How to write the shifted Chebyshev polynomials (the first kind) in mathematica?












2












$begingroup$


Is it possible to write the shifted Chebyshev polynomials (the first kind) in Mathematica? The formula is:



$$P_n(x)=sum_{k=0}^{leftlfloortfrac{n}{2}rightrfloor}(-1)^k 2^{n-2k-1}frac{n}{n-k}binom{n-k}{k}(2x-1)^{n-2k}$$



The output should be in a vector {..., ..., ..., ... etc.}
Thanks!!






special-functions polynomials







share|improve this question











$endgroup$












  • $begingroup$
    By "vector", do you mean a vector of the first few polynomials?
    $endgroup$
    – J. M. is away♦
    Apr 1 at 8:36










  • $begingroup$
    yes please depend on n
    $endgroup$
    – user62716
    Apr 1 at 8:38
















2












$begingroup$


Is it possible to write the shifted Chebyshev polynomials (the first kind) in Mathematica? The formula is:



$$P_n(x)=sum_{k=0}^{leftlfloortfrac{n}{2}rightrfloor}(-1)^k 2^{n-2k-1}frac{n}{n-k}binom{n-k}{k}(2x-1)^{n-2k}$$



The output should be in a vector {..., ..., ..., ... etc.}
Thanks!!






special-functions polynomials







share|improve this question











$endgroup$












  • $begingroup$
    By "vector", do you mean a vector of the first few polynomials?
    $endgroup$
    – J. M. is away♦
    Apr 1 at 8:36










  • $begingroup$
    yes please depend on n
    $endgroup$
    – user62716
    Apr 1 at 8:38














2












2








2





$begingroup$


Is it possible to write the shifted Chebyshev polynomials (the first kind) in Mathematica? The formula is:



$$P_n(x)=sum_{k=0}^{leftlfloortfrac{n}{2}rightrfloor}(-1)^k 2^{n-2k-1}frac{n}{n-k}binom{n-k}{k}(2x-1)^{n-2k}$$



The output should be in a vector {..., ..., ..., ... etc.}
Thanks!!






special-functions polynomials







share|improve this question











$endgroup$




Is it possible to write the shifted Chebyshev polynomials (the first kind) in Mathematica? The formula is:



$$P_n(x)=sum_{k=0}^{leftlfloortfrac{n}{2}rightrfloor}(-1)^k 2^{n-2k-1}frac{n}{n-k}binom{n-k}{k}(2x-1)^{n-2k}$$



The output should be in a vector {..., ..., ..., ... etc.}
Thanks!!






special-functions polynomials




special-functions polynomials






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Apr 1 at 8:44









J. M. is away♦

98.9k10311467




98.9k10311467










asked Apr 1 at 8:23









user62716user62716

555




555












  • $begingroup$
    By "vector", do you mean a vector of the first few polynomials?
    $endgroup$
    – J. M. is away♦
    Apr 1 at 8:36










  • $begingroup$
    yes please depend on n
    $endgroup$
    – user62716
    Apr 1 at 8:38


















  • $begingroup$
    By "vector", do you mean a vector of the first few polynomials?
    $endgroup$
    – J. M. is away♦
    Apr 1 at 8:36










  • $begingroup$
    yes please depend on n
    $endgroup$
    – user62716
    Apr 1 at 8:38
















$begingroup$
By "vector", do you mean a vector of the first few polynomials?
$endgroup$
– J. M. is away♦
Apr 1 at 8:36




$begingroup$
By "vector", do you mean a vector of the first few polynomials?
$endgroup$
– J. M. is away♦
Apr 1 at 8:36












$begingroup$
yes please depend on n
$endgroup$
– user62716
Apr 1 at 8:38




$begingroup$
yes please depend on n
$endgroup$
– user62716
Apr 1 at 8:38










1 Answer
1






active

oldest

votes


















5












$begingroup$

With[{m = 100}, 

    ChebyshevT[Range[m], 2 x - 1] == 

    Table[Sum[(-1)^k 2^(n - 2 k - 1) Binomial[n - k, k] (2 x - 1)^(n - 2 k) n/(n - k),

              {k, 0, Quotient[n, 2]}], {n, m}] // Expand]

   True


Comparing your formula with formula 22.3.6 in Abramowitz and Stegun, we find that all you need to do is



ChebyshevT[Range[0, m - 1], 2 x - 1]


if you need an entire pile of these shifted polynomials.






share|improve this answer











$endgroup$













  • $begingroup$
    Many thanks, put the output just word true not terms? I need few terms in vector depend on n, what about [n/2] is it greatest integer number?
    $endgroup$
    – user62716
    Apr 1 at 9:00










  • $begingroup$
    The first snippet just shows that the simple expression is the same as your explicit sum. The second snippet is what you want for evaluation. Assign some positive integer to m and evaluate ChebyshevT[Range[0, m - 1], 2 x - 1] (e.g. ChebyshevT[Range[0, 100 - 1], 2 x - 1] to get a hundred of them).
    $endgroup$
    – J. M. is away♦
    Apr 1 at 9:01












  • $begingroup$
    Dear, 1) You mean I have to use ChebyshevT[Range[0, m - 1], 2 x - 1].?2) the first code With[{m = 100}, ChebyshevT[Range[m], 2 x - 1] == Table[Sum[(-1)^k 2^(n - 2 k - 1) Binomial[n - k, k] (2 x - 1)^(n - 2 k) n/(n - k), {k, 0, Quotient[n, 2]}], {n, m}] // Expand] for what then?3) [n/2] is it greatest integer number?
    $endgroup$
    – user62716
    Apr 1 at 9:09












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

With[{m = 100}, 

    ChebyshevT[Range[m], 2 x - 1] == 

    Table[Sum[(-1)^k 2^(n - 2 k - 1) Binomial[n - k, k] (2 x - 1)^(n - 2 k) n/(n - k),

              {k, 0, Quotient[n, 2]}], {n, m}] // Expand]

   True


Comparing your formula with formula 22.3.6 in Abramowitz and Stegun, we find that all you need to do is



ChebyshevT[Range[0, m - 1], 2 x - 1]


if you need an entire pile of these shifted polynomials.






share|improve this answer











$endgroup$













  • $begingroup$
    Many thanks, put the output just word true not terms? I need few terms in vector depend on n, what about [n/2] is it greatest integer number?
    $endgroup$
    – user62716
    Apr 1 at 9:00










  • $begingroup$
    The first snippet just shows that the simple expression is the same as your explicit sum. The second snippet is what you want for evaluation. Assign some positive integer to m and evaluate ChebyshevT[Range[0, m - 1], 2 x - 1] (e.g. ChebyshevT[Range[0, 100 - 1], 2 x - 1] to get a hundred of them).
    $endgroup$
    – J. M. is away♦
    Apr 1 at 9:01












  • $begingroup$
    Dear, 1) You mean I have to use ChebyshevT[Range[0, m - 1], 2 x - 1].?2) the first code With[{m = 100}, ChebyshevT[Range[m], 2 x - 1] == Table[Sum[(-1)^k 2^(n - 2 k - 1) Binomial[n - k, k] (2 x - 1)^(n - 2 k) n/(n - k), {k, 0, Quotient[n, 2]}], {n, m}] // Expand] for what then?3) [n/2] is it greatest integer number?
    $endgroup$
    – user62716
    Apr 1 at 9:09
















5












$begingroup$

With[{m = 100}, 

    ChebyshevT[Range[m], 2 x - 1] == 

    Table[Sum[(-1)^k 2^(n - 2 k - 1) Binomial[n - k, k] (2 x - 1)^(n - 2 k) n/(n - k),

              {k, 0, Quotient[n, 2]}], {n, m}] // Expand]

   True


Comparing your formula with formula 22.3.6 in Abramowitz and Stegun, we find that all you need to do is



ChebyshevT[Range[0, m - 1], 2 x - 1]


if you need an entire pile of these shifted polynomials.






share|improve this answer











$endgroup$













  • $begingroup$
    Many thanks, put the output just word true not terms? I need few terms in vector depend on n, what about [n/2] is it greatest integer number?
    $endgroup$
    – user62716
    Apr 1 at 9:00










  • $begingroup$
    The first snippet just shows that the simple expression is the same as your explicit sum. The second snippet is what you want for evaluation. Assign some positive integer to m and evaluate ChebyshevT[Range[0, m - 1], 2 x - 1] (e.g. ChebyshevT[Range[0, 100 - 1], 2 x - 1] to get a hundred of them).
    $endgroup$
    – J. M. is away♦
    Apr 1 at 9:01












  • $begingroup$
    Dear, 1) You mean I have to use ChebyshevT[Range[0, m - 1], 2 x - 1].?2) the first code With[{m = 100}, ChebyshevT[Range[m], 2 x - 1] == Table[Sum[(-1)^k 2^(n - 2 k - 1) Binomial[n - k, k] (2 x - 1)^(n - 2 k) n/(n - k), {k, 0, Quotient[n, 2]}], {n, m}] // Expand] for what then?3) [n/2] is it greatest integer number?
    $endgroup$
    – user62716
    Apr 1 at 9:09














5












5








5





$begingroup$

With[{m = 100}, 

    ChebyshevT[Range[m], 2 x - 1] == 

    Table[Sum[(-1)^k 2^(n - 2 k - 1) Binomial[n - k, k] (2 x - 1)^(n - 2 k) n/(n - k),

              {k, 0, Quotient[n, 2]}], {n, m}] // Expand]

   True


Comparing your formula with formula 22.3.6 in Abramowitz and Stegun, we find that all you need to do is



ChebyshevT[Range[0, m - 1], 2 x - 1]


if you need an entire pile of these shifted polynomials.






share|improve this answer











$endgroup$



With[{m = 100}, 

    ChebyshevT[Range[m], 2 x - 1] == 

    Table[Sum[(-1)^k 2^(n - 2 k - 1) Binomial[n - k, k] (2 x - 1)^(n - 2 k) n/(n - k),

              {k, 0, Quotient[n, 2]}], {n, m}] // Expand]

   True


Comparing your formula with formula 22.3.6 in Abramowitz and Stegun, we find that all you need to do is



ChebyshevT[Range[0, m - 1], 2 x - 1]


if you need an entire pile of these shifted polynomials.







share|improve this answer














share|improve this answer



share|improve this answer








answered Apr 1 at 8:44


























community wiki





J. M. is away













  • $begingroup$
    Many thanks, put the output just word true not terms? I need few terms in vector depend on n, what about [n/2] is it greatest integer number?
    $endgroup$
    – user62716
    Apr 1 at 9:00










  • $begingroup$
    The first snippet just shows that the simple expression is the same as your explicit sum. The second snippet is what you want for evaluation. Assign some positive integer to m and evaluate ChebyshevT[Range[0, m - 1], 2 x - 1] (e.g. ChebyshevT[Range[0, 100 - 1], 2 x - 1] to get a hundred of them).
    $endgroup$
    – J. M. is away♦
    Apr 1 at 9:01












  • $begingroup$
    Dear, 1) You mean I have to use ChebyshevT[Range[0, m - 1], 2 x - 1].?2) the first code With[{m = 100}, ChebyshevT[Range[m], 2 x - 1] == Table[Sum[(-1)^k 2^(n - 2 k - 1) Binomial[n - k, k] (2 x - 1)^(n - 2 k) n/(n - k), {k, 0, Quotient[n, 2]}], {n, m}] // Expand] for what then?3) [n/2] is it greatest integer number?
    $endgroup$
    – user62716
    Apr 1 at 9:09


















  • $begingroup$
    Many thanks, put the output just word true not terms? I need few terms in vector depend on n, what about [n/2] is it greatest integer number?
    $endgroup$
    – user62716
    Apr 1 at 9:00










  • $begingroup$
    The first snippet just shows that the simple expression is the same as your explicit sum. The second snippet is what you want for evaluation. Assign some positive integer to m and evaluate ChebyshevT[Range[0, m - 1], 2 x - 1] (e.g. ChebyshevT[Range[0, 100 - 1], 2 x - 1] to get a hundred of them).
    $endgroup$
    – J. M. is away♦
    Apr 1 at 9:01












  • $begingroup$
    Dear, 1) You mean I have to use ChebyshevT[Range[0, m - 1], 2 x - 1].?2) the first code With[{m = 100}, ChebyshevT[Range[m], 2 x - 1] == Table[Sum[(-1)^k 2^(n - 2 k - 1) Binomial[n - k, k] (2 x - 1)^(n - 2 k) n/(n - k), {k, 0, Quotient[n, 2]}], {n, m}] // Expand] for what then?3) [n/2] is it greatest integer number?
    $endgroup$
    – user62716
    Apr 1 at 9:09
















$begingroup$
Many thanks, put the output just word true not terms? I need few terms in vector depend on n, what about [n/2] is it greatest integer number?
$endgroup$
– user62716
Apr 1 at 9:00




$begingroup$
Many thanks, put the output just word true not terms? I need few terms in vector depend on n, what about [n/2] is it greatest integer number?
$endgroup$
– user62716
Apr 1 at 9:00












$begingroup$
The first snippet just shows that the simple expression is the same as your explicit sum. The second snippet is what you want for evaluation. Assign some positive integer to m and evaluate ChebyshevT[Range[0, m - 1], 2 x - 1] (e.g. ChebyshevT[Range[0, 100 - 1], 2 x - 1] to get a hundred of them).
$endgroup$
– J. M. is away♦
Apr 1 at 9:01






$begingroup$
The first snippet just shows that the simple expression is the same as your explicit sum. The second snippet is what you want for evaluation. Assign some positive integer to m and evaluate ChebyshevT[Range[0, m - 1], 2 x - 1] (e.g. ChebyshevT[Range[0, 100 - 1], 2 x - 1] to get a hundred of them).
$endgroup$
– J. M. is away♦
Apr 1 at 9:01














$begingroup$
Dear, 1) You mean I have to use ChebyshevT[Range[0, m - 1], 2 x - 1].?2) the first code With[{m = 100}, ChebyshevT[Range[m], 2 x - 1] == Table[Sum[(-1)^k 2^(n - 2 k - 1) Binomial[n - k, k] (2 x - 1)^(n - 2 k) n/(n - k), {k, 0, Quotient[n, 2]}], {n, m}] // Expand] for what then?3) [n/2] is it greatest integer number?
$endgroup$
– user62716
Apr 1 at 9:09




$begingroup$
Dear, 1) You mean I have to use ChebyshevT[Range[0, m - 1], 2 x - 1].?2) the first code With[{m = 100}, ChebyshevT[Range[m], 2 x - 1] == Table[Sum[(-1)^k 2^(n - 2 k - 1) Binomial[n - k, k] (2 x - 1)^(n - 2 k) n/(n - k), {k, 0, Quotient[n, 2]}], {n, m}] // Expand] for what then?3) [n/2] is it greatest integer number?
$endgroup$
– user62716
Apr 1 at 9:09


















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