Mixing
Approximation of continuous function by linear combination of exponentials
Let $a, b in mathbb{R}, a<b$, and let $f:[a,b]to mathbb{R}$ be a continuous function. Given $varepsilon>0$, show that there exists $alpha_1, alpha_2, dots, alpha_n in mathbb{R}$ and $m_1, dots, m_n$ non-negative integers such that
$$|f(x) - sum_{i=1}^{n} alpha_ie^{m_ix}| leq varepsilon, forall x in [a, b].$$
This question provided a couple of hints for a slightly different version of mine, with $m_1, dots, m_n$ are negative integers, but I couldn't adapt them to the problem at hand or come up with any new tricks. I'm looking for a detailed solution, as I'm also on the process of learning uniform convergence and Berstein polynomials (which might be a way out of this).
real-analysis exponential-function uniform-convergence weierstrass-approximation
asked Nov 20 '18 at 13:55
user71487
948
add a comment |
Let $a, b in mathbb{R}, a<b$, and let $f:[a,b]to mathbb{R}$ be a continuous function. Given $varepsilon>0$, show that there exists $alpha_1, alpha_2, dots, alpha_n in mathbb{R}$ and $m_1, dots, m_n$ non-negative integers such that
$$|f(x) - sum_{i=1}^{n} alpha_ie^{m_ix}| leq varepsilon, forall x in [a, b].$$
This question provided a couple of hints for a slightly different version of mine, with $m_1, dots, m_n$ are negative integers, but I couldn't adapt them to the problem at hand or come up with any new tricks. I'm looking for a detailed solution, as I'm also on the process of learning uniform convergence and Berstein polynomials (which might be a way out of this).
real-analysis exponential-function uniform-convergence weierstrass-approximation
asked Nov 20 '18 at 13:55
user71487
948
1
Negative integers were necessary because the domain was not bounded. Here you can follow a very similar reasoning as the one you pointd out in the answer, taking into account that on your domain, the exponential are bounded.
– Martigan
Nov 20 '18 at 14:04
add a comment |
Let $a, b in mathbb{R}, a<b$, and let $f:[a,b]to mathbb{R}$ be a continuous function. Given $varepsilon>0$, show that there exists $alpha_1, alpha_2, dots, alpha_n in mathbb{R}$ and $m_1, dots, m_n$ non-negative integers such that
$$|f(x) - sum_{i=1}^{n} alpha_ie^{m_ix}| leq varepsilon, forall x in [a, b].$$
This question provided a couple of hints for a slightly different version of mine, with $m_1, dots, m_n$ are negative integers, but I couldn't adapt them to the problem at hand or come up with any new tricks. I'm looking for a detailed solution, as I'm also on the process of learning uniform convergence and Berstein polynomials (which might be a way out of this).
real-analysis exponential-function uniform-convergence weierstrass-approximation
asked Nov 20 '18 at 13:55
user71487
948
Let $a, b in mathbb{R}, a<b$, and let $f:[a,b]to mathbb{R}$ be a continuous function. Given $varepsilon>0$, show that there exists $alpha_1, alpha_2, dots, alpha_n in mathbb{R}$ and $m_1, dots, m_n$ non-negative integers such that
$$|f(x) - sum_{i=1}^{n} alpha_ie^{m_ix}| leq varepsilon, forall x in [a, b].$$
This question provided a couple of hints for a slightly different version of mine, with $m_1, dots, m_n$ are negative integers, but I couldn't adapt them to the problem at hand or come up with any new tricks. I'm looking for a detailed solution, as I'm also on the process of learning uniform convergence and Berstein polynomials (which might be a way out of this).
real-analysis exponential-function uniform-convergence weierstrass-approximation
real-analysis exponential-function uniform-convergence weierstrass-approximation
asked Nov 20 '18 at 13:55
user71487
948
asked Nov 20 '18 at 13:55
user71487
948
edited Nov 20 '18 at 14:57
asked Nov 20 '18 at 13:55
user71487
948
asked Nov 20 '18 at 13:55
user71487
948
asked Nov 20 '18 at 13:55
user71487
948
948
1
Negative integers were necessary because the domain was not bounded. Here you can follow a very similar reasoning as the one you pointd out in the answer, taking into account that on your domain, the exponential are bounded.
– Martigan
Nov 20 '18 at 14:04
add a comment |
1
Negative integers were necessary because the domain was not bounded. Here you can follow a very similar reasoning as the one you pointd out in the answer, taking into account that on your domain, the exponential are bounded.
– Martigan
Nov 20 '18 at 14:04
1
1
Negative integers were necessary because the domain was not bounded. Here you can follow a very similar reasoning as the one you pointd out in the answer, taking into account that on your domain, the exponential are bounded.
– Martigan
Nov 20 '18 at 14:04
Negative integers were necessary because the domain was not bounded. Here you can follow a very similar reasoning as the one you pointd out in the answer, taking into account that on your domain, the exponential are bounded.
– Martigan
Nov 20 '18 at 14:04
add a comment |
2 Answers
2
active
oldest
votes
This is a direct consequence of Stone-Weierstrass theorem, considering the subalgebra
$$
left{sum_{i=1}^nalpha_ie^{m_ix} : alpha_iinmathbb R,m_igeq0right}
$$
of $C([a,b])$.
answered Nov 20 '18 at 14:23
Federico
4,679514
I understand informally that if the subalgebra above is dense in $C(X, mathbb(R))$, $X$ a compact set, then I can approximate any continuous real-valued function defined on $X$ by a polynomial in the subalgebra, but formally why is density equivalent to uniform convergence? I ask it as a clarification of why proving the density is enough to answer my question.
– user71487
Nov 20 '18 at 22:12
1
Density of $mathcal A$ in $C([a,b])$, with respect to the $|,cdot,|_infty$ norm, means that for every $fin C([a,b])$ and every $varepsilon>0$ you can find $ginmathcal A$ such that $|f-g|_infty<varepsilon$. This is precisely what you need to get uniform convergence. Indeed, you can find for instance $g_ninmathcal A$ such that $|f-g_n|_infty<1/n$, hence $g_nto f$ uniformly.
– Federico
Nov 21 '18 at 15:48
1
Also, "I can approximate any continuous real-valued function defined on X by a polynomial in the subalgebra", there are no polynomials in the subalgebra above.
– Federico
Nov 21 '18 at 15:51
1
Uniform convergence is precisely the convergence induced by the norm $|,cdot,|_infty$. Being able to converge uniformly to any continuous function with a sequence in $mathcal A$ is equivalent to saying that $mathcal A$ is dense.
– Federico
Nov 21 '18 at 15:53
add a comment |
Consider the function $phi(y) = f(log(y))$ on the domain $[e^a,e^b]$.
Find a polynomial $p(y) = sum_k alpha_k y^{m_k}$ such that $sup_{y in [e^a,e^b]}|phi(y)-p(y)| < epsilon$.
Then $sup_{y in [e^a,e^b]}|phi(y)-p(y)| = sup_{x in [a,b]}|f(x)-p(e^x)| < epsilon$.
answered Nov 20 '18 at 14:43
copper.hat
126k559159
Is there a standard way to do it? I've attempted the Gauss function approach to a dead end.
– user71487
Nov 20 '18 at 20:22
1
The polynomials are dense in $C[a,b]$.
– copper.hat
Nov 21 '18 at 0:59
add a comment |
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2 Answers
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active
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votes
2 Answers
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votes
This is a direct consequence of Stone-Weierstrass theorem, considering the subalgebra
$$
left{sum_{i=1}^nalpha_ie^{m_ix} : alpha_iinmathbb R,m_igeq0right}
$$
of $C([a,b])$.
answered Nov 20 '18 at 14:23
Federico
4,679514
I understand informally that if the subalgebra above is dense in $C(X, mathbb(R))$, $X$ a compact set, then I can approximate any continuous real-valued function defined on $X$ by a polynomial in the subalgebra, but formally why is density equivalent to uniform convergence? I ask it as a clarification of why proving the density is enough to answer my question.
– user71487
Nov 20 '18 at 22:12
1
Density of $mathcal A$ in $C([a,b])$, with respect to the $|,cdot,|_infty$ norm, means that for every $fin C([a,b])$ and every $varepsilon>0$ you can find $ginmathcal A$ such that $|f-g|_infty<varepsilon$. This is precisely what you need to get uniform convergence. Indeed, you can find for instance $g_ninmathcal A$ such that $|f-g_n|_infty<1/n$, hence $g_nto f$ uniformly.
– Federico
Nov 21 '18 at 15:48
1
Also, "I can approximate any continuous real-valued function defined on X by a polynomial in the subalgebra", there are no polynomials in the subalgebra above.
– Federico
Nov 21 '18 at 15:51
1
Uniform convergence is precisely the convergence induced by the norm $|,cdot,|_infty$. Being able to converge uniformly to any continuous function with a sequence in $mathcal A$ is equivalent to saying that $mathcal A$ is dense.
– Federico
Nov 21 '18 at 15:53
add a comment |
This is a direct consequence of Stone-Weierstrass theorem, considering the subalgebra
$$
left{sum_{i=1}^nalpha_ie^{m_ix} : alpha_iinmathbb R,m_igeq0right}
$$
of $C([a,b])$.
answered Nov 20 '18 at 14:23
Federico
4,679514
I understand informally that if the subalgebra above is dense in $C(X, mathbb(R))$, $X$ a compact set, then I can approximate any continuous real-valued function defined on $X$ by a polynomial in the subalgebra, but formally why is density equivalent to uniform convergence? I ask it as a clarification of why proving the density is enough to answer my question.
– user71487
Nov 20 '18 at 22:12
1
Density of $mathcal A$ in $C([a,b])$, with respect to the $|,cdot,|_infty$ norm, means that for every $fin C([a,b])$ and every $varepsilon>0$ you can find $ginmathcal A$ such that $|f-g|_infty<varepsilon$. This is precisely what you need to get uniform convergence. Indeed, you can find for instance $g_ninmathcal A$ such that $|f-g_n|_infty<1/n$, hence $g_nto f$ uniformly.
– Federico
Nov 21 '18 at 15:48
1
Also, "I can approximate any continuous real-valued function defined on X by a polynomial in the subalgebra", there are no polynomials in the subalgebra above.
– Federico
Nov 21 '18 at 15:51
1
Uniform convergence is precisely the convergence induced by the norm $|,cdot,|_infty$. Being able to converge uniformly to any continuous function with a sequence in $mathcal A$ is equivalent to saying that $mathcal A$ is dense.
– Federico
Nov 21 '18 at 15:53
add a comment |
This is a direct consequence of Stone-Weierstrass theorem, considering the subalgebra
$$
left{sum_{i=1}^nalpha_ie^{m_ix} : alpha_iinmathbb R,m_igeq0right}
$$
of $C([a,b])$.
answered Nov 20 '18 at 14:23
Federico
4,679514
This is a direct consequence of Stone-Weierstrass theorem, considering the subalgebra
$$
left{sum_{i=1}^nalpha_ie^{m_ix} : alpha_iinmathbb R,m_igeq0right}
$$
of $C([a,b])$.
answered Nov 20 '18 at 14:23
Federico
4,679514
answered Nov 20 '18 at 14:23
Federico
4,679514
answered Nov 20 '18 at 14:23
Federico
4,679514
answered Nov 20 '18 at 14:23
Federico
4,679514
4,679514
I understand informally that if the subalgebra above is dense in $C(X, mathbb(R))$, $X$ a compact set, then I can approximate any continuous real-valued function defined on $X$ by a polynomial in the subalgebra, but formally why is density equivalent to uniform convergence? I ask it as a clarification of why proving the density is enough to answer my question.
– user71487
Nov 20 '18 at 22:12
1
Density of $mathcal A$ in $C([a,b])$, with respect to the $|,cdot,|_infty$ norm, means that for every $fin C([a,b])$ and every $varepsilon>0$ you can find $ginmathcal A$ such that $|f-g|_infty<varepsilon$. This is precisely what you need to get uniform convergence. Indeed, you can find for instance $g_ninmathcal A$ such that $|f-g_n|_infty<1/n$, hence $g_nto f$ uniformly.
– Federico
Nov 21 '18 at 15:48
1
Also, "I can approximate any continuous real-valued function defined on X by a polynomial in the subalgebra", there are no polynomials in the subalgebra above.
– Federico
Nov 21 '18 at 15:51
1
Uniform convergence is precisely the convergence induced by the norm $|,cdot,|_infty$. Being able to converge uniformly to any continuous function with a sequence in $mathcal A$ is equivalent to saying that $mathcal A$ is dense.
– Federico
Nov 21 '18 at 15:53
add a comment |
I understand informally that if the subalgebra above is dense in $C(X, mathbb(R))$, $X$ a compact set, then I can approximate any continuous real-valued function defined on $X$ by a polynomial in the subalgebra, but formally why is density equivalent to uniform convergence? I ask it as a clarification of why proving the density is enough to answer my question.
– user71487
Nov 20 '18 at 22:12
1
Density of $mathcal A$ in $C([a,b])$, with respect to the $|,cdot,|_infty$ norm, means that for every $fin C([a,b])$ and every $varepsilon>0$ you can find $ginmathcal A$ such that $|f-g|_infty<varepsilon$. This is precisely what you need to get uniform convergence. Indeed, you can find for instance $g_ninmathcal A$ such that $|f-g_n|_infty<1/n$, hence $g_nto f$ uniformly.
– Federico
Nov 21 '18 at 15:48
1
Also, "I can approximate any continuous real-valued function defined on X by a polynomial in the subalgebra", there are no polynomials in the subalgebra above.
– Federico
Nov 21 '18 at 15:51
1
Uniform convergence is precisely the convergence induced by the norm $|,cdot,|_infty$. Being able to converge uniformly to any continuous function with a sequence in $mathcal A$ is equivalent to saying that $mathcal A$ is dense.
– Federico
Nov 21 '18 at 15:53
I understand informally that if the subalgebra above is dense in $C(X, mathbb(R))$, $X$ a compact set, then I can approximate any continuous real-valued function defined on $X$ by a polynomial in the subalgebra, but formally why is density equivalent to uniform convergence? I ask it as a clarification of why proving the density is enough to answer my question.
– user71487
Nov 20 '18 at 22:12
I understand informally that if the subalgebra above is dense in $C(X, mathbb(R))$, $X$ a compact set, then I can approximate any continuous real-valued function defined on $X$ by a polynomial in the subalgebra, but formally why is density equivalent to uniform convergence? I ask it as a clarification of why proving the density is enough to answer my question.
– user71487
Nov 20 '18 at 22:12
1
1
Density of $mathcal A$ in $C([a,b])$, with respect to the $|,cdot,|_infty$ norm, means that for every $fin C([a,b])$ and every $varepsilon>0$ you can find $ginmathcal A$ such that $|f-g|_infty<varepsilon$. This is precisely what you need to get uniform convergence. Indeed, you can find for instance $g_ninmathcal A$ such that $|f-g_n|_infty<1/n$, hence $g_nto f$ uniformly.
– Federico
Nov 21 '18 at 15:48
Density of $mathcal A$ in $C([a,b])$, with respect to the $|,cdot,|_infty$ norm, means that for every $fin C([a,b])$ and every $varepsilon>0$ you can find $ginmathcal A$ such that $|f-g|_infty<varepsilon$. This is precisely what you need to get uniform convergence. Indeed, you can find for instance $g_ninmathcal A$ such that $|f-g_n|_infty<1/n$, hence $g_nto f$ uniformly.
– Federico
Nov 21 '18 at 15:48
1
1
Also, "I can approximate any continuous real-valued function defined on X by a polynomial in the subalgebra", there are no polynomials in the subalgebra above.
– Federico
Nov 21 '18 at 15:51
Also, "I can approximate any continuous real-valued function defined on X by a polynomial in the subalgebra", there are no polynomials in the subalgebra above.
– Federico
Nov 21 '18 at 15:51
1
1
Uniform convergence is precisely the convergence induced by the norm $|,cdot,|_infty$. Being able to converge uniformly to any continuous function with a sequence in $mathcal A$ is equivalent to saying that $mathcal A$ is dense.
– Federico
Nov 21 '18 at 15:53
Uniform convergence is precisely the convergence induced by the norm $|,cdot,|_infty$. Being able to converge uniformly to any continuous function with a sequence in $mathcal A$ is equivalent to saying that $mathcal A$ is dense.
– Federico
Nov 21 '18 at 15:53
add a comment |
Consider the function $phi(y) = f(log(y))$ on the domain $[e^a,e^b]$.
Find a polynomial $p(y) = sum_k alpha_k y^{m_k}$ such that $sup_{y in [e^a,e^b]}|phi(y)-p(y)| < epsilon$.
Then $sup_{y in [e^a,e^b]}|phi(y)-p(y)| = sup_{x in [a,b]}|f(x)-p(e^x)| < epsilon$.
answered Nov 20 '18 at 14:43
copper.hat
126k559159
Is there a standard way to do it? I've attempted the Gauss function approach to a dead end.
– user71487
Nov 20 '18 at 20:22
1
The polynomials are dense in $C[a,b]$.
– copper.hat
Nov 21 '18 at 0:59
add a comment |
Consider the function $phi(y) = f(log(y))$ on the domain $[e^a,e^b]$.
Find a polynomial $p(y) = sum_k alpha_k y^{m_k}$ such that $sup_{y in [e^a,e^b]}|phi(y)-p(y)| < epsilon$.
Then $sup_{y in [e^a,e^b]}|phi(y)-p(y)| = sup_{x in [a,b]}|f(x)-p(e^x)| < epsilon$.
answered Nov 20 '18 at 14:43
copper.hat
126k559159
Is there a standard way to do it? I've attempted the Gauss function approach to a dead end.
– user71487
Nov 20 '18 at 20:22
1
The polynomials are dense in $C[a,b]$.
– copper.hat
Nov 21 '18 at 0:59
add a comment |
Consider the function $phi(y) = f(log(y))$ on the domain $[e^a,e^b]$.
Find a polynomial $p(y) = sum_k alpha_k y^{m_k}$ such that $sup_{y in [e^a,e^b]}|phi(y)-p(y)| < epsilon$.
Then $sup_{y in [e^a,e^b]}|phi(y)-p(y)| = sup_{x in [a,b]}|f(x)-p(e^x)| < epsilon$.
answered Nov 20 '18 at 14:43
copper.hat
126k559159
Consider the function $phi(y) = f(log(y))$ on the domain $[e^a,e^b]$.
Find a polynomial $p(y) = sum_k alpha_k y^{m_k}$ such that $sup_{y in [e^a,e^b]}|phi(y)-p(y)| < epsilon$.
Then $sup_{y in [e^a,e^b]}|phi(y)-p(y)| = sup_{x in [a,b]}|f(x)-p(e^x)| < epsilon$.
answered Nov 20 '18 at 14:43
copper.hat
126k559159
answered Nov 20 '18 at 14:43
copper.hat
126k559159
answered Nov 20 '18 at 14:43
copper.hat
126k559159
answered Nov 20 '18 at 14:43
copper.hat
126k559159
126k559159
Is there a standard way to do it? I've attempted the Gauss function approach to a dead end.
– user71487
Nov 20 '18 at 20:22
1
The polynomials are dense in $C[a,b]$.
– copper.hat
Nov 21 '18 at 0:59
add a comment |
Is there a standard way to do it? I've attempted the Gauss function approach to a dead end.
– user71487
Nov 20 '18 at 20:22
1
The polynomials are dense in $C[a,b]$.
– copper.hat
Nov 21 '18 at 0:59
Is there a standard way to do it? I've attempted the Gauss function approach to a dead end.
– user71487
Nov 20 '18 at 20:22
Is there a standard way to do it? I've attempted the Gauss function approach to a dead end.
– user71487
Nov 20 '18 at 20:22
1
1
The polynomials are dense in $C[a,b]$.
– copper.hat
Nov 21 '18 at 0:59
The polynomials are dense in $C[a,b]$.
– copper.hat
Nov 21 '18 at 0:59
add a comment |
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Negative integers were necessary because the domain was not bounded. Here you can follow a very similar reasoning as the one you pointd out in the answer, taking into account that on your domain, the exponential are bounded.
– Martigan
Nov 20 '18 at 14:04