这个教程会以简单易懂的方式推导泰勒展开公式。
0. 变限积分求解
$$\large \int_{0}^{x} \int_{0}^{x} \int_{0}^{x} 1 \, dx \, dx , dx = , ?$$
$$\text{先解最内层变限积分：}$$
$$\large \int_{0}^{x} 1 , du = u \Big|_{0}^{x} = x - 0 = x $$
$$\text{将最内层变限积分的解带入第二层积分：}$$
$$\large \int_{0}^{x} \int_{0}^{x} x , dx , dx = , ?$$
$$\text{再解第二层变限积分：}$$
$$\large \int_{0}^{x} x , dx = \frac{1}{2!}x^2\Big|_{0}^{x} $$
$$\text{把积分符号内的x看作u: }$$
$$\large \int_{0}^{x} u , du = \frac{1}{2!}u^2\Big|_{0}^{x} $$
$$\text{求解：}$$
\large \frac{1}{2!}u^2\Big|_{0}^{x} = \frac{1}{2!}x^2 - \frac{1}{2!}(0)^2
$$\text{得出第二层变限积分结果为：}$$
$$\large \frac{1}{2!}x^2 $$
$$\text{将第二层变限积分的解带入最外层积分：}$$
$$\large \int_{0}^{x} \frac{1}{2!}x^2 , dx = , ?$$
$$\text{解最外层变限积分：}$$
$$\large \int_{0}^{x} \frac{1}{2!}x^2 , dx = \frac{1}{3!}x^3\Big|{0}^{x} $$
$$\text{把积分符号内的x看作u: }$$
$$\large \int{0}^{x} \frac{1}{2!}u^2 , du = \frac{1}{3!}u^3\Big|_{0}^{x} $$
$$\text{得出最外层变限积分结果为：}$$
$$\large \frac{1}{3!}x^3 - \frac{1}{3!}(0)^3 = \frac{1}{3!}x^3 $$
1. 泰勒公式推导
$$\large f(x) = \int_{c}^{x} f’(x) , dx + f(a)$$
$$\large f’(x) = \int_{c}^{x} f’’(x) , dx + f’(a)$$
$$\large f’’(x) = \int_{c}^{x} f’’’(x) , dx + f’’(a)$$
$$\large f’’’(x) = \int_{c}^{x} f^{(4)}(x) , dx + f’’’(a)$$
f(x) = ∫_{c}^{x} f’(x) , dx + f(a)
f’(x) = ∫_{c}^{x} f’’(x) , dx + f’(a)
f’’(x) = ∫_{c}^{x} f’’’(x) , dx + f’’(a)
f’’’(x) = ∫_{c}^{x} f^{(4)}(x) , dx + f’’’(a)
原函数的信息=导函数的信息积分+初始值
$$\large f(x) = \int_{c}^{x} \left( \int_{c}^{x} \left( \int_{c}^{x} \left( \int_{c}^{x} f^{(4)}(x) , dx + f’’’(a) \right) , dx + f’’(a) \right) , dx + f’(a) \right) , dx + f(a)$$
$$\large f(x) = \int_{c}^{x} \left( \int_{c}^{x} \left( \int_{c}^{x} \left( \int_{c}^{x} f^{(4)}(x) , dx + f’’’(a) \right) , dx \right) , dx + f’’(a) \right) + \frac{f’(a)}{1!} \cdot (x-c)^1 + f(a)$$
$$\large f(x) = \int_{c}^{x} \left( \int_{c}^{x} \left( \int_{c}^{x} \left( \int_{c}^{x} f^{(4)}(x) , dx \right) , dx + f’’’(a) \right) , dx + \frac{f’’(a)}{2!} \cdot (x-c)^2 + \frac{f’(a)}{1!} \cdot (x-c)^1 + f(a) \right) , dx$$
$$\large f(x) = \int_{c}^{x} \left( \int_{c}^{x} \left( \int_{c}^{x} \left( \int_{c}^{x} f^{(4)}(x) , dx \right) , dx + \frac{f’’’(a)}{3!} \cdot (x-c)^3 + \frac{f’’(a)}{2!} \cdot (x-c)^2 + \frac{f’(a)}{1!} \cdot (x-c)^1 + f(a) \right) , dx \right)$$
$$\text{以此类推，前面剩余的积分为余项，后面为泰勒展开公式。}$$
$$\large f(x) = f(a) + \frac{f’(a)}{1!} \cdot (x-c)^1 + \frac{f’’(a)}{2!} \cdot (x-c)^2 + \frac{f’’’(a)}{3!} \cdot (x-c)^3 + \ldots$$
End.
如内容有错误或不明确的地方欢迎联系我进行更正！
$$\large f(x) = \int_{c}^{x} f’(x) , dx + f(a)
$$
$$
\large f’(x) = \int_{c}^{x} f’’(x) , dx + f’(a)
$$
$$
\large f’’(x) = \int_{c}^{x} f’’’(x) , dx + f’’(a)$$
$$\large f’’’(x) = \int_{c}^{x} f^{(4)}(x) , dx + f’’’(a)$$
$x^2+y^2=z^2$
$$ \huge \int_{0}^{x} \int_{0}^{x} \int_{0}^{x} 1 , dx , dx , dx = , ? $$