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Determine the interval of convergence of a given power series and show that it satisfies a differential equation

Consider the function defined by the power series

Determine the interval of convergence for the power series and show that satisfies the differential equation

First, we use the ratio test to determine the interval of convergence,

Hence, the series converges for all . Then, to show that it satisfies the given differential equation we take the first two derivatives,

Then we have

Therefore, indeed is a solution of the given differential equation.

Now, to find the sum we first need to get the general form of the solutions for the differential equation

First, we find the general form of the solutions of the homogeneous equation

In this case we have an equation of the form where and . From this we can compute and . Therefore, the general form of the solutions is

Then, we can find a particular solution of the given equation by inspection since

is a solution. Therefore, the general solution to the given inhomogeneous equation is

Now, in the particular case we also have the initial condition and so we have

Furthermore, since is an odd function we must have

Therefore, we conclude and . And so,

Determine the interval of convergence for a given power series and show that it satisfies a given differential equation

Consider the function defined by the power series

Determine the interval of convergence for and show that it satisfies the differential equation

First, to determine the interval of convergence for the power series we use the ratio test

Hence, the series converges for all . Next, to show that it satisfies the given differential equation we take the first two derivatives,

Then, we have

Thus, indeed satisfies the given differential equation.

Further, in a previous exercise (Section 8.14, Exercise #2) that the solution of the differential equation are all of the form

For this problem we also have so

Finally, we know this function is an even function (since for all because we have inside the sum is the only term). This means we must have . Hence, we must have

Determine the interval of convergence of ∑ xn / n! and show that it satisfies y′ = x + y

Consider the function defined by the power series

Determine the interval of convergence for and show that it satisfies the differential equation

(We might notice that this is almost the power series expansion for the exponential function and deduce the interval of convergence and the differential equation from properties of the exponential that we already know. We can do it from scratch just as easily though.)

First, we apply the ratio test

Hence, the series converges for all . Next, we take a derivative

Then we have

Now, to compute the sum we can solve the given differential equation

This is a first order linear differential equation of the form with and . We also know that ; therefore, this equation has a unique solution satisfying the given initial condition which is given by

Where and

Therefore, we have

Determine the interval of convergence of ∑ x2n / n! and show it satisfies a given differential equation

Consider the function defined by the power series

Determine the interval of convergence for and show that it satisfies the differential equation

First, to determine the interval of convergence we use the ratio test,

Hence, the series converges for all .

To show that it satisfies the given differential equation, we first take the derivative,

Then, it satisfies the given differential equation since

Then, since the given differential equation is a first-order linear differential equation of the form

with we know that the solutions are uniquely determined by the formula

Since we have the initial condition (by plugging in to the power series expansion for ) we have and the unique solution of this differential equation is

Determine the interval of convergence of a given power series and show that it satisfies a given differential equation

Consider the function defined by the power series

Determine the interval of convergence for and show that it satisfies the differential equation

Find and .

First, to determine the interval of convergence for we use the ratio test,

Therefore, the series converges for all . Next, to show that it satisfies the given differential equation, we take the first two derivatives,

Then, we have the differential equation ,

But, for this equation to hold we must have (since there is no constant term in on the left) and we must also have since there the coefficient of on the left is 1 and the only possible term on the right is if . Using these values of and we verify that the given differential equation is satisfied since we have

Hence, we indeed have .

Determine the interval of convergence of ∑xn / (n!)2 and show that it satisfies a given differential equation

Consider the function f(x) defined by the power series,

Determine the interval of convergence of f(x) and show that f(x) satisfies the differential equation

First, to determine the radius of convergence we use the ratio test,

Therefore, the series converges for all (or the radius of convergence is ). Next, to show that satisfies the given differential equation we take the first two derivatives,

Plugging this into the given differential equation we have

Hence, indeed satisfies the given differential equation.

Determine the interval of convergence of ∑ x4n / (4n)! and show that it satisfies a given differential equation

Consider the function defined by the power series,

Determine the interval of convergence of and show that satisfies the differential equation

First, to determine the radius of convergence we use the ratio test

Therefore, converges for all (equivalently, ). Next, to show that satisfies the differential equation we take the first four derivatives,

But, reindexing this expression for the fourth derivative we have

Thus, satisfies the given differential equation.

Use the method of undetermined coefficients to solve (1-x2)y′′ – 2xy′ + 6y = 0

Consider the differential equation

The solution to this differential equation has a power-series expansion

Using the method of undetermined coefficients obtain a recursion formula relating the terms to the terms . Give an explicit formula for for each and find the sum of the series.

First, we differentiate twice,

From the initial conditions and we have

Now, we plug the expressions for , and back into the given differential equation,

Then, we use the fact from above that and to get

Since this sum is equal to 0, we know that every coefficient of every power of must be equal to 0. First, we solve for and ,

Then, we establish the recursive relationship between and ,

for all . Then since we have for all (since for every odd integer the formula for has is multiplied by , but each of these will be 0). For the even terms we have for ,

This means all of the remaining even terms will be 0 as well. So we have

Hence,

Determine the coefficients in the Taylor series of (2 + x2)5/2

Consider the function . Determine the first five coefficients () in the Taylor series expansion of at .

First, we take some derivatives,

Then, since the coefficients in the Taylor series at are of the form we can compute as follows:

Determine the 98th coefficient in the power series of sin (2x + (1/4)π)

Consider the power series expansion

Find the coefficient in this expansion.

First, we recall the identity for the sine of a sum,

So we have

Then, we know the expansions of and are

Therefore, we have

(Where denotes the greatest integer less than or equal to .) So, the 98th coefficient is given by