Checking for real roots of an even-th degree polynomial $ (a-x)^2n+1+x^2n+1-b=0 $ The Next CEO of Stack OverflowFind roots of polynomial with degree $ge 5$Given roots (real and complex), find the polynomialRules for whether an $n$ degree polynomial is an $n$ degree powerNumber of real roots of polynomial derivativePolynomial with odd number of real rootsLow degree polynomials over finite fields with no rootsWhat is the maximum number of real roots a polynomial of any degree can have?Finding the real roots and complex roots to a polynomialGet all roots of a 6th degree polynom numericalFind the number of distinct real roots of a polynomial
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Checking for real roots of an even-th degree polynomial $ (a-x)^2n+1+x^2n+1-b=0 $
The Next CEO of Stack OverflowFind roots of polynomial with degree $ge 5$Given roots (real and complex), find the polynomialRules for whether an $n$ degree polynomial is an $n$ degree powerNumber of real roots of polynomial derivativePolynomial with odd number of real rootsLow degree polynomials over finite fields with no rootsWhat is the maximum number of real roots a polynomial of any degree can have?Finding the real roots and complex roots to a polynomialGet all roots of a 6th degree polynom numericalFind the number of distinct real roots of a polynomial
$begingroup$
How should one check for real solutions of an even-th degree polynomial similar to this $ (a-x)^2n+1+x^2n+1-b=0 $ ? a, b and n are constant non-zero natural numbers
roots
asked Mar 19 at 19:40
Matei RaduMatei Radu
82
$endgroup$
add a comment |
$begingroup$
How should one check for real solutions of an even-th degree polynomial similar to this $ (a-x)^2n+1+x^2n+1-b=0 $ ? a, b and n are constant non-zero natural numbers
roots
asked Mar 19 at 19:40
Matei RaduMatei Radu
82
$endgroup$
add a comment |
$begingroup$
How should one check for real solutions of an even-th degree polynomial similar to this $ (a-x)^2n+1+x^2n+1-b=0 $ ? a, b and n are constant non-zero natural numbers
roots
asked Mar 19 at 19:40
Matei RaduMatei Radu
82
$endgroup$
How should one check for real solutions of an even-th degree polynomial similar to this $ (a-x)^2n+1+x^2n+1-b=0 $ ? a, b and n are constant non-zero natural numbers
roots
roots
asked Mar 19 at 19:40
Matei RaduMatei Radu
82
asked Mar 19 at 19:40
Matei RaduMatei Radu
82
edited Mar 19 at 19:46
Matei Radu
asked Mar 19 at 19:40
Matei RaduMatei Radu
82
asked Mar 19 at 19:40
Matei RaduMatei Radu
82
asked Mar 19 at 19:40
Matei RaduMatei Radu
82
82
add a comment |
add a comment |
1 Answer
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In general, the number of distinct real roots of a polynomial can be computed using Sturm's theorem.
In your case, if $P(x) = (a - x)^2n+1 + x^2n+1 - b$, it is not hard to show that $P$ is convex and symmetric around $x=a/2$. Thus we have one real root if $b = 2 (a/2)^2n+1$, none if $b < 2 (a/2)^2n+1$, two if $b > 2 (a/2)^2n+1$.
answered Mar 19 at 20:10
Robert IsraelRobert Israel
330k23218473
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1 Answer
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1 Answer
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active
oldest
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active
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active
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$begingroup$
In general, the number of distinct real roots of a polynomial can be computed using Sturm's theorem.
In your case, if $P(x) = (a - x)^2n+1 + x^2n+1 - b$, it is not hard to show that $P$ is convex and symmetric around $x=a/2$. Thus we have one real root if $b = 2 (a/2)^2n+1$, none if $b < 2 (a/2)^2n+1$, two if $b > 2 (a/2)^2n+1$.
answered Mar 19 at 20:10
Robert IsraelRobert Israel
330k23218473
$endgroup$
add a comment |
$begingroup$
In general, the number of distinct real roots of a polynomial can be computed using Sturm's theorem.
In your case, if $P(x) = (a - x)^2n+1 + x^2n+1 - b$, it is not hard to show that $P$ is convex and symmetric around $x=a/2$. Thus we have one real root if $b = 2 (a/2)^2n+1$, none if $b < 2 (a/2)^2n+1$, two if $b > 2 (a/2)^2n+1$.
answered Mar 19 at 20:10
Robert IsraelRobert Israel
330k23218473
$endgroup$
add a comment |
$begingroup$
In general, the number of distinct real roots of a polynomial can be computed using Sturm's theorem.
In your case, if $P(x) = (a - x)^2n+1 + x^2n+1 - b$, it is not hard to show that $P$ is convex and symmetric around $x=a/2$. Thus we have one real root if $b = 2 (a/2)^2n+1$, none if $b < 2 (a/2)^2n+1$, two if $b > 2 (a/2)^2n+1$.
answered Mar 19 at 20:10
Robert IsraelRobert Israel
330k23218473
$endgroup$
In general, the number of distinct real roots of a polynomial can be computed using Sturm's theorem.
In your case, if $P(x) = (a - x)^2n+1 + x^2n+1 - b$, it is not hard to show that $P$ is convex and symmetric around $x=a/2$. Thus we have one real root if $b = 2 (a/2)^2n+1$, none if $b < 2 (a/2)^2n+1$, two if $b > 2 (a/2)^2n+1$.
answered Mar 19 at 20:10
Robert IsraelRobert Israel
330k23218473
answered Mar 19 at 20:10
Robert IsraelRobert Israel
330k23218473
answered Mar 19 at 20:10
Robert IsraelRobert Israel
330k23218473
answered Mar 19 at 20:10
Robert IsraelRobert Israel
330k23218473
330k23218473
add a comment |
add a comment |
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