Problem 31
Question
Use a fourth-degree Taylor approximation for \(e^{h},\) for \(h\) near \(0,\) to evaluate the following limits. Would your answer be different if you used a Taylor polynomial of higher degree? (a) \(\lim _{h \rightarrow 0} \frac{e^{h}-1-h}{h^{2}}\) (b) \(\lim _{h \rightarrow 0} \frac{e^{h}-1-h-\frac{h^{2}}{2}}{h^{3}}\)
Step-by-Step Solution
Verified Answer
(a) \(\frac{1}{2}\); (b) \(\frac{1}{6}\). The answer would be the same with a higher-degree series.
1Step 1: Recall Taylor Series Expansion
The Taylor series for a function about 0 is given by: \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \ldots \). For \(e^h\), the derivatives at 0 are all 1, so the expansion is \(e^h = 1 + h + \frac{h^2}{2!} + \frac{h^3}{3!} + \frac{h^4}{4!} + \ldots\).
2Step 2: Simplify Expression for (a) Using the Taylor Series
Substitute the series expansion into \( e^h - 1 - h \):\[ e^h - 1 - h = \left(1 + h + \frac{h^2}{2} + \frac{h^3}{6} + \frac{h^4}{24}\right) - 1 - h = \frac{h^2}{2} + \frac{h^3}{6} + \frac{h^4}{24} \]The expression becomes \( \frac{\frac{h^2}{2} + \frac{h^3}{6} + \frac{h^4}{24}}{h^2} \).
3Step 3: Evaluate the Limit for (a)
Divide each term by \( h^2 \):\[ \frac{\frac{h^2}{2} + \frac{h^3}{6} + \frac{h^4}{24}}{h^2} = \frac{1}{2} + \frac{h}{6} + \frac{h^2}{24} \]As \( h \to 0 \), the higher-order terms vanish, so the limit is \( \frac{1}{2} \).
4Step 4: Simplify Expression for (b) Using the Taylor Series
Substitute the series expansion into \( e^h - 1 - h - \frac{h^2}{2} \):\[ e^h - 1 - h - \frac{h^2}{2} = \left(1 + h + \frac{h^2}{2} + \frac{h^3}{6} + \frac{h^4}{24}\right) - 1 - h - \frac{h^2}{2} = \frac{h^3}{6} + \frac{h^4}{24} \]The expression becomes \( \frac{\frac{h^3}{6} + \frac{h^4}{24}}{h^3} \).
5Step 5: Evaluate the Limit for (b)
Divide each term by \( h^3 \):\[ \frac{\frac{h^3}{6} + \frac{h^4}{24}}{h^3} = \frac{1}{6} + \frac{h}{24} \]As \( h \to 0 \), the higher-order terms vanish, so the limit is \( \frac{1}{6} \).
6Step 6: Consider Higher Degree Taylor Series Impact
Using a Taylor series with a degree higher than 4 would not change the results for the given limits, as higher-degree terms vanish when divided by the respective powers of \( h \) as \( h \to 0 \). The contributions of higher-degree terms become negligible in the limit.
Key Concepts
Taylor approximationLimits in CalculusExponential functionPolynomial approximation
Taylor approximation
The Taylor approximation is a powerful tool in calculus for estimating the values of functions that might be too complex to compute directly. Essentially, it involves using a polynomial to approximate the function. This is done by matching the function's value and several of its derivatives at a particular point, most commonly 0 for simplicity, which is then called a Maclaurin series.
- The Taylor series for a function is expressed as an infinite sum of terms, each involving a derivative of the function evaluated at a specific point, divided by a factorial term.
- The approximation's accuracy depends on the number of terms used; more terms mean higher accuracy.
- For the exponential function, the series converges quickly, providing very good approximations even with a few terms.
Limits in Calculus
Limits are a cornerstone concept in calculus, providing a method to determine how a function behaves as its input approaches a specific value. Limits are essential for understanding continuous functions and are the foundational elements for defining derivatives and integrals.
- In the context of Taylor approximations, limits help determine the behavior of the function as an independent variable becomes infinitely small or large.
- This is particularly useful when examining the differences between a function and its Taylor approximation, as the small parameters in the expansion diminish.
Exponential function
The exponential function, denoted as \( e^x \), is one of the most important functions in mathematics. It is the function that describes continuous growth or decay and is defined as the limit of \( (1 + \frac{x}{n})^n \) as \( n \to \infty \). This function has unique properties that make it central in calculus.
- The derivative of the exponential function is the function itself, i.e., \( \frac{d}{dx} e^x = e^x \), which simplifies calculations significantly.
- At \( x = 0 \), \( e^0 = 1 \), which provides a convenient base point for Taylor expansions.
Polynomial approximation
Polynomial approximation involves representing a complex function with a polynomial, making it easier to analyze and compute. The most common form of polynomial approximation in calculus is the Taylor series, where a function is approximated by a series of its derivatives at a specific point.
- The Taylor polynomial creates a tangent approximation that locally closely matches the function.
- The degree of the polynomial indicates the number of derivatives included and thus influences the approximation's accuracy.
- Lower degree polynomials provide rough approximations, while higher degree polynomials improve precision.
Other exercises in this chapter
Problem 31
Decide if the statements are true or false. Assume that the Taylor series for a function converges to that function. Give an explanation for your answer. If \(f
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Suppose that \(f(x)\) is a periodic function with period \(b\) Show that (a) \(g(t)=f(b t / 2 \pi)\) is periodic with period \(2 \pi\) and \(f(x)=g(2 \pi x / b)
View solution Problem 31
Using the Taylor series for \(f(x)=e^{x}\) around \(0,\) compute the following limit: $$\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}$$
View solution Problem 32
Decide if the statements are true or false. Assume that the Taylor series for a function converges to that function. Give an explanation for your answer. If \(f
View solution