Does this mean I should show $L(x)$ is well-defined? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is this a valid way to show that the recursive sequence $x_n = x_n-1 + frac1x_n-1^2$ is unbounded?Is this functional well defined?If, $lim x_n$ exists and finite then there is a function $f$ that is continuousconvolution is well-defined and differentiable for continuous $f$ and differentiable $g$ with compact supportShow $x^alpha $ is well defined.Show for $f$ defined on $mathbb Q cap [0,1]$ uniformly continuous, exists extension $g$ defined on $[0,1]$ that is continuousShow that $T(X) = (X_(1), X_(n))$ is sufficient.Show operator is well-definedShow operator is well definedMetrics well defined
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Does this mean I should show $L(x)$ is well-defined?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is this a valid way to show that the recursive sequence $x_n = x_n-1 + frac1x_n-1^2$ is unbounded?Is this functional well defined?If, $lim x_n$ exists and finite then there is a function $f$ that is continuousconvolution is well-defined and differentiable for continuous $f$ and differentiable $g$ with compact supportShow $x^alpha $ is well defined.Show for $f$ defined on $mathbb Q cap [0,1]$ uniformly continuous, exists extension $g$ defined on $[0,1]$ that is continuousShow that $T(X) = (X_(1), X_(n))$ is sufficient.Show operator is well-definedShow operator is well definedMetrics well defined
$begingroup$
I need to show that $L(x) = int^x_1 frac1t dt$ is a function $L:(0,infty) to mathbbR$.
Do I just need to say that if $x = y$, then $L(x) = L(y)$ for $x,y in (0,infty)$? I'm not sure what this is asking exactly. Any help?
real-analysis
asked Apr 9 at 4:17
user439126user439126
1726
$endgroup$
add a comment |
$begingroup$
I need to show that $L(x) = int^x_1 frac1t dt$ is a function $L:(0,infty) to mathbbR$.
Do I just need to say that if $x = y$, then $L(x) = L(y)$ for $x,y in (0,infty)$? I'm not sure what this is asking exactly. Any help?
real-analysis
asked Apr 9 at 4:17
user439126user439126
1726
$endgroup$
2
$begingroup$
I would guess that you need to prove that the integral exists.
$endgroup$
– Ted
Apr 9 at 5:35
add a comment |
$begingroup$
I need to show that $L(x) = int^x_1 frac1t dt$ is a function $L:(0,infty) to mathbbR$.
Do I just need to say that if $x = y$, then $L(x) = L(y)$ for $x,y in (0,infty)$? I'm not sure what this is asking exactly. Any help?
real-analysis
asked Apr 9 at 4:17
user439126user439126
1726
$endgroup$
I need to show that $L(x) = int^x_1 frac1t dt$ is a function $L:(0,infty) to mathbbR$.
Do I just need to say that if $x = y$, then $L(x) = L(y)$ for $x,y in (0,infty)$? I'm not sure what this is asking exactly. Any help?
real-analysis
real-analysis
asked Apr 9 at 4:17
user439126user439126
1726
asked Apr 9 at 4:17
user439126user439126
1726
asked Apr 9 at 4:17
user439126user439126
1726
asked Apr 9 at 4:17
user439126user439126
1726
asked Apr 9 at 4:17
user439126user439126
1726
1726
2
$begingroup$
I would guess that you need to prove that the integral exists.
$endgroup$
– Ted
Apr 9 at 5:35
add a comment |
2
$begingroup$
I would guess that you need to prove that the integral exists.
$endgroup$
– Ted
Apr 9 at 5:35
2
2
$begingroup$
I would guess that you need to prove that the integral exists.
$endgroup$
– Ted
Apr 9 at 5:35
$begingroup$
I would guess that you need to prove that the integral exists.
$endgroup$
– Ted
Apr 9 at 5:35
add a comment |
1 Answer
1
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$begingroup$
We must show that the integral exists. Notice that the function $f(t) = frac1t$ is continuous on $ (0,+infty)$ and is therefore also continuous on $[1,x]$ ($x>1$) and $[x,1]$ for $1 > x > 0$. Since $f$ is continuous on $[1,x]$, $[x,1]$ it is also Riemann-integrable on $[1,x]$, $[x,1]$ and the function
$$ L : mathbbR_0^+ rightarrow mathbbR : x mapsto int_1^x frac1t dt $$
is well defined.
answered Apr 9 at 8:55
DigitalisDigitalis
624216
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add a comment |
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$begingroup$
We must show that the integral exists. Notice that the function $f(t) = frac1t$ is continuous on $ (0,+infty)$ and is therefore also continuous on $[1,x]$ ($x>1$) and $[x,1]$ for $1 > x > 0$. Since $f$ is continuous on $[1,x]$, $[x,1]$ it is also Riemann-integrable on $[1,x]$, $[x,1]$ and the function
$$ L : mathbbR_0^+ rightarrow mathbbR : x mapsto int_1^x frac1t dt $$
is well defined.
answered Apr 9 at 8:55
DigitalisDigitalis
624216
$endgroup$
add a comment |
$begingroup$
We must show that the integral exists. Notice that the function $f(t) = frac1t$ is continuous on $ (0,+infty)$ and is therefore also continuous on $[1,x]$ ($x>1$) and $[x,1]$ for $1 > x > 0$. Since $f$ is continuous on $[1,x]$, $[x,1]$ it is also Riemann-integrable on $[1,x]$, $[x,1]$ and the function
$$ L : mathbbR_0^+ rightarrow mathbbR : x mapsto int_1^x frac1t dt $$
is well defined.
answered Apr 9 at 8:55
DigitalisDigitalis
624216
$endgroup$
add a comment |
$begingroup$
We must show that the integral exists. Notice that the function $f(t) = frac1t$ is continuous on $ (0,+infty)$ and is therefore also continuous on $[1,x]$ ($x>1$) and $[x,1]$ for $1 > x > 0$. Since $f$ is continuous on $[1,x]$, $[x,1]$ it is also Riemann-integrable on $[1,x]$, $[x,1]$ and the function
$$ L : mathbbR_0^+ rightarrow mathbbR : x mapsto int_1^x frac1t dt $$
is well defined.
answered Apr 9 at 8:55
DigitalisDigitalis
624216
$endgroup$
We must show that the integral exists. Notice that the function $f(t) = frac1t$ is continuous on $ (0,+infty)$ and is therefore also continuous on $[1,x]$ ($x>1$) and $[x,1]$ for $1 > x > 0$. Since $f$ is continuous on $[1,x]$, $[x,1]$ it is also Riemann-integrable on $[1,x]$, $[x,1]$ and the function
$$ L : mathbbR_0^+ rightarrow mathbbR : x mapsto int_1^x frac1t dt $$
is well defined.
answered Apr 9 at 8:55
DigitalisDigitalis
624216
answered Apr 9 at 8:55
DigitalisDigitalis
624216
answered Apr 9 at 8:55
DigitalisDigitalis
624216
answered Apr 9 at 8:55
DigitalisDigitalis
624216
624216
add a comment |
add a comment |
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I would guess that you need to prove that the integral exists.
$endgroup$
– Ted
Apr 9 at 5:35