Exponential Approximation
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I was recently reading a text book on random walk. In the proof of a local central limit theorem the book used the following step: $e^{-2k^2/(n+k)} = e^{-2k^2/n} space exp{frac{2k^3}{n^2} +O(frac{k^4}{n^3})}$. And similarly, there is: $(1-frac{k^2}{n^2})^{-1/2} = exp{O(frac{1}{n}+frac{k^4}{n^3})}$, and the latter case is deduced using the fact that $k^2/n^2 leq max{(1/n),(k^4/n^3)}$. I think the expanding of complex logarithm with power series is used in both cases, but I'm still quite confused how these equalities are deduced.
Could anyone help me with it?
The big O notation above is defined as follows:"write $f(n,x) = g(n,x) +O(h(n))$ to mean that there exists a constant c such that for all n $|f(n,x)-g(n,x)| leq c|h(n)|$, where c denotes a positive constant depending on the increment distribution"
complex-analysis logarithms power-series taylor-expansion random-walk
asked Nov 24 '18 at 1:40
HomeomorphismHomeomorphism
83
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add a comment |
$begingroup$
I was recently reading a text book on random walk. In the proof of a local central limit theorem the book used the following step: $e^{-2k^2/(n+k)} = e^{-2k^2/n} space exp{frac{2k^3}{n^2} +O(frac{k^4}{n^3})}$. And similarly, there is: $(1-frac{k^2}{n^2})^{-1/2} = exp{O(frac{1}{n}+frac{k^4}{n^3})}$, and the latter case is deduced using the fact that $k^2/n^2 leq max{(1/n),(k^4/n^3)}$. I think the expanding of complex logarithm with power series is used in both cases, but I'm still quite confused how these equalities are deduced.
Could anyone help me with it?
The big O notation above is defined as follows:"write $f(n,x) = g(n,x) +O(h(n))$ to mean that there exists a constant c such that for all n $|f(n,x)-g(n,x)| leq c|h(n)|$, where c denotes a positive constant depending on the increment distribution"
complex-analysis logarithms power-series taylor-expansion random-walk
asked Nov 24 '18 at 1:40
HomeomorphismHomeomorphism
83
$endgroup$
add a comment |
$begingroup$
I was recently reading a text book on random walk. In the proof of a local central limit theorem the book used the following step: $e^{-2k^2/(n+k)} = e^{-2k^2/n} space exp{frac{2k^3}{n^2} +O(frac{k^4}{n^3})}$. And similarly, there is: $(1-frac{k^2}{n^2})^{-1/2} = exp{O(frac{1}{n}+frac{k^4}{n^3})}$, and the latter case is deduced using the fact that $k^2/n^2 leq max{(1/n),(k^4/n^3)}$. I think the expanding of complex logarithm with power series is used in both cases, but I'm still quite confused how these equalities are deduced.
Could anyone help me with it?
The big O notation above is defined as follows:"write $f(n,x) = g(n,x) +O(h(n))$ to mean that there exists a constant c such that for all n $|f(n,x)-g(n,x)| leq c|h(n)|$, where c denotes a positive constant depending on the increment distribution"
complex-analysis logarithms power-series taylor-expansion random-walk
asked Nov 24 '18 at 1:40
HomeomorphismHomeomorphism
83
$endgroup$
I was recently reading a text book on random walk. In the proof of a local central limit theorem the book used the following step: $e^{-2k^2/(n+k)} = e^{-2k^2/n} space exp{frac{2k^3}{n^2} +O(frac{k^4}{n^3})}$. And similarly, there is: $(1-frac{k^2}{n^2})^{-1/2} = exp{O(frac{1}{n}+frac{k^4}{n^3})}$, and the latter case is deduced using the fact that $k^2/n^2 leq max{(1/n),(k^4/n^3)}$. I think the expanding of complex logarithm with power series is used in both cases, but I'm still quite confused how these equalities are deduced.
Could anyone help me with it?
The big O notation above is defined as follows:"write $f(n,x) = g(n,x) +O(h(n))$ to mean that there exists a constant c such that for all n $|f(n,x)-g(n,x)| leq c|h(n)|$, where c denotes a positive constant depending on the increment distribution"
complex-analysis logarithms power-series taylor-expansion random-walk
complex-analysis logarithms power-series taylor-expansion random-walk
asked Nov 24 '18 at 1:40
HomeomorphismHomeomorphism
83
asked Nov 24 '18 at 1:40
HomeomorphismHomeomorphism
83
edited Nov 24 '18 at 1:47
Homeomorphism
asked Nov 24 '18 at 1:40
HomeomorphismHomeomorphism
83
asked Nov 24 '18 at 1:40
HomeomorphismHomeomorphism
83
asked Nov 24 '18 at 1:40
HomeomorphismHomeomorphism
83
83
add a comment |
add a comment |
1 Answer
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Consider $$a_n=e^{-frac{2 k^2}{k+n}}implies log(a_n)=-frac{2 k^2}{k+n}$$ Now, if $n gg k$, use Taylor expansion
$$log(a_n)=-frac{2 k^2}{n}+frac{2 k^3}{n^2}+Oleft(frac{1}{n^3}right)$$ Exponentiate to get
$$a_n=e^{-frac{2 k^2}{n}}expleft(frac{2 k^3}{n^2}+Oleft(frac{1}{n^3}right) right)$$
answered Nov 26 '18 at 10:19
Claude LeiboviciClaude Leibovici
119k1157132
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1 Answer
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1 Answer
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$begingroup$
Consider $$a_n=e^{-frac{2 k^2}{k+n}}implies log(a_n)=-frac{2 k^2}{k+n}$$ Now, if $n gg k$, use Taylor expansion
$$log(a_n)=-frac{2 k^2}{n}+frac{2 k^3}{n^2}+Oleft(frac{1}{n^3}right)$$ Exponentiate to get
$$a_n=e^{-frac{2 k^2}{n}}expleft(frac{2 k^3}{n^2}+Oleft(frac{1}{n^3}right) right)$$
answered Nov 26 '18 at 10:19
Claude LeiboviciClaude Leibovici
119k1157132
$endgroup$
add a comment |
$begingroup$
Consider $$a_n=e^{-frac{2 k^2}{k+n}}implies log(a_n)=-frac{2 k^2}{k+n}$$ Now, if $n gg k$, use Taylor expansion
$$log(a_n)=-frac{2 k^2}{n}+frac{2 k^3}{n^2}+Oleft(frac{1}{n^3}right)$$ Exponentiate to get
$$a_n=e^{-frac{2 k^2}{n}}expleft(frac{2 k^3}{n^2}+Oleft(frac{1}{n^3}right) right)$$
answered Nov 26 '18 at 10:19
Claude LeiboviciClaude Leibovici
119k1157132
$endgroup$
add a comment |
$begingroup$
Consider $$a_n=e^{-frac{2 k^2}{k+n}}implies log(a_n)=-frac{2 k^2}{k+n}$$ Now, if $n gg k$, use Taylor expansion
$$log(a_n)=-frac{2 k^2}{n}+frac{2 k^3}{n^2}+Oleft(frac{1}{n^3}right)$$ Exponentiate to get
$$a_n=e^{-frac{2 k^2}{n}}expleft(frac{2 k^3}{n^2}+Oleft(frac{1}{n^3}right) right)$$
answered Nov 26 '18 at 10:19
Claude LeiboviciClaude Leibovici
119k1157132
$endgroup$
Consider $$a_n=e^{-frac{2 k^2}{k+n}}implies log(a_n)=-frac{2 k^2}{k+n}$$ Now, if $n gg k$, use Taylor expansion
$$log(a_n)=-frac{2 k^2}{n}+frac{2 k^3}{n^2}+Oleft(frac{1}{n^3}right)$$ Exponentiate to get
$$a_n=e^{-frac{2 k^2}{n}}expleft(frac{2 k^3}{n^2}+Oleft(frac{1}{n^3}right) right)$$
answered Nov 26 '18 at 10:19
Claude LeiboviciClaude Leibovici
119k1157132
answered Nov 26 '18 at 10:19
Claude LeiboviciClaude Leibovici
119k1157132
answered Nov 26 '18 at 10:19
Claude LeiboviciClaude Leibovici
119k1157132
answered Nov 26 '18 at 10:19
Claude LeiboviciClaude Leibovici
119k1157132
119k1157132
add a comment |
add a comment |
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