Example: Shifting the Index of Summation
We’ll show how to shift the index of summation by one and how this process requires adjusting both the terms inside the summation and the bounds of summation.
-
Original Summation (N from 1 to 3): \(\Sigma_{N=1}^{3} N = 1 + 2 + 3\)
This is the starting expression, which sums the values from N = 1 to N = 3. -
Choose a Shift: We want to shift the index by one, so we replace \(N\) with \(N+1\).
This shift means each term in the summation will be increased by 1, so we’ll need to adjust the bounds accordingly. -
Adjust the Lower Bound: Since we’re increasing each term by 1, the lower bound must decrease by 1. The original lower bound is 1, so it becomes 0.
The lower bound changes from 1 to 0 to compensate for the shift in the index. -
Adjust the Upper Bound: The upper bound also needs to be adjusted. The original upper bound is 3, so it becomes 2.
The upper bound changes from 3 to 2, keeping the number of terms in the summation the same as before the shift. -
New Summation (N from 0 to 2): \(\Sigma_{N=0}^{2} (N+1) = (0+1) + (1+1) + (2+1) = 1 + 2 + 3\)
This is the new expression after shifting the index and adjusting the bounds. The sum remains the same as in the original expression.
This step-by-step example demonstrates how to shift the index of summation and how to adjust the bounds to ensure the sum stays the same. By carefully handling both the terms inside the summation and the bounds, we’ve created an equivalent expression representing the same value.
Example: Shifting the Index of Summation by Subtracting One
We’ll show how to shift the index of summation by subtracting one, and how this process requires adjusting both the terms inside the summation and the bounds of summation.
-
Original Summation (N from 1 to 3): \(\Sigma_{N=1}^{3} N = 1 + 2 + 3\)
This is the starting expression, which sums the values from N = 1 to N = 3. -
Choose a Shift: We want to shift the index by subtracting one, so we replace \(N\) with \(N-1\).
This shift means each term in the summation will be decreased by 1, so we’ll need to adjust the bounds accordingly. -
Adjust the Lower Bound: Since we’re decreasing each term by 1, the lower bound must increase by 1. The original lower bound is 1, so it becomes 2.
The lower bound changes from 1 to 2 to compensate for the shift in the index. -
Adjust the Upper Bound: The upper bound also needs to be adjusted. The original upper bound is 3, so it becomes 4.
The upper bound changes from 3 to 4, keeping the number of terms in the summation the same as before the shift. -
New Summation (N from 2 to 4): \(\Sigma_{N=2}^{4} (N-1) = (2-1) + (3-1) + (4-1) = 1 + 2 + 3\)
This is the new expression after shifting the index and adjusting the bounds. The sum remains the same as in the original expression.
This example illustrates a different way to shift the index of summation by subtracting one from each term. By carefully adjusting the bounds, we’ve shown that the sum remains the same, demonstrating the flexibility and mathematical rigor of summation notation.
Example: Shifting the Index of Summation by Adding One
We’ll show how to shift the index of summation by adding one in the series \(x + x^2 + x^3 + \ldots\) and how this process requires adjusting both the terms inside the summation and the bounds of summation.
-
Original Summation (k from 1 to N): \(\Sigma_{k=1}^{N} x^k = x + x^2 + x^3 + \ldots + x^N\)
This is the starting expression, which sums the powers of \(x\) from \(k = 1\) to \(k = N\). -
Choose a Shift: We want to shift the index by adding one, so we replace \(k\) with \(k+1\).
This shift means each term in the summation will be increased by 1 in the exponent, so we’ll need to adjust the bounds accordingly. -
Adjust the Lower Bound: Since we’re increasing each term by 1 in the exponent, the lower bound must decrease by 1. The original lower bound is 1, so it becomes 0.
The lower bound changes from 1 to 0 to compensate for the shift in the index. -
Adjust the Upper Bound: The upper bound also needs to be adjusted. The original upper bound is \(N\), so it becomes \(N-1\).
The upper bound changes from \(N\) to \(N-1\), keeping the number of terms in the summation the same as before the shift. -
New Summation (k from 0 to N-1): \(\Sigma_{k=0}^{N-1} x^{k+1} = x^1 + x^2 + x^3 + \ldots + x^N\)
This is the new expression after shifting the index and adjusting the bounds. The sum remains the same as in the original expression.
This example illustrates how to shift the index of summation by adding one to each term in the series \(x + x^2 + x^3 + \ldots\). By carefully adjusting the bounds, we’ve shown that the sum remains the same, providing insight into the nature of summation notation.
Example: Aligning Summations with Different Indexes
We will look at two summations and adjust them so they can be combined. We’ll take very small steps to explain the process:
-
Start with Two Different Summations:
\(\Sigma_{k=0}^{N} (k + x) + \Sigma_{j=1}^{N+1} j\)
Notice how the first summation starts at 0 and the second starts at 1. We need to align them. -
Identify the Differences:
The first summation uses index \(k\) and starts at 0, while the second uses index \(j\) and starts at 1. -
Plan the Alignment:
We’ll increase the lower bound of the first summation by 1 and adjust the terms inside accordingly. -
Adjust the First Summation:
\(\Sigma_{k=1}^{N+1} ((k-1) + x)\)
We increased the bounds by 1 and subtracted 1 from the \(k\) term to keep the values the same. -
Show Both Summations Aligned:
\(\Sigma_{k=1}^{N+1} ((k-1) + x) + \Sigma_{j=1}^{N+1} j\)
Now both summations have the same bounds, from 1 to \(N+1\). -
Combine into One Summation:
\(\Sigma_{k=1}^{N+1} ((k-1) + x + k)\)
We combine the summations by adding the terms inside. The bounds remain the same. -
Expand to Show the Pattern:
\(1 + x + 2 + x + 3 + x + \ldots + N + x + N + 1\)
Finally, we expand the summation to show the explicit series.
This careful step-by-step process helps us understand how to align and combine different summations, a useful technique in many mathematical contexts.
1. Given Differential Equation
\( y’ = y \): We have a first-order differential equation, where the rate of change of the function y with respect to x is equal to the value of the function itself.
2. Assumption of Power Series Solution
We assume that the solution can be represented as a power series. This is a common method for solving differential equations, where we look for solutions that can be expressed as infinite polynomials.
3. Expressing the Power Series
\( y(x) = \sum_{n=0}^{\infty} a_nx^n \): This is the general form of a power series. The series goes on forever, and the coefficients \( a_n \) might vary with n.
4. Explanation of Summation Notation
The symbol \( \sum \) means we’re adding up an infinite number of terms, where each term follows a pattern described by \( a_nx^n \).
5. Expanded Form of Power Series
\( y(x) = a_0 + a_1x + a_2x^2 + \ldots \): Here, we’ve expanded the series into its individual terms. We can’t write all the terms since there are infinitely many, but this gives a sense of the pattern.
6. Differentiation of Power Series
We take the derivative of both sides with respect to x. This involves applying basic rules of differentiation to each term of the series separately.
7. Detailed Differentiation
\( y'(x) = \sum_{n=0}^{\infty} na_nx^{n-1} \): By using the power rule for differentiation, we find that each term \( a_nx^n \) becomes \( na_nx^{n-1} \) after differentiation.
8. Handling the First Term After Differentiation
The first term (where n=0) becomes zero, so we adjust the sum to start from n=1: \( y'(x) = \sum_{n=1}^{\infty} na_nx^{n-1} \).
9. Further Expansion of the Derivative
\( y'(x) = a_1 + 2a_2x + 3a_3x^2 + \ldots \): This is the derivative of the series. We see how the power of x decreases by one in each term and the coefficient is multiplied by the original power.
10. Original Differential Equation
Remembering that our given equation is \( y’ = y \), we now have expressions for both \( y’ \) and \( y \) and can equate them to find the coefficients.
11. Equating the Expressions for y and y’
\( a_1 + 2a_2x + 3a_3x^2 + \ldots = a_0 + a_1x + a_2x^2 + \ldots \): We have equated the expressions for \( y’ \) and \( y \), using the given equation \( y’ = y \).
12. Matching Coefficients of Powers of x
By comparing the coefficients of corresponding powers of x, we can find relationships between the coefficients.
13. Detailed Relationships Between Coefficients
\( a_1 = a_0, \, a_1 = 2a_2, \, a_2 = 3a_3, \, \ldots \): This gives us a recursive relationship between the coefficients, allowing us to find each coefficient in terms of the previous one.
14. General Recursive Relationship
\( a_{n+1} = (n+1)a_{n+1}, \, n \geq 1 \): We can express the pattern we found earlier in a general form that holds for all n greater than or equal to 1.
15. Explanation of Recursive Relationship
This recursive relationship tells us how each coefficient is related to the next one. It gives us a systematic way to find all the coefficients.
16. Finding a General Expression for the Coefficients
\( a_n = \frac{a_0}{n!} \): By solving the recursive relationship, we find a general expression for any coefficient in terms of the first one, \( a_0 \).
17. Explanation of Factorial Notation
The symbol \( n! \) denotes the factorial of n, which is the product of all positive integers up to and including n.
18. Final Expression for y(x)
\( y(x) = a_0 \left(1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots\right) \): Substituting the general expression for the coefficients back into the series, we get this expression.
19. Recognizing the Exponential Function
\( y(x) = a_0e^x \): The series inside the parentheses is the Taylor expansion of the exponential function \( e^x \), so we can write the solution in this compact form.
20. Conclusion and Interpretation
This expression represents the general solution to the differential equation \( y’ = y \), with \( a_0 \) as an arbitrary constant. It tells us that any function of the form \( y(x) = Ce^x \) (where C is a constant) is a solution to the given equation.