Math Problem Statement

Question content area top Part 1 Find the power series representation for g centered at 0 by differentiating or integrating the power series for f. Give the interval of convergence for the resulting series. ​g(x)equalsStartFraction 21 Over left parenthesis 1 minus 21 x right parenthesis squared EndFraction ​, ​f(x)equalsStartFraction 1 Over 1 minus 21 x EndFraction Question content area bottom Part 1 The power series representation for g is Summation from k equals 1 to infinity left parenthesis nothing right parenthesis    .

Solution

To find the power series representation for g(x)g(x), we start with the given functions:

  1. f(x)=1121xf(x) = \frac{1}{1 - 21x}
  2. g(x)=21(121x)2g(x) = \frac{21}{(1 - 21x)^2}

Step 1: Power Series for f(x)f(x)

The power series for f(x)f(x) is derived from the geometric series formula:

11r=k=0rk(for r<1).\frac{1}{1 - r} = \sum_{k=0}^{\infty} r^k \quad \text{(for } |r| < 1 \text{)}.

Substitute r=21xr = 21x: f(x)=1121x=k=0(21x)k=k=021kxk,for 21x<1 or x<121.f(x) = \frac{1}{1 - 21x} = \sum_{k=0}^{\infty} (21x)^k = \sum_{k=0}^{\infty} 21^k x^k, \quad \text{for } |21x| < 1 \text{ or } |x| < \frac{1}{21}.

Step 2: Derivative of f(x)f(x) to Obtain g(x)g(x)

The derivative of f(x)f(x) with respect to xx is: f(x)=ddx(1121x)=21(121x)2.f'(x) = \frac{d}{dx} \left( \frac{1}{1 - 21x} \right) = \frac{21}{(1 - 21x)^2}.

Since g(x)=21f(x)g(x) = 21 \cdot f'(x), we have: g(x)=2121(121x)2=21(121x)2.g(x) = 21 \cdot \frac{21}{(1 - 21x)^2} = \frac{21}{(1 - 21x)^2}.

Step 3: Power Series for g(x)g(x)

To find the power series for g(x)g(x), differentiate the power series for f(x)f(x): f(x)=k=021kxk.f(x) = \sum_{k=0}^{\infty} 21^k x^k.

Differentiating term by term: f(x)=k=1k21kxk1.f'(x) = \sum_{k=1}^{\infty} k \cdot 21^k x^{k-1}.

Multiply by 21 to account for g(x)=21f(x)g(x) = 21 \cdot f'(x): g(x)=21f(x)=k=1k21k+1xk1.g(x) = 21 \cdot f'(x) = \sum_{k=1}^{\infty} k \cdot 21^{k+1} x^{k-1}.

Rewrite the power series by shifting the index kk+1k \to k+1: g(x)=k=1k21k+1xk.g(x) = \sum_{k=1}^{\infty} k \cdot 21^{k+1} x^k.

Thus, the power series representation for g(x)g(x) is: g(x)=k=1k21kxk.g(x) = \sum_{k=1}^{\infty} k \cdot 21^k x^k.

Step 4: Interval of Convergence

The power series for g(x)g(x) converges for the same x<121|x| < \frac{1}{21}, since it was derived from f(x)f(x) by differentiation.

Final Answer

The power series representation for g(x)g(x) is: g(x)=k=1k21kxk,g(x) = \sum_{k=1}^{\infty} k \cdot 21^k x^k, and the interval of convergence is: x<121.|x| < \frac{1}{21}.

Would you like additional explanations or derivations for any part?


5 Related Questions:

  1. How does differentiation affect the interval of convergence of a power series?
  2. What is the general formula for the nn-th derivative of a geometric series?
  3. Can the power series for g(x)g(x) be integrated to recover f(x)f(x)?
  4. What is the significance of the radius of convergence in applications of power series?
  5. How do you find the nth term of a power series derived from a given function?

Tip:

When differentiating or integrating a power series, always carefully adjust the index of summation and confirm the interval of convergence remains unchanged.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Power Series
Differentiation
Interval of Convergence
Geometric Series

Formulas

Geometric Series: 1 / (1 - r) = Σ r^k for |r| < 1
Derivative of a power series term-by-term: d/dx Σ a_k x^k = Σ k * a_k x^(k-1)

Theorems

Geometric Series Theorem
Differentiation of Power Series

Suitable Grade Level

College Level Calculus