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Chapter 9: Problem 3

How do you find the coefficients of the Taylor series for \(f\) centered at \(a ?\)

Short Answer

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Question: Find the coefficients of the Taylor series for the following function centered at point a: \(f(x) = e^x\), with \(a = 0\). Step 1: Determine the derivatives of the function at point a. The first few derivatives of the function \(f(x) = e^x\) are: \(f'(x) = e^x\) \(f''(x) = e^x\) \(f'''(x) = e^x\) And so on. Now, evaluate these derivatives at \(a = 0\): \(f(0) = e^0 = 1\) \(f'(0) = e^0 = 1\) \(f''(0) = e^0 = 1\) \(f'''(0) = e^0 = 1\) The pattern continues for higher derivatives as well. Step 2: Write down the general definition of the coefficients. The general coefficient \(c_n\) of the Taylor series is given by: $$c_n = \frac{f^{(n)}(0)}{n!}$$ Step 3: Substitute the derivatives evaluated at a into the coefficient definition formula. Since all derivatives of the function \(f(x) = e^x\) evaluated at a (= 0) are equal to 1, we can substitute this value into the formula: $$c_n = \frac{1}{n!}$$ Step 4: Find the coefficients of the Taylor series expansion. Using the equation obtained in Step 3, we can find the coefficients of the Taylor series for the function \(f(x) = e^x\) centered at \(a = 0\): \(c_0 = \frac{1}{0!} = 1\) \(c_1 = \frac{1}{1!} = 1\) \(c_2 = \frac{1}{2!} = \frac{1}{2}\) \(c_3 = \frac{1}{3!} = \frac{1}{6}\) And so on. The coefficients of the Taylor series follow a pattern: \(c_n = \frac{1}{n!}\) for \(n = 0, 1, 2, 3, ...\).

Step by step solution

01

Determine the derivatives of the function at point a

Calculate the first few derivatives of the function \(f(x)\) and evaluate them at the point \(a\). Depending on the function, it might be easier to find a pattern in the derivatives evaluated at \(a\), which will help us in finding the coefficients of the Taylor series.
02

Write down the general definition of the coefficients

Recall the formula for the general coefficient \(c_n\) of the Taylor series expansion: $$c_n = \frac{f^{(n)}(a)}{n!}$$
03

Substitute the derivatives evaluated at a into the coefficient definition formula

Substitute the results from Step 1, which were the derivatives of the function evaluated at the point \(a\), into the coefficient definition formula we wrote down in Step 2. This will allow us to find the coefficients \(c_n\) of the Taylor series expansion.
04

Find the coefficients of the Taylor series expansion

Use the equations obtained in Step 3 to calculate the coefficients of the Taylor series for the function \(f(x)\) centered at \(a\).

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Key Concepts

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

Coefficients in Taylor Series
To understand a Taylor series, we must focus on calculating its coefficients. In mathematical terms, these coefficients provide the weight that each term carries in the series. The general formula for finding a coefficient, denoted as \(c_n\) in a Taylor series, is \(c_n = \frac{f^{(n)}(a)}{n!}\).
This formula tells us to take the \(n\)-th derivative of the function \(f\), evaluate it at the point \(a\), and then divide by \(n!\) (n factorial). Here, \(n!\) is the product of all positive integers up to \(n\), which normalizes the coefficients.
  • The \(n\)-th derivative \(f^{(n)}(a)\) provides specific values of the function's behavior at point \(a\).
  • This helps in approximating the function within the vicinity of \(a\) for better convergence.
  • Coefficients are crucial because they directly influence the series' accuracy and convergence rate.
Understanding coefficients is vital as they form the basis of constructing the Taylor series itself.
Derivatives in Taylor Series
Derivatives are an essential part of forming a Taylor series. Each term in the series includes a derivative of the function being analyzed. The more derivatives we take, the better the approximation of the function we achieve.
To find these, take the following steps:
  • Start from the first derivative of \(f(x)\) at the point of interest \(a\).
  • Proceed to higher-order derivatives as required for the series expansion.
  • Evaluate each derivative at point \(a\) to find its specific value.
Taking derivatives helps depict how the function behaves locally around the point \(a\). Precise and orderly computation of these derivatives is necessary to develop accurate Taylor series coefficients.
Function Evaluation at a Point
Evaluating the function and its derivatives at a specific point, denoted as \(a\), is a cornerstone in forming Taylor series. This involves calculating the exact values of the function and its successive derivatives at \(a\) to incorporate them into the Taylor series formula.
Why is it important?
  • These values provide insights into the function's behavior close to \(a\).
  • They form the backbone of the Taylor series coefficients which define each term's contribution.
  • Accurate function evaluation can significantly enhance the series' precision near \(a\).
By leveraging these evaluations, we begin constructing the series that mirrors the function's nature around its center point \(a\).
Expansion of a Function Using Taylor Series
Expansion is the process of expressing a function as an infinite sum of terms calculated from the values of its derivatives at a particular point. This concept is pivotal when using Taylor series to approximate functions.
The general expression for the Taylor series expansion around \(a\) is represented as:
\[\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n\] Here is why expansion is critical:
  • It translates complex functions into manageable infinite series.
  • The expansion helps estimate function values that are otherwise difficult to compute directly.
  • It provides varying degrees of approximation flexibility based on the number of terms used.
An effective expansion brings a function closer to its actual behavior, especially in regions near \(a\). Concluding, expansion through Taylor series is a powerful tool in both theoretical and applied mathematics.

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Most popular questions from this chapter

Find the remainder in the Taylor series centered at the point a for the following functions. Then show that \(\lim _{n \rightarrow \infty} R_{n}(x)=0\) for all \(x\) in the interval of convergence. $$f(x)=\sin x, a=0$$

Explain why or why not ,Determine whether the following statements are true and give an explanation or counterexample. a. The interval of convergence of the power series \(\Sigma c_{k}(x-3)^{k}\) could be (-2,8) b. The series \(\sum(-2 x)^{k}\) converges on the interval \(-\frac{1}{2}

Elliptic integrals The period of a pendulum is given by $$ T=4 \sqrt{\frac{\ell}{g}} \int_{0}^{\pi / 2} \frac{d \theta}{\sqrt{1-k^{2} \sin ^{2} \theta}}=4 \sqrt{\frac{\ell}{g}} F(k) $$ where \(\ell\) is the length of the pendulum, \(g \approx 9.8 \mathrm{m} / \mathrm{s}^{2}\) is the acceleration due to gravity, \(k=\sin \left(\theta_{0} / 2\right),\) and \(\theta_{0}\) is the initial angular displacement of the pendulum (in radians). The integral in this formula \(F(k)\) is called an elliptic integral, and it cannot be evaluated analytically. a. Approximate \(F(0.1)\) by expanding the integrand in a Taylor (binomial) series and integrating term by term. b. How many terms of the Taylor series do you suggest using to obtain an approximation to \(F(0.1)\) with an error less than \(10^{-3} ?\) c. Would you expect to use fewer or more terms (than in part (b)) to approximate \(F(0.2)\) to the same accuracy? Explain.

Summation notation Write the following power series in summation (sigma) notation. $$-\frac{x^{2}}{1 !}+\frac{x^{4}}{2 !}-\frac{x^{6}}{3 !}+\frac{x^{8}}{4 !}-\cdots$$

Use a Taylor series to approximate the following definite integrals. Retain as many terms as needed to ensure the error is less than \(10^{-4}.\) $$\int_{0}^{0.35} \tan ^{-1} x d x$$
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