# Power Series/series solutions near a point

• dp182
In summary, the student attempted to find two independent solutions for a(n+2), but was not able to find a solution with the given equation. They attempted to find the first four terms of the solutions using a recurrence relation, but were not successful. They then tried to find the solutions using individual coefficients, but were not successful. They finally solved the equation for an+2 and substituted in values for n to get terms for a2...a10.
dp182

## Homework Statement

given the equation (2+x2)y''-xy'+4y determine the first four terms of the coresponding independent solutions (a(0),a(1))
given the solution is $\Sigma$a(n)xn

## Homework Equations

when I calculated the series i got a(n+2)= (n(n-2)+4)/2(n+2)(n+1)

## The Attempt at a Solution

so when I used values of 0-7 for n i got a(0),a(1)/4,...ect I'm not sure if my series equations is incorrect any help would be great

There should be 2 independent solutions as it is a 2nd order DE. These can be presented in terms of the first coefficients a(0) and a(1)

Generally you should find a general recurrence relation for general n, however you may need specific cases for the first few terms

There should be 2 independent solutions as it is a 2nd order DE. These can be presented in terms of the first coefficients a(0) and a(1)

Generally you should find a general recurrence relation for general n, however you may need specific cases for the first few terms

I find two series with a(0) and a(1) but its wrong and no matter how many times I try to equate my equation for a(n+2) i get the same equation I posted so I'm not sure where I am screwing up

I can't be sure either without seeing what you've tried... ;) if you show your attempt I can have a look though

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dp182 said:

## Homework Statement

given the equation (2+x2)y''-xy'+4y determine the first four terms of the coresponding independent solutions (a(0),a(1))
given the solution is $\Sigma$a(n)xn

also that's not quite an equation, do you mean
(2+x2)y''-xy'+4y=0?

from the beginning:
since the answer is of the form y=$\Sigma$anxn so the first and second order derivatives are y'=$\Sigma$nanxn-1 and y''=$\Sigma$n(n-1)anxn-2 so plugging those into the original equation we get
(2+x2)$\Sigma$n(n-1)anxn-2-x$\Sigma$nanxn-1+4$\Sigma$anxn and from that equation you can see that the zero terms for the first order and second order for x2 will be zero so you won't need to change the order, and for the 2y'' term setting the sigma to 0 gives the form
2$\Sigma$(n+2)(n+1)an+2xn
so you can then factor out the $\Sigma$ and the xn terms so what's left i used to formulate the equation for an+2 and then subbed in values for n to get terms for a2...a10 in terms of a0 and a1
thats how I did it and it follows my instructors example completely but I am unable to get the correct answer hope this is helpful

ok so following along (note you can write a whole equation in tex)

first in these problems its worth including the sum as terms will be a little different at low n as the derivative of a constant is zero
$$y = \sum_{n=0}a_nx^n$$
$$y' = \sum_{n=1}a_nnx^{n-1}$$
$$y'' = \sum_{n=2}a_nn(n-1)x^{n-2}$$

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the sums and powers can get a little confusing so here's a reasonably foolproof recipe for putting it all together

now calculating each term in the DE
$$4y = 4\sum_{n=0}a_nx^n$$
$$xy' = x\sum_{n=1}a_n n x^{n-1} = \sum_{n=1}a_n n x^{n}$$
$$2y'' = 2\sum_{n=2}a_n n (n-1)x^{n-2}$$
$$x^2y'' = x^2\sum_{n=2}a_nn (n-1)x^{n-2}= \sum_{n=2}a_nn (n-1)x^{n}$$

the next trick I find useful is to re-write each sum so the sum has the power of x expressed in the same form and change the indexes to align.

so we can group three terms together, those with x to the power of n, and these need no change but we change m to n for clarity
$$m = n$$
$$4y = 4\sum_{m=0}a_m x^m$$
$$xy'= \sum_{m=1}a_m m x^{m}$$
$$x^2y'' = \sum_{m=2}a_m m (m-1) x^{m}$$

now for the n-2 power term we do the following shift
$$m = n-2\implies n=m+2$$
$$2y'' = 2\sum_{n=2}a_nn(n-1)x^{n-2}= 2\sum_{m=0}a_{m+2}(m+2)(m+1)x^{m}$$

from here it should be a simple case to substitute into the original and compare terms,

note the sums start at different m values so you may want to write out the terms for m<2 explicitly and have sums form m=2 up

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That is the same as what I did except there are no constants when you took the derivative's of the series how come you don't add any new terms when you took your derivatives

sorry i don't understand the question?
$$\frac{d}{dx} (a_0x^0) = \frac{d}{dx} (a_0.1) = 0$$

when you took the answer to the ode
y=$\sum$anxn and then derived it to get
y'=$\sum$anxn-1 shouldn't you bring down the n as a constant? and if not then wouldn't your solution for am+2=am?

woops that one slipped through cutting & pasting tex, sorry, have corrected above

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so then the equation for am+2 would be
am+2=(-m(m-1)+m-4)am/(m+2)(m+1) which would be the same one I got before

Nvm I am an idiot lol thanks a lot you've really helped

yeah so i think you got it but it should become
$$(2+x^2)y''-xy'+4y=0$$
$$=2\sum_{m=0}a_{m+2}(m+2)(m+1)x^{m}+ \sum_{m=2}a_m m (m-1) x^{m}+ \sum_{m=1}a_m m x^{m}+ 4\sum_{m=0}a_m x^m$$
$$=2\sum_{m=0}a_{m+2}(m+2)(m+1)x^{m}+ \sum_{m=2}a_m m (m-1) x^{m}+ \sum_{m=1}a_m m x^{m}+ 4\sum_{m=0}a_m x^m$$
$$=2a_{2}+6a_{3}x+ a_1 x+ a_0 +a_1 x+ +\sum_{m=2}(a_{m+2}(m+2)(m+1)+a_m m (m-1)+a_m m +a_m )x^{m} =0$$

## 1. What is a power series and how is it used in finding series solutions near a point?

A power series is an infinite series of the form Σ an(x-c)n, where c is a constant and an is a coefficient. It is used in finding series solutions near a point by representing a function as an infinite sum of polynomials, making it easier to manipulate and approximate the function.

## 2. What is the radius of convergence for a power series and why is it important?

The radius of convergence for a power series is the distance from the center point c within which the series will converge. It is important because it determines the interval of values for which the series solution will be valid and accurate.

## 3. How do you determine the interval of convergence for a power series?

The interval of convergence can be determined by using the ratio test, which compares the absolute value of the ratio of successive terms in the series to a limiting value. If the limiting value is less than 1, the series will converge within that interval. However, the endpoints of the interval should also be checked separately for convergence.

## 4. Can a power series be used to approximate functions near a point that is not the center of the series?

Yes, a power series can be used to approximate functions near a point that is not the center of the series. This is known as a Taylor series, which is a power series centered at a different point and can be used to approximate a function near that point.

## 5. What are some common applications of series solutions near a point in real-world problems?

Series solutions near a point are commonly used in physics and engineering to model physical phenomena, such as the motion of a pendulum or the behavior of an electric circuit. They are also used in economics and finance to model complex financial systems. Additionally, series solutions are used in computer science and data analysis for approximation and prediction purposes.

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