Uniform convergence of a sequence on limited t range
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Hi I know the following sequence converges pointwise to $t$ from the limit but I am not sure if it converges uniformly for $t in [0,1]$ and $n≥1$
$$f_n(t)= frac {nt}{n+t}$$
I have found $|f_n(t)-f(t)|$ to be $$frac {-t^2}{n+t}$$
But I am not sure how to get to uniform convergence from here.
I believe it converges to $t$ uniformly, but could someone please clarify?
sequences-and-series convergence uniform-convergence
edited Dec 12 '18 at 13:45
Arnaud D.
16k52443
asked Dec 12 '18 at 13:19
MathExchange123MathExchange123
61
$endgroup$
add a comment |
$begingroup$
Hi I know the following sequence converges pointwise to $t$ from the limit but I am not sure if it converges uniformly for $t in [0,1]$ and $n≥1$
$$f_n(t)= frac {nt}{n+t}$$
I have found $|f_n(t)-f(t)|$ to be $$frac {-t^2}{n+t}$$
But I am not sure how to get to uniform convergence from here.
I believe it converges to $t$ uniformly, but could someone please clarify?
sequences-and-series convergence uniform-convergence
edited Dec 12 '18 at 13:45
Arnaud D.
16k52443
asked Dec 12 '18 at 13:19
MathExchange123MathExchange123
61
$endgroup$
$begingroup$
You recieved an answer to your question. Is it what you needed? If so, you should upvote and accept it.
$endgroup$
– 5xum
Jan 15 at 8:50
add a comment |
$begingroup$
Hi I know the following sequence converges pointwise to $t$ from the limit but I am not sure if it converges uniformly for $t in [0,1]$ and $n≥1$
$$f_n(t)= frac {nt}{n+t}$$
I have found $|f_n(t)-f(t)|$ to be $$frac {-t^2}{n+t}$$
But I am not sure how to get to uniform convergence from here.
I believe it converges to $t$ uniformly, but could someone please clarify?
sequences-and-series convergence uniform-convergence
edited Dec 12 '18 at 13:45
Arnaud D.
16k52443
asked Dec 12 '18 at 13:19
MathExchange123MathExchange123
61
$endgroup$
Hi I know the following sequence converges pointwise to $t$ from the limit but I am not sure if it converges uniformly for $t in [0,1]$ and $n≥1$
$$f_n(t)= frac {nt}{n+t}$$
I have found $|f_n(t)-f(t)|$ to be $$frac {-t^2}{n+t}$$
But I am not sure how to get to uniform convergence from here.
I believe it converges to $t$ uniformly, but could someone please clarify?
sequences-and-series convergence uniform-convergence
sequences-and-series convergence uniform-convergence
edited Dec 12 '18 at 13:45
Arnaud D.
16k52443
asked Dec 12 '18 at 13:19
MathExchange123MathExchange123
61
edited Dec 12 '18 at 13:45
Arnaud D.
16k52443
asked Dec 12 '18 at 13:19
MathExchange123MathExchange123
61
edited Dec 12 '18 at 13:45
Arnaud D.
16k52443
edited Dec 12 '18 at 13:45
Arnaud D.
16k52443
edited Dec 12 '18 at 13:45
Arnaud D.
16k52443
16k52443
asked Dec 12 '18 at 13:19
MathExchange123MathExchange123
61
asked Dec 12 '18 at 13:19
MathExchange123MathExchange123
61
asked Dec 12 '18 at 13:19
MathExchange123MathExchange123
61
61
$begingroup$
You recieved an answer to your question. Is it what you needed? If so, you should upvote and accept it.
$endgroup$
– 5xum
Jan 15 at 8:50
add a comment |
$begingroup$
You recieved an answer to your question. Is it what you needed? If so, you should upvote and accept it.
$endgroup$
– 5xum
Jan 15 at 8:50
$begingroup$
You recieved an answer to your question. Is it what you needed? If so, you should upvote and accept it.
$endgroup$
– 5xum
Jan 15 at 8:50
$begingroup$
You recieved an answer to your question. Is it what you needed? If so, you should upvote and accept it.
$endgroup$
– 5xum
Jan 15 at 8:50
add a comment |
1 Answer
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$begingroup$
First of all, a minor mistake, your calculation $$|f_n(t)-f(t)|=frac{-t^2}{n+t}$$ is wrong. For example, for $n=1,t=1$, the left side equals
$$left|frac{1cdot 1}{1+1} - 1right| = left|frac12right|=frac12$$ while the right side equals $$frac{-1^2}{1+1}=-frac12$$ Similar, but not the same :)
You are almost there. Remember, convergence is uniform if, for all $epsilon$, there exists some $N$ such that $sup_{tin[0,1]} |f_n(t)-f(t)| < epsilon$ for all $n>N$.
In your case, the supremum becomes a maximum, and you already (almost) calculated $|f_n(t)-f(t)|$. The fact that $$frac{t^2}{n+t} leq frac{1}{n+t}leq frac1n$$ should help you through the final steps.
answered Dec 12 '18 at 13:24
5xum5xum
91.1k394161
$endgroup$
$begingroup$
Does $ frac {-t^2}{n+t}$ being negative affect this? Also I'm not quite sure how to find the supremum on [0,1] or pick a suitable n. Could you please help me with this?
$endgroup$
– MathExchange123
Dec 12 '18 at 13:28
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@MathExchange123 Please, carefully read the first part of my answer.
$endgroup$
– 5xum
Dec 12 '18 at 13:28
$begingroup$
Thank you, I missed that the first time.
$endgroup$
– MathExchange123
Dec 12 '18 at 13:44
add a comment |
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1 Answer
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1 Answer
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$begingroup$
First of all, a minor mistake, your calculation $$|f_n(t)-f(t)|=frac{-t^2}{n+t}$$ is wrong. For example, for $n=1,t=1$, the left side equals
$$left|frac{1cdot 1}{1+1} - 1right| = left|frac12right|=frac12$$ while the right side equals $$frac{-1^2}{1+1}=-frac12$$ Similar, but not the same :)
You are almost there. Remember, convergence is uniform if, for all $epsilon$, there exists some $N$ such that $sup_{tin[0,1]} |f_n(t)-f(t)| < epsilon$ for all $n>N$.
In your case, the supremum becomes a maximum, and you already (almost) calculated $|f_n(t)-f(t)|$. The fact that $$frac{t^2}{n+t} leq frac{1}{n+t}leq frac1n$$ should help you through the final steps.
answered Dec 12 '18 at 13:24
5xum5xum
91.1k394161
$endgroup$
$begingroup$
Does $ frac {-t^2}{n+t}$ being negative affect this? Also I'm not quite sure how to find the supremum on [0,1] or pick a suitable n. Could you please help me with this?
$endgroup$
– MathExchange123
Dec 12 '18 at 13:28
$begingroup$
@MathExchange123 Please, carefully read the first part of my answer.
$endgroup$
– 5xum
Dec 12 '18 at 13:28
$begingroup$
Thank you, I missed that the first time.
$endgroup$
– MathExchange123
Dec 12 '18 at 13:44
add a comment |
$begingroup$
First of all, a minor mistake, your calculation $$|f_n(t)-f(t)|=frac{-t^2}{n+t}$$ is wrong. For example, for $n=1,t=1$, the left side equals
$$left|frac{1cdot 1}{1+1} - 1right| = left|frac12right|=frac12$$ while the right side equals $$frac{-1^2}{1+1}=-frac12$$ Similar, but not the same :)
You are almost there. Remember, convergence is uniform if, for all $epsilon$, there exists some $N$ such that $sup_{tin[0,1]} |f_n(t)-f(t)| < epsilon$ for all $n>N$.
In your case, the supremum becomes a maximum, and you already (almost) calculated $|f_n(t)-f(t)|$. The fact that $$frac{t^2}{n+t} leq frac{1}{n+t}leq frac1n$$ should help you through the final steps.
answered Dec 12 '18 at 13:24
5xum5xum
91.1k394161
$endgroup$
$begingroup$
Does $ frac {-t^2}{n+t}$ being negative affect this? Also I'm not quite sure how to find the supremum on [0,1] or pick a suitable n. Could you please help me with this?
$endgroup$
– MathExchange123
Dec 12 '18 at 13:28
$begingroup$
@MathExchange123 Please, carefully read the first part of my answer.
$endgroup$
– 5xum
Dec 12 '18 at 13:28
$begingroup$
Thank you, I missed that the first time.
$endgroup$
– MathExchange123
Dec 12 '18 at 13:44
add a comment |
$begingroup$
First of all, a minor mistake, your calculation $$|f_n(t)-f(t)|=frac{-t^2}{n+t}$$ is wrong. For example, for $n=1,t=1$, the left side equals
$$left|frac{1cdot 1}{1+1} - 1right| = left|frac12right|=frac12$$ while the right side equals $$frac{-1^2}{1+1}=-frac12$$ Similar, but not the same :)
You are almost there. Remember, convergence is uniform if, for all $epsilon$, there exists some $N$ such that $sup_{tin[0,1]} |f_n(t)-f(t)| < epsilon$ for all $n>N$.
In your case, the supremum becomes a maximum, and you already (almost) calculated $|f_n(t)-f(t)|$. The fact that $$frac{t^2}{n+t} leq frac{1}{n+t}leq frac1n$$ should help you through the final steps.
answered Dec 12 '18 at 13:24
5xum5xum
91.1k394161
$endgroup$
First of all, a minor mistake, your calculation $$|f_n(t)-f(t)|=frac{-t^2}{n+t}$$ is wrong. For example, for $n=1,t=1$, the left side equals
$$left|frac{1cdot 1}{1+1} - 1right| = left|frac12right|=frac12$$ while the right side equals $$frac{-1^2}{1+1}=-frac12$$ Similar, but not the same :)
You are almost there. Remember, convergence is uniform if, for all $epsilon$, there exists some $N$ such that $sup_{tin[0,1]} |f_n(t)-f(t)| < epsilon$ for all $n>N$.
In your case, the supremum becomes a maximum, and you already (almost) calculated $|f_n(t)-f(t)|$. The fact that $$frac{t^2}{n+t} leq frac{1}{n+t}leq frac1n$$ should help you through the final steps.
answered Dec 12 '18 at 13:24
5xum5xum
91.1k394161
edited Dec 12 '18 at 13:29
answered Dec 12 '18 at 13:24
5xum5xum
91.1k394161
answered Dec 12 '18 at 13:24
5xum5xum
91.1k394161
answered Dec 12 '18 at 13:24
5xum5xum
91.1k394161
91.1k394161
$begingroup$
Does $ frac {-t^2}{n+t}$ being negative affect this? Also I'm not quite sure how to find the supremum on [0,1] or pick a suitable n. Could you please help me with this?
$endgroup$
– MathExchange123
Dec 12 '18 at 13:28
$begingroup$
@MathExchange123 Please, carefully read the first part of my answer.
$endgroup$
– 5xum
Dec 12 '18 at 13:28
$begingroup$
Thank you, I missed that the first time.
$endgroup$
– MathExchange123
Dec 12 '18 at 13:44
add a comment |
$begingroup$
Does $ frac {-t^2}{n+t}$ being negative affect this? Also I'm not quite sure how to find the supremum on [0,1] or pick a suitable n. Could you please help me with this?
$endgroup$
– MathExchange123
Dec 12 '18 at 13:28
$begingroup$
@MathExchange123 Please, carefully read the first part of my answer.
$endgroup$
– 5xum
Dec 12 '18 at 13:28
$begingroup$
Thank you, I missed that the first time.
$endgroup$
– MathExchange123
Dec 12 '18 at 13:44
$begingroup$
Does $ frac {-t^2}{n+t}$ being negative affect this? Also I'm not quite sure how to find the supremum on [0,1] or pick a suitable n. Could you please help me with this?
$endgroup$
– MathExchange123
Dec 12 '18 at 13:28
$begingroup$
Does $ frac {-t^2}{n+t}$ being negative affect this? Also I'm not quite sure how to find the supremum on [0,1] or pick a suitable n. Could you please help me with this?
$endgroup$
– MathExchange123
Dec 12 '18 at 13:28
$begingroup$
@MathExchange123 Please, carefully read the first part of my answer.
$endgroup$
– 5xum
Dec 12 '18 at 13:28
$begingroup$
@MathExchange123 Please, carefully read the first part of my answer.
$endgroup$
– 5xum
Dec 12 '18 at 13:28
$begingroup$
Thank you, I missed that the first time.
$endgroup$
– MathExchange123
Dec 12 '18 at 13:44
$begingroup$
Thank you, I missed that the first time.
$endgroup$
– MathExchange123
Dec 12 '18 at 13:44
add a comment |
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$begingroup$
You recieved an answer to your question. Is it what you needed? If so, you should upvote and accept it.
$endgroup$
– 5xum
Jan 15 at 8:50