Does a taylor polynomial always have a non-zero remainder? [closed]
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A bit of a silly question but it isn't really covered in my book and can't seem to find this question anywhere.
analysis
asked Dec 13 '18 at 11:50
Job EijgenhuijsenJob Eijgenhuijsen
62
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closed as unclear what you're asking by amWhy, user10354138, Jyrki Lahtonen, Daniel Moskovich, onurcanbektas Dec 13 '18 at 16:13
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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$begingroup$
A bit of a silly question but it isn't really covered in my book and can't seem to find this question anywhere.
analysis
asked Dec 13 '18 at 11:50
Job EijgenhuijsenJob Eijgenhuijsen
62
$endgroup$
closed as unclear what you're asking by amWhy, user10354138, Jyrki Lahtonen, Daniel Moskovich, onurcanbektas Dec 13 '18 at 16:13
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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What do you mean? Do you mean a non-zero remainder? But of course the Taylor Series for a polynomial is exact.
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– lulu
Dec 13 '18 at 11:51
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@lulu yes sorry, that is what I meant. A non-zero remainder.
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– Job Eijgenhuijsen
Dec 13 '18 at 11:55
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Well, like I say...that first order Taylor polynomial for $f(x)=x$ is $x$ which is exact.
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– lulu
Dec 13 '18 at 11:57
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I think you should look at the definitions of the Taylor polynomial and the Taylor series of a function. A Taylor polynomial contains a finite number of terms since it is a full-fledged polynomial. And thus there is no remainder.
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– Bernard Massé
Dec 13 '18 at 12:31
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$begingroup$
A bit of a silly question but it isn't really covered in my book and can't seem to find this question anywhere.
analysis
asked Dec 13 '18 at 11:50
Job EijgenhuijsenJob Eijgenhuijsen
62
$endgroup$
A bit of a silly question but it isn't really covered in my book and can't seem to find this question anywhere.
analysis
analysis
asked Dec 13 '18 at 11:50
Job EijgenhuijsenJob Eijgenhuijsen
62
asked Dec 13 '18 at 11:50
Job EijgenhuijsenJob Eijgenhuijsen
62
edited Dec 13 '18 at 11:56
Job Eijgenhuijsen
asked Dec 13 '18 at 11:50
Job EijgenhuijsenJob Eijgenhuijsen
62
asked Dec 13 '18 at 11:50
Job EijgenhuijsenJob Eijgenhuijsen
62
asked Dec 13 '18 at 11:50
Job EijgenhuijsenJob Eijgenhuijsen
62
62
closed as unclear what you're asking by amWhy, user10354138, Jyrki Lahtonen, Daniel Moskovich, onurcanbektas Dec 13 '18 at 16:13
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by amWhy, user10354138, Jyrki Lahtonen, Daniel Moskovich, onurcanbektas Dec 13 '18 at 16:13
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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What do you mean? Do you mean a non-zero remainder? But of course the Taylor Series for a polynomial is exact.
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– lulu
Dec 13 '18 at 11:51
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@lulu yes sorry, that is what I meant. A non-zero remainder.
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– Job Eijgenhuijsen
Dec 13 '18 at 11:55
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Well, like I say...that first order Taylor polynomial for $f(x)=x$ is $x$ which is exact.
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– lulu
Dec 13 '18 at 11:57
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I think you should look at the definitions of the Taylor polynomial and the Taylor series of a function. A Taylor polynomial contains a finite number of terms since it is a full-fledged polynomial. And thus there is no remainder.
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– Bernard Massé
Dec 13 '18 at 12:31
add a comment |
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What do you mean? Do you mean a non-zero remainder? But of course the Taylor Series for a polynomial is exact.
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– lulu
Dec 13 '18 at 11:51
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@lulu yes sorry, that is what I meant. A non-zero remainder.
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– Job Eijgenhuijsen
Dec 13 '18 at 11:55
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Well, like I say...that first order Taylor polynomial for $f(x)=x$ is $x$ which is exact.
$endgroup$
– lulu
Dec 13 '18 at 11:57
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I think you should look at the definitions of the Taylor polynomial and the Taylor series of a function. A Taylor polynomial contains a finite number of terms since it is a full-fledged polynomial. And thus there is no remainder.
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– Bernard Massé
Dec 13 '18 at 12:31
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What do you mean? Do you mean a non-zero remainder? But of course the Taylor Series for a polynomial is exact.
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– lulu
Dec 13 '18 at 11:51
$begingroup$
What do you mean? Do you mean a non-zero remainder? But of course the Taylor Series for a polynomial is exact.
$endgroup$
– lulu
Dec 13 '18 at 11:51
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@lulu yes sorry, that is what I meant. A non-zero remainder.
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– Job Eijgenhuijsen
Dec 13 '18 at 11:55
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@lulu yes sorry, that is what I meant. A non-zero remainder.
$endgroup$
– Job Eijgenhuijsen
Dec 13 '18 at 11:55
$begingroup$
Well, like I say...that first order Taylor polynomial for $f(x)=x$ is $x$ which is exact.
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– lulu
Dec 13 '18 at 11:57
$begingroup$
Well, like I say...that first order Taylor polynomial for $f(x)=x$ is $x$ which is exact.
$endgroup$
– lulu
Dec 13 '18 at 11:57
$begingroup$
I think you should look at the definitions of the Taylor polynomial and the Taylor series of a function. A Taylor polynomial contains a finite number of terms since it is a full-fledged polynomial. And thus there is no remainder.
$endgroup$
– Bernard Massé
Dec 13 '18 at 12:31
$begingroup$
I think you should look at the definitions of the Taylor polynomial and the Taylor series of a function. A Taylor polynomial contains a finite number of terms since it is a full-fledged polynomial. And thus there is no remainder.
$endgroup$
– Bernard Massé
Dec 13 '18 at 12:31
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1 Answer
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Assuming that we are dealing with a smooth function $f$ here, if $ninmathbb N$ and $ain D_f$, then the Taylor polynomial of $f$ at $a$ has a non-zero remainder if and only if $f$ is not a polynomial function or if it is a polynomial function whose degree is greater than $n$.
answered Dec 13 '18 at 12:21
José Carlos SantosJosé Carlos Santos
160k22126232
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Assuming that we are dealing with a smooth function $f$ here, if $ninmathbb N$ and $ain D_f$, then the Taylor polynomial of $f$ at $a$ has a non-zero remainder if and only if $f$ is not a polynomial function or if it is a polynomial function whose degree is greater than $n$.
answered Dec 13 '18 at 12:21
José Carlos SantosJosé Carlos Santos
160k22126232
$endgroup$
add a comment |
$begingroup$
Assuming that we are dealing with a smooth function $f$ here, if $ninmathbb N$ and $ain D_f$, then the Taylor polynomial of $f$ at $a$ has a non-zero remainder if and only if $f$ is not a polynomial function or if it is a polynomial function whose degree is greater than $n$.
answered Dec 13 '18 at 12:21
José Carlos SantosJosé Carlos Santos
160k22126232
$endgroup$
add a comment |
$begingroup$
Assuming that we are dealing with a smooth function $f$ here, if $ninmathbb N$ and $ain D_f$, then the Taylor polynomial of $f$ at $a$ has a non-zero remainder if and only if $f$ is not a polynomial function or if it is a polynomial function whose degree is greater than $n$.
answered Dec 13 '18 at 12:21
José Carlos SantosJosé Carlos Santos
160k22126232
$endgroup$
Assuming that we are dealing with a smooth function $f$ here, if $ninmathbb N$ and $ain D_f$, then the Taylor polynomial of $f$ at $a$ has a non-zero remainder if and only if $f$ is not a polynomial function or if it is a polynomial function whose degree is greater than $n$.
answered Dec 13 '18 at 12:21
José Carlos SantosJosé Carlos Santos
160k22126232
answered Dec 13 '18 at 12:21
José Carlos SantosJosé Carlos Santos
160k22126232
answered Dec 13 '18 at 12:21
José Carlos SantosJosé Carlos Santos
160k22126232
answered Dec 13 '18 at 12:21
José Carlos SantosJosé Carlos Santos
160k22126232
160k22126232
add a comment |
add a comment |
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What do you mean? Do you mean a non-zero remainder? But of course the Taylor Series for a polynomial is exact.
$endgroup$
– lulu
Dec 13 '18 at 11:51
$begingroup$
@lulu yes sorry, that is what I meant. A non-zero remainder.
$endgroup$
– Job Eijgenhuijsen
Dec 13 '18 at 11:55
$begingroup$
Well, like I say...that first order Taylor polynomial for $f(x)=x$ is $x$ which is exact.
$endgroup$
– lulu
Dec 13 '18 at 11:57
$begingroup$
I think you should look at the definitions of the Taylor polynomial and the Taylor series of a function. A Taylor polynomial contains a finite number of terms since it is a full-fledged polynomial. And thus there is no remainder.
$endgroup$
– Bernard Massé
Dec 13 '18 at 12:31