테일러급수에서 2nd symmetry derivative 유도

WooSeongkyun·2023년 3월 16일
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  • f(t)=i=0f(n)(x)n!(tx)nf(t)=\displaystyle\sum_{i=0}^{\infty }{\displaystyle\frac{f ^{(n)}(x)}{n!}}(t-x) ^{n}
    - f(ϵ)=i=0f(n)(x)n!(ϵx)nf(\epsilon)=\displaystyle\sum_{i=0}^{\infty }{\displaystyle\frac{f ^{(n)}(x)}{n!}}(\epsilon-x) ^{n}
    - f(x+ϵ)=i=0f(n)(x)n!ϵn=f(x)+f(x)ϵ+12f(x)ϵ2+η(ϵ)f(x+\epsilon)=\displaystyle\sum_{i=0}^{\infty }{\displaystyle\frac{f ^{(n)}(x)}{n!}}\epsilon ^{n}=f(x)+f'(x)\epsilon+\displaystyle\frac{1}{2}f''(x)\epsilon ^{2}+\eta(\epsilon)
    - f(xϵ)=i=0f(n)(x)n!(ϵ)n=f(x)f(x)ϵ+12f(x)ϵ2+η(ϵ)f(x-\epsilon)=\displaystyle\sum_{i=0}^{\infty }{\displaystyle\frac{f ^{(n)}(x)}{n!}}(-\epsilon) ^{n}=f(x)-f'(x)\epsilon+\displaystyle\frac{1}{2}f''(x)\epsilon ^{2}+\eta(-\epsilon)
    - 12f(x)ϵ2=f(x+ϵ)f(x)f(x)ϵη(ϵ)=f(xϵ)f(x)+f(x)ϵη(ϵ)\displaystyle\frac{1}{2}f''(x)\epsilon ^{2}=f(x+\epsilon)-f(x)-f'(x)\epsilon-\eta(\epsilon)=f(x-\epsilon)-f(x)+f'(x)\epsilon-\eta(-\epsilon)
    - f(x)=1ϵ2[f(x+ϵ)+f(xϵ)2f(x)η(ϵ)η(ϵ)]f''(x)= \displaystyle\frac{1}{\epsilon ^{2}}[f(x+\epsilon)+f(x-\epsilon)-2f(x)-\eta(\epsilon)-\eta(-\epsilon)]
    - 이는 ϵ\epsilon 이 무슨 값이든 참이여야 하므로 ϵ0\epsilon \to 0 으로 보내면 limϵ0η(ϵ)ϵ2=η(ϵ)ϵ2=0\lim_{\epsilon \to 0}{\displaystyle\frac{\eta(\epsilon)}{\epsilon ^{2}}}=\displaystyle\frac{\eta(-\epsilon)}{\epsilon ^{2}}=0 이되고,
    - f(x)=limϵ0f(x+ϵ)+f(xϵ)2f(x)ϵ2f''(x)=\lim_{\epsilon \to 0}{\displaystyle\frac{f(x+\epsilon)+f(x-\epsilon)-2f(x)}{\epsilon ^{2}}}
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