 z ¯ ) ¯ , and also we have |G(z)|=|H(z)| for |z|=k.

Applying Lemma 10 to the polynomials H(z) and G(z), we have

for every real or complex number α i with | α i |k, i=1,2,,t, and for any real or complex number β with |β|1 and |z|=1.

On substituting G(z) and H(z) in the above inequality, we obtain the following for every real or complex number α i with | α i |k, i=1,2,,t, and for any real or complex number β with |β|1 and |z|=1,

(32)

Since all the zeros of Q(z) lie in |z|k and |P(z)|=|Q(z)| for |z|=k, therefore, by applying Lemma 7 to Q(z), we have

Then, for an appropriate choice of the argument of λ, we get

(33)

for |z|=1.

Then combining the right-hand side of (32) and (33), we rewrite (32) as

Equivalently,

As |λ|1, we have

It implies that for every real or complex number β with |β|1 and |z|=1,

This in conjunction with Lemma 9 gives, for |β|1 and |z|=1,

This completes the proof of the theorem.

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GS and AL studied the related literature under the supervision of WMS and jointly developed the idea and drafted the manuscript. GS made the text file in Latex and communicated the manuscript. GS also revised it as per the directions of the referee under the guidance of WMS. All three authors read and approved the final manuscript.

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Singh, G., Shah, W.M. & Liman, A. A generalized inequality for the polar derivative of a polynomial. J Inequal Appl 2013, 183 (2013). https://doi.org/10.1186/1029-242X-2013-183

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Keywords

  • polynomial
  • inequality
  • zeros
  • polar derivative