Power series solutions to ODEs
Polynomial functions have the form
A power series is an infinite sum of power terms
which is determined by an infinite series of coefficient numbers
If a power series is given with either:
- (1) the general term
as a function of in terms of an algebraic formula, - (2) the
term as a function of all previous terms in a recursion relation,
then a computer can quickly produce the coefficients of a power series to any degree
Polynomials do approximate everything
Any continuous function
on an interval can be approximated by a sequence of polynomials with increasing degree.
However, such approximating polynomials might not form a power series, meaning that their initial coefficients do not get locked in, or in other words, the sequence of polynomials have ever differing initial coefficients.
Power series can not approximate everything
Many important functions in mathematics cannot be approximated by power series.
When a function is approximated by a power series, it is called analytic or (to distinguish from the complex case) real analytic.
Example
Smooth, non-analytic function
The function
cannot be approximated by a power series near
, even though it can be approximated by a sequence of polynomials there. Moreover, it does have all of its higher derivatives, and thus it is smooth and yet non-analytic. Near
, the sequence of polynomials gives a great approximation to . However, any initial coefficient becomes zero, whereas any single power series would have to set some coefficient to be non-zero.
One perspective on power series is the approximation perspective, according to which our goal is to approximate a class of potentially relevant functions using some standardized functions, and power series give a powerful type of such standardized functions. Even though many important functions are non-analytic, many other important ones are indeed analytic.
Another perspective on power series is the transcendental functions perspective, according to which our goal is to understand certain functions that are given to us and studied in the form of converging power series. This perspective takes an interest in higher transcendental functions. These functions are usually determined as power series in terms of an integral, an ODE (a more complex integral), or an asymptotic expansion of some kind. The so-called “special functions” like Bessel functions and Legendre functions and Chebyshev functions are of this kind.
Review of power series
A power series of the form
is said to be expanded around
is said to be expanded around
Radius of convergence
Power series have the interesting property of converging inside intervals centered at the expansion point
Radius of convergence
The reason that power series converge inside intervals / disks that are symmetric about the expansion point
is the root test, which is really a “comparison” to the geometric series that converges for and diverges for . Inside the disk of convergence , the quantities are (eventually) all smaller than one, so the
term of the power series is (eventually) smaller in absolute value than the term of a converging geometric series , namely with . Outside the disk of convergence, , the individual terms of the series do not even converge to zero, so the series itself surely cannot converge. Of course, it is important to note that the behavior of the series on the boundary may be non-symmetric. For example,
converges on and diverges outside it. Remember, also, that the radius of convergence might be .
Power series define functions on their domains of convergence. These functions are fairly convenient to handle using algebra and calculus. Algebra Power series can be added, subtracted, and multiplied:
where
Calculus It is easy to differentiate and integrate power series functions. This is one of their primary advantages. Derivatives:
Integrals:
The coefficients of a power series function can be extracted by evaluating the function and its derivatives:
This fact entails that the coefficients of a power series are unique, and so any two power series with at least one differing coefficient will necessarily have differing values somewhere.
Power series have intrinsic merit, not just of other functions
We treat power series as functions on their interval of convergence.
Some functions are not necessarily initially defined using power series, and yet they “have” power series, for example
and . This means that they are equal to a power series on some domain (possibly everywhere). This is an equality of two definitions. For the most part, it is good to think of such a function as simply (defined by) its power series, and its other definition (involving geometry or integration or whatever) is actually a property of said power series, not a definition of said function.
Familiar examples
Question 10-01
Power series calculus
Use calculus and the first familiar example
to find the power series that equals each of:
Question 10-02
Bessel function
Show that the series
satisfies the ODE
.
Power series solutions: basic technique
The fundamental idea is to convert an ODE into an algebraic system relating the coefficients of a power series function that solves the ODE, and then to solve this system to produce formulas for the coefficients.
Example
Defining the exponential function as a power series
Consider the ODE
. Suppose the series solves this ODE. Plug in the series and obtain: Now we manipulate the summation expressions using two important techniques: (1) reindex summations to align the powers, (2) break off extra terms until the summations begin at the same number.
Starting with (1):
Fortunately, (2) is already done. Now add these termwise:
Assuming the series on the left converges, this function is zero only when all coefficients are zero. So we arrive at the recursive relation
. Assuming a starting value of , this relation determines every using: The power series solution is therefore
, as we expected. Notice! The starting coefficient
becomes the parameter constant giving the complete family of solutions of the ODE.
Example
Newton’s Binomial Expansion of
The binomial theorem is a formula for the coefficients of the expansion of
as a polynomial, where is a natural number. Isaac Newton found an analogous formula for where . We will show how to derive his formula. It is actually sufficient to discover a formula for
, because and we can therefore find the expansion of
for , and then multiply every coefficient by to obtain the expansion of . To see the overall strategy, notice that
solves the ODE given by . The method of power series solutions applied to this ODE will give us the desired series expansion. Let us write the ODE as
, and substitute : Now we (1) reindex the first summation, and (2) break off the
terms from the first and third, and (3) collect the summations: Since all coefficients of
(including ) must be zero, we know , and Therefore, the power series function is:
Compare this to the formula for the expansion when
: In other words, the series for
terminates, and the resulting polynomial (finite series) is exactly the binomial expansion polynomial. That series is generalized to non-integral powers by interpreting the formula for coefficients, , which a priori makes sense only for integers, as the specific fraction , in which we can allow to take non-integral values. Actually this numerator goes by its own name, the Pochhammer symbol: . Using this notation, Newton’s formula becomes:
Question 10-03
Method of power series
Find a power series solution to
. Then solve the equation algebraically using prior methods we have learned, and compare your answers.
Question 10-04
Method of power series
Find a power series solution to
. Then solve the equation algebraically using prior methods we have learned, and compare your answers.
Exercise 10-01
Method of power series: series terminates
Look for a power series solution to
. Then solve the equation algebraically using prior methods we have learned, and compare your answers.
Exercise 10-02
Method of power series: encountering problems
Look for a power series solution to
. Then solve the equation algebraically using prior methods we have learned, and compare your answers.
Exercise 10-03
Method of power series: one free coefficient
Find a power series solution to
. Then solve the equation algebraically using prior methods we have learned, and compare your answers.
Exercise 10-04
Method of power series: missing negative power
Find a power series solution to
. Then solve the equation algebraically using prior methods we have learned, and compare your answers.
Second-order linear equations
A number of famous special transcendental functions are determined by second-order differential equations
having the property that
Example
Trig functions
Consider the ODE given by
. Suppose there is a power series function that solves this ODE. By plugging in, reindexing the summation to achieve the same powers, and factoring out the powers, we find: We conclude that
This equation determines the even coefficients and the odd coefficients separately. There is no restriction on
and , but once these are set, the formula above determines all others. The complete family of solutions in terms of the and is then: Notice: we could recognize the trig power series and write instead:
.
Now for a more difficult example. (See the textbook for more detailed steps!)
Example
Legendre equation, Legendre series, Legendre polynomials
We seek the solutions to Legendre’s Equation:
Here
is any real constant. Notice that and do not become unbounded around , so is an ordinary point. To solve this ODE using power series, it is convenient to write it without denominators:
Plug in the power series
: Now (1) reindex the first term so it shows
, (2) break off separately the terms with and , and (3) collect under one summation and factor out : We set each coefficient equal to zero, perform some algebra, and deduce the equations:
Some additional algebra (which is a bit tricky to discover, but easy to verify) gives:
From this last formulation, we see that
and are free and they become the two constants in our complete family of solutions. Once they are set, all remaining constants are determined by iterating the formulas. As in the previous example, the even and odd terms are controlled separately from the two starting values and . Notice! When
is an even integer, then at some point the even coefficients will encounter a zero in the numerator, and remain zero ever thereafter. Similarly, when is an odd integer, then at some point the odd coefficients will encounter a zero and then remain zero. In either case, the power series is then a polynomial (because the terms are eventually just zeros). These polynomials are called Legendre polynomials, and they are very important in mathematics.
Theorem: power series solutions at ordinary points
Suppose
is an ordinary point of . Then there is some neighborhood of on which this equation has a power series solution. Any initial values may be given, and the radius of convergence of the solution is at least as big as the smaller of the radii of convergence of and .
Problems due Monday 25 Mar 2024 by 11:59pm
Easier Problems
Problem 10-01
Power series solution of second-order ODE
Find the complete family of power series solutions to
.
Problem 10-02
Power series solution of second-order ODE
Find the complete family of power series solutions to
.
Problem 10-03
Power series solution of second-order ODE
Find the complete family of power series solutions to
.
Harder Problems
Problem 10-04
Chebyshev polynomials
Consider the Chebyshev ODE given by
, where is some constant. Find two independent power series solutions to this ODE (expanded around ), and verify that when is an integer, then these series terminate, thus giving polynomials.
Problem 10-05
Alternative power series method: using initial conditions
Consider the IVP given by
, , . Notice that the equation directly determines . Now differentiate the equation:
. Notice that this equation now determines in terms of the initial conditions plus the information of . We can iterate the process to compute every
. Recall that . We therefore have a new procedure to calculate every . Follow this procedure to compute the first 7 terms .
Problem 10-06
Power series: using series multiplication
Find the first 6 terms of a power series solution to the IVP given by
, . [Hint: substitute in the power series for .]