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수식 랜더링 test

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선형대수학에서 벡터공간은 매우 중요한 개념입니다. 본문 내용 쏼라쏼라…

$f(x)$

테일러 시리즈

$\text{Let}x,t,a\in \mathbb{R},f(x)\text{be a real valued function s.t. infinite times differentiable}$ ¹

and let

$D^{n}f(x)=\frac{d^{n}f}{d{x}^n},j_n(t)=\frac{(-1{)}^{n+1}(x-t{)}^n}{n!}$

by solving this simple integral, we get 

$${\int }_a^x(x-t)D^{2}f(t)dt=f(x)-f(a)-f^{\prime }(a)(x-a)$$

and using integration-by-parts method, we are able to observe that

$${\int }_a^x(x-t)D^{2}f(t)dt$$

$$={[j_2(t)D^{2}f(t)]}_a^x-{\int }_a^x{j}_2(t)D^{3}f(t)dt$$

$$={[j_2(t)D^{2}f(t)]}_a^x-{[j_3(t)D^{3}f(t)]}_a^x+{\int }_a^x{j}_3(t)D^{4}f(t)dt$$

$$={[j_2(t)D^{2}f(t)]}_a^x-{[j_3(t)D^{3}f(t)]}_a^x+{[j_4(t)D^{4}f(t)]}_a^x-{\int }_a^x{j}_4(t)D^{5}f(t)dt\cdots$$ ²

since $j_n(x)=0.j_n(a)=\frac{(x-a{)}^n(-1{)}^{n+1}}{n!}$ the equation above becomes 

$$=\frac{(x-a{)}^2{D}^{2}f(a)}{2!}+\frac{(x-a{)}^3{D}^{3}f(a)}{3!}+\frac{(x-a{)}^4{D}^{4}f(a)}{4!}\cdots$$

thus, if

$$\underset{n\to \mathrm{\infty }}{lim}{\int }_a^x{j}_n(t)D^{n+1}f(t)dt=0$$

then,³

$${\int }_a^x(x-t)D^{2}f(t)dt=f(x)-f(a)-f^{\prime }(a)(x-a)$$ 

$$=\frac{(x-a{)}^2{D}^{2}f(a)}{2!}+\frac{(x-a{)}^3{D}^{3}f(a)}{3!}+\frac{(x-a{)}^4{D}^{4}f(a)}{4!}\dots$$

$$\therefore f(x)=f(a)+\frac{(x-a)D^{1}f(a)}{1!}+\frac{(x-a{)}^2{D}^{2}f(a)}{2!}+\frac{(x-a{)}^3{D}^{3}f(a)}{3!}\cdots$$

$$\therefore f(x)=\underset{n\to \mathrm{\infty }}{lim}\sum _{k=0}^n\frac{(x-a{)}^k{D}^{k}f(a)}{k!}$$

$(\cos (\frac{\pi }{2}+\theta ),\sin (\frac{\pi }{2}+\theta ))$

$(1+0i)\times (\cos (\theta )+i\sin (\theta ))=(\cos (\theta )+i\sin (\theta )$

$(0+i)\times (\cos (\theta )+i\sin (\theta ))=i\cos (\theta )-\sin (\theta )=i\sin (\frac{\pi }{2}+\theta )+\cos (\frac{\pi }{2}+\theta )$

$a+bi\times (\cos (\theta +i\sin (\theta ))$

$(\cos (n\theta )+i\sin (n\theta ))=(\cos (\theta )+i\sin (\theta ){)}^n$

$x=\cos \left(\frac{2k\pi }{n}\right)+i\sin \left(\frac{2k\pi }{n}\right)$

각주

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1. 테일러 시리즈이다↩︎

2. $ff(x)-e^{3u}$ 이라고 할수 있고, 이는 저것의 내용인 그렇다 $$x=\cos \left(\frac{2k\pi }{n}\right)+i\sin \left(\frac{2k\pi }{n}\right)-{\displaystyle x=\cos \left(\frac{2k\pi }{n}\right)+i\sin \left(\frac{2k\pi }{n}\right)+r\cos (x-\int e^{x}dx-100+f(x-f(x+f(x)))\prod x-1000=\sin (2x)\exp (e^{x^2})+1000)}$$↩︎

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