Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A trigonometric function (such as sine or cosine), or some combination thereof, can be the solution of a first order differential equation with constant coefficients. But the solution of a higher order differential equation with non-constant coefficients is often a power series.

Why is that? Is it because the coefficients are themselves a function of the differentials? Or do the power series in question tend to converge toward simpler trigonometric functions?

share|improve this question
    
Let $P \in \mathbb C[[t][x_0, x_1, \dots, x_n]$ be a polynomial in $n+1$ variables and coefficients in the ring of complex formal power series. If a solution to $P(f,f', \dots, f^{(n)}) = 0$ exists, then it is a power series. –  Alexander Thumm Sep 2 '12 at 17:56
1  
I'm not sure I understand the question. A trigonometric function is also a power series. –  Qiaochu Yuan Sep 2 '12 at 19:08
    
@QiaochuYuan: This might have been addressed in the answer below; the more complicated equations have power series of complicated functions, while the ordinary equations can be solved by simple functions (that, of course, have power series). –  Tom Au Sep 2 '12 at 19:13

3 Answers 3

up vote 3 down vote accepted

Here is the intuitive story behind Robert Israel's answer:

The following general principle has been invented by Newton: Any reasonable given or unknown function $x\mapsto f(x)$ defined in a neighborhood of $x=0$, in particular any "analytical expression" like, e.g., $e^x$, can be developed into a power series of the form $\sum_{k=0}^\infty a_k x^k$, where the $a_k$ are real (or complex) constants. This means that there is a $\rho>0$ such that $$f(x)=\sum_{k=0}^\infty a_kx^k\qquad\bigl(|x|<\rho\bigr)\ .$$ These power series behave in a simple way under addition and multiplication (particularly by polynomials in $x$), and above all, under differentiation. Even composition ("plugging" one series into an other) can be handled, however the computations get more involved in this case.

When an initial value problem $$y'(x)=F\bigl(x,y(x)\bigr),\quad y(0)=y_0\qquad(*)$$ is given, where the right side is some "analytical expression" $F(x,y)$ in the two variables $x$ and $y$, then only in rare cases it is possible to guess and subsequently verify some expression $x\mapsto y(x)$ that solves this problem. But it is always allowed to develop everything in sight into a power series in $x$, the unknown solution $y(x)$ with undetermined coefficients $a_k$, and in the majority of cases one then obtains a recursion formula for the $a_k$, with starting value $a_0=y_0$.

In the end one has the solution of $(*)$ in the form $y(x)=\sum_{k=0}^\infty a_kx^k$, and this series is hopefully convergent for all $x$ with $|x|<\rho$ for some $\rho>0$. In this way one has an at least numerically feasible hold on the solution, even though there may be no "elementary" expression for this solution available.

share|improve this answer
1  
And even if the series is not convergent almost everywhere (such as $\sum n!x^n$), the series can sometimes be manipulated to give useful information. –  marty cohen Sep 3 '12 at 0:38

The power series converge to complicated special functions. For uncomplicated differential equations with uncomplicated power series we have names for the functions like "sin" "cos" "exp" etc. An example of a more complicated special function that is not as well known as cosine is http://en.wikipedia.org/wiki/Fox_H-function .

share|improve this answer
    
exactly, power series also define functions. They may just lack a name that we know. –  James S. Cook Sep 2 '12 at 17:53
    
Welcome to the site. An upvote to get you going. –  Tom Au Sep 2 '12 at 18:10

For an initial value problem, say $\dfrac{d^n y}{dx^n} = F\left(x, y, \dfrac{dy}{dx}, \ldots, \dfrac{d^{n-1} y}{dx^{n-1}}\right)$ with $y(x_0) = y_0, \ldots, y^{(n-1)}(x_0) = y_{n-1} $, if $F$ is analytic in a neighbourhood of $(x_0, y_0, \ldots, y_{n-1})$, then there is a solution that is analytic in a neighbourhood of $x_0$. You can find arbitrarily many of the Taylor coefficients by writing $y = y_0 + c_1 (x - x_0) + c_2 (x - x_0)^2 + \ldots$, expanding $\dfrac{d^n y}{dx^n} - F\left(x, y, \dfrac{dy}{dx}, \ldots, \dfrac{d^{n-1} y}{dx^{n-1}}\right)$ in powers of $x$, and solving equations for each power of $x$. Moreover, in the case of a homogeneous linear equation with coefficients that are polynomials in $x$, you get a linear recurrence for the coefficients of the solution. In some cases you can solve that recurrence to get an explicit formula for the coefficients.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.