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 , we start with the given functions:
Step 1: Power Series for
The power series for is derived from the geometric series formula:
Substitute :
Step 2: Derivative of to Obtain
The derivative of with respect to is:
Since , we have:
Step 3: Power Series for
To find the power series for , differentiate the power series for :
Differentiating term by term:
Multiply by 21 to account for :
Rewrite the power series by shifting the index :
Thus, the power series representation for is:
Step 4: Interval of Convergence
The power series for converges for the same , since it was derived from by differentiation.
Final Answer
The power series representation for is: and the interval of convergence is:
Would you like additional explanations or derivations for any part?
5 Related Questions:
- How does differentiation affect the interval of convergence of a power series?
- What is the general formula for the -th derivative of a geometric series?
- Can the power series for be integrated to recover ?
- What is the significance of the radius of convergence in applications of power series?
- 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.
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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
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