a. Find the first four nonzero terms of the Maclaurin series for the given function.
b. Write the power series using summation notation.
c. Determine the interval of convergence of the series.
$$
f(x)=5 e^{-5 x}
$$

Question

a. Find the first four nonzero terms of the Maclaurin series for the given function.
b. Write the power series using summation notation.
c. Determine the interval of convergence of the series.
$$
f(x)=5 e^{-5 x}
$$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the first four nonzero terms of the Maclaurin series for $ f(x) = 5e^{-5x} $, we will use the Maclaurin series expansion for $ e^x $, which is $ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} $. ‖ ‖ step 2 ⋮ Substituting $ -5x $ for $ x $ in the series expansion, we get $ e^{-5x} = \sum_{n=0}^{\infty} \frac{(-5x)^n}{n!} $. ‖ ‖ step 3 ⋮ Multiplying the series by 5, we obtain the series for $ 5e^{-5x} $: $ 5e^{-5x} = 5 \sum_{n=0}^{\infty} \frac{(-5x)^n}{n!} $. ‖ ‖ step 4 ⋮ The first four nonzero terms of the series are given by the terms for $ n = 0, 1, 2, 3 $: $ 5, -25x, \frac{125x^2}{2!}, -\frac{625x^3}{3!} $. ‖ ∻Answer∻ ⚹ The first four nonzero terms of the Maclaurin series for $ f(x) = 5e^{-5x} $ are $ 5, -25x, \frac{125x^2}{2}, -\frac{625x^3}{6} $. ⚹ ∻Key Concept∻ ⚹Maclaurin Series Expansion⚹ ∻Explanation∻ ⚹The Maclaurin series is a special case of the Taylor series centered at $ x = 0 $. For the exponential function $ e^x $, the series expansion is $ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} $. By substituting $ -5x $ into this formula and multiplying by 5, we obtain the series for $ 5e^{-5x} $.⚹ --- ‖ step 1 ⋮ To represent $ 5e^{-5x} $ as a power series using summation notation, we start with the general form of the Maclaurin series for $ e^x $: $ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} $. ‖ ‖ step 2 ⋮ We substitute $ -5x $ for $ x $ to get the series for $ e^{-5x} $: $ e^{-5x} = \sum_{n=0}^{\infty} \frac{(-5x)^n}{n!} $. ‖ ‖ step 3 ⋮ Multiplying the series by 5 gives us the power series for $ 5e^{-5x} $: $ 5e^{-5x} = 5 \sum_{n=0}^{\infty} \frac{(-5x)^n}{n!} $. ‖ ∻Answer∻ ⚹ The power series for $ f(x) = 5e^{-5x} $ using summation notation is $ 5 \sum_{n=0}^{\infty} \frac{(-5x)^n}{n!} $. ⚹ ∻Key Concept∻ ⚹Power Series Representation⚹ ∻Explanation∻ ⚹A power series is an infinite series of the form $ \sum_{n=0}^{\infty} a_n(x - c)^n $, where $ a_n $ are the coefficients and $ c $ is the center of the series. For $ 5e^{-5x} $, the power series is centered at $ x = 0 $ and has coefficients derived from the Maclaurin series expansion of $ e^x $.⚹ --- ‖ step 1 ⋮ To find the interval of convergence for the power series of $ 5e^{-5x} $, we can use the ratio test, which states that a series $ \sum a_n $ converges if $ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 $. ‖ ‖ step 2 ⋮ Applying the ratio test to the series $ \sum_{n=0}^{\infty} \frac{(-5x)^n}{n!} $, we get $ \lim_{n \to \infty} \left| \frac{(-5x)^{n+1}/(n+1)!}{(-5x)^n/n!} \right| = \lim_{n \to \infty} \left| \frac{-5x}{n+1} \right| $. ‖ ‖ step 3 ⋮ Since $ \lim_{n \to \infty} \left| \frac{-5x}{n+1} \right| = 0 $ for all $ x $, the ratio is less than 1 for all $ x $. ‖ ‖ step 4 ⋮ Therefore, the interval of convergence for the power series of $ 5e^{-5x} $ is all real numbers, $ (-\infty, \infty) $. ‖ ∻Answer∻ ⚹ The interval of convergence for the power series of $ f(x) = 5e^{-5x} $ is $ (-\infty, \infty) $. ⚹ ∻Key Concept∻ ⚹Interval of Convergence⚹ ∻Explanation∻ ⚹The interval of convergence for a power series is the set of all $ x $ values for which the series converges. The exponential function $ e^x $ has a power series that converges for all $ x $, and thus the power series for $ 5e^{-5x} $ also converges for all $ x $.⚹◊What is the formula for finding the nth term in a Maclaurin series?⍭ Generate me a similar question◊

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