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f(x) 导数计算

創建於：2024年12月8日

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問題

设f(x)=1/(x^2-3x+2), 则fn=?. （[]代表上标）

思考

Confirming the notation

I’m piecing together the meaning of [n] in Chinese mathematical notation. Assuming it refers to the n-th derivative, I’m factoring the denominator of f(x) = 1 / (x^2 - 3x + 2) as (x-1)(x-2).

Uncovering partial fractions

I determined the partial fractions for f(x) as -1/(x-1) + 1/(x-2). Using the formula for derivatives of 1/(x-a), I found f^{(n)}(x) = (-1)^n n! / (x - a)^{n+1}.

Verifying calculations

I confirmed that f^{(n)}(0) is -n! / a^{n+1} and applied it to both components of f(x), ensuring accuracy for each partial fraction in evaluating the n-th derivative at 0.

Reevaluating the formula

I›m re-evaluating f^{(n)}(0) for f(x) = -1/(x-1) + 1/(x-2), ensuring accuracy by testing different forms and simplifying expressions.

Verifying partial fractions

I’m examining the partial fractions for f(x) at x=0. The first component gives -n!, while the second simplifies to -n! / 2^{n+1}, confirming their accuracy.

Confirming the expression

OK, let me see. I’m ensuring fn represents the n-th derivative and verifying that n! [1 - 1 / 2^{n+1}] is the correct representation.

Mapping out iterates

I’m noting that f(f(f(...f(0)))) for f(x) = 1/(x^2-3x+2) doesn’t form a simple pattern. This suggests fn likely represents the n-th derivative at 0.

Rewriting as series

I’m expressing f(x) as a power series around 0, showing that the coefficient of x^n is 1 - 1/2^{n+1}, aligning with earlier findings.

Thought for 16 seconds

回答

设 f(x)=1x2−3x+2f(x) = \frac{1}{x^2 - 3x + 2}，我们需要计算 f(n)(0)f^{(n)}(0)，即 f(x)f(x) 在 x=0x = 0 处的第 nn 阶导数。

步骤如下：

  1. 因式分解分母：

    x2−3x+2=(x−1)(x−2)x^2 - 3x + 2 = (x - 1)(x - 2)

    所以，

    f(x)=1(x−1)(x−2)=−1x−1+1x−2f(x) = \frac{1}{(x - 1)(x - 2)} = \frac{-1}{x - 1} + \frac{1}{x - 2}

  2. 求导数公式： 对于函数 1x−a\frac{1}{x - a}，其第 nn 阶导数为：

    dndxn(1x−a)=(−1)nn!(x−a)n+1\frac{d^n}{dx^n} \left( \frac{1}{x - a} \right) = \frac{(-1)^n n!}{(x - a)^{n+1}}

  3. 计算 f(n)(0)f^{(n)}(0)：

    f(n)(x)=−(−1)nn!(x−1)n+1+(−1)nn!(x−2)n+1f^{(n)}(x) = -\frac{(-1)^n n!}{(x - 1)^{n+1}} + \frac{(-1)^n n!}{(x - 2)^{n+1}}

    代入 x=0x = 0：

    f(n)(0)=−(−1)nn!(−1)n+1+(−1)nn!(−2)n+1f^{(n)}(0) = -\frac{(-1)^n n!}{(-1)^{n+1}} + \frac{(-1)^n n!}{(-2)^{n+1}}

    化简得到：

    f(n)(0)=n!(1−12n+1)f^{(n)}(0) = n! \left( 1 - \frac{1}{2^{n+1}} \right)

结论：

f(n)(0)=n!(1−12n+1)f^{(n)}(0) = n! \left(1 - \frac{1}{2^{n+1}}\right)

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