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







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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







    share|cite|improve this question



























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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







      share|cite|improve this question















      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






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      edited Nov 24 at 1:47

























      asked Nov 24 at 1:40









      Homeomorphism

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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
          $$a_n=e^{-frac{2 k^2}{n}}expleft(frac{2 k^3}{n^2}+Oleft(frac{1}{n^3}right) right)$$






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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
            $$a_n=e^{-frac{2 k^2}{n}}expleft(frac{2 k^3}{n^2}+Oleft(frac{1}{n^3}right) right)$$






            share|cite|improve this 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
              $$a_n=e^{-frac{2 k^2}{n}}expleft(frac{2 k^3}{n^2}+Oleft(frac{1}{n^3}right) right)$$






              share|cite|improve this 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
                $$a_n=e^{-frac{2 k^2}{n}}expleft(frac{2 k^3}{n^2}+Oleft(frac{1}{n^3}right) right)$$






                share|cite|improve this answer












                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
                $$a_n=e^{-frac{2 k^2}{n}}expleft(frac{2 k^3}{n^2}+Oleft(frac{1}{n^3}right) right)$$







                share|cite|improve this answer












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                answered Nov 26 at 10:19









                Claude Leibovici

                118k1156132




                118k1156132






























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