How to write the shifted Chebyshev polynomials (the first kind) in mathematica?Compose two special power series expansionsHow can I plot a Chebyshev spiral?Plotting a partial sumConfusion regarding the incomplete elliptic integral of the first kindExpand power of a polynomialExpansion for Modified Bessel Function Around InfinityEvaluate a precision-sensitive functionHow to deal with the loss of significant digits in this expression with Fresnel integrals?The set of polynomials under the action by a symmetric groupUsing Mathematica to find series expansions for partial derivatives of the generalized Riemann zeta function
Do UK voters know if their MP will be the Speaker of the House?
Is it inappropriate for a student to attend their mentor's dissertation defense?
Why is this clock signal connected to a capacitor to gnd?
How seriously should I take size and weight limits of hand luggage?
What type of content (depth/breadth) is expected for a short presentation for Asst Professor interview in the UK?
Avoiding the "not like other girls" trope?
Can we compute the area of a quadrilateral with one right angle when we only know the lengths of any three sides?
ssTTsSTtRrriinInnnnNNNIiinngg
What is the idiomatic way to say "clothing fits"?
Is "remove commented out code" correct English?
One verb to replace 'be a member of' a club
What are some good books on Machine Learning and AI like Krugman, Wells and Graddy's "Essentials of Economics"
Determining Impedance With An Antenna Analyzer
Is it possible to create a QR code using text?
Why didn't Boeing produce its own regional jet?
What's the in-universe reasoning behind sorcerers needing material components?
CAST throwing error when run in stored procedure but not when run as raw query
What is the most common color to indicate the input-field is disabled?
Unable to supress ligatures in headings which are set in Caps
What do you call someone who asks many questions?
Is there a hemisphere-neutral way of specifying a season?
Is there an expression that means doing something right before you will need it rather than doing it in case you might need it?
How to show a landlord what we have in savings?
I would say: "You are another teacher", but she is a woman and I am a man
How to write the shifted Chebyshev polynomials (the first kind) in mathematica?
Compose two special power series expansionsHow can I plot a Chebyshev spiral?Plotting a partial sumConfusion regarding the incomplete elliptic integral of the first kindExpand power of a polynomialExpansion for Modified Bessel Function Around InfinityEvaluate a precision-sensitive functionHow to deal with the loss of significant digits in this expression with Fresnel integrals?The set of polynomials under the action by a symmetric groupUsing Mathematica to find series expansions for partial derivatives of the generalized Riemann zeta function
$begingroup$
Is it possible to write the shifted Chebyshev polynomials (the first kind) in Mathematica? The formula is:
$$P_n(x)=sum_k=0^leftlfloortfracn2rightrfloor(-1)^k 2^n-2k-1fracnn-kbinomn-kk(2x-1)^n-2k$$
The output should be in a vector ..., ..., ..., ... etc.
Thanks!!
special-functions polynomials
edited 2 days ago
J. M. is slightly pensive♦
99k10311467
asked 2 days ago
user62716user62716
455
$endgroup$
add a comment |
$begingroup$
Is it possible to write the shifted Chebyshev polynomials (the first kind) in Mathematica? The formula is:
$$P_n(x)=sum_k=0^leftlfloortfracn2rightrfloor(-1)^k 2^n-2k-1fracnn-kbinomn-kk(2x-1)^n-2k$$
The output should be in a vector ..., ..., ..., ... etc.
Thanks!!
special-functions polynomials
edited 2 days ago
J. M. is slightly pensive♦
99k10311467
asked 2 days ago
user62716user62716
455
$endgroup$
$begingroup$
By "vector", do you mean a vector of the first few polynomials?
$endgroup$
– J. M. is slightly pensive♦
2 days ago
$begingroup$
yes please depend on n
$endgroup$
– user62716
2 days ago
add a comment |
$begingroup$
Is it possible to write the shifted Chebyshev polynomials (the first kind) in Mathematica? The formula is:
$$P_n(x)=sum_k=0^leftlfloortfracn2rightrfloor(-1)^k 2^n-2k-1fracnn-kbinomn-kk(2x-1)^n-2k$$
The output should be in a vector ..., ..., ..., ... etc.
Thanks!!
special-functions polynomials
edited 2 days ago
J. M. is slightly pensive♦
99k10311467
asked 2 days ago
user62716user62716
455
$endgroup$
Is it possible to write the shifted Chebyshev polynomials (the first kind) in Mathematica? The formula is:
$$P_n(x)=sum_k=0^leftlfloortfracn2rightrfloor(-1)^k 2^n-2k-1fracnn-kbinomn-kk(2x-1)^n-2k$$
The output should be in a vector ..., ..., ..., ... etc.
Thanks!!
special-functions polynomials
special-functions polynomials
edited 2 days ago
J. M. is slightly pensive♦
99k10311467
asked 2 days ago
user62716user62716
455
edited 2 days ago
J. M. is slightly pensive♦
99k10311467
asked 2 days ago
user62716user62716
455
edited 2 days ago
J. M. is slightly pensive♦
99k10311467
edited 2 days ago
J. M. is slightly pensive♦
99k10311467
edited 2 days ago
J. M. is slightly pensive♦
99k10311467
99k10311467
asked 2 days ago
user62716user62716
455
asked 2 days ago
user62716user62716
455
asked 2 days ago
user62716user62716
455
455
$begingroup$
By "vector", do you mean a vector of the first few polynomials?
$endgroup$
– J. M. is slightly pensive♦
2 days ago
$begingroup$
yes please depend on n
$endgroup$
– user62716
2 days ago
add a comment |
$begingroup$
By "vector", do you mean a vector of the first few polynomials?
$endgroup$
– J. M. is slightly pensive♦
2 days ago
$begingroup$
yes please depend on n
$endgroup$
– user62716
2 days ago
$begingroup$
By "vector", do you mean a vector of the first few polynomials?
$endgroup$
– J. M. is slightly pensive♦
2 days ago
$begingroup$
By "vector", do you mean a vector of the first few polynomials?
$endgroup$
– J. M. is slightly pensive♦
2 days ago
$begingroup$
yes please depend on n
$endgroup$
– user62716
2 days ago
$begingroup$
yes please depend on n
$endgroup$
– user62716
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
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
2 days ago
$begingroup$
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 tomand evaluateChebyshevT[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 slightly pensive♦
2 days ago
$begingroup$
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
2 days ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "387"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f194359%2fhow-to-write-the-shifted-chebyshev-polynomials-the-first-kind-in-mathematica%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
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
2 days ago
$begingroup$
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 tomand evaluateChebyshevT[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 slightly pensive♦
2 days ago
$begingroup$
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
2 days ago
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
2 days ago
$begingroup$
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 tomand evaluateChebyshevT[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 slightly pensive♦
2 days ago
$begingroup$
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
2 days ago
add a comment |
$begingroup$
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.
$endgroup$
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.
answered 2 days ago
community wiki
J. M. is slightly pensive
$begingroup$
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
2 days ago
$begingroup$
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 tomand evaluateChebyshevT[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 slightly pensive♦
2 days ago
$begingroup$
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
2 days ago
add a comment |
$begingroup$
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
2 days ago
$begingroup$
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 tomand evaluateChebyshevT[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 slightly pensive♦
2 days ago
$begingroup$
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
2 days ago
$begingroup$
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
2 days ago
$begingroup$
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
2 days ago
$begingroup$
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 slightly pensive♦
2 days ago
$begingroup$
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 slightly pensive♦
2 days ago
$begingroup$
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
2 days ago
$begingroup$
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
2 days ago
add a comment |
Thanks for contributing an answer to Mathematica Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f194359%2fhow-to-write-the-shifted-chebyshev-polynomials-the-first-kind-in-mathematica%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
By "vector", do you mean a vector of the first few polynomials?
$endgroup$
– J. M. is slightly pensive♦
2 days ago
$begingroup$
yes please depend on n
$endgroup$
– user62716
2 days ago