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--- 배규호:눈금_바꿈_가설 [2017/07/12 08:47] – bekuho
+++ 배규호:눈금_바꿈_가설 [2023/09/05 15:46] (current) – external edit 127.0.0.1
@@ Line 10: / Line 10: @@
 $\xi$ 가 발산할 때 $G(k)$를 위 변수의 함수로 나타낸다면
-\begin{equation*}
-\begin{split*}
-G(k) &= f(k\xi,b_1/\xi,b_2/\xi, \dots)  \\
+$$G(k) = f(k\xi,b_1\xi,b_2\xi, \dots) $$
-&= f(k\xi) + \sum_{i=1}\frac{\partial{f(k\xi,b_1/\xi,b_2/\xi,...)}}{\partial{(b_i/\xi)}}(b_i/\xi) + \sum_{i=1}\frac{\partial^{2}{f(k\xi,b_1/\xi,b_2/\xi,...)}}{\partial{(b_i/\xi)^2}}(b_i/\xi)^2 + higher\: orders \: of \left (b_i/\xi) \\
+$$= f(k\xi) + \sum_{i=1}\frac{\partial{f(k\xi,b_1\xi,b_2\xi,\dots)}}{\partial{(b_i\xi)}} (b_i \xi) + \sum_{i=1}\frac{\partial^{2}{f(k\xi,b_1 \xi,b_2\xi,\dots)}}{\partial{(b_i\xi)^2}}(b_i\xi)^2 + \text{higher orders of }\left(b_i\xi \right) $$
-&= f(k\xi) + c_{b_1, x_1}(b_1/\xi)^{x_1} + c_{b_1, x_1-1}(b_1/\xi)^{{x_1}-1} + \dots + c_{b_2, x_2}(b_2/\xi)^{x_2} + \dots  \\
+$$= f(k\xi) + c_{b_1, x_1}(b_1\xi)^{x_1} + c_{b_1, x_1-1}(b_1\xi)^{{x_1}-1} + \dots + c_{b_2, x_2}(b_2\xi)^{x_2} + \dots  $$
-& = \xi^{y}(g(k\xi) + \: higher \: powers \: of \: \xi^{-1}) \\
+$$= \xi^{y}(g(k\xi) + \text{higher powers of }\xi^{-1}) $$
-& \approx \xi^{y}g(k\xi)
+$$ \approx \xi^{y}g(k\xi)$$
-\end{split*}
-\end{equation*}
 다음을 유도할 때 2번째 줄에서는 $b_i/\xi$ 에 대해 급수전개 하였고 3번쨰 줄에서는 $-y = x_1 + x_2 +\dots$ 를 이용하였다.