# Why same initial value in power series

• shanepitts
In summary, the conversation discusses finding a power series representation for the function (1+x)/(1-x)^2 and determining the radius of convergence. The relevant equation is shown in the attached image and the attempt at a solution involves making the starting values equal. However, it is noted that this is the goal, which is achieved in the fourth line. The formula for the power series representation is also mentioned.
shanepitts

## Homework Statement

Find a power series representation for the function below & determine the radius of convergence.

f (x)=(1+x)/(1-x)2

2.Relevant equation

Shown in attached image below which is the solution the problem.

3.The attempt at a solution

I'm trying to fathom the solution here.

I am pretty sure the initial value is the value of n. If so, it does not seem that the starting values were made equal here. Unless it has something to do with the 1+ ∑ that shows up on the fourth line?

shanepitts said:

## Homework Statement

Find a power series representation for the function below & determine the radius of convergence.

f (x)=(1+x)/(1-x)2

2.Relevant equation

Shown in attached image below which is the solution the problem.

3.The attempt at a solution

View attachment 84215
I'm trying to fathom the solution here.

I am pretty sure the initial value is the value of n. If so, it does not seem that the starting values were made equal here. Unless it has something to do with the 1+ ∑ that shows up on the fourth line?
The starting values are not equal in the third line. The comment to this effect is not that the starting values were made equal, but rather that this is the goal, which happens in the 4th line.

Note that ##\sum_{n = 0}^{\infty} (n + 1)x^n = 1x^0 + \sum_{n = 1}^{\infty} (n + 1)x^n = 1 + \sum_{n = 1}^{\infty} (n + 1)x^n##. Is that what you're asking about?

shanepitts
Mark44 said:
The starting values are not equal in the third line. The comment to this effect is not that the starting values were made equal, but rather that this is the goal, which happens in the 4th line.

Note that ##\sum_{n = 0}^{\infty} (n + 1)x^n = 1x^0 + \sum_{n = 1}^{\infty} (n + 1)x^n = 1 + \sum_{n = 1}^{\infty} (n + 1)x^n##. Is that what you're asking about?

Thanks for your quick response. I fathom now.

## 1. Why is the initial value important in a power series?

The initial value in a power series is important because it serves as the starting point for the series. It allows us to determine the behavior of the series and how it changes as the power increases. Without a specific initial value, the series would not have a clear starting point and would not be able to accurately represent a function.

## 2. Can the initial value in a power series be any number?

No, the initial value in a power series must be a fixed value that is defined by the function being represented. It cannot be any arbitrary number. For example, if the function is f(x) = x^2, the initial value must be 0 since the series starts at x^0, which is equal to 1. Changing the initial value in this case would result in a different function being represented.

## 3. How does changing the initial value affect the power series?

Changing the initial value in a power series will shift the series horizontally on a graph. For example, if the initial value is increased, the graph will shift to the left, and if the initial value is decreased, the graph will shift to the right. This is because the initial value determines the starting point of the series, and any change will affect the entire series.

## 4. Can the initial value of a power series be negative?

Yes, the initial value of a power series can be negative if the function being represented has a negative value at the starting point. It is important to note that the initial value does not necessarily reflect the behavior of the entire series, as the series can have both positive and negative terms.

## 5. Is the initial value the same as the y-intercept of the function?

No, the initial value in a power series is not always the same as the y-intercept of the function. The initial value is the starting point of the series, while the y-intercept is the point where the function crosses the y-axis. In some cases, the initial value may coincide with the y-intercept, but this is not always the case.

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