Why is it true that $sqrt{frac{a}{2x}+O(x^{-2})} = frac{a}{4x} + O(x^{-2})$ for $|x|<1$
Simplify $sqrt{frac{a}{2x}+O(x^{-2})}$ for $|x|<1$. This problem uses "big-O" notation where $f(x) = O(g(x))$ as $xrightarrow a$ if and only if there exists positive numbers $M$ and $delta$ such that $|f(x)|leq Mg(x)$ when $0<|x-a|<delta$.
I am trying to validate a claim that this expression is equal to $frac{a}{4x} + O(x^{-2})$. I don't know where to start because I have no idea how to evaluate this square root. My initial idea was to use the binomial expansion for the square root of two terms, which leads me to $$sqrt{frac{a}{2x}} + frac{1}{2}*frac{1}{(frac{a}{2})^{frac{1}{2}}}O(x^{-2}) -frac{1}{4}*frac{1}{(frac{a}{2x})^{frac{3}{2}}})O^2(x^{-2}):+:...$$ Next I would need to find a way to round off all the remaining terms with some $O(g(x))$, but I also don't know how to do this.
asymptotics
asked Nov 29 at 20:51
Ryan Greyling
1707
add a comment |
Simplify $sqrt{frac{a}{2x}+O(x^{-2})}$ for $|x|<1$. This problem uses "big-O" notation where $f(x) = O(g(x))$ as $xrightarrow a$ if and only if there exists positive numbers $M$ and $delta$ such that $|f(x)|leq Mg(x)$ when $0<|x-a|<delta$.
I am trying to validate a claim that this expression is equal to $frac{a}{4x} + O(x^{-2})$. I don't know where to start because I have no idea how to evaluate this square root. My initial idea was to use the binomial expansion for the square root of two terms, which leads me to $$sqrt{frac{a}{2x}} + frac{1}{2}*frac{1}{(frac{a}{2})^{frac{1}{2}}}O(x^{-2}) -frac{1}{4}*frac{1}{(frac{a}{2x})^{frac{3}{2}}})O^2(x^{-2}):+:...$$ Next I would need to find a way to round off all the remaining terms with some $O(g(x))$, but I also don't know how to do this.
asymptotics
asked Nov 29 at 20:51
Ryan Greyling
1707
This isn’t correct
– Clayton
Nov 29 at 20:55
1
Your big O is near which point?
– hamam_Abdallah
Nov 29 at 21:01
The big-oh expressions only make sense as $x to infty$. The problem is wrong.
– marty cohen
Nov 29 at 21:11
The big O is near 0 @hamam_Abdallah
– Ryan Greyling
Nov 29 at 21:22
1
These sources claim that big O notation is applicable as $xrightarrow a$ @marty cohen en.wikipedia.org/wiki/Big_O_notation under formal definition and web.mit.edu/16.070/www/lecture/big_o.pdf page 3 under related notations
– Ryan Greyling
Nov 29 at 21:26
add a comment |
Simplify $sqrt{frac{a}{2x}+O(x^{-2})}$ for $|x|<1$. This problem uses "big-O" notation where $f(x) = O(g(x))$ as $xrightarrow a$ if and only if there exists positive numbers $M$ and $delta$ such that $|f(x)|leq Mg(x)$ when $0<|x-a|<delta$.
I am trying to validate a claim that this expression is equal to $frac{a}{4x} + O(x^{-2})$. I don't know where to start because I have no idea how to evaluate this square root. My initial idea was to use the binomial expansion for the square root of two terms, which leads me to $$sqrt{frac{a}{2x}} + frac{1}{2}*frac{1}{(frac{a}{2})^{frac{1}{2}}}O(x^{-2}) -frac{1}{4}*frac{1}{(frac{a}{2x})^{frac{3}{2}}})O^2(x^{-2}):+:...$$ Next I would need to find a way to round off all the remaining terms with some $O(g(x))$, but I also don't know how to do this.
asymptotics
asked Nov 29 at 20:51
Ryan Greyling
1707
Simplify $sqrt{frac{a}{2x}+O(x^{-2})}$ for $|x|<1$. This problem uses "big-O" notation where $f(x) = O(g(x))$ as $xrightarrow a$ if and only if there exists positive numbers $M$ and $delta$ such that $|f(x)|leq Mg(x)$ when $0<|x-a|<delta$.
I am trying to validate a claim that this expression is equal to $frac{a}{4x} + O(x^{-2})$. I don't know where to start because I have no idea how to evaluate this square root. My initial idea was to use the binomial expansion for the square root of two terms, which leads me to $$sqrt{frac{a}{2x}} + frac{1}{2}*frac{1}{(frac{a}{2})^{frac{1}{2}}}O(x^{-2}) -frac{1}{4}*frac{1}{(frac{a}{2x})^{frac{3}{2}}})O^2(x^{-2}):+:...$$ Next I would need to find a way to round off all the remaining terms with some $O(g(x))$, but I also don't know how to do this.
asymptotics
asymptotics
asked Nov 29 at 20:51
Ryan Greyling
1707
asked Nov 29 at 20:51
Ryan Greyling
1707
asked Nov 29 at 20:51
Ryan Greyling
1707
asked Nov 29 at 20:51
Ryan Greyling
1707
asked Nov 29 at 20:51
Ryan Greyling
1707
1707
This isn’t correct
– Clayton
Nov 29 at 20:55
1
Your big O is near which point?
– hamam_Abdallah
Nov 29 at 21:01
The big-oh expressions only make sense as $x to infty$. The problem is wrong.
– marty cohen
Nov 29 at 21:11
The big O is near 0 @hamam_Abdallah
– Ryan Greyling
Nov 29 at 21:22
1
These sources claim that big O notation is applicable as $xrightarrow a$ @marty cohen en.wikipedia.org/wiki/Big_O_notation under formal definition and web.mit.edu/16.070/www/lecture/big_o.pdf page 3 under related notations
– Ryan Greyling
Nov 29 at 21:26
add a comment |
This isn’t correct
– Clayton
Nov 29 at 20:55
1
Your big O is near which point?
– hamam_Abdallah
Nov 29 at 21:01
The big-oh expressions only make sense as $x to infty$. The problem is wrong.
– marty cohen
Nov 29 at 21:11
The big O is near 0 @hamam_Abdallah
– Ryan Greyling
Nov 29 at 21:22
1
These sources claim that big O notation is applicable as $xrightarrow a$ @marty cohen en.wikipedia.org/wiki/Big_O_notation under formal definition and web.mit.edu/16.070/www/lecture/big_o.pdf page 3 under related notations
– Ryan Greyling
Nov 29 at 21:26
This isn’t correct
– Clayton
Nov 29 at 20:55
This isn’t correct
– Clayton
Nov 29 at 20:55
1
1
Your big O is near which point?
– hamam_Abdallah
Nov 29 at 21:01
Your big O is near which point?
– hamam_Abdallah
Nov 29 at 21:01
The big-oh expressions only make sense as $x to infty$. The problem is wrong.
– marty cohen
Nov 29 at 21:11
The big-oh expressions only make sense as $x to infty$. The problem is wrong.
– marty cohen
Nov 29 at 21:11
The big O is near 0 @hamam_Abdallah
– Ryan Greyling
Nov 29 at 21:22
The big O is near 0 @hamam_Abdallah
– Ryan Greyling
Nov 29 at 21:22
1
1
These sources claim that big O notation is applicable as $xrightarrow a$ @marty cohen en.wikipedia.org/wiki/Big_O_notation under formal definition and web.mit.edu/16.070/www/lecture/big_o.pdf page 3 under related notations
– Ryan Greyling
Nov 29 at 21:26
These sources claim that big O notation is applicable as $xrightarrow a$ @marty cohen en.wikipedia.org/wiki/Big_O_notation under formal definition and web.mit.edu/16.070/www/lecture/big_o.pdf page 3 under related notations
– Ryan Greyling
Nov 29 at 21:26
add a comment |
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This isn’t correct
– Clayton
Nov 29 at 20:55
1
Your big O is near which point?
– hamam_Abdallah
Nov 29 at 21:01
The big-oh expressions only make sense as $x to infty$. The problem is wrong.
– marty cohen
Nov 29 at 21:11
The big O is near 0 @hamam_Abdallah
– Ryan Greyling
Nov 29 at 21:22
1
These sources claim that big O notation is applicable as $xrightarrow a$ @marty cohen en.wikipedia.org/wiki/Big_O_notation under formal definition and web.mit.edu/16.070/www/lecture/big_o.pdf page 3 under related notations
– Ryan Greyling
Nov 29 at 21:26