Taylor series of product of two functionsProving an inequality with Taylor polynomialsIntuition behind Taylor/Maclaurin SeriesUse of taylor series in convergenceRunge Phenomena and Taylor ExpansionWhat is the justification for taylor series for functions with one or no critical points?Marsden's definition of Taylor SeriesFind the Taylor series of $f(x)=sum_k=0^infty frac2^-kk+1(x-1)^k$Smoothness of Taylor polynomials coefficients as function of position of expansion?Taylor series with initial value of infinityMisunderstanding about Taylor series
Java - What do constructor type arguments mean when placed *before* the type?
How will losing mobility of one hand affect my career as a programmer?
Is it okay / does it make sense for another player to join a running game of Munchkin?
Why is delta-v is the most useful quantity for planning space travel?
Why are all the doors on Ferenginar (the Ferengi home world) far shorter than the average Ferengi?
Identify a stage play about a VR experience in which participants are encouraged to simulate performing horrific activities
Science Fiction story where a man invents a machine that can help him watch history unfold
Adding empty element to declared container without declaring type of element
How to deal with or prevent idle in the test team?
How can I successfully establish a nationwide combat training program for a large country?
Does "Dominei" mean something?
Can somebody explain Brexit in a few child-proof sentences?
What to do when my ideas aren't chosen, when I strongly disagree with the chosen solution?
Stereotypical names
Should a half Jewish man be discouraged from marrying a Jewess?
Simple image editor tool to draw a simple box/rectangle in an existing image
How to color a zone in Tikz
What was required to accept "troll"?
Could solar power be utilized and substitute coal in the 19th century?
Visiting the UK as unmarried couple
Can I rely on these GitHub repository files?
I'm in charge of equipment buying but no one's ever happy with what I choose. How to fix this?
Partial sums of primes
Word describing multiple paths to the same abstract outcome
Taylor series of product of two functions
Proving an inequality with Taylor polynomialsIntuition behind Taylor/Maclaurin SeriesUse of taylor series in convergenceRunge Phenomena and Taylor ExpansionWhat is the justification for taylor series for functions with one or no critical points?Marsden's definition of Taylor SeriesFind the Taylor series of $f(x)=sum_k=0^infty frac2^-kk+1(x-1)^k$Smoothness of Taylor polynomials coefficients as function of position of expansion?Taylor series with initial value of infinityMisunderstanding about Taylor series
$begingroup$
let $f$ and $g$ be infinitley differentiable functions and $a_k = fracf^(k)(a)k!$ and $b_e = fracg^(e)(a)e!$ be cofficients of Taylor Polynomial at $a$ then what would be the coefficients of $fg$.
rather than asking my specific question I asked this general question so other can benefit too
So I think we would need to multiply the two polynomials but that's just my intuition and I don't know how to justify it and I don't think it would be as simple.
analysis taylor-expansion
asked 4 hours ago
ConorConor
556
$endgroup$
add a comment |
$begingroup$
let $f$ and $g$ be infinitley differentiable functions and $a_k = fracf^(k)(a)k!$ and $b_e = fracg^(e)(a)e!$ be cofficients of Taylor Polynomial at $a$ then what would be the coefficients of $fg$.
rather than asking my specific question I asked this general question so other can benefit too
So I think we would need to multiply the two polynomials but that's just my intuition and I don't know how to justify it and I don't think it would be as simple.
analysis taylor-expansion
asked 4 hours ago
ConorConor
556
$endgroup$
$begingroup$
Is it supposed to be $b_e=fracg^(e)(a)e!$ ? If so, look up the Cauchy Product Formula.
$endgroup$
– robjohn♦
4 hours ago
add a comment |
$begingroup$
let $f$ and $g$ be infinitley differentiable functions and $a_k = fracf^(k)(a)k!$ and $b_e = fracg^(e)(a)e!$ be cofficients of Taylor Polynomial at $a$ then what would be the coefficients of $fg$.
rather than asking my specific question I asked this general question so other can benefit too
So I think we would need to multiply the two polynomials but that's just my intuition and I don't know how to justify it and I don't think it would be as simple.
analysis taylor-expansion
asked 4 hours ago
ConorConor
556
$endgroup$
let $f$ and $g$ be infinitley differentiable functions and $a_k = fracf^(k)(a)k!$ and $b_e = fracg^(e)(a)e!$ be cofficients of Taylor Polynomial at $a$ then what would be the coefficients of $fg$.
rather than asking my specific question I asked this general question so other can benefit too
So I think we would need to multiply the two polynomials but that's just my intuition and I don't know how to justify it and I don't think it would be as simple.
analysis taylor-expansion
analysis taylor-expansion
asked 4 hours ago
ConorConor
556
asked 4 hours ago
ConorConor
556
edited 4 hours ago
Conor
asked 4 hours ago
ConorConor
556
asked 4 hours ago
ConorConor
556
asked 4 hours ago
ConorConor
556
556
$begingroup$
Is it supposed to be $b_e=fracg^(e)(a)e!$ ? If so, look up the Cauchy Product Formula.
$endgroup$
– robjohn♦
4 hours ago
add a comment |
$begingroup$
Is it supposed to be $b_e=fracg^(e)(a)e!$ ? If so, look up the Cauchy Product Formula.
$endgroup$
– robjohn♦
4 hours ago
$begingroup$
Is it supposed to be $b_e=fracg^(e)(a)e!$ ? If so, look up the Cauchy Product Formula.
$endgroup$
– robjohn♦
4 hours ago
$begingroup$
Is it supposed to be $b_e=fracg^(e)(a)e!$ ? If so, look up the Cauchy Product Formula.
$endgroup$
– robjohn♦
4 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your intuition is good.
Multiplying the series gives an n-th term coefficient of
$$c_n = a_0b_n + a_1b_n-1 + dots + a_n-1b_1 + a_nb_0= sum_i=0^n a_i b_n-i$$
which is the same as doing the Taylor series of $fg$ the long way, since
$$c_n = frac(fg)^(n)(a)n! = fracsum_i=0^n binomnif^(i)(a)g^(n-i)(a)n! = sum_i=0^n fracf^(i)(a)i! fracg^(n-i)(a)(n-i)! = sum_i=0^n a_i b_n-i$$
answered 4 hours ago
Michael BiroMichael Biro
11.3k21831
$endgroup$
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: "69"
;
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: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
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
,
noCode: 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%2fmath.stackexchange.com%2fquestions%2f3162570%2ftaylor-series-of-product-of-two-functions%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
$begingroup$
Your intuition is good.
Multiplying the series gives an n-th term coefficient of
$$c_n = a_0b_n + a_1b_n-1 + dots + a_n-1b_1 + a_nb_0= sum_i=0^n a_i b_n-i$$
which is the same as doing the Taylor series of $fg$ the long way, since
$$c_n = frac(fg)^(n)(a)n! = fracsum_i=0^n binomnif^(i)(a)g^(n-i)(a)n! = sum_i=0^n fracf^(i)(a)i! fracg^(n-i)(a)(n-i)! = sum_i=0^n a_i b_n-i$$
answered 4 hours ago
Michael BiroMichael Biro
11.3k21831
$endgroup$
add a comment |
$begingroup$
Your intuition is good.
Multiplying the series gives an n-th term coefficient of
$$c_n = a_0b_n + a_1b_n-1 + dots + a_n-1b_1 + a_nb_0= sum_i=0^n a_i b_n-i$$
which is the same as doing the Taylor series of $fg$ the long way, since
$$c_n = frac(fg)^(n)(a)n! = fracsum_i=0^n binomnif^(i)(a)g^(n-i)(a)n! = sum_i=0^n fracf^(i)(a)i! fracg^(n-i)(a)(n-i)! = sum_i=0^n a_i b_n-i$$
answered 4 hours ago
Michael BiroMichael Biro
11.3k21831
$endgroup$
add a comment |
$begingroup$
Your intuition is good.
Multiplying the series gives an n-th term coefficient of
$$c_n = a_0b_n + a_1b_n-1 + dots + a_n-1b_1 + a_nb_0= sum_i=0^n a_i b_n-i$$
which is the same as doing the Taylor series of $fg$ the long way, since
$$c_n = frac(fg)^(n)(a)n! = fracsum_i=0^n binomnif^(i)(a)g^(n-i)(a)n! = sum_i=0^n fracf^(i)(a)i! fracg^(n-i)(a)(n-i)! = sum_i=0^n a_i b_n-i$$
answered 4 hours ago
Michael BiroMichael Biro
11.3k21831
$endgroup$
Your intuition is good.
Multiplying the series gives an n-th term coefficient of
$$c_n = a_0b_n + a_1b_n-1 + dots + a_n-1b_1 + a_nb_0= sum_i=0^n a_i b_n-i$$
which is the same as doing the Taylor series of $fg$ the long way, since
$$c_n = frac(fg)^(n)(a)n! = fracsum_i=0^n binomnif^(i)(a)g^(n-i)(a)n! = sum_i=0^n fracf^(i)(a)i! fracg^(n-i)(a)(n-i)! = sum_i=0^n a_i b_n-i$$
answered 4 hours ago
Michael BiroMichael Biro
11.3k21831
answered 4 hours ago
Michael BiroMichael Biro
11.3k21831
answered 4 hours ago
Michael BiroMichael Biro
11.3k21831
answered 4 hours ago
Michael BiroMichael Biro
11.3k21831
11.3k21831
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics 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%2fmath.stackexchange.com%2fquestions%2f3162570%2ftaylor-series-of-product-of-two-functions%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
$begingroup$
Is it supposed to be $b_e=fracg^(e)(a)e!$ ? If so, look up the Cauchy Product Formula.
$endgroup$
– robjohn♦
4 hours ago