Chapter 11: Problem 48
Write out all of the terms of each series. \(\sum_{j=0}^{3}(-1)^{j} x^{3-j} y^{j}\)
Short Answer
Expert verified
The terms are \(x^{3}, -x^{2} y, x y^{2}, -y^{3}\).
Step by step solution
01
- Understand the Summation
The given summation is \(\sum_{j=0}^{3}(-1)^{j} x^{3-j} y^{j}\). This means we need to compute the sum of the terms from j=0 to j=3.
02
- Calculate for j=0
Substitute j=0 into \((-1)^{j} x^{3-j} y^{j}\). This gives \((-1)^{0} x^{3-0} y^{0} = x^{3}\).
03
- Calculate for j=1
Substitute j=1 into \((-1)^{j} x^{3-j} y^{j}\). This gives \((-1)^{1} x^{3-1} y^{1} = -x^{2} y\).
04
- Calculate for j=2
Substitute j=2 into \((-1)^{j} x^{3-j} y^{j}\). This gives \((-1)^{2} x^{3-2} y^{2} = x y^{2}\).
05
- Calculate for j=3
Substitute j=3 into \((-1)^{j} x^{3-j} y^{j}\). This gives \((-1)^{3} x^{3-3} y^{3} = -y^{3}\).
06
- Write out All Terms
The terms of the series are \(x^{3}, -x^{2} y, x y^{2}, -y^{3}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
series terms
In mathematics, a series is the sum of the terms of a sequence. The terms of the series are the individual elements that are being summed. When analyzing a series, we look at each term separately before we sum them all together.
In the given exercise, the series is represented by \(\sum_{j=0}^{3}(-1)^{j} x^{3-j} y^{j}\). This notation means we need to find the sum for each term from j=0 to j=3.
For example, when j=0, the term is \((-1)^{0} x^{3-0} y^{0} = x^{3}\). When j=1, the term adjusts to \((-1)^{1} x^{3-1} y^{1} = -x^{2} y\). This process continues until j reaches its upper limit of 3.
Understanding these individual terms helps in creating a clear picture of how the entire series sums up.
In the given exercise, the series is represented by \(\sum_{j=0}^{3}(-1)^{j} x^{3-j} y^{j}\). This notation means we need to find the sum for each term from j=0 to j=3.
For example, when j=0, the term is \((-1)^{0} x^{3-0} y^{0} = x^{3}\). When j=1, the term adjusts to \((-1)^{1} x^{3-1} y^{1} = -x^{2} y\). This process continues until j reaches its upper limit of 3.
Understanding these individual terms helps in creating a clear picture of how the entire series sums up.
substitution
Substitution is a method used to simplify expressions or functions by replacing variables with their values. In our exercise, substitution is key to finding each term of the series.
The series \(\sum_{j=0}^{3}(-1)^{j} x^{3-j} y^{j}\) requires us to substitute different values for j to determine each term.
Starting with j=0, we substitute this value into the expression \((-1)^{j} x^{3-j} y^{j}\). Thus, \((-1)^{0} x^{3-0} y^{0} = x^{3}\).
When j=1, the substitution gives us \((-1)^{1} x^{3-1} y^{1} = -x^{2} y\). Continuing this method for j=2 and j=3, we find \((-1)^{2} x^{3-2} y^{2} = x y^{2}\) and \((-1)^{3} x^{3-3} y^{3} = -y^{3}\) respectively.
By substituting these values successively, we can list all terms needed to sum up the series.
The series \(\sum_{j=0}^{3}(-1)^{j} x^{3-j} y^{j}\) requires us to substitute different values for j to determine each term.
Starting with j=0, we substitute this value into the expression \((-1)^{j} x^{3-j} y^{j}\). Thus, \((-1)^{0} x^{3-0} y^{0} = x^{3}\).
When j=1, the substitution gives us \((-1)^{1} x^{3-1} y^{1} = -x^{2} y\). Continuing this method for j=2 and j=3, we find \((-1)^{2} x^{3-2} y^{2} = x y^{2}\) and \((-1)^{3} x^{3-3} y^{3} = -y^{3}\) respectively.
By substituting these values successively, we can list all terms needed to sum up the series.
exponential functions
Exponential functions are mathematical expressions in which a variable appears in the exponent. They have the general form \(f(x) = a \cdot e^{bx}\) where e is Euler's number, a constant approximately equal to 2.71828.
In our exercise, the variables x and y are raised to different powers, creating exponential terms in the series.
For instance, the term \(x^{3}\) means x is raised to the power of 3. Similarly, \(-x^{2} y\) and \(-y^{3}\) contain variables raised to various powers.
Understanding how to work with exponential functions involves recognizing that these terms grow quickly with increasing exponents and play a critical role in the nature of the series.
In our exercise, the variables x and y are raised to different powers, creating exponential terms in the series.
For instance, the term \(x^{3}\) means x is raised to the power of 3. Similarly, \(-x^{2} y\) and \(-y^{3}\) contain variables raised to various powers.
Understanding how to work with exponential functions involves recognizing that these terms grow quickly with increasing exponents and play a critical role in the nature of the series.