Math Problem Statement

Use the following information to complete parts a. and b. below.

f left parenthesis x right parenthesis equals 5 ln xf(x)=5lnx​,

aequals=44

Question content area bottom

Part 1

a. Find the first four nonzero terms of the Taylor series for the given function centered at a.

A.

The first four nonzero terms are

5 ln 4 plus five fourths left parenthesis x minus 4 right parenthesis minus StartFraction 5 Over 32 EndFraction left parenthesis x minus 4 right parenthesis squared plus StartFraction 5 Over 192 EndFraction left parenthesis x minus 4 right parenthesis cubed5ln 4+54(x−4)−532(x−4)2+5192(x−4)3.

B.

The first four nonzero terms are

5 ln 4 minus five fourths left parenthesis x minus 4 right parenthesis minus five sixteenths left parenthesis x minus 4 right parenthesis squared minus StartFraction 5 Over 32 EndFraction left parenthesis x minus 4 right parenthesis cubed5ln 4−54(x−4)−516(x−4)2−532(x−4)3.

C.

The first four nonzero terms are

negative five fourths plus StartFraction 5 Over 32 EndFraction left parenthesis x minus 4 right parenthesis minus StartFraction 5 Over 32 EndFraction left parenthesis x minus 4 right parenthesis squared plus StartFraction 15 Over 64 EndFraction left parenthesis x minus 4 right parenthesis cubed−54+532(x−4)−532(x−4)2+1564(x−4)3.

D.

The first four nonzero terms are

negative five fourths plus StartFraction 5 Over 32 EndFraction left parenthesis x minus 4 right parenthesis minus StartFraction 5 Over 192 EndFraction left parenthesis x minus 4 right parenthesis squared plus StartFraction 5 Over 1024 EndFraction left parenthesis x minus 4 right parenthesis cubed−54+532(x−4)−5192(x−4)2+51024(x−4)3.

Part 2

b. Write the power series using summation notation.

A.5 ln 4 plus Summation from k equals 0 to infinity StartFraction 5 left parenthesis negative 1 right parenthesis Superscript k plus 1 Over 4 Superscript k Baseline left parenthesis x minus 4 right parenthesis Superscript k EndFraction

5 ln 45ln 4plus+Summation from k equals 0 to infinity StartFraction 5 left parenthesis negative 1 right parenthesis Superscript k plus 1 Over 4 Superscript k Baseline left parenthesis x minus 4 right parenthesis Superscript k EndFraction∑k=0∞5(−1)k+14k(x−4)k

B.5 ln 4 plus Summation from k equals 1 to infinity left parenthesis negative 1 right parenthesis Superscript k plus 1 Baseline times StartFraction 5 Over k times 4 Superscript k EndFraction left parenthesis x minus 4 right parenthesis Superscript k

5 ln 45ln 4plus+Summation from k equals 1 to infinity left parenthesis negative 1 right parenthesis Superscript k plus 1 Baseline times StartFraction 5 Over k times 4 Superscript k EndFraction left parenthesis x minus 4 right parenthesis Superscript k∑k=1∞(−1)k+1•5k•4k(x−4)k

C.5 ln 4 plus Summation from k equals 0 to infinity left parenthesis negative 1 right parenthesis Superscript k Baseline times StartFraction 5 Over 4 Superscript k Baseline left parenthesis x minus 4 right parenthesis Superscript k EndFraction

5 ln 45ln 4plus+Summation from k equals 0 to infinity left parenthesis negative 1 right parenthesis Superscript k Baseline times StartFraction 5 Over 4 Superscript k Baseline left parenthesis x minus 4 right parenthesis Superscript k EndFraction∑k=0∞(−1)k•54k(x−4)k

D.5 ln 4 plus Summation from k equals 1 to infinity StartFraction 5 left parenthesis negative 1 right parenthesis Superscript k plus 1 Baseline left parenthesis x minus 4 right parenthesis Superscript k Over 4 Superscript k EndFraction

5 ln 45ln 4plus+Summation from k equals 1 to infinity StartFraction 5 left parenthesis negative 1 right parenthesis Superscript k plus 1 Baseline left parenthesis x minus 4 right parenthesis Superscript k Over 4 Superscript k EndFraction∑k=1∞5(−1)k+1(x−4)k4k

Solution

Part 1: Find the first four nonzero terms of the Taylor series

To find the Taylor series for f(x)=5ln⁡xf(x) = 5 \ln x centered at a=4a = 4, we use the formula for the Taylor series expansion: f(x)=f(a)+f′(a)(x−a)+f′′(a)2!(x−a)2+f′′′(a)3!(x−a)3+…f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \dots where f(a)f(a), f′(a)f'(a), f′′(a)f''(a), etc., are derivatives of f(x)f(x) evaluated at x=ax = a.

Step 1: Compute the derivatives of f(x)=5ln⁡xf(x) = 5 \ln x

  1. f(x)=5ln⁡xf(x) = 5 \ln x
  2. f′(x)=5xf'(x) = \frac{5}{x}
  3. f′′(x)=−5x2f''(x) = -\frac{5}{x^2}
  4. f′′′(x)=10x3f'''(x) = \frac{10}{x^3}
  5. f(4)(x)=−30x4f^{(4)}(x) = -\frac{30}{x^4}

Step 2: Evaluate the derivatives at x=4x = 4

  1. f(4)=5ln⁡4f(4) = 5 \ln 4
  2. f′(4)=54f'(4) = \frac{5}{4}
  3. f′′(4)=−516f''(4) = -\frac{5}{16}
  4. f′′′(4)=1064=532f'''(4) = \frac{10}{64} = \frac{5}{32}
  5. f(4)(4)=−30256=−15128f^{(4)}(4) = -\frac{30}{256} = -\frac{15}{128}

Step 3: Write the first four terms of the Taylor series

f(x)=f(4)+f′(4)(x−4)+f′′(4)2!(x−4)2+f′′′(4)3!(x−4)3+…f(x) = f(4) + f'(4)(x-4) + \frac{f''(4)}{2!}(x-4)^2 + \frac{f'''(4)}{3!}(x-4)^3 + \dots Substituting the values: f(x)=5ln⁡4+54(x−4)−516⋅2(x−4)2+532⋅6(x−4)3f(x) = 5 \ln 4 + \frac{5}{4}(x-4) - \frac{5}{16 \cdot 2}(x-4)^2 + \frac{5}{32 \cdot 6}(x-4)^3 Simplify: f(x)=5ln⁡4+54(x−4)−532(x−4)2+5192(x−4)3f(x) = 5 \ln 4 + \frac{5}{4}(x-4) - \frac{5}{32}(x-4)^2 + \frac{5}{192}(x-4)^3

The first four nonzero terms are: 5ln⁡4+54(x−4)−532(x−4)2+5192(x−4)3\boxed{5 \ln 4 + \frac{5}{4}(x-4) - \frac{5}{32}(x-4)^2 + \frac{5}{192}(x-4)^3} This corresponds to Option A.

Part 2: Write the power series using summation notation

The Taylor series can also be written in summation form: f(x)=5ln⁡4+∑k=1∞(−1)k+15k⋅4k(x−4)kf(x) = 5 \ln 4 + \sum_{k=1}^\infty (-1)^{k+1} \frac{5}{k \cdot 4^k}(x-4)^k Here:

  • (−1)k+1(-1)^{k+1} alternates the sign of each term.
  • 5k⋅4k\frac{5}{k \cdot 4^k} is the coefficient of the kk-th term.

The correct summation form is: 5ln⁡4+∑k=1∞(−1)k+15k⋅4k(x−4)k\boxed{5 \ln 4 + \sum_{k=1}^\infty (-1)^{k+1} \frac{5}{k \cdot 4^k}(x-4)^k} This corresponds to Option B.

Final Answers:

  1. Part (a): 5ln⁡4+54(x−4)−532(x−4)2+5192(x−4)3\boxed{5 \ln 4 + \frac{5}{4}(x-4) - \frac{5}{32}(x-4)^2 + \frac{5}{192}(x-4)^3} (Option A)
  2. Part (b): 5ln⁡4+∑k=1∞(−1)k+15k⋅4k(x−4)k\boxed{5 \ln 4 + \sum_{k=1}^\infty (-1)^{k+1} \frac{5}{k \cdot 4^k}(x-4)^k} (Option B)

Do you want a breakdown of any specific step or further clarification?
Here are some related questions to explore:

  1. How do Taylor series help approximate functions?
  2. Can the Taylor series of ln⁡x\ln x converge for all values of xx?
  3. How is the radius of convergence for a Taylor series determined?
  4. How does the factorial in the denominator affect the series' terms?
  5. What are real-world applications of logarithmic Taylor expansions?

Tip: Always test the convergence of a series to ensure it is valid for your range of interest!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Taylor Series
Logarithmic Functions
Power Series Representation

Formulas

Taylor series formula: f(x) = f(a) + f'(a)(x-a) + (f''(a)/2!)(x-a)^2 + ...
Logarithmic derivative: d(ln x)/dx = 1/x

Theorems

Taylor Series Expansion Theorem
Convergence of Taylor Series

Suitable Grade Level

Undergraduate Calculus (or Advanced High School Calculus)

Related Recommendation

Find the Taylor Series for f(x) centered at a = 2

Taylor Series for ln(x) Centered at 1

Power Series Expansion of 5x * ln(1+2x) and Coefficients c_0 to c_4

Taylor Series Approximation for Natural Logarithms: 1.05ln(1.05) and 2ln(2)

Taylor Series for ln((1+x)/(1-x)) and Approximation of ln(3) with 4 Terms