Help with manipulating power series.

In summary, the conversation discusses factoring a power series and finding the general formula for the partial sums. The method involves finding the terms with the same exponent and combining them to get the desired result. This can be achieved by playing around with the summation notation and looking for patterns.
  • #1
MidgetDwarf
1,471
615
So i am given (1+x)/(1-x)^2 and I have to put it into a power series. I know that 1/(1-x)= 1+x+x^2+x^3+...=∑x^n from 0 to infinity. I am having problems factoring series.

I differentiate 1/(1-x).

I get, 1/(1-x)^2= 1+2x+3x^2+...= ∑nX^(n-1) the sum from 1 to infinity.

rewriting this equation.

1/(1-x)^2=∑(n+1)X^n. The sum from 0 to infinity.

multiply both sides by (1+x)

I get (x+1)/(1-x)^2 = (1+x)∑(n+1)X^(n) , from 0 to infinity.

Then I distribute 1+x.

∑(n+1)X^(n) + ∑(n+1)X^(n+1) both sums have an index of 0.

My problem is that I have having trouble factoring using summation notation and I am not sure how my book factored these 2 sums into 1 to get the answer

∑(2n+1)X^n with the index of 0.
When I factored my previous work I got. ∑(n+1)(X^(n)[1+X]

which I cannot seem to to put into the form the book has it.
 
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  • #2
Pick an exponent n and look at the terms with xn. From the first and second summation, respectively, they are (n+1)xn and nxn. Sum them to get the book answer.
 
  • #3
FactChecker said:
Pick an exponent n and look at the terms with xn. From the first and second summation, respectively, they are (n+1)xn and nxn. Sum them to get the book answer.

hmm. Do you mean by finding a general formula for the partial sums? Ie. n=0 +n=1+n=3+...+n= whatever? from here look at a pattern and construct the formula?

What i am thinking: Is it kind of the way of how we play around with a telescopic series?
 
  • #4
Make the exponents of the ##x## equal and add.
 
  • #5
MidgetDwarf said:
hmm. Do you mean by finding a general formula for the partial sums? Ie. n=0 +n=1+n=3+...+n= whatever? from here look at a pattern and construct the formula?
No. Nothing fancy. Just find the term in each summation that has the same xn and add them together. Combine like terms.
 

1. What is a power series?

A power series is a mathematical series that represents a function as an infinite sum of monomials (terms with a variable raised to a power). It is typically written in the form of Σanxn, where an represents the coefficient of each term and x represents the variable.

2. How can I manipulate a power series?

There are several ways to manipulate a power series, depending on the desired outcome. Some common techniques include multiplying or dividing by a constant, factoring out common terms, and reindexing the series. Additionally, there are specific operations, such as differentiation and integration, that can be applied to power series.

3. What is the purpose of manipulating a power series?

The purpose of manipulating a power series is to make it easier to solve problems involving the function it represents. By rearranging and simplifying the series, we can often find patterns and relationships that make it possible to evaluate the function at different values or to approximate its behavior.

4. What are some applications of power series?

Power series have many applications in mathematics, physics, and engineering. They are used to represent functions, such as trigonometric functions and exponential functions, that are difficult to work with using traditional methods. Power series also play a key role in calculus, where they are used to approximate and solve problems involving derivatives and integrals.

5. Are there any limitations to manipulating power series?

While power series can be a powerful tool for solving problems, there are some limitations to their use. One limitation is that they can only represent functions that are continuous and differentiable within a certain interval. Additionally, the accuracy of a power series approximation depends on the size of the interval and the number of terms used, so it may not always provide an exact solution.

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