Given $f(x)=o(1), g(x)=O(h(x)) text{ as } xto a $, show that $f(x)g(x)=o(h(x))$
$begingroup$
We are given that $f(x)=o(1)$ as $ xto a $, $g(x)=O(h(x))$ as $ xto a $. It is required to prove or show that $f(x)g(x)=o(h(x))$ as $ xto a $
I just multiplied the two functions to get: $$f(x)g(x)=o(1)O(h(x))=color{red}{o(1)O(1)h(x) = o(1)h(x)=o(h(x))} text{ as $xto a$ } $$
My main question is whether or not the part in red is a valid use of the properties of asymptotic expressions.
calculus limits discrete-mathematics asymptotics
asked Dec 14 '18 at 10:27
E.NoleE.Nole
113112
$endgroup$
add a comment |
$begingroup$
We are given that $f(x)=o(1)$ as $ xto a $, $g(x)=O(h(x))$ as $ xto a $. It is required to prove or show that $f(x)g(x)=o(h(x))$ as $ xto a $
I just multiplied the two functions to get: $$f(x)g(x)=o(1)O(h(x))=color{red}{o(1)O(1)h(x) = o(1)h(x)=o(h(x))} text{ as $xto a$ } $$
My main question is whether or not the part in red is a valid use of the properties of asymptotic expressions.
calculus limits discrete-mathematics asymptotics
asked Dec 14 '18 at 10:27
E.NoleE.Nole
113112
$endgroup$
add a comment |
$begingroup$
We are given that $f(x)=o(1)$ as $ xto a $, $g(x)=O(h(x))$ as $ xto a $. It is required to prove or show that $f(x)g(x)=o(h(x))$ as $ xto a $
I just multiplied the two functions to get: $$f(x)g(x)=o(1)O(h(x))=color{red}{o(1)O(1)h(x) = o(1)h(x)=o(h(x))} text{ as $xto a$ } $$
My main question is whether or not the part in red is a valid use of the properties of asymptotic expressions.
calculus limits discrete-mathematics asymptotics
asked Dec 14 '18 at 10:27
E.NoleE.Nole
113112
$endgroup$
We are given that $f(x)=o(1)$ as $ xto a $, $g(x)=O(h(x))$ as $ xto a $. It is required to prove or show that $f(x)g(x)=o(h(x))$ as $ xto a $
I just multiplied the two functions to get: $$f(x)g(x)=o(1)O(h(x))=color{red}{o(1)O(1)h(x) = o(1)h(x)=o(h(x))} text{ as $xto a$ } $$
My main question is whether or not the part in red is a valid use of the properties of asymptotic expressions.
calculus limits discrete-mathematics asymptotics
calculus limits discrete-mathematics asymptotics
asked Dec 14 '18 at 10:27
E.NoleE.Nole
113112
asked Dec 14 '18 at 10:27
E.NoleE.Nole
113112
asked Dec 14 '18 at 10:27
E.NoleE.Nole
113112
asked Dec 14 '18 at 10:27
E.NoleE.Nole
113112
asked Dec 14 '18 at 10:27
E.NoleE.Nole
113112
113112
add a comment |
add a comment |
1 Answer
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$begingroup$
Recall that by definition as $xto a$
$$f(x)=o(1)iff f(x)=1cdot omega_1(x) quad omega_1(x)to 0$$
$$g(x)=O(h(x)) iff g(x)=h(x)cdot omega_2(x) quad limsup|omega_2(x)|in mathbb R$$
then
$$f(x)g(x)=h(x)cdotomega_1(x)cdotomega_2(x)=h(x)cdot omega_3(x) quad omega_3(x)to 0$$
indeed we have that $|omega_2(x)|le M$ and by squeeze theorem
$$|omega_3(x)|=|omega_1(x)cdotomega_2(x)|le M|omega_1(x)to 0$$
In the same way we can see that
$O(h(x))=h(x)cdot O(1)$
$o(1)cdot O(1)=o(1)$
answered Dec 14 '18 at 10:30
gimusigimusi
1
$endgroup$
$begingroup$
I was given slightly different definitions so I'd like to clarify, How is $g(x)=O(h(x)) iff g(x)=h(x)cdot omega_2(x) quad limsup|omega_2(x)|in mathbb R$? The limsup part is where I'm particularly lost. The definition I know is that $g(x)=O(h(x)) iff lim_{xrightarrow a} frac{g(x)}{h(x)} = C < infty$
$endgroup$
– E.Nole
Dec 14 '18 at 15:40
1
$begingroup$
Your definition $$lim_{xrightarrow a} frac{g(x)}{h(x)} = C < infty$$ is uncorrect we should have $$limsup_{xrightarrow a} left|frac{g(x)}{h(x)}right| = C < infty$$ Refer to formal definition. The definition I used is more general since include also te case with $h(x)=0$.
$endgroup$
– gimusi
Dec 14 '18 at 15:50
add a comment |
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1 Answer
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active
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1 Answer
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active
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active
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active
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$begingroup$
Recall that by definition as $xto a$
$$f(x)=o(1)iff f(x)=1cdot omega_1(x) quad omega_1(x)to 0$$
$$g(x)=O(h(x)) iff g(x)=h(x)cdot omega_2(x) quad limsup|omega_2(x)|in mathbb R$$
then
$$f(x)g(x)=h(x)cdotomega_1(x)cdotomega_2(x)=h(x)cdot omega_3(x) quad omega_3(x)to 0$$
indeed we have that $|omega_2(x)|le M$ and by squeeze theorem
$$|omega_3(x)|=|omega_1(x)cdotomega_2(x)|le M|omega_1(x)to 0$$
In the same way we can see that
$O(h(x))=h(x)cdot O(1)$
$o(1)cdot O(1)=o(1)$
answered Dec 14 '18 at 10:30
gimusigimusi
1
$endgroup$
$begingroup$
I was given slightly different definitions so I'd like to clarify, How is $g(x)=O(h(x)) iff g(x)=h(x)cdot omega_2(x) quad limsup|omega_2(x)|in mathbb R$? The limsup part is where I'm particularly lost. The definition I know is that $g(x)=O(h(x)) iff lim_{xrightarrow a} frac{g(x)}{h(x)} = C < infty$
$endgroup$
– E.Nole
Dec 14 '18 at 15:40
1
$begingroup$
Your definition $$lim_{xrightarrow a} frac{g(x)}{h(x)} = C < infty$$ is uncorrect we should have $$limsup_{xrightarrow a} left|frac{g(x)}{h(x)}right| = C < infty$$ Refer to formal definition. The definition I used is more general since include also te case with $h(x)=0$.
$endgroup$
– gimusi
Dec 14 '18 at 15:50
add a comment |
$begingroup$
Recall that by definition as $xto a$
$$f(x)=o(1)iff f(x)=1cdot omega_1(x) quad omega_1(x)to 0$$
$$g(x)=O(h(x)) iff g(x)=h(x)cdot omega_2(x) quad limsup|omega_2(x)|in mathbb R$$
then
$$f(x)g(x)=h(x)cdotomega_1(x)cdotomega_2(x)=h(x)cdot omega_3(x) quad omega_3(x)to 0$$
indeed we have that $|omega_2(x)|le M$ and by squeeze theorem
$$|omega_3(x)|=|omega_1(x)cdotomega_2(x)|le M|omega_1(x)to 0$$
In the same way we can see that
$O(h(x))=h(x)cdot O(1)$
$o(1)cdot O(1)=o(1)$
answered Dec 14 '18 at 10:30
gimusigimusi
1
$endgroup$
$begingroup$
I was given slightly different definitions so I'd like to clarify, How is $g(x)=O(h(x)) iff g(x)=h(x)cdot omega_2(x) quad limsup|omega_2(x)|in mathbb R$? The limsup part is where I'm particularly lost. The definition I know is that $g(x)=O(h(x)) iff lim_{xrightarrow a} frac{g(x)}{h(x)} = C < infty$
$endgroup$
– E.Nole
Dec 14 '18 at 15:40
1
$begingroup$
Your definition $$lim_{xrightarrow a} frac{g(x)}{h(x)} = C < infty$$ is uncorrect we should have $$limsup_{xrightarrow a} left|frac{g(x)}{h(x)}right| = C < infty$$ Refer to formal definition. The definition I used is more general since include also te case with $h(x)=0$.
$endgroup$
– gimusi
Dec 14 '18 at 15:50
add a comment |
$begingroup$
Recall that by definition as $xto a$
$$f(x)=o(1)iff f(x)=1cdot omega_1(x) quad omega_1(x)to 0$$
$$g(x)=O(h(x)) iff g(x)=h(x)cdot omega_2(x) quad limsup|omega_2(x)|in mathbb R$$
then
$$f(x)g(x)=h(x)cdotomega_1(x)cdotomega_2(x)=h(x)cdot omega_3(x) quad omega_3(x)to 0$$
indeed we have that $|omega_2(x)|le M$ and by squeeze theorem
$$|omega_3(x)|=|omega_1(x)cdotomega_2(x)|le M|omega_1(x)to 0$$
In the same way we can see that
$O(h(x))=h(x)cdot O(1)$
$o(1)cdot O(1)=o(1)$
answered Dec 14 '18 at 10:30
gimusigimusi
1
$endgroup$
Recall that by definition as $xto a$
$$f(x)=o(1)iff f(x)=1cdot omega_1(x) quad omega_1(x)to 0$$
$$g(x)=O(h(x)) iff g(x)=h(x)cdot omega_2(x) quad limsup|omega_2(x)|in mathbb R$$
then
$$f(x)g(x)=h(x)cdotomega_1(x)cdotomega_2(x)=h(x)cdot omega_3(x) quad omega_3(x)to 0$$
indeed we have that $|omega_2(x)|le M$ and by squeeze theorem
$$|omega_3(x)|=|omega_1(x)cdotomega_2(x)|le M|omega_1(x)to 0$$
In the same way we can see that
$O(h(x))=h(x)cdot O(1)$
$o(1)cdot O(1)=o(1)$
answered Dec 14 '18 at 10:30
gimusigimusi
1
edited Dec 14 '18 at 10:45
answered Dec 14 '18 at 10:30
gimusigimusi
1
answered Dec 14 '18 at 10:30
gimusigimusi
1
answered Dec 14 '18 at 10:30
gimusigimusi
1
1
$begingroup$
I was given slightly different definitions so I'd like to clarify, How is $g(x)=O(h(x)) iff g(x)=h(x)cdot omega_2(x) quad limsup|omega_2(x)|in mathbb R$? The limsup part is where I'm particularly lost. The definition I know is that $g(x)=O(h(x)) iff lim_{xrightarrow a} frac{g(x)}{h(x)} = C < infty$
$endgroup$
– E.Nole
Dec 14 '18 at 15:40
1
$begingroup$
Your definition $$lim_{xrightarrow a} frac{g(x)}{h(x)} = C < infty$$ is uncorrect we should have $$limsup_{xrightarrow a} left|frac{g(x)}{h(x)}right| = C < infty$$ Refer to formal definition. The definition I used is more general since include also te case with $h(x)=0$.
$endgroup$
– gimusi
Dec 14 '18 at 15:50
add a comment |
$begingroup$
I was given slightly different definitions so I'd like to clarify, How is $g(x)=O(h(x)) iff g(x)=h(x)cdot omega_2(x) quad limsup|omega_2(x)|in mathbb R$? The limsup part is where I'm particularly lost. The definition I know is that $g(x)=O(h(x)) iff lim_{xrightarrow a} frac{g(x)}{h(x)} = C < infty$
$endgroup$
– E.Nole
Dec 14 '18 at 15:40
1
$begingroup$
Your definition $$lim_{xrightarrow a} frac{g(x)}{h(x)} = C < infty$$ is uncorrect we should have $$limsup_{xrightarrow a} left|frac{g(x)}{h(x)}right| = C < infty$$ Refer to formal definition. The definition I used is more general since include also te case with $h(x)=0$.
$endgroup$
– gimusi
Dec 14 '18 at 15:50
$begingroup$
I was given slightly different definitions so I'd like to clarify, How is $g(x)=O(h(x)) iff g(x)=h(x)cdot omega_2(x) quad limsup|omega_2(x)|in mathbb R$? The limsup part is where I'm particularly lost. The definition I know is that $g(x)=O(h(x)) iff lim_{xrightarrow a} frac{g(x)}{h(x)} = C < infty$
$endgroup$
– E.Nole
Dec 14 '18 at 15:40
$begingroup$
I was given slightly different definitions so I'd like to clarify, How is $g(x)=O(h(x)) iff g(x)=h(x)cdot omega_2(x) quad limsup|omega_2(x)|in mathbb R$? The limsup part is where I'm particularly lost. The definition I know is that $g(x)=O(h(x)) iff lim_{xrightarrow a} frac{g(x)}{h(x)} = C < infty$
$endgroup$
– E.Nole
Dec 14 '18 at 15:40
1
1
$begingroup$
Your definition $$lim_{xrightarrow a} frac{g(x)}{h(x)} = C < infty$$ is uncorrect we should have $$limsup_{xrightarrow a} left|frac{g(x)}{h(x)}right| = C < infty$$ Refer to formal definition. The definition I used is more general since include also te case with $h(x)=0$.
$endgroup$
– gimusi
Dec 14 '18 at 15:50
$begingroup$
Your definition $$lim_{xrightarrow a} frac{g(x)}{h(x)} = C < infty$$ is uncorrect we should have $$limsup_{xrightarrow a} left|frac{g(x)}{h(x)}right| = C < infty$$ Refer to formal definition. The definition I used is more general since include also te case with $h(x)=0$.
$endgroup$
– gimusi
Dec 14 '18 at 15:50
add a comment |
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