Chapter 10: Problem 28
Find the Taylor series generated by \(f\) at \(x=a.\) \(f(x)=1 /(1-x)^{3}, \quad a=0\)
Short Answer
Expert verified
The Taylor series is \( \sum_{n=0}^{\infty} \frac{n(n+1)}{2} x^n \).
Step by step solution
01
Identify the General Form for Taylor Series
The Taylor series of a function \( f(x) \) about \( x = a \) is given by: \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots \] For this exercise, we need to find the series about \( x = 0 \) (hence \( a = 0 \)).
02
Find the First Few Derivatives
First calculate a few derivatives of \( f(x) = \frac{1}{(1-x)^3} \):- \( f(x) = (1 - x)^{-3} \)- \( f'(x) = 3(1-x)^{-4} \)- \( f''(x) = 12(1-x)^{-5} \)- \( f'''(x) = 60(1-x)^{-6} \)
03
Evaluate Derivatives at \( x = 0 \)
Substitute \( x = 0 \) into each derivative:- \( f(0) = (1-0)^{-3} = 1 \)- \( f'(0) = 3(1-0)^{-4} = 3 \)- \( f''(0) = 12(1-0)^{-5} = 12 \)- \( f'''(0) = 60(1-0)^{-6} = 60 \)
04
Construct the Taylor Series
Substitute the derivatives evaluated at \( x = 0 \) into the Taylor series formula: \[ f(x) = 1 + 3x + \frac{12}{2!}x^2 + \frac{60}{3!}x^3 + \cdots\]Simplifying each term gives: \[ f(x) = 1 + 3x + 6x^2 + 10x^3 + \cdots\]
05
Recognize the Pattern and Generalize
Notice the coefficients \(1, 3, 6, 10, \cdots\) which are the so-called "triangular numbers." The general pattern is given by the formula for the nth triangular number: \( T_n = \frac{n(n+1)}{2} \).Thus, the Taylor series can be written as:\[ \sum_{n=0}^{\infty} T_n x^n = \sum_{n=0}^{\infty} \frac{n(n+1)}{2} x^n \] where \( T_n \) is the nth triangular number.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Triangular Numbers
Triangular numbers are a sequence of numbers that form an equilateral triangle when arranged in a dot pattern. To visualize, consider lining up dots to form successive rows, with each row containing one more dot than the previous. For example, the first few triangular numbers are:
- 1 (a single dot)
- 3 (one row of two dots above a single dot)
- 6 (one row of three dots above a row of two dots above a single dot)
Derivatives
Derivatives are fundamentally about finding the rate of change of a function. In calculus, when we speak about a derivative, we refer to the slope of a function at any given point. Mathematically, the derivative of a function \(f(x)\) is denoted as \(f'(x)\).
Understanding derivatives involves:
Understanding derivatives involves:
- The Power Rule: If \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\).
- The Chain Rule: For composite functions, the derivative is the product of the outer function's derivative and the derivative of the inner function.
Maclaurin Series
A Maclaurin series is a special case of the Taylor series where we expand the function around the point \(x=0\). For a given function \(f(x)\), the Maclaurin series is:\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots \]Maclaurin series provide a polynomial approximation of the function that can be infinitely extended to represent the function accurately over a wide range of values near zero. This makes them incredibly useful for simplifying complex functions and aiding in computations.When constructing the Maclaurin series for \(f(x) = \frac{1}{(1-x)^3}\), the calculated derivatives are substituted into this series expansion. As a result, each term contributes to building an approximation of the function based on the coefficients derived from the function and its derivatives evaluated at zero.