Uniform convergence of $ U_n(x) = sum_{n=0}^{+ infty} (-1)^n ln ( 1 + frac{x}{1+ nx} ) $.
$begingroup$
We consider the series of functions:
$$U_n(x) = sum_{k=0}^{n} (-1)^k ln left( 1 + frac{x}{1+ kx} right) ,~ x geq0.$$
Prove that $U_n$ is convergent.
Study the uniform convergence of $U_n$.
Study the normal convergence.
We consider $U(x) = sum_{n=0}^{+ infty} (-1)^n ln ( 1 + frac{x}{1+ nx} )$. Prove that $U(x)$ is of class $C^1$.
Compute $U'(x)$.
I have problem with question 2 and 4.
For question 2. I do not know how to answer this question since I do not know the value of the sum $U_n (x)$ to compute : $lim sup |U_n(x) - l| $. How to know if the series have uniform convergence?
For question 4. $U(x)$ is of class $C^1$ means it is differentiable and its derivative is continuous. If $ sum U_n$ converges uniformaly, $U$ will have the same properties, but how can I can prove that $U$ is of $C^1$ without computing $U'(x)$ first? I am confused because in the last question I am asked to compute $U'(x)$. Is there a way to deduce that from uniform convergence?
real-analysis sequences-and-series
edited Dec 28 '18 at 23:56
Larry
2,41131129
asked Dec 28 '18 at 14:08
Zouhair El YaagoubiZouhair El Yaagoubi
538411
$endgroup$
add a comment |
$begingroup$
We consider the series of functions:
$$U_n(x) = sum_{k=0}^{n} (-1)^k ln left( 1 + frac{x}{1+ kx} right) ,~ x geq0.$$
Prove that $U_n$ is convergent.
Study the uniform convergence of $U_n$.
Study the normal convergence.
We consider $U(x) = sum_{n=0}^{+ infty} (-1)^n ln ( 1 + frac{x}{1+ nx} )$. Prove that $U(x)$ is of class $C^1$.
Compute $U'(x)$.
I have problem with question 2 and 4.
For question 2. I do not know how to answer this question since I do not know the value of the sum $U_n (x)$ to compute : $lim sup |U_n(x) - l| $. How to know if the series have uniform convergence?
For question 4. $U(x)$ is of class $C^1$ means it is differentiable and its derivative is continuous. If $ sum U_n$ converges uniformaly, $U$ will have the same properties, but how can I can prove that $U$ is of $C^1$ without computing $U'(x)$ first? I am confused because in the last question I am asked to compute $U'(x)$. Is there a way to deduce that from uniform convergence?
real-analysis sequences-and-series
edited Dec 28 '18 at 23:56
Larry
2,41131129
asked Dec 28 '18 at 14:08
Zouhair El YaagoubiZouhair El Yaagoubi
538411
$endgroup$
add a comment |
$begingroup$
We consider the series of functions:
$$U_n(x) = sum_{k=0}^{n} (-1)^k ln left( 1 + frac{x}{1+ kx} right) ,~ x geq0.$$
Prove that $U_n$ is convergent.
Study the uniform convergence of $U_n$.
Study the normal convergence.
We consider $U(x) = sum_{n=0}^{+ infty} (-1)^n ln ( 1 + frac{x}{1+ nx} )$. Prove that $U(x)$ is of class $C^1$.
Compute $U'(x)$.
I have problem with question 2 and 4.
For question 2. I do not know how to answer this question since I do not know the value of the sum $U_n (x)$ to compute : $lim sup |U_n(x) - l| $. How to know if the series have uniform convergence?
For question 4. $U(x)$ is of class $C^1$ means it is differentiable and its derivative is continuous. If $ sum U_n$ converges uniformaly, $U$ will have the same properties, but how can I can prove that $U$ is of $C^1$ without computing $U'(x)$ first? I am confused because in the last question I am asked to compute $U'(x)$. Is there a way to deduce that from uniform convergence?
real-analysis sequences-and-series
edited Dec 28 '18 at 23:56
Larry
2,41131129
asked Dec 28 '18 at 14:08
Zouhair El YaagoubiZouhair El Yaagoubi
538411
$endgroup$
We consider the series of functions:
$$U_n(x) = sum_{k=0}^{n} (-1)^k ln left( 1 + frac{x}{1+ kx} right) ,~ x geq0.$$
Prove that $U_n$ is convergent.
Study the uniform convergence of $U_n$.
Study the normal convergence.
We consider $U(x) = sum_{n=0}^{+ infty} (-1)^n ln ( 1 + frac{x}{1+ nx} )$. Prove that $U(x)$ is of class $C^1$.
Compute $U'(x)$.
I have problem with question 2 and 4.
For question 2. I do not know how to answer this question since I do not know the value of the sum $U_n (x)$ to compute : $lim sup |U_n(x) - l| $. How to know if the series have uniform convergence?
For question 4. $U(x)$ is of class $C^1$ means it is differentiable and its derivative is continuous. If $ sum U_n$ converges uniformaly, $U$ will have the same properties, but how can I can prove that $U$ is of $C^1$ without computing $U'(x)$ first? I am confused because in the last question I am asked to compute $U'(x)$. Is there a way to deduce that from uniform convergence?
real-analysis sequences-and-series
real-analysis sequences-and-series
edited Dec 28 '18 at 23:56
Larry
2,41131129
asked Dec 28 '18 at 14:08
Zouhair El YaagoubiZouhair El Yaagoubi
538411
edited Dec 28 '18 at 23:56
Larry
2,41131129
asked Dec 28 '18 at 14:08
Zouhair El YaagoubiZouhair El Yaagoubi
538411
edited Dec 28 '18 at 23:56
Larry
2,41131129
edited Dec 28 '18 at 23:56
Larry
2,41131129
edited Dec 28 '18 at 23:56
Larry
2,41131129
2,41131129
asked Dec 28 '18 at 14:08
Zouhair El YaagoubiZouhair El Yaagoubi
538411
asked Dec 28 '18 at 14:08
Zouhair El YaagoubiZouhair El Yaagoubi
538411
asked Dec 28 '18 at 14:08
Zouhair El YaagoubiZouhair El Yaagoubi
538411
538411
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
This is a series with alternating sign terms, so a good idea would be to split it into even and odd terms. For even $n$, say $n=2m$, we have
$$U_{2m-1}(x) = sum_{k=0}^m ln left( 1+ frac{x}{1+2kx} right) - ln left( 1+ frac{x}{1+(2k+1)x} right)$$ which (after some calculations) turns out to be
$$sum_{k=0}^m ln left( 1+ frac{x^2}{(1+2kx)(1+2(k+1)x)} right)$$
As $m to + infty$ this series is normally convergent, thus $U_{2m-1}(x)$ is uniformly convergent on $Bbb R$. Now you should study the series $U_{2m}$ and conclude similarly.
Thus $U_n$ is uniformly convergent, although it is not normally convergent (since it is not absolutely convergent).
As for differentiability of $U(x)$, you have to apply this limit .
answered Dec 28 '18 at 15:17
CrostulCrostul
27.9k22352
$endgroup$
$begingroup$
In the order of the problem, I am asked about the uniform convergence first. You have started with the normal convergence. Although, I found that: $sum sup |U_n(x)| = sum ln (1 + frac{1}{n}) = + infty $, which is not normally convergent. For your last sentence, I want to know about the continuity of $U'(x)$ not the differentiability. I need to prove that $U(x) in C^1$
$endgroup$
– Zouhair El Yaagoubi
Dec 28 '18 at 16:45
$begingroup$
Indeed what I proved is that the series converges uniformly. After that, you have to study the uniform convergence of $U'_n(x)$ in a similar way: since those are continuous, they will converge to a continuous $U'(x)$.
$endgroup$
– Crostul
Dec 28 '18 at 18:04
add a comment |
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1 Answer
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1 Answer
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votes
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votes
$begingroup$
This is a series with alternating sign terms, so a good idea would be to split it into even and odd terms. For even $n$, say $n=2m$, we have
$$U_{2m-1}(x) = sum_{k=0}^m ln left( 1+ frac{x}{1+2kx} right) - ln left( 1+ frac{x}{1+(2k+1)x} right)$$ which (after some calculations) turns out to be
$$sum_{k=0}^m ln left( 1+ frac{x^2}{(1+2kx)(1+2(k+1)x)} right)$$
As $m to + infty$ this series is normally convergent, thus $U_{2m-1}(x)$ is uniformly convergent on $Bbb R$. Now you should study the series $U_{2m}$ and conclude similarly.
Thus $U_n$ is uniformly convergent, although it is not normally convergent (since it is not absolutely convergent).
As for differentiability of $U(x)$, you have to apply this limit .
answered Dec 28 '18 at 15:17
CrostulCrostul
27.9k22352
$endgroup$
$begingroup$
In the order of the problem, I am asked about the uniform convergence first. You have started with the normal convergence. Although, I found that: $sum sup |U_n(x)| = sum ln (1 + frac{1}{n}) = + infty $, which is not normally convergent. For your last sentence, I want to know about the continuity of $U'(x)$ not the differentiability. I need to prove that $U(x) in C^1$
$endgroup$
– Zouhair El Yaagoubi
Dec 28 '18 at 16:45
$begingroup$
Indeed what I proved is that the series converges uniformly. After that, you have to study the uniform convergence of $U'_n(x)$ in a similar way: since those are continuous, they will converge to a continuous $U'(x)$.
$endgroup$
– Crostul
Dec 28 '18 at 18:04
add a comment |
$begingroup$
This is a series with alternating sign terms, so a good idea would be to split it into even and odd terms. For even $n$, say $n=2m$, we have
$$U_{2m-1}(x) = sum_{k=0}^m ln left( 1+ frac{x}{1+2kx} right) - ln left( 1+ frac{x}{1+(2k+1)x} right)$$ which (after some calculations) turns out to be
$$sum_{k=0}^m ln left( 1+ frac{x^2}{(1+2kx)(1+2(k+1)x)} right)$$
As $m to + infty$ this series is normally convergent, thus $U_{2m-1}(x)$ is uniformly convergent on $Bbb R$. Now you should study the series $U_{2m}$ and conclude similarly.
Thus $U_n$ is uniformly convergent, although it is not normally convergent (since it is not absolutely convergent).
As for differentiability of $U(x)$, you have to apply this limit .
answered Dec 28 '18 at 15:17
CrostulCrostul
27.9k22352
$endgroup$
$begingroup$
In the order of the problem, I am asked about the uniform convergence first. You have started with the normal convergence. Although, I found that: $sum sup |U_n(x)| = sum ln (1 + frac{1}{n}) = + infty $, which is not normally convergent. For your last sentence, I want to know about the continuity of $U'(x)$ not the differentiability. I need to prove that $U(x) in C^1$
$endgroup$
– Zouhair El Yaagoubi
Dec 28 '18 at 16:45
$begingroup$
Indeed what I proved is that the series converges uniformly. After that, you have to study the uniform convergence of $U'_n(x)$ in a similar way: since those are continuous, they will converge to a continuous $U'(x)$.
$endgroup$
– Crostul
Dec 28 '18 at 18:04
add a comment |
$begingroup$
This is a series with alternating sign terms, so a good idea would be to split it into even and odd terms. For even $n$, say $n=2m$, we have
$$U_{2m-1}(x) = sum_{k=0}^m ln left( 1+ frac{x}{1+2kx} right) - ln left( 1+ frac{x}{1+(2k+1)x} right)$$ which (after some calculations) turns out to be
$$sum_{k=0}^m ln left( 1+ frac{x^2}{(1+2kx)(1+2(k+1)x)} right)$$
As $m to + infty$ this series is normally convergent, thus $U_{2m-1}(x)$ is uniformly convergent on $Bbb R$. Now you should study the series $U_{2m}$ and conclude similarly.
Thus $U_n$ is uniformly convergent, although it is not normally convergent (since it is not absolutely convergent).
As for differentiability of $U(x)$, you have to apply this limit .
answered Dec 28 '18 at 15:17
CrostulCrostul
27.9k22352
$endgroup$
This is a series with alternating sign terms, so a good idea would be to split it into even and odd terms. For even $n$, say $n=2m$, we have
$$U_{2m-1}(x) = sum_{k=0}^m ln left( 1+ frac{x}{1+2kx} right) - ln left( 1+ frac{x}{1+(2k+1)x} right)$$ which (after some calculations) turns out to be
$$sum_{k=0}^m ln left( 1+ frac{x^2}{(1+2kx)(1+2(k+1)x)} right)$$
As $m to + infty$ this series is normally convergent, thus $U_{2m-1}(x)$ is uniformly convergent on $Bbb R$. Now you should study the series $U_{2m}$ and conclude similarly.
Thus $U_n$ is uniformly convergent, although it is not normally convergent (since it is not absolutely convergent).
As for differentiability of $U(x)$, you have to apply this limit .
answered Dec 28 '18 at 15:17
CrostulCrostul
27.9k22352
answered Dec 28 '18 at 15:17
CrostulCrostul
27.9k22352
answered Dec 28 '18 at 15:17
CrostulCrostul
27.9k22352
answered Dec 28 '18 at 15:17
CrostulCrostul
27.9k22352
27.9k22352
$begingroup$
In the order of the problem, I am asked about the uniform convergence first. You have started with the normal convergence. Although, I found that: $sum sup |U_n(x)| = sum ln (1 + frac{1}{n}) = + infty $, which is not normally convergent. For your last sentence, I want to know about the continuity of $U'(x)$ not the differentiability. I need to prove that $U(x) in C^1$
$endgroup$
– Zouhair El Yaagoubi
Dec 28 '18 at 16:45
$begingroup$
Indeed what I proved is that the series converges uniformly. After that, you have to study the uniform convergence of $U'_n(x)$ in a similar way: since those are continuous, they will converge to a continuous $U'(x)$.
$endgroup$
– Crostul
Dec 28 '18 at 18:04
add a comment |
$begingroup$
In the order of the problem, I am asked about the uniform convergence first. You have started with the normal convergence. Although, I found that: $sum sup |U_n(x)| = sum ln (1 + frac{1}{n}) = + infty $, which is not normally convergent. For your last sentence, I want to know about the continuity of $U'(x)$ not the differentiability. I need to prove that $U(x) in C^1$
$endgroup$
– Zouhair El Yaagoubi
Dec 28 '18 at 16:45
$begingroup$
Indeed what I proved is that the series converges uniformly. After that, you have to study the uniform convergence of $U'_n(x)$ in a similar way: since those are continuous, they will converge to a continuous $U'(x)$.
$endgroup$
– Crostul
Dec 28 '18 at 18:04
$begingroup$
In the order of the problem, I am asked about the uniform convergence first. You have started with the normal convergence. Although, I found that: $sum sup |U_n(x)| = sum ln (1 + frac{1}{n}) = + infty $, which is not normally convergent. For your last sentence, I want to know about the continuity of $U'(x)$ not the differentiability. I need to prove that $U(x) in C^1$
$endgroup$
– Zouhair El Yaagoubi
Dec 28 '18 at 16:45
$begingroup$
In the order of the problem, I am asked about the uniform convergence first. You have started with the normal convergence. Although, I found that: $sum sup |U_n(x)| = sum ln (1 + frac{1}{n}) = + infty $, which is not normally convergent. For your last sentence, I want to know about the continuity of $U'(x)$ not the differentiability. I need to prove that $U(x) in C^1$
$endgroup$
– Zouhair El Yaagoubi
Dec 28 '18 at 16:45
$begingroup$
Indeed what I proved is that the series converges uniformly. After that, you have to study the uniform convergence of $U'_n(x)$ in a similar way: since those are continuous, they will converge to a continuous $U'(x)$.
$endgroup$
– Crostul
Dec 28 '18 at 18:04
$begingroup$
Indeed what I proved is that the series converges uniformly. After that, you have to study the uniform convergence of $U'_n(x)$ in a similar way: since those are continuous, they will converge to a continuous $U'(x)$.
$endgroup$
– Crostul
Dec 28 '18 at 18:04
add a comment |
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