Exponential Approximation
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 at 1:40
Homeomorphism
83
add a comment |
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 at 1:40
Homeomorphism
83
add a comment |
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 at 1:40
Homeomorphism
83
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 at 1:40
Homeomorphism
83
asked Nov 24 at 1:40
Homeomorphism
83
edited Nov 24 at 1:47
asked Nov 24 at 1:40
Homeomorphism
83
asked Nov 24 at 1:40
Homeomorphism
83
asked Nov 24 at 1:40
Homeomorphism
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 at 10:19
Claude Leibovici
118k1156132
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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 at 10:19
Claude Leibovici
118k1156132
add a comment |
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 at 10:19
Claude Leibovici
118k1156132
add a comment |
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 at 10:19
Claude Leibovici
118k1156132
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 at 10:19
Claude Leibovici
118k1156132
answered Nov 26 at 10:19
Claude Leibovici
118k1156132
answered Nov 26 at 10:19
Claude Leibovici
118k1156132
answered Nov 26 at 10:19
Claude Leibovici
118k1156132
118k1156132
add a comment |
add a comment |
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