The conditions under which the Taylor polynomial $P_n(x)$ will converge to $f(x)$
$begingroup$
Background
It is well-known that for the following function:
$$f(x) = begin{cases}
e^{frac{-1}{x^2}} & text{if} & x neq 0 \
0 & text{if} & x = 0
end{cases} $$
the Taylor polynomial does not converge to $f(x)$ as $nto infty$, although $f^{(k)}(0)$ exists and it is 0 for all $kgeq 0$. The problem is with the general conditions under which a Taylor polynomial will converge to the function.
Problem
In general, what conditions exist (or do they exist at all) such that the Taylor polynomial of a function will converge to the function as $nto infty$?
real-analysis analysis taylor-expansion exponential-function
asked Dec 5 '18 at 11:01
hephaeshephaes
1709
$endgroup$
add a comment |
$begingroup$
Background
It is well-known that for the following function:
$$f(x) = begin{cases}
e^{frac{-1}{x^2}} & text{if} & x neq 0 \
0 & text{if} & x = 0
end{cases} $$
the Taylor polynomial does not converge to $f(x)$ as $nto infty$, although $f^{(k)}(0)$ exists and it is 0 for all $kgeq 0$. The problem is with the general conditions under which a Taylor polynomial will converge to the function.
Problem
In general, what conditions exist (or do they exist at all) such that the Taylor polynomial of a function will converge to the function as $nto infty$?
real-analysis analysis taylor-expansion exponential-function
asked Dec 5 '18 at 11:01
hephaeshephaes
1709
$endgroup$
$begingroup$
The $n$-th derivative of any polynomial function is another polynomial function and therefore a continuous function.
$endgroup$
– José Carlos Santos
Dec 5 '18 at 11:03
$begingroup$
@JoséCarlosSantos Thank you for your answer. I understand it now that my question should now change a bit.
$endgroup$
– hephaes
Dec 5 '18 at 11:29
add a comment |
$begingroup$
Background
It is well-known that for the following function:
$$f(x) = begin{cases}
e^{frac{-1}{x^2}} & text{if} & x neq 0 \
0 & text{if} & x = 0
end{cases} $$
the Taylor polynomial does not converge to $f(x)$ as $nto infty$, although $f^{(k)}(0)$ exists and it is 0 for all $kgeq 0$. The problem is with the general conditions under which a Taylor polynomial will converge to the function.
Problem
In general, what conditions exist (or do they exist at all) such that the Taylor polynomial of a function will converge to the function as $nto infty$?
real-analysis analysis taylor-expansion exponential-function
asked Dec 5 '18 at 11:01
hephaeshephaes
1709
$endgroup$
Background
It is well-known that for the following function:
$$f(x) = begin{cases}
e^{frac{-1}{x^2}} & text{if} & x neq 0 \
0 & text{if} & x = 0
end{cases} $$
the Taylor polynomial does not converge to $f(x)$ as $nto infty$, although $f^{(k)}(0)$ exists and it is 0 for all $kgeq 0$. The problem is with the general conditions under which a Taylor polynomial will converge to the function.
Problem
In general, what conditions exist (or do they exist at all) such that the Taylor polynomial of a function will converge to the function as $nto infty$?
real-analysis analysis taylor-expansion exponential-function
real-analysis analysis taylor-expansion exponential-function
asked Dec 5 '18 at 11:01
hephaeshephaes
1709
asked Dec 5 '18 at 11:01
hephaeshephaes
1709
edited Dec 5 '18 at 11:29
hephaes
asked Dec 5 '18 at 11:01
hephaeshephaes
1709
asked Dec 5 '18 at 11:01
hephaeshephaes
1709
asked Dec 5 '18 at 11:01
hephaeshephaes
1709
1709
$begingroup$
The $n$-th derivative of any polynomial function is another polynomial function and therefore a continuous function.
$endgroup$
– José Carlos Santos
Dec 5 '18 at 11:03
$begingroup$
@JoséCarlosSantos Thank you for your answer. I understand it now that my question should now change a bit.
$endgroup$
– hephaes
Dec 5 '18 at 11:29
add a comment |
$begingroup$
The $n$-th derivative of any polynomial function is another polynomial function and therefore a continuous function.
$endgroup$
– José Carlos Santos
Dec 5 '18 at 11:03
$begingroup$
@JoséCarlosSantos Thank you for your answer. I understand it now that my question should now change a bit.
$endgroup$
– hephaes
Dec 5 '18 at 11:29
$begingroup$
The $n$-th derivative of any polynomial function is another polynomial function and therefore a continuous function.
$endgroup$
– José Carlos Santos
Dec 5 '18 at 11:03
$begingroup$
The $n$-th derivative of any polynomial function is another polynomial function and therefore a continuous function.
$endgroup$
– José Carlos Santos
Dec 5 '18 at 11:03
$begingroup$
@JoséCarlosSantos Thank you for your answer. I understand it now that my question should now change a bit.
$endgroup$
– hephaes
Dec 5 '18 at 11:29
$begingroup$
@JoséCarlosSantos Thank you for your answer. I understand it now that my question should now change a bit.
$endgroup$
– hephaes
Dec 5 '18 at 11:29
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
One such condition is given by Bernstein's theorem: if $(forall xin D_f)(forall ninmathbb{N}):f^{(n)}(x)geqslant0$, then the Taylor series of $f$ converges pointwise to $f(x)$ in the neighborhood of every point of $D_f$.
answered Dec 5 '18 at 11:35
José Carlos SantosJosé Carlos Santos
164k22132235
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
One such condition is given by Bernstein's theorem: if $(forall xin D_f)(forall ninmathbb{N}):f^{(n)}(x)geqslant0$, then the Taylor series of $f$ converges pointwise to $f(x)$ in the neighborhood of every point of $D_f$.
answered Dec 5 '18 at 11:35
José Carlos SantosJosé Carlos Santos
164k22132235
$endgroup$
add a comment |
$begingroup$
One such condition is given by Bernstein's theorem: if $(forall xin D_f)(forall ninmathbb{N}):f^{(n)}(x)geqslant0$, then the Taylor series of $f$ converges pointwise to $f(x)$ in the neighborhood of every point of $D_f$.
answered Dec 5 '18 at 11:35
José Carlos SantosJosé Carlos Santos
164k22132235
$endgroup$
add a comment |
$begingroup$
One such condition is given by Bernstein's theorem: if $(forall xin D_f)(forall ninmathbb{N}):f^{(n)}(x)geqslant0$, then the Taylor series of $f$ converges pointwise to $f(x)$ in the neighborhood of every point of $D_f$.
answered Dec 5 '18 at 11:35
José Carlos SantosJosé Carlos Santos
164k22132235
$endgroup$
One such condition is given by Bernstein's theorem: if $(forall xin D_f)(forall ninmathbb{N}):f^{(n)}(x)geqslant0$, then the Taylor series of $f$ converges pointwise to $f(x)$ in the neighborhood of every point of $D_f$.
answered Dec 5 '18 at 11:35
José Carlos SantosJosé Carlos Santos
164k22132235
answered Dec 5 '18 at 11:35
José Carlos SantosJosé Carlos Santos
164k22132235
answered Dec 5 '18 at 11:35
José Carlos SantosJosé Carlos Santos
164k22132235
answered Dec 5 '18 at 11:35
José Carlos SantosJosé Carlos Santos
164k22132235
164k22132235
add a comment |
add a comment |
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$begingroup$
The $n$-th derivative of any polynomial function is another polynomial function and therefore a continuous function.
$endgroup$
– José Carlos Santos
Dec 5 '18 at 11:03
$begingroup$
@JoséCarlosSantos Thank you for your answer. I understand it now that my question should now change a bit.
$endgroup$
– hephaes
Dec 5 '18 at 11:29