Uniform convergence of a sequence on limited t range












1












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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?










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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
















1












$begingroup$


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?










share|cite|improve this question











$endgroup$












  • $begingroup$
    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














1












1








1


0



$begingroup$


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?










share|cite|improve this question











$endgroup$




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






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share|cite|improve this question













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edited Dec 12 '18 at 13:45









Arnaud D.

16k52443




16k52443










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


















  • $begingroup$
    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










1 Answer
1






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oldest

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1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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1 Answer
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1 Answer
1






active

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active

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1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 12 '18 at 13:29

























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


















  • $begingroup$
    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


















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