Problem 19
Question
In Problems 19-24, find the Taylor series in \(x-\) a through the term \((x-a)^{3}\). $$ e^{x}, a=1 $$
Step-by-Step Solution
Verified Answer
The Taylor series for \(e^x\) at \(x=1\) up to \((x-1)^3\) is \(e + e(x-1) + \frac{e}{2}(x-1)^2 + \frac{e}{6}(x-1)^3\).
1Step 1: Taylor Series Formula
The Taylor series of a function \(f(x)\) about \(x = a\) is given by:\[f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots\]We need the series through the \((x-a)^3\) term.
2Step 2: Determine the Function and Derivatives at \(a=1\)
Given \(f(x) = e^x\), compute \(f(a), f'(a), f''(a), f'''(a)\) at \(a=1\):- \(f(1) = e^1 = e\)- \(f'(x) = e^x\) so, \(f'(1) = e\)- \(f''(x) = e^x\) so, \(f''(1) = e\)- \(f'''(x) = e^x\) so, \(f'''(1) = e\)
3Step 3: Substitute Values into Taylor Series Formula
Substitute the derivatives into the Taylor series formula:\[e^x = e + e(x-1) + \frac{e}{2}(x-1)^2 + \frac{e}{6}(x-1)^3\]This provides each term up to \((x-1)^3\).
Key Concepts
Exponential FunctionDerivativesPower SeriesPolynomial Approximation
Exponential Function
The exponential function is a fundamental mathematical concept often denoted by the notation \( e^x \). It's a function where the constant \( e \) (approximately 2.71828) is raised to the power of \( x \). This function has the unique property of being equal to its own derivative and is frequently used in various fields such as calculus, complex analysis, and differential equations.
The exponential function is used to model growth or decay processes, like population growth and radioactive decay. It's particularly important because of its continuous growth rate. In calculus, the function \( e^x \) serves as an example where functions maintain the same shape even as we apply derivatives.
The exponential function is used to model growth or decay processes, like population growth and radioactive decay. It's particularly important because of its continuous growth rate. In calculus, the function \( e^x \) serves as an example where functions maintain the same shape even as we apply derivatives.
- Always results in positive values since \( e^x > 0 \) for all \( x \).
- Changes rapidly with changes in \( x \), thus reflecting exponentially growing phenomena.
Derivatives
Derivatives are tools in calculus that describe the rate at which a function changes. For any given function, the derivative is a function that tells us the slope of the tangent line at any point. This makes derivatives vital for understanding the behavior of functions.
In the case of the exponential function, \( f(x) = e^x \), the derivative is quite simple. Its derivative, \( f'(x) \), is the same \( e^x \). This characteristic is unique to the exponential function, as most functions have derivatives that differ from the original function.
Finding derivatives involves applying differentiation rules:
In the case of the exponential function, \( f(x) = e^x \), the derivative is quite simple. Its derivative, \( f'(x) \), is the same \( e^x \). This characteristic is unique to the exponential function, as most functions have derivatives that differ from the original function.
Finding derivatives involves applying differentiation rules:
- The derivative of a constant times a function is the constant times the derivative of the function.
- The chain rule and product rule may be used for more complex functions involving exponential terms.
Power Series
A power series is a series of the form \( \sum_{n=0}^{\infty} a_n (x-c)^n \), where \( a_n \) are coefficients and \( c \) is the center of the series. Power series extend the idea of a polynomial to potentially infinitely many terms, capturing the behavior of functions across intervals.
In the context of the Taylor series, power series are used to represent functions like \( e^x \) as an infinite sum of terms. The Taylor series at a certain point \( a \) utilizes the function's derivatives to determine the coefficients. This provides an efficient way to approximate complex functions by polynomials.
In the context of the Taylor series, power series are used to represent functions like \( e^x \) as an infinite sum of terms. The Taylor series at a certain point \( a \) utilizes the function's derivatives to determine the coefficients. This provides an efficient way to approximate complex functions by polynomials.
- Power series converge within a certain radius around the center \( c \).
- The rate of convergence depends on the function and the chosen center.
Polynomial Approximation
Polynomial approximation involves replacing a complex function with a polynomial to simplify analysis and computations. The Taylor series provides one way to approximate functions using polynomials.
Through polynomial approximation, we can estimate the value of functions like \( e^x \) by a finite sum of terms. The more terms included, the more accurate the approximation becomes. This is particularly useful in computing and numerical methods where exact calculation of functions isn't feasible.
Through polynomial approximation, we can estimate the value of functions like \( e^x \) by a finite sum of terms. The more terms included, the more accurate the approximation becomes. This is particularly useful in computing and numerical methods where exact calculation of functions isn't feasible.
- The Taylor series up to \((x-a)^3\) approximates \( e^x \) at \( a = 1 \), as seen with the expression: \( e + e(x-1) + \frac{e}{2}(x-1)^2 + \frac{e}{6}(x-1)^3 \).
- Higher degree polynomials provide closer approximations.
Other exercises in this chapter
Problem 19
In Problems 9-28, find the convergence set for the given power series. Hint: First find a formula for the nth term; then use the Absolute Ratio Test. $$ 1-\frac
View solution Problem 19
In Problems 1-20, an explicit formula for \(a_{n}\) is given. Write the first five terms of \(\left\\{a_{n}\right\\}\), determine whether the sequence converges
View solution Problem 19
In Problems 13-22, use any test developed so far, including any from Section 9.2, to decide about the convergence or divergence of the series. Give a reason for
View solution Problem 19
\(\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}+\cdots\) Hint: \(a_{n}=\frac{1}{n(n+1)}\)
View solution