Integrate the binomial series for to obtain the power-series expansion

**Incomplete.**

Skip to content
#
Stumbling Robot

A Fraction of a Dot
#
Tag: power series

#
Using the binomial series obtain a power-series expansion for *arcsin x*

#
Compute some things using the binomial series

#
Compute the coefficients of a power series with coefficients satisfying a given identity

#
Compute the coefficients of a given power series

#
Assume *y′′ + xy′ + y = 0* has a power-series solution and determine the coefficient *a*_{n}

#
Assume *y′′ = xy* has a power-series solution and determine the coefficient *a*_{n}

#
Assume *y′ = α y* has a power series solution and determine the coefficient *a*_{n}

#
Find the first four nonzero terms of the power series solution of *y′ = x + y*^{2}

#
Find the first four nonzero terms of the power series solution of *y′ = 1 + xy*^{2}

#
Find the first four nonzero terms of the power series solution of *y′ = x*^{2} + y^{2}

Integrate the binomial series for to obtain the power-series expansion

**Incomplete.**

- Show that the first six terms of the binomial series for are
- For let be the th coefficient in the binomial series and let denote the remainder after terms, i.e.,
for . Prove that

- Prove the validity of the identity
Use this identity to compute the first ten decimal places of .

**Incomplete.**

Consider the power series

whose coefficients satisfy the identity

Compute the coefficients and determine the value of .

First, we know the power-series expansion of is given by

So, we have

Equating like powers of we can compute the coefficients and ,

Then to compute the value of we solve the differential equation

If we have

Therefore, . If then we can divide the above differential equation by to obtain the first order linear differential equation

Now, we can solve this differential equation as follows,

Where is an arbitrary constant.

(**Incomplete.** Judging by the answer in the back of the book, Apostol computes this constant as . I’m not sure how to get that though. I think we need some kind of initial condition to determine the constant, and so get a unique solution for . Maybe we can assume this must be continuous at 0 and then take a limit as ? I do think that would get us to , but I don’t know why can assume is continuous at 0. Leave a comment if you have any suggestions.)

Consider the power series

with coefficients determined by the identity

Compute the coefficients and determine the sum of the series.

We know the power-series expansion for is given by

Starting with and the given identity we can compute the coefficients by equating the coefficients of like powers of ,

Then from the identity for the coefficients (and noting that the series converges absolutely for all real so we may split the sum into separate sums without worry),

This is a first order linear differential equation of the form . Furthermore, the initial condition implies that when . Therefore, the solution is

where

So, we have

Assume that the differential equation

has a power-series solution and find a formula for the coefficient .

First, we have

So, from the given differential equation we have

Since each coefficient of must equal 0 for this equation to hold we have

By induction we then have

The coefficients and are arbitrary and we denote them by and respectively. Then we have

Assume that the differential equation

has a power-series solution and find a formula for the coefficient .

First, we have

Therefore, we have

Equating like powers of , we have a recursive relation for when given by

Furthermore, we have that and that and are arbitrary constants, say and , respectively. Then by induction we establish

Therefore,

Assume that the differential equation

has a power-series solution and find a formula for the coefficient .

First, we have

Therefore, we have

Equating like powers of we obtain the recursive relation

By induction, we then have

Therefore,

Consider the differential equation

with initial conditions when . Assume this differential equation has a power-series solution and compute the first four nonzero terms of the expansion.

Let

be the power-series solution of the differential equation. Then we must have

From the initial conditions we know . Then, equating like powers of we can solve for the first four nonzero terms in the power series expansion:

(**Note:** I think the solution in the back of Apostol is wrong on this. Apostol has , , and . I’m going to mark this as errata until someone convinces me Apostol is actually correct.)

Therefore, we have

Consider the differential equation

with initial conditions when . Assume this differential equation has a power-series solution and compute the first four nonzero terms of the expansion.

Let

be the power series solution of the differential equation. Then we must have

From the initial condition when we know . Therefore, equating like powers of we have

(**Note:** The book gives the value . I think the answer we have above is correct. I’m marking this as errata unless someone convinces me that Apostol is correct.)

Therefore, we have

Consider the differential equation

with initial conditions when . Assume this differential equation has a power-series solution and compute the first four nonzero terms of the expansion.

Let

be the power-series solution of the differential equation. Then we must have

From the initial condition when we have . Therefore, equating like powers of we have the following equations

Therefore, we have