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Why does the limit of the power rule yield the natural log?
Clash Royale CLAN TAG#URR8PPP
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For any $a>0$ and any real number $r neq -1$, we have
$$int_a^x t^r dt = fracx^r+1r+1-fraca^r+1r+1$$
Now we can’t plug in $r=-1$ into the formula above, because we’d get zero in the denominator. But we can take the limit:
$$lim_r rightarrow -1int_a^x t^r , dt = lim_r rightarrow -1fracx^r+1-a^r+1r+1=ln(x)-ln(a) = int_a^x t^-1 dt = int_a^x lim_r rightarrow -1 t^r , dt$$
My question is, why was it possible to interchange the limit and integral here?
Is $t^r$ uniformly convergent to $t^-1$ as $r$ goes to $-1$? That would be one possible cause of being able to interchange limit and integral (though there are others).
calculus real-analysis integration convergence uniform-convergence
edited Jul 31 at 14:44
md2perpe
5,6391921
asked Jul 31 at 14:29
Keshav Srinivasan
1,74511338
add a comment |Â
up vote
3
down vote
favorite
For any $a>0$ and any real number $r neq -1$, we have
$$int_a^x t^r dt = fracx^r+1r+1-fraca^r+1r+1$$
Now we can’t plug in $r=-1$ into the formula above, because we’d get zero in the denominator. But we can take the limit:
$$lim_r rightarrow -1int_a^x t^r , dt = lim_r rightarrow -1fracx^r+1-a^r+1r+1=ln(x)-ln(a) = int_a^x t^-1 dt = int_a^x lim_r rightarrow -1 t^r , dt$$
My question is, why was it possible to interchange the limit and integral here?
Is $t^r$ uniformly convergent to $t^-1$ as $r$ goes to $-1$? That would be one possible cause of being able to interchange limit and integral (though there are others).
calculus real-analysis integration convergence uniform-convergence
edited Jul 31 at 14:44
md2perpe
5,6391921
asked Jul 31 at 14:29
Keshav Srinivasan
1,74511338
@InterstellarProbe Yeah, I already know how to find $$lim_r rightarrow -1fracx^r+1-a^r+1r+1$$
– Keshav Srinivasan
Jul 31 at 14:47
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
For any $a>0$ and any real number $r neq -1$, we have
$$int_a^x t^r dt = fracx^r+1r+1-fraca^r+1r+1$$
Now we can’t plug in $r=-1$ into the formula above, because we’d get zero in the denominator. But we can take the limit:
$$lim_r rightarrow -1int_a^x t^r , dt = lim_r rightarrow -1fracx^r+1-a^r+1r+1=ln(x)-ln(a) = int_a^x t^-1 dt = int_a^x lim_r rightarrow -1 t^r , dt$$
My question is, why was it possible to interchange the limit and integral here?
Is $t^r$ uniformly convergent to $t^-1$ as $r$ goes to $-1$? That would be one possible cause of being able to interchange limit and integral (though there are others).
calculus real-analysis integration convergence uniform-convergence
edited Jul 31 at 14:44
md2perpe
5,6391921
asked Jul 31 at 14:29
Keshav Srinivasan
1,74511338
For any $a>0$ and any real number $r neq -1$, we have
$$int_a^x t^r dt = fracx^r+1r+1-fraca^r+1r+1$$
Now we can’t plug in $r=-1$ into the formula above, because we’d get zero in the denominator. But we can take the limit:
$$lim_r rightarrow -1int_a^x t^r , dt = lim_r rightarrow -1fracx^r+1-a^r+1r+1=ln(x)-ln(a) = int_a^x t^-1 dt = int_a^x lim_r rightarrow -1 t^r , dt$$
My question is, why was it possible to interchange the limit and integral here?
Is $t^r$ uniformly convergent to $t^-1$ as $r$ goes to $-1$? That would be one possible cause of being able to interchange limit and integral (though there are others).
calculus real-analysis integration convergence uniform-convergence
edited Jul 31 at 14:44
md2perpe
5,6391921
asked Jul 31 at 14:29
Keshav Srinivasan
1,74511338
edited Jul 31 at 14:44
md2perpe
5,6391921
edited Jul 31 at 14:44
md2perpe
5,6391921
edited Jul 31 at 14:44
md2perpe
5,6391921
5,6391921
asked Jul 31 at 14:29
Keshav Srinivasan
1,74511338
asked Jul 31 at 14:29
Keshav Srinivasan
1,74511338
asked Jul 31 at 14:29
Keshav Srinivasan
1,74511338
1,74511338
@InterstellarProbe Yeah, I already know how to find $$lim_r rightarrow -1fracx^r+1-a^r+1r+1$$
– Keshav Srinivasan
Jul 31 at 14:47
add a comment |Â
@InterstellarProbe Yeah, I already know how to find $$lim_r rightarrow -1fracx^r+1-a^r+1r+1$$
– Keshav Srinivasan
Jul 31 at 14:47
@InterstellarProbe Yeah, I already know how to find $$lim_r rightarrow -1fracx^r+1-a^r+1r+1$$
– Keshav Srinivasan
Jul 31 at 14:47
@InterstellarProbe Yeah, I already know how to find $$lim_r rightarrow -1fracx^r+1-a^r+1r+1$$
– Keshav Srinivasan
Jul 31 at 14:47
add a comment |Â
2 Answers
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0
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I think you can apply the Arzela-Ascoli theorem here. Take a function sequence $f_n$ with $ f_n(t) = t^r_n $ defined on $[a,x]$ with $ r_n rightarrow -1 $ monotonly. Obviously, it is uniformly bounded. So are the derivatives $f_n'(t)$ which means that $f_n$ is equicontinous. Due to Arzela-Ascoli we then get uniform convergence of at least a subsequence of $f_n$. However, one should easily be able to show the uniform convergence of the whole sequence using the monotony of the values $ f_n(t) _n in mathbbN $ at each point $ t in [a,x] $.
Herefor, you can interchange integral and limit if you restrict to a concrete sequence $r_n$ as above. The generalization to continous $r$ should be possible by again exploiting the monotony properties w.r.t. $r$.
Edit: Details to the continuous case
Let $r_n rightarrow -1$ be monotonly decreasing and wlog $r_n<0$ . Then for each $t in [a,x]$, $f_n(t) = t^r_n$ is monotonly increasing. So for an arbitrary $r in (-1,r_1)$ there is an $n$ such that $r_n >=r >= r_n+1 $. Therefore, we get $f_n(t) > t^r > f_n+1 $ and hence
$ int_a^x f_n(t) dt > int_a^x t^r dt > int_a^x f_n+1 dt $
which implies the continuous convergence for $r>-1$. Analogously, this can be done for $r<-1$.
answered Jul 31 at 15:01
til
694
If you can flesh out the details (in the continuous $r$ case), I’d be happy to accept your answer.
– Keshav Srinivasan
Jul 31 at 15:04
@KeshavSrinivasan I tried, but its a bit ugly to carry out concretely i think ... :-)
– til
Jul 31 at 15:45
add a comment |Â
up vote
0
down vote
It suffices to show that $f:(r,t) mapsto exp(rlogt)$ is continuous on $ [a,x] times [-1-varepsilon,-1+varepsilon] $ for fixed $0<a<x$: then $f$ is uniformly continuous since this set is compact. So given any $varepsilon>0$, there is one $delta$ so that $lvert f(r,t)-f(r',t') rvert<varepsilon$ for any pairs with $d((r,t),(r',t'))<delta$ for, e.g. the Euclidean distance on the product of intervals.
We want to show that $f(r,t) to f(r_0,t)$ uniformly as $r to r_0$. But for any $varepsilon>0$, there is a single $delta$ so that $lvert r-r_0 rvert=d((r,t),(r_0,t))<delta$ implies that $lvert f(r,t)-f(r_0,t)rvert<varepsilon$, for any $t$; this is just a specialisation of the condition of being uniformly continuous, so we are done.
Why is $f$ continuous? It's a composition of continuous functions
$$ exp circ operatornamemult circ (operatornameid,log), $$
where mult is the function that sends $(a,b)mapsto ab$.
answered Jul 31 at 16:17
Chappers
54.9k74190
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I think you can apply the Arzela-Ascoli theorem here. Take a function sequence $f_n$ with $ f_n(t) = t^r_n $ defined on $[a,x]$ with $ r_n rightarrow -1 $ monotonly. Obviously, it is uniformly bounded. So are the derivatives $f_n'(t)$ which means that $f_n$ is equicontinous. Due to Arzela-Ascoli we then get uniform convergence of at least a subsequence of $f_n$. However, one should easily be able to show the uniform convergence of the whole sequence using the monotony of the values $ f_n(t) _n in mathbbN $ at each point $ t in [a,x] $.
Herefor, you can interchange integral and limit if you restrict to a concrete sequence $r_n$ as above. The generalization to continous $r$ should be possible by again exploiting the monotony properties w.r.t. $r$.
Edit: Details to the continuous case
Let $r_n rightarrow -1$ be monotonly decreasing and wlog $r_n<0$ . Then for each $t in [a,x]$, $f_n(t) = t^r_n$ is monotonly increasing. So for an arbitrary $r in (-1,r_1)$ there is an $n$ such that $r_n >=r >= r_n+1 $. Therefore, we get $f_n(t) > t^r > f_n+1 $ and hence
$ int_a^x f_n(t) dt > int_a^x t^r dt > int_a^x f_n+1 dt $
which implies the continuous convergence for $r>-1$. Analogously, this can be done for $r<-1$.
answered Jul 31 at 15:01
til
694
If you can flesh out the details (in the continuous $r$ case), I’d be happy to accept your answer.
– Keshav Srinivasan
Jul 31 at 15:04
@KeshavSrinivasan I tried, but its a bit ugly to carry out concretely i think ... :-)
– til
Jul 31 at 15:45
add a comment |Â
up vote
0
down vote
I think you can apply the Arzela-Ascoli theorem here. Take a function sequence $f_n$ with $ f_n(t) = t^r_n $ defined on $[a,x]$ with $ r_n rightarrow -1 $ monotonly. Obviously, it is uniformly bounded. So are the derivatives $f_n'(t)$ which means that $f_n$ is equicontinous. Due to Arzela-Ascoli we then get uniform convergence of at least a subsequence of $f_n$. However, one should easily be able to show the uniform convergence of the whole sequence using the monotony of the values $ f_n(t) _n in mathbbN $ at each point $ t in [a,x] $.
Herefor, you can interchange integral and limit if you restrict to a concrete sequence $r_n$ as above. The generalization to continous $r$ should be possible by again exploiting the monotony properties w.r.t. $r$.
Edit: Details to the continuous case
Let $r_n rightarrow -1$ be monotonly decreasing and wlog $r_n<0$ . Then for each $t in [a,x]$, $f_n(t) = t^r_n$ is monotonly increasing. So for an arbitrary $r in (-1,r_1)$ there is an $n$ such that $r_n >=r >= r_n+1 $. Therefore, we get $f_n(t) > t^r > f_n+1 $ and hence
$ int_a^x f_n(t) dt > int_a^x t^r dt > int_a^x f_n+1 dt $
which implies the continuous convergence for $r>-1$. Analogously, this can be done for $r<-1$.
answered Jul 31 at 15:01
til
694
If you can flesh out the details (in the continuous $r$ case), I’d be happy to accept your answer.
– Keshav Srinivasan
Jul 31 at 15:04
@KeshavSrinivasan I tried, but its a bit ugly to carry out concretely i think ... :-)
– til
Jul 31 at 15:45
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I think you can apply the Arzela-Ascoli theorem here. Take a function sequence $f_n$ with $ f_n(t) = t^r_n $ defined on $[a,x]$ with $ r_n rightarrow -1 $ monotonly. Obviously, it is uniformly bounded. So are the derivatives $f_n'(t)$ which means that $f_n$ is equicontinous. Due to Arzela-Ascoli we then get uniform convergence of at least a subsequence of $f_n$. However, one should easily be able to show the uniform convergence of the whole sequence using the monotony of the values $ f_n(t) _n in mathbbN $ at each point $ t in [a,x] $.
Herefor, you can interchange integral and limit if you restrict to a concrete sequence $r_n$ as above. The generalization to continous $r$ should be possible by again exploiting the monotony properties w.r.t. $r$.
Edit: Details to the continuous case
Let $r_n rightarrow -1$ be monotonly decreasing and wlog $r_n<0$ . Then for each $t in [a,x]$, $f_n(t) = t^r_n$ is monotonly increasing. So for an arbitrary $r in (-1,r_1)$ there is an $n$ such that $r_n >=r >= r_n+1 $. Therefore, we get $f_n(t) > t^r > f_n+1 $ and hence
$ int_a^x f_n(t) dt > int_a^x t^r dt > int_a^x f_n+1 dt $
which implies the continuous convergence for $r>-1$. Analogously, this can be done for $r<-1$.
answered Jul 31 at 15:01
til
694
I think you can apply the Arzela-Ascoli theorem here. Take a function sequence $f_n$ with $ f_n(t) = t^r_n $ defined on $[a,x]$ with $ r_n rightarrow -1 $ monotonly. Obviously, it is uniformly bounded. So are the derivatives $f_n'(t)$ which means that $f_n$ is equicontinous. Due to Arzela-Ascoli we then get uniform convergence of at least a subsequence of $f_n$. However, one should easily be able to show the uniform convergence of the whole sequence using the monotony of the values $ f_n(t) _n in mathbbN $ at each point $ t in [a,x] $.
Herefor, you can interchange integral and limit if you restrict to a concrete sequence $r_n$ as above. The generalization to continous $r$ should be possible by again exploiting the monotony properties w.r.t. $r$.
Edit: Details to the continuous case
Let $r_n rightarrow -1$ be monotonly decreasing and wlog $r_n<0$ . Then for each $t in [a,x]$, $f_n(t) = t^r_n$ is monotonly increasing. So for an arbitrary $r in (-1,r_1)$ there is an $n$ such that $r_n >=r >= r_n+1 $. Therefore, we get $f_n(t) > t^r > f_n+1 $ and hence
$ int_a^x f_n(t) dt > int_a^x t^r dt > int_a^x f_n+1 dt $
which implies the continuous convergence for $r>-1$. Analogously, this can be done for $r<-1$.
answered Jul 31 at 15:01
til
694
edited Jul 31 at 15:45
answered Jul 31 at 15:01
til
694
answered Jul 31 at 15:01
til
694
answered Jul 31 at 15:01
til
694
694
If you can flesh out the details (in the continuous $r$ case), I’d be happy to accept your answer.
– Keshav Srinivasan
Jul 31 at 15:04
@KeshavSrinivasan I tried, but its a bit ugly to carry out concretely i think ... :-)
– til
Jul 31 at 15:45
add a comment |Â
If you can flesh out the details (in the continuous $r$ case), I’d be happy to accept your answer.
– Keshav Srinivasan
Jul 31 at 15:04
@KeshavSrinivasan I tried, but its a bit ugly to carry out concretely i think ... :-)
– til
Jul 31 at 15:45
If you can flesh out the details (in the continuous $r$ case), I’d be happy to accept your answer.
– Keshav Srinivasan
Jul 31 at 15:04
If you can flesh out the details (in the continuous $r$ case), I’d be happy to accept your answer.
– Keshav Srinivasan
Jul 31 at 15:04
@KeshavSrinivasan I tried, but its a bit ugly to carry out concretely i think ... :-)
– til
Jul 31 at 15:45
@KeshavSrinivasan I tried, but its a bit ugly to carry out concretely i think ... :-)
– til
Jul 31 at 15:45
add a comment |Â
up vote
0
down vote
It suffices to show that $f:(r,t) mapsto exp(rlogt)$ is continuous on $ [a,x] times [-1-varepsilon,-1+varepsilon] $ for fixed $0<a<x$: then $f$ is uniformly continuous since this set is compact. So given any $varepsilon>0$, there is one $delta$ so that $lvert f(r,t)-f(r',t') rvert<varepsilon$ for any pairs with $d((r,t),(r',t'))<delta$ for, e.g. the Euclidean distance on the product of intervals.
We want to show that $f(r,t) to f(r_0,t)$ uniformly as $r to r_0$. But for any $varepsilon>0$, there is a single $delta$ so that $lvert r-r_0 rvert=d((r,t),(r_0,t))<delta$ implies that $lvert f(r,t)-f(r_0,t)rvert<varepsilon$, for any $t$; this is just a specialisation of the condition of being uniformly continuous, so we are done.
Why is $f$ continuous? It's a composition of continuous functions
$$ exp circ operatornamemult circ (operatornameid,log), $$
where mult is the function that sends $(a,b)mapsto ab$.
answered Jul 31 at 16:17
Chappers
54.9k74190
add a comment |Â
up vote
0
down vote
It suffices to show that $f:(r,t) mapsto exp(rlogt)$ is continuous on $ [a,x] times [-1-varepsilon,-1+varepsilon] $ for fixed $0<a<x$: then $f$ is uniformly continuous since this set is compact. So given any $varepsilon>0$, there is one $delta$ so that $lvert f(r,t)-f(r',t') rvert<varepsilon$ for any pairs with $d((r,t),(r',t'))<delta$ for, e.g. the Euclidean distance on the product of intervals.
We want to show that $f(r,t) to f(r_0,t)$ uniformly as $r to r_0$. But for any $varepsilon>0$, there is a single $delta$ so that $lvert r-r_0 rvert=d((r,t),(r_0,t))<delta$ implies that $lvert f(r,t)-f(r_0,t)rvert<varepsilon$, for any $t$; this is just a specialisation of the condition of being uniformly continuous, so we are done.
Why is $f$ continuous? It's a composition of continuous functions
$$ exp circ operatornamemult circ (operatornameid,log), $$
where mult is the function that sends $(a,b)mapsto ab$.
answered Jul 31 at 16:17
Chappers
54.9k74190
add a comment |Â
up vote
0
down vote
up vote
0
down vote
It suffices to show that $f:(r,t) mapsto exp(rlogt)$ is continuous on $ [a,x] times [-1-varepsilon,-1+varepsilon] $ for fixed $0<a<x$: then $f$ is uniformly continuous since this set is compact. So given any $varepsilon>0$, there is one $delta$ so that $lvert f(r,t)-f(r',t') rvert<varepsilon$ for any pairs with $d((r,t),(r',t'))<delta$ for, e.g. the Euclidean distance on the product of intervals.
We want to show that $f(r,t) to f(r_0,t)$ uniformly as $r to r_0$. But for any $varepsilon>0$, there is a single $delta$ so that $lvert r-r_0 rvert=d((r,t),(r_0,t))<delta$ implies that $lvert f(r,t)-f(r_0,t)rvert<varepsilon$, for any $t$; this is just a specialisation of the condition of being uniformly continuous, so we are done.
Why is $f$ continuous? It's a composition of continuous functions
$$ exp circ operatornamemult circ (operatornameid,log), $$
where mult is the function that sends $(a,b)mapsto ab$.
answered Jul 31 at 16:17
Chappers
54.9k74190
It suffices to show that $f:(r,t) mapsto exp(rlogt)$ is continuous on $ [a,x] times [-1-varepsilon,-1+varepsilon] $ for fixed $0<a<x$: then $f$ is uniformly continuous since this set is compact. So given any $varepsilon>0$, there is one $delta$ so that $lvert f(r,t)-f(r',t') rvert<varepsilon$ for any pairs with $d((r,t),(r',t'))<delta$ for, e.g. the Euclidean distance on the product of intervals.
We want to show that $f(r,t) to f(r_0,t)$ uniformly as $r to r_0$. But for any $varepsilon>0$, there is a single $delta$ so that $lvert r-r_0 rvert=d((r,t),(r_0,t))<delta$ implies that $lvert f(r,t)-f(r_0,t)rvert<varepsilon$, for any $t$; this is just a specialisation of the condition of being uniformly continuous, so we are done.
Why is $f$ continuous? It's a composition of continuous functions
$$ exp circ operatornamemult circ (operatornameid,log), $$
where mult is the function that sends $(a,b)mapsto ab$.
answered Jul 31 at 16:17
Chappers
54.9k74190
answered Jul 31 at 16:17
Chappers
54.9k74190
answered Jul 31 at 16:17
Chappers
54.9k74190
answered Jul 31 at 16:17
Chappers
54.9k74190
54.9k74190
add a comment |Â
add a comment |Â
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@InterstellarProbe Yeah, I already know how to find $$lim_r rightarrow -1fracx^r+1-a^r+1r+1$$
– Keshav Srinivasan
Jul 31 at 14:47