这个教程会以简单易懂的方式推导泰勒展开公式。

0. 变现积分求解

[\int_{0}^{x} \int_{0}^{x} \int_{0}^{x} 1 , dx , dx , dx = , ?]

先解最内层变现积分：[\int_{0}^{x} 1 , dx = x]

将最内层变现积分的解带入第二层积分：[\int_{0}^{x} \int_{0}^{x} x , dx , dx = , ?]

再解第二层变现积分：[\int_{0}^{x} x = \frac{1}{2!}x^2\bigg|_{0}^{x}]

把积分符号内的x看作u: [\int_{0}^{x} u = \frac{1}{2!}u^2\bigg|_{0}^{x}]

求解：[\frac{1}{2!}u^2\bigg|_{0}^{x} = \frac{1}{2!}x^2 - \frac{1}{2!}(0)^2,]

得出第二层变现积分结果为：[\frac{1}{2!}x^2]

将第二层变现积分的解带入第外层积分：[\int_{0}^{x} \frac{1}{2!}x^2 , dx = , ?]

解最外层变现积分：[\int_{0}^{x} \frac{1}{2!}x^2 , dx = \frac{1}{3!}x^3\bigg|_0^x]

把积分符号内的x看作u: [\int_{0}^{x} \frac{1}{2!}u^2 , dx = \frac{1}{3!}u^3\bigg|_0^x]

得出最外层变现积分结果为：[\frac{1}{3!}x^3 - \frac{1}{3!}(0)^3 = \frac{1}{3!}x^3]

1. 泰勒公式推导

[f(x) = \int_{c}^{x} f’(x) , dx + f(a)]

[f’(x) = \int_{c}^{x} f’’(x) , dx + f’(a)]

[f’’(x) = \int_{c}^{x} f’’’(x) , dx + f’’(a)]

[f’’’(x) = \int_{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

[f(x) = \int_{c}^{x} \left( \int_{c}^{x} \left( \int_{c}^{x} \left( \int_{c}^{x} f^{(4)}(a) , dx + f’’’(a) \right) , dx + f’’(a) \right) , dx + f’(a) \right) , dx + f(a)]

[f(x) = \int_{c}^{x} \left( \int_{c}^{x} \left( \int_{c}^{x} \left( \int_{c}^{x} f^{(4)}(a) , dx + f’’’(a) \right) , dx \right) , dx + f’’(a) \right) + \frac{f’(a)}{1!} \cdot (x-c)^1 + f(a)]

[f(x) = \int_{c}^{x} \left( \int_{c}^{x} \left( \int_{c}^{x} \left( \int_{c}^{x} f^{(4)}(a) , 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 \right)]

[f(x) = \int_{c}^{x} \left( \int_{c}^{x} \left( \int_{c}^{x} \left( \int_{c}^{x} f^{(4)}(a) , da \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)]

以此类推，前面剩余的积分为余项，后面为泰勒展开公式。

[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.

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