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Chapter 3: Problem 67

Find a formula for the \(n\) th derivative of \(f\), for \(n \geq 1 .\) $$ f(x)=1 /(1-x) $$

Short Answer

Expert verified
The \( n \)th derivative of \( f(x) = \frac{1}{1-x} \) is \( n!(1-x)^{-n-1} \).

Step by step solution

01

Recognize the Function Type

The function given is \( f(x) = \frac{1}{1-x} \), which is a geometric series. In this form, it resembles the sum of an infinite series \( \sum_{k=0}^{\infty} x^k \) with a common ratio of \( x \) when expanded.
02

Recall the Series Expansion

The function \( f(x) = \frac{1}{1-x} \) can be expanded into the series \( 1 + x + x^2 + x^3 + \cdots \). This expansion corresponds to the series \( \sum_{k=0}^{\infty} x^k \) for \( |x| < 1 \).
03

Derivative of the Geometric Series Term

Consider a single term in the series, \( x^k \). The derivative of \( x^k \) with respect to \( x \) is \( kx^{k-1} \). Differentiating multiple times, the \(n\)th derivative is \( \frac{d^n}{dx^n} x^k = k(k-1)\cdots(k-n+1)x^{k-n} \).
04

Apply to All Terms

The \(n\)th derivative of the sum \( \sum_{k=0}^{\infty} x^k \) becomes \( \sum_{k=n}^{\infty} k(k-1)\cdots(k-n+1) x^{k-n} \) with the \(n\) leading terms disappearing due to differentiation.
05

Simplify the Expression

Recognize that this expression is equivalent to \( n! \sum_{k=n}^{\infty} \frac{x^{k-n}}{(k-n)!} \). Changing the index of summation \((j = k-n)\), this series simplifies further to \( n!(1+x+x^2+\cdots) = n!(1-x)^{-n-1}\).
06

Conclusion on the Derivative Formula

Thus, the \( n \)th derivative of \( f(x) = \frac{1}{1-x} \) is given by \( \frac{d^n}{dx^n} f(x) = n!(1-x)^{-n-1} \). This handles all derivatives using the factorial function and the power series property.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Geometric Series
A geometric series is a sum of terms where each term is found by multiplying the previous term by a fixed, non-zero number called the common ratio. In the function \( f(x) = \frac{1}{1-x} \), this is identified as a geometric series with a common ratio of \( x \). This specific form, \( \sum_{k=0}^{\infty} x^k \), represents the sum of an infinite series. This means:
  • The first term is 1 (when \( k = 0 \))
  • Each subsequent term is multiplied by \( x \)
  • The series continues infinitely as long as \( |x| < 1 \)
Geometric series have numerous applications in mathematics, especially in convergence tests and series summation techniques. This particular series converges due to the condition \( |x| < 1 \), allowing us to express functions in terms of an infinite sum.
Power Series Expansion
Power series expansion is a way to express a function as an infinite sum of terms in the form \( a_n (x - c)^n \), where \( a_n \) denotes the coefficients. For the function \( f(x) = \frac{1}{1-x} \), the power series expansion is \( 1 + x + x^2 + x^3 + \cdots \). Some key points include:
  • The expansion converges for \( |x| < 1 \)
  • Every function expanded in this way has a so-called 'radius of convergence鈥�
  • In this problem, the center is \( c = 0 \), making it a simple Taylor series
Power series ties into many aspects of calculus because it simplifies complex functions into manageable pieces. It鈥檚 particularly useful for integration and differentiation since each term can be handled individually.
Calculus Differentiation
Calculus differentiation involves finding the rate at which a function changes at any point. For the given function \( f(x) = \frac{1}{1-x} \), differentiation is applied repeatedly to find the nth derivative. This process leverages the power series expansion, as each term \( x^k \) has its derivative \( kx^{k-1} \) when differentiated once. Some differentiation rules to remember:
  • The derivative of \( x^k \) is \( kx^{k-1} \)
  • The nth derivative of \( x^k \) involves multiplying by additional constants as seen in factorial expressions
  • When differentiated extensively, constant terms vanish, affecting the sum of series
The ultimate derivative obtained, \( n!(1-x)^{-n-1} \), demonstrates the power of utilizing series and differentiation rules to simplify complex derivatives.

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Most popular questions from this chapter

Use the results of this section to find the derivative of the given function at the given numbers. $$ f(x)=x^{3} ; a=-\frac{1}{4} $$

Use (6) to find the derivative of the function. $$ f(x)=3 x-7 $$

Use the Newton-Raphson method to find an approximate solution of the given equation in the given interval. Use the method until successive approximations obtained by calculator are identical. \(x^{3}-3 x-1=0 ;[1,2]\) (This equation is related to the study of inscribing a nine-sided polygon in a circle.)

Use the results of this section to find the derivative of the given function at the given numbers. $$ f(x)=x^{5} ; a=-2 $$

The outside surface of a hemispherical dome of radius 20 feet is to be given a coat of paint, \(\frac{1}{100}\) inch thick. Use differentials to approximate the volume of paint needed for the job. (Hint: Approximate the change in the volume of a hemisphere when the radius increases from 20 feet to \(20 \frac{1}{1200}\) feet.)
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