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Expand \(f(x)\) in powers of \(x\) $$f(x)=\left(x^{2}+x\right) \ln (1+x)$$

Short Answer

Expert verified
Short Answer: The expanded form of \(f(x)=(x^2+x)\ln(1+x)\) in powers of \(x\) is given by: \[f(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}x^{n+2} + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}x^{n+1}\]

Step by step solution

01

Expanding the natural logarithm as a power series

The natural logarithm, \(\ln (1+x)\), can be expanded as an infinite power series using the formula: \[\ln (1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} x^n\] where the summation runs from \(n=1\) to infinity.
02

Multiplying each term of the power series by \((x^2 + x)\)

Now, we will multiply each term of the series by \((x^2 + x)\). Thus, we have: \(f(x) = (x^2 + x) \ln (1+x)\) \(f(x) = (x^2 + x) \cdot \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} x^n\) To compute this multiplication, we distribute the terms inside the brackets like follows: \[f(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(x^3 + x^2)x^{n-1}\]
03

Simplifying the result

Now, we have to simplify the expression obtained in the last step by combining terms and arranging the summation in a standard form. First, let's distribute the \(x^n\) term inside the parentheses: \[f(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(x^{n+2} + x^{n+1})\] Now, let's split the sum into two parts and rearrange the terms: \[f(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}x^{n+2} + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}x^{n+1}\] Thus, the expression \(f(x)\) has been expanded in powers of \(x\) as: \[f(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}x^{n+2} + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}x^{n+1}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Natural Logarithm
The natural logarithm function, denoted as \(\ln(1+x)\), is a fundamental concept in calculus and mathematical analysis. It represents the logarithm where the base is the mathematical constant \(e\), approximately equal to 2.71828. Its power series expansion is particularly useful for finding more straightforward representations of complex expressions.
To expand \(\ln(1+x)\) as a power series, we use the formula:
  • \[ \ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} x^n \]
This equation tells us that the logarithm can be represented as an infinite sum of terms. The series begins from \(n=1\) and proceeds to infinity. Each term involves a fraction \(\frac{1}{n}\), adjusted alternately by a negative sign \((-1)^{n+1}\).
In essence, by using this power series, we can approximate the natural logarithm for small values of \(x\) and perform mathematical operations like differentiation or integration more seamlessly.
Infinite Series
An infinite series represents a sum of infinitely many terms. In mathematics, infinite series are used to express functions, such as the sine, cosine, or exponential functions, as sums of simpler terms.
The general form of an infinite series is represented by:
  • \[ \sum_{n=1}^{\infty} a_n \]
which means adding up all terms \(a_n\) from \(n=1\) to infinity. In the context of this problem, we expand \(\ln(1+x)\) into its infinite series form, involving terms \(x^n\). By writing it this way, we allow easier manipulation, like multiplication with polynomials.
This approach is beneficial in calculus and advanced mathematics, providing a more straightforward approach to evaluate or approximate functions that otherwise seem complex.
Polynomial Multiplication
Polynomial multiplication involves distributing and combining terms in two or more polynomial expressions. This fundamental arithmetic operation is not only essential for straightforward calculations but is crucial in higher-level mathematics.
For this exercise, we've multiplied the power series expansion of \(\ln(1+x)\) by a simple polynomial \((x^2 + x)\). Each term in \((x^2 + x)\) multiplies every term in the power series, leading to new powers of \(x\).
Here's how it works:
  • Take each term from the polynomial, \(x^2 + x\), and multiply it by every term in the equation
  • After distribution, you'll arrive at new, combined polynomial expressions
  • These new expressions are finally summed up to get the expanded polynomial expression in terms of \(x\)
Polynomial multiplication is key to simplifying expressions and solving algebraic equations effectively.

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Most popular questions from this chapter

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