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Magazine Chapter 9: Problem 27
Let \(f(x)\) be the sum of the series. Find \(f^{\prime}(x)\) and \(\int_{0}^{x} f(t) d t\). $$ \sum_{n=1}^{\infty}(n+1) x^{n} $$
Short Answer
Expert verified
\(f'(x) = \frac{(1+x)}{(1-x)^3} - \frac{1}{(1-x)^2}\) and \(\int_{0}^{x} f(t) dt = -\frac{x + \log(1-x)}{(1-x)^2}\)."
Step by step solution
01
Understand the Series
The given series is \( \sum_{n=1}^{\infty}(n+1) x^{n} \). We need to find two things: the derivative \( f^{\prime}(x) \) and the integral \( \int_{0}^{x} f(t) dt \). First, we need to recognize that this is a power series and determine its sum.
02
Express the Function as a Series
The series expression is \( f(x) = \sum_{n=1}^{\infty} (n+1)x^n \). If we write out the first few terms, it looks like: \( 2x + 3x^2 + 4x^3 + \cdots \). This series converges for \( |x| < 1 \).
03
Find a Closed Form for the Series
We can find a closed form for the series \( \sum_{n=1}^{\infty}(n+1)x^n \) using the formula for the sum of a geometric series and some algebraic manipulation. Recall the geometric series sum formula: \( \sum_{n=0}^{\infty} x^n = \frac{1}{1-x} \), for \( |x| < 1 \).
04
Calculate f(x)
The function \( f(x) \) can be approached by using existing series manipulation techniques, such as deriving known series. Differentiate \( \sum_{n=0}^{\infty} x^n \) to get \( \sum_{n=1}^{\infty} nx^{n-1} = \frac{1}{(1-x)^2} \) and integrate \( x \sum_{n=1}^{\infty} nx^{n-1} = x\cdot \frac{1}{(1-x)^2} \). After appropriate algebra, the result is the derived function: \( f(x) = \frac{x}{(1-x)^2} - \sum_{n=0}^{\infty} x^n = \frac{x}{(1-x)^2} - \frac{1}{1-x} \), for \( |x| < 1 \).
05
Derive f'(x)
To find the derivative \( f^{\prime}(x) \), we take the derivative of \( f(x) = \sum_{n=1}^{\infty} (n+1)x^n \). Use: \( \frac{d}{dx} \left( \frac{x}{(1-x)^2} \right) - \frac{d}{dx} \left( \frac{1}{1-x} \right) \). Compute to get \( f^{\prime}(x) = \frac{(1+x)}{(1-x)^3} - \frac{1}{(1-x)^2} \).
06
Integrate the Series
The integral \( \int_{0}^{x} f(t) dt \) is found by integrating the expression derived in \( f(t) \). Integrate directly: \( \int_{0}^{x} \frac{t}{(1-t)^2} dt - \int_{0}^{x} \frac{1}{1-t} dt = \left[ -\frac{t + \log(1-t)}{(1-t)^2} \right]_0^x \). Compute the definite integral to find the result of the integration.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative of a Series
The derivative of a series is a critical concept in calculus, helping us understand the behavior of functions represented as infinite sums. When dealing with a power series like \( \sum_{n=1}^{\infty} (n+1)x^n \), taking the derivative involves applying the basic rules of differentiation to each term in the series.
For a term \( (n+1)x^n \), the derivative with respect to \( x \) is found using the power rule, which states that if you have a term \( ax^n \), its derivative is \( anx^{n-1} \). Applying this rule,
After computing, the series can be re-summarized, and sometimes, as in this exercise, expressed in a closed-form formula to get a neat succinct answer. The derivative hence derived reveals how the series changes at every point \( x \).
For a term \( (n+1)x^n \), the derivative with respect to \( x \) is found using the power rule, which states that if you have a term \( ax^n \), its derivative is \( anx^{n-1} \). Applying this rule,
- differentiate \( (n+1)x^n \) to get \( (n+1)nx^{n-1} \).
After computing, the series can be re-summarized, and sometimes, as in this exercise, expressed in a closed-form formula to get a neat succinct answer. The derivative hence derived reveals how the series changes at every point \( x \).
Integral of a Series
To find the integral of a series means to determine the area under the curve represented by the series. It can help us grasp the accumulation effect of the function over an interval. When given a function like \( f(t) = \sum_{n=1}^{\infty} (n+1)t^n \) and asked to integrate from 0 to \( x \), the task is to integrate each term separately.
This process involves reversing differentiation. For a term \( (n+1)t^n \), integrate using the rule: \( \frac{t^{n+1}}{n+1} \), plus any necessary bounds or simplifications.
This process involves reversing differentiation. For a term \( (n+1)t^n \), integrate using the rule: \( \frac{t^{n+1}}{n+1} \), plus any necessary bounds or simplifications.
- Integrate \( (n+1)t^n \) from 0 to \( x \).
- Account for convergence where \( |x| < 1 \).
Sum of a Geometric Series
The concept of a geometric series is foundational in calculus when dealing with infinite series. A geometric series has the form \( \sum_{n=0}^{\infty} ar^n \), where \( a \) is the first term and \( r \) is the common ratio. Understanding it involves recognizing how this series adds up its terms and converges.
When \( |r| < 1 \), the series converges to a sum \( \frac{a}{1-r} \). This formula is handy since many complex series can often be rearranged or differentiated to form a geometric series.
When \( |r| < 1 \), the series converges to a sum \( \frac{a}{1-r} \). This formula is handy since many complex series can often be rearranged or differentiated to form a geometric series.
- The series \( \sum_{n=0}^{\infty} x^n \) converges to \( \frac{1}{1-x} \).
- This series formula is a starting key in deriving and summing various series encountered in calculus, allowing manipulation into recognizable patterns for easier computation.
