Mixing
Understanding Taylor series error function and Lagrange error bound
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I am in high school and find math interesting so lately I have been trying to learn as much about it as I can. I recently began studying Taylor Series as it pertains to the research areas that I am currently following and have two questions. I am trying to conceptualize Lagrange's error bound and the remainder function.
I have read that:
$f(x)=f(c)+f'(c)(x-c)+...+frac{f^{(n)}(c)}{n!}(x-c)^{n}+R_n(x)$
where $R_n(x)$ is the remainder given by:
$frac{f^{(n+1)}(z)}{(n+1)!}(x-c)^{n+1}$.
From what I have read, I understand that the remainder gives the difference between a given Taylor polynomial and its original function, $f(x)$. I have also read about La Grange's error bound which differs from the remainder.
My first question is, if Lagrange's error bound gives the maximum "remainder" for a given $x$, how does it differ from the remainder function $R_n(x)$? Are they equivalent? If a function can be found from its Taylor polynomial and Remainder term or any combination vice-versa, what is the purpose of Lagrange's error bound?
My second question is, how is the error term function, $R_n(x)$ derived? I understand that it can be found from the difference between a function and its Taylor polynomial but I am not sure where it originates from in general form.
Any help with conceptualizing this is greatly appreciated.
calculus taylor-expansion error-function
asked Jan 9 at 0:08
GnumbertesterGnumbertester
495111
$endgroup$
add a comment |
$begingroup$
I am in high school and find math interesting so lately I have been trying to learn as much about it as I can. I recently began studying Taylor Series as it pertains to the research areas that I am currently following and have two questions. I am trying to conceptualize Lagrange's error bound and the remainder function.
I have read that:
$f(x)=f(c)+f'(c)(x-c)+...+frac{f^{(n)}(c)}{n!}(x-c)^{n}+R_n(x)$
where $R_n(x)$ is the remainder given by:
$frac{f^{(n+1)}(z)}{(n+1)!}(x-c)^{n+1}$.
From what I have read, I understand that the remainder gives the difference between a given Taylor polynomial and its original function, $f(x)$. I have also read about La Grange's error bound which differs from the remainder.
My first question is, if Lagrange's error bound gives the maximum "remainder" for a given $x$, how does it differ from the remainder function $R_n(x)$? Are they equivalent? If a function can be found from its Taylor polynomial and Remainder term or any combination vice-versa, what is the purpose of Lagrange's error bound?
My second question is, how is the error term function, $R_n(x)$ derived? I understand that it can be found from the difference between a function and its Taylor polynomial but I am not sure where it originates from in general form.
Any help with conceptualizing this is greatly appreciated.
calculus taylor-expansion error-function
asked Jan 9 at 0:08
GnumbertesterGnumbertester
495111
$endgroup$
add a comment |
$begingroup$
I am in high school and find math interesting so lately I have been trying to learn as much about it as I can. I recently began studying Taylor Series as it pertains to the research areas that I am currently following and have two questions. I am trying to conceptualize Lagrange's error bound and the remainder function.
I have read that:
$f(x)=f(c)+f'(c)(x-c)+...+frac{f^{(n)}(c)}{n!}(x-c)^{n}+R_n(x)$
where $R_n(x)$ is the remainder given by:
$frac{f^{(n+1)}(z)}{(n+1)!}(x-c)^{n+1}$.
From what I have read, I understand that the remainder gives the difference between a given Taylor polynomial and its original function, $f(x)$. I have also read about La Grange's error bound which differs from the remainder.
My first question is, if Lagrange's error bound gives the maximum "remainder" for a given $x$, how does it differ from the remainder function $R_n(x)$? Are they equivalent? If a function can be found from its Taylor polynomial and Remainder term or any combination vice-versa, what is the purpose of Lagrange's error bound?
My second question is, how is the error term function, $R_n(x)$ derived? I understand that it can be found from the difference between a function and its Taylor polynomial but I am not sure where it originates from in general form.
Any help with conceptualizing this is greatly appreciated.
calculus taylor-expansion error-function
asked Jan 9 at 0:08
GnumbertesterGnumbertester
495111
$endgroup$
I am in high school and find math interesting so lately I have been trying to learn as much about it as I can. I recently began studying Taylor Series as it pertains to the research areas that I am currently following and have two questions. I am trying to conceptualize Lagrange's error bound and the remainder function.
I have read that:
$f(x)=f(c)+f'(c)(x-c)+...+frac{f^{(n)}(c)}{n!}(x-c)^{n}+R_n(x)$
where $R_n(x)$ is the remainder given by:
$frac{f^{(n+1)}(z)}{(n+1)!}(x-c)^{n+1}$.
From what I have read, I understand that the remainder gives the difference between a given Taylor polynomial and its original function, $f(x)$. I have also read about La Grange's error bound which differs from the remainder.
My first question is, if Lagrange's error bound gives the maximum "remainder" for a given $x$, how does it differ from the remainder function $R_n(x)$? Are they equivalent? If a function can be found from its Taylor polynomial and Remainder term or any combination vice-versa, what is the purpose of Lagrange's error bound?
My second question is, how is the error term function, $R_n(x)$ derived? I understand that it can be found from the difference between a function and its Taylor polynomial but I am not sure where it originates from in general form.
Any help with conceptualizing this is greatly appreciated.
calculus taylor-expansion error-function
calculus taylor-expansion error-function
asked Jan 9 at 0:08
GnumbertesterGnumbertester
495111
asked Jan 9 at 0:08
GnumbertesterGnumbertester
495111
edited Jan 9 at 2:55
Gnumbertester
asked Jan 9 at 0:08
GnumbertesterGnumbertester
495111
asked Jan 9 at 0:08
GnumbertesterGnumbertester
495111
asked Jan 9 at 0:08
GnumbertesterGnumbertester
495111
495111
add a comment |
add a comment |
1 Answer
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$R_n(x)$ is defined to be the difference between the function and its $n$th order Taylor polynomial (not series).
It is an important consequence of the Mean Value Theorem that there exists $z$ between $x$ and $c$ such that the remainder can be written as $R_n(x) = frac{f^{(n+1)}(z)}{(n+1)!} (x-c)^{n+1}$. However in general we do not know more about the exact value of $z$ (or $f^{(n+1)}(z)$) beyond the fact that it lies between $x$ and $c$.
Since we often cannot compute the remainder explicitly, we would like to bound it. For example, it would be nice to say $|R_n(x)| le 0.01$; this would tell us that the value of the $n$th order Taylor polynomial at $x$ is within $0.01$ of the actual function value at $x$. Error bounds use assumptions on $f$ to bound $|R_n(x)|$.
answered Jan 9 at 0:18
angryavianangryavian
40.6k23380
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$begingroup$
$R_n(x)$ is defined to be the difference between the function and its $n$th order Taylor polynomial (not series).
It is an important consequence of the Mean Value Theorem that there exists $z$ between $x$ and $c$ such that the remainder can be written as $R_n(x) = frac{f^{(n+1)}(z)}{(n+1)!} (x-c)^{n+1}$. However in general we do not know more about the exact value of $z$ (or $f^{(n+1)}(z)$) beyond the fact that it lies between $x$ and $c$.
Since we often cannot compute the remainder explicitly, we would like to bound it. For example, it would be nice to say $|R_n(x)| le 0.01$; this would tell us that the value of the $n$th order Taylor polynomial at $x$ is within $0.01$ of the actual function value at $x$. Error bounds use assumptions on $f$ to bound $|R_n(x)|$.
answered Jan 9 at 0:18
angryavianangryavian
40.6k23380
$endgroup$
add a comment |
$begingroup$
$R_n(x)$ is defined to be the difference between the function and its $n$th order Taylor polynomial (not series).
It is an important consequence of the Mean Value Theorem that there exists $z$ between $x$ and $c$ such that the remainder can be written as $R_n(x) = frac{f^{(n+1)}(z)}{(n+1)!} (x-c)^{n+1}$. However in general we do not know more about the exact value of $z$ (or $f^{(n+1)}(z)$) beyond the fact that it lies between $x$ and $c$.
Since we often cannot compute the remainder explicitly, we would like to bound it. For example, it would be nice to say $|R_n(x)| le 0.01$; this would tell us that the value of the $n$th order Taylor polynomial at $x$ is within $0.01$ of the actual function value at $x$. Error bounds use assumptions on $f$ to bound $|R_n(x)|$.
answered Jan 9 at 0:18
angryavianangryavian
40.6k23380
$endgroup$
add a comment |
$begingroup$
$R_n(x)$ is defined to be the difference between the function and its $n$th order Taylor polynomial (not series).
It is an important consequence of the Mean Value Theorem that there exists $z$ between $x$ and $c$ such that the remainder can be written as $R_n(x) = frac{f^{(n+1)}(z)}{(n+1)!} (x-c)^{n+1}$. However in general we do not know more about the exact value of $z$ (or $f^{(n+1)}(z)$) beyond the fact that it lies between $x$ and $c$.
Since we often cannot compute the remainder explicitly, we would like to bound it. For example, it would be nice to say $|R_n(x)| le 0.01$; this would tell us that the value of the $n$th order Taylor polynomial at $x$ is within $0.01$ of the actual function value at $x$. Error bounds use assumptions on $f$ to bound $|R_n(x)|$.
answered Jan 9 at 0:18
angryavianangryavian
40.6k23380
$endgroup$
$R_n(x)$ is defined to be the difference between the function and its $n$th order Taylor polynomial (not series).
It is an important consequence of the Mean Value Theorem that there exists $z$ between $x$ and $c$ such that the remainder can be written as $R_n(x) = frac{f^{(n+1)}(z)}{(n+1)!} (x-c)^{n+1}$. However in general we do not know more about the exact value of $z$ (or $f^{(n+1)}(z)$) beyond the fact that it lies between $x$ and $c$.
Since we often cannot compute the remainder explicitly, we would like to bound it. For example, it would be nice to say $|R_n(x)| le 0.01$; this would tell us that the value of the $n$th order Taylor polynomial at $x$ is within $0.01$ of the actual function value at $x$. Error bounds use assumptions on $f$ to bound $|R_n(x)|$.
answered Jan 9 at 0:18
angryavianangryavian
40.6k23380
answered Jan 9 at 0:18
angryavianangryavian
40.6k23380
answered Jan 9 at 0:18
angryavianangryavian
40.6k23380
answered Jan 9 at 0:18
angryavianangryavian
40.6k23380
40.6k23380
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