Solving Derivation Problem: \sum_{j=\kappa}^{\inf} \varphi^{2j-\kappa}\sigma^2

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In summary, the equation given is a geometric series with reparameterized index and can be simplified to \frac{\sigma^2}{1-\sigma^2} \varphi^\kappa. The exponent of \sigma is related to the index j in the series. The sum of a geometric series is also used in the derivation. There is no need for a negative exponent in the reparameterized series.
  • #1
roadworx
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Hi,

I have [tex]\sum_{j=\kappa}^{\inf} \varphi^{2j-\kappa}\sigma^2[/tex]

I know the answer is [tex] = \frac{\sigma^2}{1-\sigma^2} \varphi^\kappa [/tex]

Can someone explain the mathematics involved in this derivation?

Thanks.
 
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  • #2
Does [itex]\sigma[/itex] have an exponent related to j? Otherwise, the sum is a straightforward geometric series, once you reparameterize the index of summation. Note that
[tex]\sum_{j=\kappa}^\infty \varphi^{2j-\kappa} = \varphi^\kappa + \varphi^{\kappa + 2} + \varphi^{\kappa + 6} + \cdots = \varphi^{\kappa}\sum_{n=0}^\infty \varphi^{2n} = \varphi^{\kappa}\sum_{n=0}^\infty (\varphi^2)^n[/tex]
Do you know the sum of a geometric series?
 
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  • #3
slider142 said:
Does [itex]\sigma[/itex] have an exponent related to j? Otherwise, the sum is a straightforward geometric series, once you reparameterize the index of summation. Note that
[tex]\sum_{j=\kappa}^\infty \varphi^{2j-\kappa} = \varphi^\kappa + \varphi^{\kappa + 2} + \varphi^{\kappa + 6} + \cdots = \varphi^{\kappa}\sum_{n=0}^\infty \varphi^{2n} = \varphi^{\kappa}\sum_{n=0}^\infty (\varphi^2)^n[/tex]
Do you know the sum of a geometric series?

Thanks for that. I have a question on your reparameterization. Isn't [tex]\varphi^{\kappa}[/tex] to the negative index, and therefore shouldn't it be [tex] \frac{1}{\varphi^{\kappa}}[/tex] when you write out the series? Or did I miss something?
 
  • #4
[itex]\kappa[/itex] is the same as that in the original sum. When j starts at [itex]\kappa[/itex], we get 2K - K = K, 2(K + 1) - K = K + 2, 2(K + 2) - K = K + 4, and so on.
 

1. What is the purpose of solving derivation problems?

The purpose of solving derivation problems is to find the derivative of a given function, which represents the rate of change or slope of the function at any given point. This is useful in many areas of science, including physics, engineering, and economics.

2. What is the meaning of the notation \sum_{j=\kappa}^{\inf} \varphi^{2j-\kappa}\sigma^2?

The notation \sum_{j=\kappa}^{\inf} \varphi^{2j-\kappa}\sigma^2 represents a summation, where j is the index variable and \kappa is the starting value. The variable \varphi is being raised to the power of 2j-\kappa, and then multiplied by \sigma^2. The summation is performed from j=\kappa to infinity.

3. How do I solve a derivation problem using the summation notation?

To solve a derivation problem using the summation notation, you need to first identify the function for which you need to find the derivative. Then, you can use the rules of differentiation to simplify the function and express it in terms of the summation notation. Finally, you can use known formulas and techniques to evaluate the summation and find the derivative.

4. What are some common techniques for solving derivation problems?

Some common techniques for solving derivation problems include using the power rule, product rule, quotient rule, and chain rule. Other techniques include using known derivatives of common functions, using logarithmic differentiation, and applying the rules of differentiation to the summation notation.

5. How can solving derivation problems be applied in real life?

Solving derivation problems has many practical applications in real life. For example, in physics, the derivative of a position function can be used to find the velocity and acceleration of an object. In economics, derivatives can be used to analyze changes in supply and demand. In engineering, derivatives are used to optimize designs and analyze systems. Overall, solving derivation problems allows us to better understand and model the world around us.

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