Math Problem Statement

nth derivative of 1/(1-5x+6x2)

Solution

The problem asks us to find the nnth derivative of the function:

f(x)=115x+6x2.f(x) = \frac{1}{1 - 5x + 6x^2}.

Let's break down the steps:

1. Factor the Denominator

First, note that the denominator is a quadratic expression. Let's factor it:

15x+6x2=(12x)(13x).1 - 5x + 6x^2 = (1 - 2x)(1 - 3x).

So, the function can be rewritten as:

f(x)=1(12x)(13x).f(x) = \frac{1}{(1 - 2x)(1 - 3x)}.

2. Express as a Partial Fraction

We can express the function as a sum of partial fractions:

f(x)=A12x+B13x,f(x) = \frac{A}{1 - 2x} + \frac{B}{1 - 3x},

where AA and BB are constants to be determined. To find AA and BB, we equate:

1(12x)(13x)=A12x+B13x.\frac{1}{(1 - 2x)(1 - 3x)} = \frac{A}{1 - 2x} + \frac{B}{1 - 3x}.

Multiplying through by the denominator gives:

1=A(13x)+B(12x).1 = A(1 - 3x) + B(1 - 2x).

Expanding this expression:

1=A3Ax+B2Bx.1 = A - 3Ax + B - 2Bx.

Grouping terms gives:

1=(A+B)+(3A2B)x.1 = (A + B) + (-3A - 2B)x.

For this to hold for all xx, the coefficients of xx and the constant term must be equal:

  1. A+B=1A + B = 1.
  2. 3A2B=0-3A - 2B = 0.

Solving these equations simultaneously:

From 3A2B=0-3A - 2B = 0, we get B=32AB = -\frac{3}{2}A.

Substituting into A+B=1A + B = 1:

A32A=1    12A=1    A=2.A - \frac{3}{2}A = 1 \implies -\frac{1}{2}A = 1 \implies A = -2.

Then, B=32(2)=3B = -\frac{3}{2}(-2) = 3.

So, the partial fraction decomposition is:

f(x)=212x+313x.f(x) = \frac{-2}{1 - 2x} + \frac{3}{1 - 3x}.

3. Express as a Series

Each term can be expressed as a geometric series:

11ax=k=0(ax)k.\frac{1}{1 - ax} = \sum_{k=0}^{\infty} (ax)^k.

Applying this to our expression:

f(x)=2k=0(2x)k+3k=0(3x)k.f(x) = -2 \sum_{k=0}^{\infty} (2x)^k + 3 \sum_{k=0}^{\infty} (3x)^k.

Simplifying:

f(x)=k=0[22k+33k]xk.f(x) = \sum_{k=0}^{\infty} \left[ -2 \cdot 2^k + 3 \cdot 3^k \right] x^k.

4. Differentiate to Find the nnth Derivative

The nnth derivative of xkx^k is given by:

\frac{k!}{(k-n)!} x^{k-n} & \text{if } k \geq n, \\ 0 & \text{if } k < n. \end{cases}$$ So, the $$n$$th derivative of $$f(x)$$ is: $$f^{(n)}(x) = \sum_{k=n}^{\infty} \left[ -2 \cdot 2^k + 3 \cdot 3^k \right] \frac{k!}{(k-n)!} x^{k-n}.$$ Evaluating this at $$x = 0$$ gives: $$f^{(n)}(0) = \sum_{k=n}^{\infty} \left[ -2 \cdot 2^k + 3 \cdot 3^k \right] \frac{k!}{(k-n)!}.$$ This expression gives the $$n$$th derivative of the function at $$x = 0$$. Would you like more details or have any other questions? Here are some related questions: 1. How do you compute the second derivative of a function? 2. What is the radius of convergence for a geometric series? 3. How do you find the partial fraction decomposition of a function? 4. How do you compute the series expansion for a function? 5. What are the applications of geometric series in calculus? **Tip:** In partial fractions, always check for factors in the denominator that can simplify your problem before expanding.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Partial fractions
Series expansion
Geometric series

Formulas

Geometric series sum formula
Derivative of a power function

Theorems

Binomial theorem

Suitable Grade Level

Advanced Undergraduate