Chapter 2 - Series Solutions of Linear Ordinary Differential Equations Flashcards
What is an elementary function?
-an elementary function is one that can be expressed as a power series expansion
How are power series related to differential equations?
- we can calculate power series solutions of differential equations which have no elementary functions as solutions
- they are also an alternative method to solving differential equations which do have a known function as their solution
List the three well known types of elementary equation
1) algebraic functions
2) the exponential function e^x
3) trigonometric functions
Algebraic Functions
-solutions y(x) of any equation of the form Pn(x)y^n + … + P1(x)y + Po(x) = 0
-polynomials which correspond to n=1, P1=-1
=> y = Po(x)
-rational functions which correspond to n=1 and general P1(x)
=> y = - Po(x) / P1(x)
-roots of rational functions e.g. n=2, P1(x)=0
=> y = √[-Po(x)/P2(x)
Combining to Form Elementary Functions
- algebraic combinations of other types,
e. g. tan(x) = sin(x)/cos(x) - function of a function
e. g. sin(x*e^x) - inverse functions
e. g. log(x) or tan^(-1)(x) - -but only a finite number of such operations
Taylor Expansion of a function f(x) about a point xo
f(x) = Σ f(‘n)(xo)/n! * (x-xo)^m
- where f(‘n)(xo) is the nth derivative of f evaluated at xo
- and the sum is taken from 0 to infinity
Radius of Convergence
-the Taylor series about a point xo is only valid within a radius of convergence of that point i.e. valid for:
|x-xo| < rc (rc>0)
rc = lim |an / an+1|
-where the limit is taken as n tends to infinity and the n and n+1 are subscripts
-if rc tends to infinity then we say that the series converges for all x
Term by Term
- the Taylor series can be differentiated term by term
- in fact it is infinitely differentiable and each of the resulting power series has the SAME radius of convergence rc
Ratio Test
-consider the power series: Σ an * x^n -where the sum is taken between 0 and infinity -for x=0 it evidently has the sum ao -for x≠0 we use the ration test: lim |an+1*x^(n+1) / an*x^n| = lim |an+1 / an | * |x| = L -where the limit is taken as n tends to infinity and the n and n+1 following a are subscripts -if |L|<1 the series converges -if |L|>1 the series diverges -if |L|=1 it is unknown
Power Series
e^x
e^x = Σ x^n/n!
-where the sum is taken from n=0 to n=infinity
e^x = 1 + x + x²/2 + x^3/3 + …
-for all x
Power Series
sinx
sinx = x - x^3/3! + x^5/5! + …
-for all x
Power Series
1/(1-x)
1/1-x = Σ x^n = 1 + x + x² + x^3 +…
-where the sum is taken from n=0 to infinity
Where can an elementary function be represented as a power series?
- given an elementary function we can represent it as a power series wherever the function is analytic
- e.g. for √x , the function is not analytic at xo=0 so has no Taylor series there, but at xo=1 a Taylor series can be defined
Analytic Solutions of ODEs
Steps
y’’ + p(x)y’ + q(x)y = 0
-start by supposing that the solution y takes the form:
y = Σanx^n = ao + a1x + a2x² +…
-where the sum is taken between 0 and infinity
-differentiate to get an expression for y’
-differentiate again for an expression for y’’ (if necessary)
-sub into the original differential equation
-collect terms by power of x e.g. a1-ao + (2a2-a1)x + (3a3-a2)x² + …. + ((n+1)an+1-an)x^n+…
-the only way for a series to be identically zero is for all coefficients to be equal to zero
-setting each coefficient equal to zero write the other coefficients in terms of constants ao, a1, a2
-sub these expression into the original
y = Σanx^n
-to obtain the general solution
When can an analytic solution for the following equation be found?
y’’ + p(x)y’ + q(x)y = 0
-in order to build analytic solutions to equations of this form we must demand that functions p and q are not only continuous but also analytic
-most examples will be derived from:
S(x)y’’ + P(x)y’ + Q(x)y = 0
-where S, P and Q are polynomials with no common factor
-so p(x) = P(x) / S(x)
-and q(x) = Q(x) / S(x)
-p and q are analytic if they can be written as a power series