Show that there exists $T>0$ such that...
Let $alphain,]0,1[$. I want to show that there exists $T>0$ such that the positive function
$F(t):=e^{-t}int_{0}^{t}frac{e^{xi}}{xi^{1-alpha}}dxi$ decays monotonically to zero.
I calculated
$$lim_{trightarrow infty} F(t)=0.$$
Also, $$F^{prime}(t)=frac{f(t)}{e^t}.$$
It remains then to show that there exists $T>0$ such that $f(t):=frac{e^{t}}{t^{1-alpha}}-int_{0}^{t}frac{e^{xi}}{xi^{1-alpha}}dxi<0$ for all $tgeq T$.
Notice that
$$f^{prime}(t)=-(1-alpha)frac{e^t}{t^{2-alpha}}<0.$$
So, $f$ is decreasing.
Therefore, it suffices to show that there exists $T>0$ such that
$$f(T)<0.$$
Testing the function $f$ numerically, it seems that there does exist $T=T(alpha)$ such that $f(T(alpha))<0$ for all $alphain,]0,1[.$
real-analysis calculus analysis multivariable-calculus inequality
asked Dec 9 at 4:55
Medo
611213
add a comment |
Let $alphain,]0,1[$. I want to show that there exists $T>0$ such that the positive function
$F(t):=e^{-t}int_{0}^{t}frac{e^{xi}}{xi^{1-alpha}}dxi$ decays monotonically to zero.
I calculated
$$lim_{trightarrow infty} F(t)=0.$$
Also, $$F^{prime}(t)=frac{f(t)}{e^t}.$$
It remains then to show that there exists $T>0$ such that $f(t):=frac{e^{t}}{t^{1-alpha}}-int_{0}^{t}frac{e^{xi}}{xi^{1-alpha}}dxi<0$ for all $tgeq T$.
Notice that
$$f^{prime}(t)=-(1-alpha)frac{e^t}{t^{2-alpha}}<0.$$
So, $f$ is decreasing.
Therefore, it suffices to show that there exists $T>0$ such that
$$f(T)<0.$$
Testing the function $f$ numerically, it seems that there does exist $T=T(alpha)$ such that $f(T(alpha))<0$ for all $alphain,]0,1[.$
real-analysis calculus analysis multivariable-calculus inequality
asked Dec 9 at 4:55
Medo
611213
add a comment |
Let $alphain,]0,1[$. I want to show that there exists $T>0$ such that the positive function
$F(t):=e^{-t}int_{0}^{t}frac{e^{xi}}{xi^{1-alpha}}dxi$ decays monotonically to zero.
I calculated
$$lim_{trightarrow infty} F(t)=0.$$
Also, $$F^{prime}(t)=frac{f(t)}{e^t}.$$
It remains then to show that there exists $T>0$ such that $f(t):=frac{e^{t}}{t^{1-alpha}}-int_{0}^{t}frac{e^{xi}}{xi^{1-alpha}}dxi<0$ for all $tgeq T$.
Notice that
$$f^{prime}(t)=-(1-alpha)frac{e^t}{t^{2-alpha}}<0.$$
So, $f$ is decreasing.
Therefore, it suffices to show that there exists $T>0$ such that
$$f(T)<0.$$
Testing the function $f$ numerically, it seems that there does exist $T=T(alpha)$ such that $f(T(alpha))<0$ for all $alphain,]0,1[.$
real-analysis calculus analysis multivariable-calculus inequality
asked Dec 9 at 4:55
Medo
611213
Let $alphain,]0,1[$. I want to show that there exists $T>0$ such that the positive function
$F(t):=e^{-t}int_{0}^{t}frac{e^{xi}}{xi^{1-alpha}}dxi$ decays monotonically to zero.
I calculated
$$lim_{trightarrow infty} F(t)=0.$$
Also, $$F^{prime}(t)=frac{f(t)}{e^t}.$$
It remains then to show that there exists $T>0$ such that $f(t):=frac{e^{t}}{t^{1-alpha}}-int_{0}^{t}frac{e^{xi}}{xi^{1-alpha}}dxi<0$ for all $tgeq T$.
Notice that
$$f^{prime}(t)=-(1-alpha)frac{e^t}{t^{2-alpha}}<0.$$
So, $f$ is decreasing.
Therefore, it suffices to show that there exists $T>0$ such that
$$f(T)<0.$$
Testing the function $f$ numerically, it seems that there does exist $T=T(alpha)$ such that $f(T(alpha))<0$ for all $alphain,]0,1[.$
real-analysis calculus analysis multivariable-calculus inequality
real-analysis calculus analysis multivariable-calculus inequality
asked Dec 9 at 4:55
Medo
611213
asked Dec 9 at 4:55
Medo
611213
edited Dec 9 at 11:02
asked Dec 9 at 4:55
Medo
611213
asked Dec 9 at 4:55
Medo
611213
asked Dec 9 at 4:55
Medo
611213
611213
add a comment |
add a comment |
1 Answer
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For $t>0$ we have $f’’(t)<0$, so $f’(t)$ is decreasing. Then for each $tge 1$ the value $$f(t)=f(1)+int_{1}^t f’(s)dsle f(1)+f’(1)(t-1)$$ is negative for all sufficiently large $t$.
answered Dec 9 at 12:40
Alex Ravsky
38.9k32079
1
Great. But with a little care because $f^{primeprime}(t)= (1-alpha),frac{e^{t}}{t^{2-alpha}},(-1+frac{2-alpha}{t})$ which is negative precisely when $t>2-alpha$. Thank you for you smart answer.
– Medo
Dec 9 at 16:49
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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votes
For $t>0$ we have $f’’(t)<0$, so $f’(t)$ is decreasing. Then for each $tge 1$ the value $$f(t)=f(1)+int_{1}^t f’(s)dsle f(1)+f’(1)(t-1)$$ is negative for all sufficiently large $t$.
answered Dec 9 at 12:40
Alex Ravsky
38.9k32079
1
Great. But with a little care because $f^{primeprime}(t)= (1-alpha),frac{e^{t}}{t^{2-alpha}},(-1+frac{2-alpha}{t})$ which is negative precisely when $t>2-alpha$. Thank you for you smart answer.
– Medo
Dec 9 at 16:49
add a comment |
For $t>0$ we have $f’’(t)<0$, so $f’(t)$ is decreasing. Then for each $tge 1$ the value $$f(t)=f(1)+int_{1}^t f’(s)dsle f(1)+f’(1)(t-1)$$ is negative for all sufficiently large $t$.
answered Dec 9 at 12:40
Alex Ravsky
38.9k32079
1
Great. But with a little care because $f^{primeprime}(t)= (1-alpha),frac{e^{t}}{t^{2-alpha}},(-1+frac{2-alpha}{t})$ which is negative precisely when $t>2-alpha$. Thank you for you smart answer.
– Medo
Dec 9 at 16:49
add a comment |
For $t>0$ we have $f’’(t)<0$, so $f’(t)$ is decreasing. Then for each $tge 1$ the value $$f(t)=f(1)+int_{1}^t f’(s)dsle f(1)+f’(1)(t-1)$$ is negative for all sufficiently large $t$.
answered Dec 9 at 12:40
Alex Ravsky
38.9k32079
For $t>0$ we have $f’’(t)<0$, so $f’(t)$ is decreasing. Then for each $tge 1$ the value $$f(t)=f(1)+int_{1}^t f’(s)dsle f(1)+f’(1)(t-1)$$ is negative for all sufficiently large $t$.
answered Dec 9 at 12:40
Alex Ravsky
38.9k32079
answered Dec 9 at 12:40
Alex Ravsky
38.9k32079
answered Dec 9 at 12:40
Alex Ravsky
38.9k32079
answered Dec 9 at 12:40
Alex Ravsky
38.9k32079
38.9k32079
1
Great. But with a little care because $f^{primeprime}(t)= (1-alpha),frac{e^{t}}{t^{2-alpha}},(-1+frac{2-alpha}{t})$ which is negative precisely when $t>2-alpha$. Thank you for you smart answer.
– Medo
Dec 9 at 16:49
add a comment |
1
Great. But with a little care because $f^{primeprime}(t)= (1-alpha),frac{e^{t}}{t^{2-alpha}},(-1+frac{2-alpha}{t})$ which is negative precisely when $t>2-alpha$. Thank you for you smart answer.
– Medo
Dec 9 at 16:49
1
1
Great. But with a little care because $f^{primeprime}(t)= (1-alpha),frac{e^{t}}{t^{2-alpha}},(-1+frac{2-alpha}{t})$ which is negative precisely when $t>2-alpha$. Thank you for you smart answer.
– Medo
Dec 9 at 16:49
Great. But with a little care because $f^{primeprime}(t)= (1-alpha),frac{e^{t}}{t^{2-alpha}},(-1+frac{2-alpha}{t})$ which is negative precisely when $t>2-alpha$. Thank you for you smart answer.
– Medo
Dec 9 at 16:49
add a comment |
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