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17-18. Find the Taylor series at \(x=0\) for each function in two ways: a. By calculating derivatives and using the definition of Taylor series b. By modifying a known Taylor series. \(e^{x / 5}\)

Short Answer

Expert verified
The Taylor series for \( e^{x/5} \) is: \( 1 + \frac{x}{5} + \frac{x^2}{2!(5^2)} + \frac{x^3}{3!(5^3)} + \cdots \).

Step by step solution

01

Definition of Taylor Series

The Taylor series of a function \( f(x) \) centered at \( x=0 \) is given by: \[ f(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \] where \( f^n(0) \) denotes the \( n \)-th derivative of \( f \) evaluated at \( x=0 \).
02

Calculating Derivatives

Let \( f(x) = e^{x/5} \). First, find the derivatives at \( x=0 \):\[ f(x) = e^{x/5} \]\[ f'(x) = \frac{1}{5}e^{x/5}, \, f'(0) = \frac{1}{5} \]\[ f''(x) = \left(\frac{1}{5}\right)^2 e^{x/5}, \, f''(0) = \left(\frac{1}{5}\right)^2 \]\[ f'''(x) = \left(\frac{1}{5}\right)^3 e^{x/5}, \, f'''(0) = \left(\frac{1}{5}\right)^3 \]And so on for higher derivatives.
03

Constructing the Taylor Series using Derivatives

Substitute the derivatives into the Taylor series formula:\[ e^{x/5} \approx 1 + \frac{1}{5}x + \frac{1}{2!}\left(\frac{1}{5}\right)^2 x^2 + \frac{1}{3!}\left(\frac{1}{5}\right)^3 x^3 + \cdots \]
04

Using Known Series to Modify

The known Taylor series for \( e^u \) about \( u=0 \) is: \[ e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \cdots \] Substitute \( u = \frac{x}{5} \) to obtain the series for \( e^{x/5} \):\[ e^{x/5} = 1 + \frac{x}{5} + \frac{\left(\frac{x}{5}\right)^2}{2!} + \frac{\left(\frac{x}{5}\right)^3}{3!} + \cdots \]
05

Final Comparison of Methods

Notice that both methods yield the same Taylor series expansion:\[ e^{x/5} = 1 + \frac{x}{5} + \frac{x^2}{2!(5^2)} + \frac{x^3}{3!(5^3)} + \cdots \] This verifies both approaches are consistent in providing the Taylor series for \( e^{x/5} \) at \( x=0 \).

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Key Concepts

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

Derivatives
In calculus, derivatives are fundamental tools for understanding how functions change. It's like finding the slope of a line that best approximates a function at a given point. The derivative of a function gives you a formula that tells you the rate of change at any point along the function. Stretching beyond this idea, we can take derivatives multiple times. Each time, it's like peeling an onion layer to reveal deeper behavior of the function.

To find the Taylor series of a function, derivatives play a crucial role. For example, if we have a function like \( e^{x/5} \), we start by finding its derivatives. At \( x=0 \):
  • The zeroth derivative (the function itself) is \( e^{0} = 1 \).
  • The first derivative \( f'(x) = \frac{1}{5}e^{x/5} \), so \( f'(0) = \frac{1}{5} \).
  • The second derivative \( f''(x) = \left(\frac{1}{5}\right)^2 e^{x/5} \), hence \( f''(0) = \left(\frac{1}{5}\right)^2 \).
  • This pattern continues, giving us deeper insights with each step.
By deriving these terms, each tells us what to include in the corresponding power of \( x \) in our series expansion.
Maclaurin Series
The Maclaurin series is simply a Taylor series centered at zero. It's a way to express a function as an infinite sum of terms based on its derivatives at a single point, specifically zero. If you've ever wondered how a smooth curve can be turned into a series of simple polynomial terms, the Maclaurin series shows you how it’s done.

For the function \( e^{x/5} \), using a Maclaurin series means we evaluate derivatives at zero and then create a series representation:
  • Start with the constant term \( f(0) \).
  • Add the first derivative times \( x \), all evaluated at zero, over \( 1! \).
  • Continue adding higher derivatives like \( \frac{f''(0)}{2!}x^2 \) and so forth.
This series is extremely useful, especially in calculus, physics, and engineering, as it transforms complex expressions into manageable polynomials.
Series Expansion
Series expansion is a mathematical method that allows us to write complex functions as the sum of simpler polynomial terms. This is like breaking a big, complicated task into smaller, more manageable tasks. The Taylor series is one of the most popular forms of series expansions.

Take, for example, how we derived the series for \( e^{x/5} \). First, using derivatives, we constructed the series step-by-step by evaluating the function at \( x=0 \). Each term in this series expansion represents a polynomial order:
  • The first term is the constant.
  • The second term involves \( x \) raised to the first power, scaled by the first derivative.
  • Subsequent terms involve higher powers of \( x \), each scaled by corresponding higher order derivatives.
Alternatively, if a standard series like \( e^u = 1 + u + \frac{u^2}{2!} + \cdots \) is known, you can substitute directly, where \( u = \frac{x}{5} \). This gives a shortcut to derive our desired series expansion without recomputing each derivative.

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Most popular questions from this chapter

a. Find the Taylor polynomial at \(x=0\) that approximates \(f(x)=\cos x\) on the interval \(-1 \leq x \leq 1\) with error less than 0.0002 b. Graph \(f(x)=\cos x\) and the Taylor polynomial on \([-2 \pi, 2 \pi]\) by \([-2,2] .\) Use TRACE to estimate the maximum difference between the two curves for \(-1 \leq x \leq 1\).

a. What linear function best approximates \(\sqrt[3]{1+x}\) near \(x=0 ?\) b. Use the error formula to find an interval \([0, x]\) on which the error is less than 0.01 c. Graph \(\sqrt[3]{1+x}\) and the linear approximation on [-2,2] by [-2,2] and use TRACE to verify that the difference is less than 0.01 for the \(x\) -values found in part (b).

BIOMEDICAL: Long-Term Population A population (of cells or people) is such that each year a number \(a\) of individuals are added (call them "immigrants"), and a proportion \(p\) of the individuals who have been there die. Therefore, the proportion that survives is \((1-p),\) so that just after an immigration the population will consist of new immigrants plus \(a(1-p)\) from the previous year's immigration plus \(a(1-p)^{2}\) from the immigration before that, and so on. In the long run, the size of the population just after an immigration will then be the sum \(a+a(1-p)+a(1-p)^{2}+a(1-p)^{3}+\cdots\) Find the long-run size of the population just after an immigration for any number of immigrants \(a\) and any survival proportion \((1-p)\). Then find the long-run size just before an immigration.

GENERAL: Chessboards It is said that the game of chess was invented by the Grand Vizier Sissa Ben Dahir for the Indian King Shirham. When the king offered him any reward, Sissa asked for one grain of wheat to be placed on the first square of the chessboard, two on the second, four on the third, and so on, doubling each time for each of the 64 squares. Estimate the total number of grains of wheat required. (It is also said that the king, upon learning that the total was many times the world production of wheat, executed Sissa for making a fool out of him.)

$$ \begin{aligned} &\text { How many terms are in the following finite geometric }\\\ &\text { series? } 2+2 \cdot 3+2 \cdot 3^{2}+\cdots+2 \cdot 3^{5280} \end{aligned} $$

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