Math Problem Statement

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)equals7 e Superscript negative 4 x Question content area bottom Part 1 a. The first nonzero term of the Maclaurin series is    7. Part 2 The second nonzero term of the Maclaurin series is    negative 28 x. Part 3 The third nonzero term of the Maclaurin series is    56 x squared. Part 4 The fourth nonzero term of the Maclaurin series is    negative StartFraction 224 Over 3 EndFraction x cubed . Part 5 b. Write the power series using summation notation. 7 e Superscript negative 4 xequalsSummation from k equals 0 to infinity left parenthesis nothing right parenthesis

Solution

Let's solve this step-by-step.

Given Function:

f(x)=7e4xf(x) = 7 e^{-4x}

The Maclaurin series is the Taylor series expanded around x=0x = 0. For f(x)=7e4xf(x) = 7 e^{-4x}, we first expand e4xe^{-4x} using its Maclaurin series:
e4x=n=0(4x)nn!=n=0(1)n(4nxn)n!.e^{-4x} = \sum_{n=0}^\infty \frac{(-4x)^n}{n!} = \sum_{n=0}^\infty \frac{(-1)^n (4^n x^n)}{n!}.
Multiplying by 7 gives:
f(x)=7n=0(1)n(4nxn)n!.f(x) = 7 \sum_{n=0}^\infty \frac{(-1)^n (4^n x^n)}{n!}.

Now we find the first four nonzero terms.


(a) First Four Nonzero Terms

To compute the first four terms, plug in n=0,1,2,3n = 0, 1, 2, 3:

  1. First term (n=0n = 0):
    f(x)=7(1)0(40x0)0!=7.f(x) = 7 \cdot \frac{(-1)^0 (4^0 x^0)}{0!} = 7.

  2. Second term (n=1n = 1):
    f(x)=7(1)1(41x1)1!=7(4x)=28x.f(x) = 7 \cdot \frac{(-1)^1 (4^1 x^1)}{1!} = 7 \cdot (-4x) = -28x.

  3. Third term (n=2n = 2):
    f(x)=7(1)2(42x2)2!=716x22=78x2=56x2.f(x) = 7 \cdot \frac{(-1)^2 (4^2 x^2)}{2!} = 7 \cdot \frac{16x^2}{2} = 7 \cdot 8x^2 = 56x^2.

  4. Fourth term (n=3n = 3):
    f(x)=7(1)3(43x3)3!=764x36=7(646x3)=2243x3.f(x) = 7 \cdot \frac{(-1)^3 (4^3 x^3)}{3!} = 7 \cdot \frac{-64x^3}{6} = 7 \cdot \left(-\frac{64}{6}x^3\right) = -\frac{224}{3}x^3.

Thus, the first four nonzero terms are:
728x+56x22243x3.7 - 28x + 56x^2 - \frac{224}{3}x^3.


(b) Power Series in Summation Notation

The general term for the series is:
an=7(1)n(4n)n!xn.a_n = 7 \cdot \frac{(-1)^n (4^n)}{n!} x^n.

Thus, the power series is:
f(x)=n=07(1)n(4n)n!xn=n=07(1)n(4n)n!xn.f(x) = \sum_{n=0}^\infty 7 \cdot \frac{(-1)^n (4^n)}{n!} x^n = \sum_{n=0}^\infty \frac{7(-1)^n (4^n)}{n!} x^n.


(c) Interval of Convergence

To determine the interval of convergence, we analyze the original series:
n=07(1)n(4n)n!xn.\sum_{n=0}^\infty \frac{7(-1)^n (4^n)}{n!} x^n.

Here, the factorial n!n! in the denominator ensures that the series converges for all xx (since the exponential series always converges for all xx).

Conclusion: The interval of convergence is:
(,).(-\infty, \infty).


Final Answers

(a) First four nonzero terms:
728x+56x22243x3.7 - 28x + 56x^2 - \frac{224}{3}x^3.

(b) Power series in summation notation:
f(x)=n=07(1)n(4n)n!xn.f(x) = \sum_{n=0}^\infty \frac{7(-1)^n (4^n)}{n!} x^n.

(c) Interval of convergence:
(,).(-\infty, \infty).


Would you like a detailed explanation of any part of this process?

Here are some related questions you might want to explore:

  1. What is the general method for finding a Maclaurin series?
  2. How does factorial growth affect the convergence of a power series?
  3. What are the properties of the exponential function's series expansion?
  4. How would the series change if the coefficient 77 were replaced by another constant?
  5. How do we test for convergence in a general series?

Tip: When working with power series, always verify convergence using the ratio test or by analyzing the growth of terms directly!

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Math Problem Analysis

Mathematical Concepts

Maclaurin Series
Power Series
Interval of Convergence
Exponential Functions

Formulas

Maclaurin series formula: f(x) = Σ (f^n(0) / n!) x^n
Exponential function expansion: e^(ax) = Σ (a^n x^n / n!)

Theorems

Maclaurin Series Expansion
Interval of Convergence for Power Series

Suitable Grade Level

Grades 11-12, College Level