There is a sequence $(P_n)$ of polynomials such that $P_n(cos x) to f(x)$ uniformly over $[0,pi]$.
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Show that for any function continuous $f:[0,pi] to mathbbR$, there is a sequence $(P_n)$ of polynomials such that
$$P_n(cos x) to f(x);textuniformly over;[0,pi].$$
The Weierstrass Theorem says that there is a sequence of polynomials $p_n$ such that $p_n to f$ uniformly. We cannot have $p_n = P_n$ if no, $P_n(cos x) to f(cos x)$, right?
I know that also, there is a sequence $q_n$ such that $q_n to cos$ uniformly. I'm trying to somehow use these two sequences $p_n,q_n$, but I'm stuck. I don't want the solution to the exercise, but I would like a hint.
real-analysis continuity weierstrass-approximation
asked Aug 14 at 9:08
Lucas Corrêa
1,173319
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up vote
0
down vote
favorite
Show that for any function continuous $f:[0,pi] to mathbbR$, there is a sequence $(P_n)$ of polynomials such that
$$P_n(cos x) to f(x);textuniformly over;[0,pi].$$
The Weierstrass Theorem says that there is a sequence of polynomials $p_n$ such that $p_n to f$ uniformly. We cannot have $p_n = P_n$ if no, $P_n(cos x) to f(cos x)$, right?
I know that also, there is a sequence $q_n$ such that $q_n to cos$ uniformly. I'm trying to somehow use these two sequences $p_n,q_n$, but I'm stuck. I don't want the solution to the exercise, but I would like a hint.
real-analysis continuity weierstrass-approximation
asked Aug 14 at 9:08
Lucas Corrêa
1,173319
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Show that for any function continuous $f:[0,pi] to mathbbR$, there is a sequence $(P_n)$ of polynomials such that
$$P_n(cos x) to f(x);textuniformly over;[0,pi].$$
The Weierstrass Theorem says that there is a sequence of polynomials $p_n$ such that $p_n to f$ uniformly. We cannot have $p_n = P_n$ if no, $P_n(cos x) to f(cos x)$, right?
I know that also, there is a sequence $q_n$ such that $q_n to cos$ uniformly. I'm trying to somehow use these two sequences $p_n,q_n$, but I'm stuck. I don't want the solution to the exercise, but I would like a hint.
real-analysis continuity weierstrass-approximation
asked Aug 14 at 9:08
Lucas Corrêa
1,173319
Show that for any function continuous $f:[0,pi] to mathbbR$, there is a sequence $(P_n)$ of polynomials such that
$$P_n(cos x) to f(x);textuniformly over;[0,pi].$$
The Weierstrass Theorem says that there is a sequence of polynomials $p_n$ such that $p_n to f$ uniformly. We cannot have $p_n = P_n$ if no, $P_n(cos x) to f(cos x)$, right?
I know that also, there is a sequence $q_n$ such that $q_n to cos$ uniformly. I'm trying to somehow use these two sequences $p_n,q_n$, but I'm stuck. I don't want the solution to the exercise, but I would like a hint.
real-analysis continuity weierstrass-approximation
asked Aug 14 at 9:08
Lucas Corrêa
1,173319
asked Aug 14 at 9:08
Lucas Corrêa
1,173319
asked Aug 14 at 9:08
Lucas Corrêa
1,173319
asked Aug 14 at 9:08
Lucas Corrêa
1,173319
1,173319
add a comment |Â
add a comment |Â
1 Answer
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votes
up vote
4
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$fcirc cos^-1$ is a continuous function on $[-1,1]$. Approximate this by polynomials. $p_n$ and that will do the trick.
answered Aug 14 at 9:11
Kavi Rama Murthy
22.2k2933
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
$fcirc cos^-1$ is a continuous function on $[-1,1]$. Approximate this by polynomials. $p_n$ and that will do the trick.
answered Aug 14 at 9:11
Kavi Rama Murthy
22.2k2933
add a comment |Â
up vote
4
down vote
accepted
$fcirc cos^-1$ is a continuous function on $[-1,1]$. Approximate this by polynomials. $p_n$ and that will do the trick.
answered Aug 14 at 9:11
Kavi Rama Murthy
22.2k2933
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
$fcirc cos^-1$ is a continuous function on $[-1,1]$. Approximate this by polynomials. $p_n$ and that will do the trick.
answered Aug 14 at 9:11
Kavi Rama Murthy
22.2k2933
$fcirc cos^-1$ is a continuous function on $[-1,1]$. Approximate this by polynomials. $p_n$ and that will do the trick.
answered Aug 14 at 9:11
Kavi Rama Murthy
22.2k2933
answered Aug 14 at 9:11
Kavi Rama Murthy
22.2k2933
answered Aug 14 at 9:11
Kavi Rama Murthy
22.2k2933
answered Aug 14 at 9:11
Kavi Rama Murthy
22.2k2933
22.2k2933
add a comment |Â
add a comment |Â
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