How to change interval [-2i, 2i] to [ai, bi] for Lucas polynomial The Next CEO of Stack OverflowSolution of a polynomial in interval $(0,1)$A polynomial identityHow to prove this polynomial expression.Polynomial identityInterval of Polynomial Root FindingExplanation of the last step in this proof of Lucas' theoremInterval of $x$ for a $7^textth$ degree polynomialChange of Basis and coordinate vector for polynomial spaces.Continuity of polynomial of $k$ variablesHow to show that power sums determines $x_1,…,x_n$
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How to change interval [-2i, 2i] to [ai, bi] for Lucas polynomial
The Next CEO of Stack OverflowSolution of a polynomial in interval $(0,1)$A polynomial identityHow to prove this polynomial expression.Polynomial identityInterval of Polynomial Root FindingExplanation of the last step in this proof of Lucas' theoremInterval of $x$ for a $7^textth$ degree polynomialChange of Basis and coordinate vector for polynomial spaces.Continuity of polynomial of $k$ variablesHow to show that power sums determines $x_1,…,x_n$
$begingroup$
On the interval [-2i, 2i], Lucas nodes are defined as
$$x_k = 2icos(2j+1)pi/2s$$, where $j=0,1,2,dots,s-1$.
Now, how we can transform Lucas nodes from [-2i,2i] to [ai,bi].
polynomials
edited Mar 28 at 12:56
Thies Heidecke
1186
$endgroup$
migrated from mathematica.stackexchange.com Mar 28 at 9:43
This question came from our site for users of Wolfram Mathematica.
add a comment |
$begingroup$
On the interval [-2i, 2i], Lucas nodes are defined as
$$x_k = 2icos(2j+1)pi/2s$$, where $j=0,1,2,dots,s-1$.
Now, how we can transform Lucas nodes from [-2i,2i] to [ai,bi].
polynomials
edited Mar 28 at 12:56
Thies Heidecke
1186
$endgroup$
migrated from mathematica.stackexchange.com Mar 28 at 9:43
This question came from our site for users of Wolfram Mathematica.
add a comment |
$begingroup$
On the interval [-2i, 2i], Lucas nodes are defined as
$$x_k = 2icos(2j+1)pi/2s$$, where $j=0,1,2,dots,s-1$.
Now, how we can transform Lucas nodes from [-2i,2i] to [ai,bi].
polynomials
edited Mar 28 at 12:56
Thies Heidecke
1186
$endgroup$
On the interval [-2i, 2i], Lucas nodes are defined as
$$x_k = 2icos(2j+1)pi/2s$$, where $j=0,1,2,dots,s-1$.
Now, how we can transform Lucas nodes from [-2i,2i] to [ai,bi].
polynomials
polynomials
edited Mar 28 at 12:56
Thies Heidecke
1186
edited Mar 28 at 12:56
Thies Heidecke
1186
edited Mar 28 at 12:56
Thies Heidecke
1186
edited Mar 28 at 12:56
Thies Heidecke
1186
edited Mar 28 at 12:56
Thies Heidecke
1186
1186
asked Mar 28 at 9:35
Nia Dutta
migrated from mathematica.stackexchange.com Mar 28 at 9:43
This question came from our site for users of Wolfram Mathematica.
migrated from mathematica.stackexchange.com Mar 28 at 9:43
This question came from our site for users of Wolfram Mathematica.
add a comment |
add a comment |
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