# Series solution up to a term, power series

• dp182
In summary, the conversation discusses finding the series solution for the initial value problem (1-x)y,,+xy,-2y=0 up to the term with x6. The attempt at a solution involves solving for an+2 using a recurrence relation involving an, an+1, and an+2. It is noted that two boundary/initial conditions are needed for a solution.
dp182

## Homework Statement

consider the initial value problem (1-x)y,,+xy,-2y=0 find the series solution up to the term with x6

## Homework Equations

(1-x)y,,+xy,-2y=0

## The Attempt at a Solution

assuming the answer has the form $\Sigma$anxn
that gives y,,=$\Sigma$nanxn-1 and y,,=$\Sigma$n(n-1)anxn-2 then plugging these back in and getting rid of the xn-2 and xn-1 you get the equation $\Sigma$(n+2)(n+1)an+2xn-$\Sigma$n(n+1)an+1xn+$\Sigma$nanxn-2$\Sigma$anxn so what I'm wondering is how do you replace the an+1 with an an so you can solve for an+2

hi dp182!
dp182 said:
… how do you replace the an+1 with an an so you can solve for an+2

just subtract 1 from everything

(eg ∑an+1xn = ∑anxn-1)

dp182 said:
how do you replace the an+1 with an an so you can solve for an+2
You don't. The recurrence relation involves an, an+1, and an+2. There's nothing wrong with that.

It makes sense that the recurrence relation for a_n+2 involves a_n+1 and a_n. It would require you to define a_0 and a_1 and Second order differential equations require two boundary/initial conditions for a solution.

So If I solve for an+2 I get an+2=n(n+1)an+1+an(2-n)/(n+2)(n+1) so to get my terms up to x6 I will just input values of n, so my series will be in terms of a0and a1?

Yup, but you made an algebra error solving for an+2 (or omitted very needed parentheses).

## What is a series solution up to a term?

A series solution up to a term is a mathematical method used to approximate a function by representing it as an infinite sum of terms. The series is often truncated at a certain term to make calculations more manageable.

## What is a power series?

A power series is a specific type of series solution up to a term, where the terms are all powers of a variable, typically x. It is represented in the form ∑n=0^∞ an(x-c)n, where c is a constant and an is the coefficient of the nth term.

## How is a series solution up to a term different from a Taylor series?

A Taylor series is a type of power series that is used to approximate a function at a specific point, while a series solution up to a term is used to approximate a function over its entire domain. Additionally, a Taylor series requires knowledge of the function's derivatives at a specific point, while a series solution up to a term can be calculated without this information.

## What are the benefits of using a series solution up to a term?

A series solution up to a term can be used to approximate functions that are difficult to solve using other methods, such as differential equations. It also allows for more accurate approximations by increasing the number of terms in the series.

## What are some applications of series solutions up to a term?

Series solutions up to a term are commonly used in physics, engineering, and other scientific fields to approximate functions that arise in real-world problems. They are also used in designing algorithms and computer programs that require efficient and accurate numerical solutions.

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