Mixing
Logic behind comparison test for integrals
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I know the comparison test for improper integrals says:
"If $f(x), g(x)$ are continuous such that $0leq g(x)leq f(x)$ for $xgeq a,$ and if $int_{a}^infty{f(x)dx}$ converges, then $int_{a}^infty{g(x)dx}$ converges.
If $int_{a}^infty{g(x)dx}$ diverges, then $int_{a}^infty{f(x)dx}$ diverges.
However, when we see that $f(x)$ diverges, we can't say anything about the convergence of $g(x).$ What are we supposed to do in this situation?
Edit: I have been told sometimes that you can set $g(x)$ as an upper bound and search for a new function $h(x) leq g(x).$ If $h(x)$ diverges, then $g(x)$ must diverge. However, I don't think this makes sense because if $h(x)$ converges, you are stuck again.
calculus convergence
edited Jan 20 at 23:32
jordan_glen
1
asked Jan 20 at 22:38
JayJay
384
$endgroup$
add a comment |
$begingroup$
I know the comparison test for improper integrals says:
"If $f(x), g(x)$ are continuous such that $0leq g(x)leq f(x)$ for $xgeq a,$ and if $int_{a}^infty{f(x)dx}$ converges, then $int_{a}^infty{g(x)dx}$ converges.
If $int_{a}^infty{g(x)dx}$ diverges, then $int_{a}^infty{f(x)dx}$ diverges.
However, when we see that $f(x)$ diverges, we can't say anything about the convergence of $g(x).$ What are we supposed to do in this situation?
Edit: I have been told sometimes that you can set $g(x)$ as an upper bound and search for a new function $h(x) leq g(x).$ If $h(x)$ diverges, then $g(x)$ must diverge. However, I don't think this makes sense because if $h(x)$ converges, you are stuck again.
calculus convergence
edited Jan 20 at 23:32
jordan_glen
1
asked Jan 20 at 22:38
JayJay
384
$endgroup$
2
$begingroup$
Short and complete answer: nothing. Use another test. The analogy is the following: "I need some food. If I have money, I can buy it at the market. What if I have no money?" Well, if you have no money, you need to find your food in some other way.
$endgroup$
– Crostul
Jan 20 at 22:40
1
$begingroup$
Basically what it's saying is that if $f$ and $g$ are continuous on $(a,b)$ with $0leq g(x)leq f(x)$ for all $xin(a,b)$ then $int_a^b0;dxleqint_a^bg(x);dxleqint_a^bf(x);dx$. The left quantity is always zero, so if the right quantity is finite, the middle one is too. On the other hand, if the middle quantity is infinite, so is the right one.
$endgroup$
– Ben W
Jan 20 at 22:46
add a comment |
$begingroup$
I know the comparison test for improper integrals says:
"If $f(x), g(x)$ are continuous such that $0leq g(x)leq f(x)$ for $xgeq a,$ and if $int_{a}^infty{f(x)dx}$ converges, then $int_{a}^infty{g(x)dx}$ converges.
If $int_{a}^infty{g(x)dx}$ diverges, then $int_{a}^infty{f(x)dx}$ diverges.
However, when we see that $f(x)$ diverges, we can't say anything about the convergence of $g(x).$ What are we supposed to do in this situation?
Edit: I have been told sometimes that you can set $g(x)$ as an upper bound and search for a new function $h(x) leq g(x).$ If $h(x)$ diverges, then $g(x)$ must diverge. However, I don't think this makes sense because if $h(x)$ converges, you are stuck again.
calculus convergence
edited Jan 20 at 23:32
jordan_glen
1
asked Jan 20 at 22:38
JayJay
384
$endgroup$
I know the comparison test for improper integrals says:
"If $f(x), g(x)$ are continuous such that $0leq g(x)leq f(x)$ for $xgeq a,$ and if $int_{a}^infty{f(x)dx}$ converges, then $int_{a}^infty{g(x)dx}$ converges.
If $int_{a}^infty{g(x)dx}$ diverges, then $int_{a}^infty{f(x)dx}$ diverges.
However, when we see that $f(x)$ diverges, we can't say anything about the convergence of $g(x).$ What are we supposed to do in this situation?
Edit: I have been told sometimes that you can set $g(x)$ as an upper bound and search for a new function $h(x) leq g(x).$ If $h(x)$ diverges, then $g(x)$ must diverge. However, I don't think this makes sense because if $h(x)$ converges, you are stuck again.
calculus convergence
calculus convergence
edited Jan 20 at 23:32
jordan_glen
1
asked Jan 20 at 22:38
JayJay
384
edited Jan 20 at 23:32
jordan_glen
1
asked Jan 20 at 22:38
JayJay
384
edited Jan 20 at 23:32
jordan_glen
1
edited Jan 20 at 23:32
jordan_glen
1
edited Jan 20 at 23:32
jordan_glen
1
1
asked Jan 20 at 22:38
JayJay
384
asked Jan 20 at 22:38
JayJay
384
asked Jan 20 at 22:38
JayJay
384
384
2
$begingroup$
Short and complete answer: nothing. Use another test. The analogy is the following: "I need some food. If I have money, I can buy it at the market. What if I have no money?" Well, if you have no money, you need to find your food in some other way.
$endgroup$
– Crostul
Jan 20 at 22:40
1
$begingroup$
Basically what it's saying is that if $f$ and $g$ are continuous on $(a,b)$ with $0leq g(x)leq f(x)$ for all $xin(a,b)$ then $int_a^b0;dxleqint_a^bg(x);dxleqint_a^bf(x);dx$. The left quantity is always zero, so if the right quantity is finite, the middle one is too. On the other hand, if the middle quantity is infinite, so is the right one.
$endgroup$
– Ben W
Jan 20 at 22:46
add a comment |
2
$begingroup$
Short and complete answer: nothing. Use another test. The analogy is the following: "I need some food. If I have money, I can buy it at the market. What if I have no money?" Well, if you have no money, you need to find your food in some other way.
$endgroup$
– Crostul
Jan 20 at 22:40
1
$begingroup$
Basically what it's saying is that if $f$ and $g$ are continuous on $(a,b)$ with $0leq g(x)leq f(x)$ for all $xin(a,b)$ then $int_a^b0;dxleqint_a^bg(x);dxleqint_a^bf(x);dx$. The left quantity is always zero, so if the right quantity is finite, the middle one is too. On the other hand, if the middle quantity is infinite, so is the right one.
$endgroup$
– Ben W
Jan 20 at 22:46
2
2
$begingroup$
Short and complete answer: nothing. Use another test. The analogy is the following: "I need some food. If I have money, I can buy it at the market. What if I have no money?" Well, if you have no money, you need to find your food in some other way.
$endgroup$
– Crostul
Jan 20 at 22:40
$begingroup$
Short and complete answer: nothing. Use another test. The analogy is the following: "I need some food. If I have money, I can buy it at the market. What if I have no money?" Well, if you have no money, you need to find your food in some other way.
$endgroup$
– Crostul
Jan 20 at 22:40
1
1
$begingroup$
Basically what it's saying is that if $f$ and $g$ are continuous on $(a,b)$ with $0leq g(x)leq f(x)$ for all $xin(a,b)$ then $int_a^b0;dxleqint_a^bg(x);dxleqint_a^bf(x);dx$. The left quantity is always zero, so if the right quantity is finite, the middle one is too. On the other hand, if the middle quantity is infinite, so is the right one.
$endgroup$
– Ben W
Jan 20 at 22:46
$begingroup$
Basically what it's saying is that if $f$ and $g$ are continuous on $(a,b)$ with $0leq g(x)leq f(x)$ for all $xin(a,b)$ then $int_a^b0;dxleqint_a^bg(x);dxleqint_a^bf(x);dx$. The left quantity is always zero, so if the right quantity is finite, the middle one is too. On the other hand, if the middle quantity is infinite, so is the right one.
$endgroup$
– Ben W
Jan 20 at 22:46
add a comment |
1 Answer
1
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$begingroup$
The comparison test for integrals can only tell you if certain integrals converge, given certain bounds. If f(x) $le$ g(x) for on some interval I, then $$displaystyleint_I f(x) dx le displaystyleint_I g(x) dx$$
So if f diverges, then g must diverge and the contrapositive is true: that is if g converges, then f converges. Those are the only conclusions to be drawn. the hypotheses are not satisfied, then you have to use a different convergence test OR different functions.
answered Jan 21 at 0:10
Joel PereiraJoel Pereira
78719
$endgroup$
add a comment |
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$begingroup$
The comparison test for integrals can only tell you if certain integrals converge, given certain bounds. If f(x) $le$ g(x) for on some interval I, then $$displaystyleint_I f(x) dx le displaystyleint_I g(x) dx$$
So if f diverges, then g must diverge and the contrapositive is true: that is if g converges, then f converges. Those are the only conclusions to be drawn. the hypotheses are not satisfied, then you have to use a different convergence test OR different functions.
answered Jan 21 at 0:10
Joel PereiraJoel Pereira
78719
$endgroup$
add a comment |
$begingroup$
The comparison test for integrals can only tell you if certain integrals converge, given certain bounds. If f(x) $le$ g(x) for on some interval I, then $$displaystyleint_I f(x) dx le displaystyleint_I g(x) dx$$
So if f diverges, then g must diverge and the contrapositive is true: that is if g converges, then f converges. Those are the only conclusions to be drawn. the hypotheses are not satisfied, then you have to use a different convergence test OR different functions.
answered Jan 21 at 0:10
Joel PereiraJoel Pereira
78719
$endgroup$
add a comment |
$begingroup$
The comparison test for integrals can only tell you if certain integrals converge, given certain bounds. If f(x) $le$ g(x) for on some interval I, then $$displaystyleint_I f(x) dx le displaystyleint_I g(x) dx$$
So if f diverges, then g must diverge and the contrapositive is true: that is if g converges, then f converges. Those are the only conclusions to be drawn. the hypotheses are not satisfied, then you have to use a different convergence test OR different functions.
answered Jan 21 at 0:10
Joel PereiraJoel Pereira
78719
$endgroup$
The comparison test for integrals can only tell you if certain integrals converge, given certain bounds. If f(x) $le$ g(x) for on some interval I, then $$displaystyleint_I f(x) dx le displaystyleint_I g(x) dx$$
So if f diverges, then g must diverge and the contrapositive is true: that is if g converges, then f converges. Those are the only conclusions to be drawn. the hypotheses are not satisfied, then you have to use a different convergence test OR different functions.
answered Jan 21 at 0:10
Joel PereiraJoel Pereira
78719
answered Jan 21 at 0:10
Joel PereiraJoel Pereira
78719
answered Jan 21 at 0:10
Joel PereiraJoel Pereira
78719
answered Jan 21 at 0:10
Joel PereiraJoel Pereira
78719
78719
add a comment |
add a comment |
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$begingroup$
Short and complete answer: nothing. Use another test. The analogy is the following: "I need some food. If I have money, I can buy it at the market. What if I have no money?" Well, if you have no money, you need to find your food in some other way.
$endgroup$
– Crostul
Jan 20 at 22:40
1
$begingroup$
Basically what it's saying is that if $f$ and $g$ are continuous on $(a,b)$ with $0leq g(x)leq f(x)$ for all $xin(a,b)$ then $int_a^b0;dxleqint_a^bg(x);dxleqint_a^bf(x);dx$. The left quantity is always zero, so if the right quantity is finite, the middle one is too. On the other hand, if the middle quantity is infinite, so is the right one.
$endgroup$
– Ben W
Jan 20 at 22:46