배규호:눈금_바꿈_가설

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배규호:눈금_바꿈_가설 [2017/07/12 08:53] – bekuho배규호:눈금_바꿈_가설 [2023/09/05 15:46] (current) – external edit 127.0.0.1
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 $$G(k) = f(k\xi,b_1\xi,b_2\xi, \dots) $$ $$G(k) = f(k\xi,b_1\xi,b_2\xi, \dots) $$
-$$= f(k\xi) + \sum_{i=1}\frac{\partial{f(k\xi,b_1\xi,b_2\xi,\dots)}}{\partial{(b_i\xi)}} (b_i \xi) + \sum_{i=1}\frac{\partial^{2}{f(k\xi,b_1 \xi,b_2\xi,\dots)}}{\partial{(b_i\xi)^2}}(b_i\xi)^2 + higher\: orders \: of \left (b_i\xi) $$+$$= f(k\xi) + \sum_{i=1}\frac{\partial{f(k\xi,b_1\xi,b_2\xi,\dots)}}{\partial{(b_i\xi)}} (b_i \xi) + \sum_{i=1}\frac{\partial^{2}{f(k\xi,b_1 \xi,b_2\xi,\dots)}}{\partial{(b_i\xi)^2}}(b_i\xi)^2 + \text{higher orders of }\left(b_i\xi \right) $$
 $$= f(k\xi) + c_{b_1, x_1}(b_1\xi)^{x_1} + c_{b_1, x_1-1}(b_1\xi)^{{x_1}-1} + \dots + c_{b_2, x_2}(b_2\xi)^{x_2} + \dots  $$  $$= f(k\xi) + c_{b_1, x_1}(b_1\xi)^{x_1} + c_{b_1, x_1-1}(b_1\xi)^{{x_1}-1} + \dots + c_{b_2, x_2}(b_2\xi)^{x_2} + \dots  $$ 
-$$= \xi^{y}(g(k\xi) + \: higher \: powers \: of \: \xi^{-1}) $$+$$= \xi^{y}(g(k\xi) + \text{higher powers of }\xi^{-1}) $$
 $$ \approx \xi^{y}g(k\xi)$$ $$ \approx \xi^{y}g(k\xi)$$
  
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  • Last modified: 2023/09/05 15:46
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