Power Series Solutions And Special Functions: Review of Power Series

Transcription

1 Power Series Solutions And Special Functions: Review of Power Series Pradeep Boggarapu Department of Mathematics BITS PILANI K K Birla Goa Campus, Goa September, 205 Pradeep Boggarapu (Dept. of Maths) Review of Power Series September, 205 / 25

2 Finding the general solution of a linear differential equation y + P(x)y + Q(x)y = R(x) depends on determining any two linear independent solutions of the homogeneous equation y + P(x)y + Q(x)y = 0 So far, we have a systematic procedure for constructing fundamental solutions (linear independent solutions of associated homogeneous equation), if the equation has constant coefficients (i.e., P(x) and Q(x) are constant functions). For the equations with variable coefficients we don t have any method to find fundamental solutions expect the case of knowing one solution where we use the known solution to find another solution via the formula y 2 (x) = v(x)y (x) with v(x) = e P(x)dx dx y 2 Pradeep Boggarapu (Dept. of Maths) Review of Power Series September, / 25

3 Examples: Let p, a, b, c and k are real constants, Bessel s Differential Equation: x 2 y + xy + (x 2 p 2 )y = 0 2 Legender s Differential Equation: ( x 2 )y 2xy + p(p )y = 0 3 Hermite Differential Equation: y 2xy + 2py = 0 4 Guass Hypergeometric Differentila Equation: x(x )y + [c (a + b + )x]y aby = 0 5 Laguerre s Differential Equations: xy + ( x)y + py = 0 6 Airy s Equation: y ± p 2 xy = 0 Pradeep Boggarapu (Dept. of Maths) Review of Power Series September, / 25

4 In most of the cases the solutions of the above differential equations are beyond the elementary functions which are called as special functions. Many of the special functions find applications in connection with the partial differential equations of mathematical physics. Thare are also important in modern pure mathematics, through the theory of orthogonal expansions. For a larger class of linear diffrerential equations with variable coefficients such as above equations, we must search for solutions beyond the familiar elementary functions of calculus. The principal tool we need is the representation of a given function by a power series. Then, we assume that the solutions y have power series representations a n(x x 0 ) n, and then determine the coefficients a n s so as to satisfy the differential equation similar to the method of undetermined coefficients. Pradeep Boggarapu (Dept. of Maths) Review of Power Series September, / 25

5 Review of Power Series Power series about the point zero: It is an infinite series of the form a n x n = a 0 + a x + a 2 x 2 + a 3 x 3 + (0.) where a 0, a, a 2,..., a n,... are real constants. Power series about the point x 0 : It is an infinite series of the form a n (x x 0 ) n = a 0 + a (x x 0 ) + a 2 (x x 0 ) 2 + a 3 (x x 0 ) 3 + (0.2) where a 0, a, a 2,..., a n,... are real constants. x n Examples: n! = + x! + x 2 2! + x 3 3! + ; x n = + x + x 2 + x 3 +. Pradeep Boggarapu (Dept. of Maths) Review of Power Series September, / 25

6 Convergence of power series: The power series is said to converge at a point x if its n-th partial sum n k=0 a kx k converges that is to say that the n limit L = lim a k x k exists. n k=0 In this case the sum of the series is the limit i.e. L = a nx n and such points x are called points of convergence. Note that x = 0 is always a point of convergence of the power series (0.). See the following examples x n n! = + x! + x 2 2! + x 3 3! + ; (0.3) x n = + x + x 2 + x 3 + ; (0.4) n!x n = + x + 2!x 2 + 3!x 3 +. (0.5) Pradeep Boggarapu (Dept. of Maths) Review of Power Series September, / 25

7 The first series converges for every value of x in R; second converges only for x < and the third series diverges for all x 0. The power series in x that behaves like third series are of no interest to us. Fact: The points of convergence of a power series (0.) (or (0.2)) form an interval. Moreover there exists 0 R such that the power series (0.) (or (0.2)) converges for all x < R (resp. x x 0 < R) and diverges for all x > R (resp. x x 0 > R). Here R is called radius of convergence. In many cases the radius of convergence can be found by using the following formulas, whenever the limits exist. R = lim n 0 a n a n+ or R = lim an n 0 n Regardless of the existence of the above limits, it is known that R always exists. Pradeep Boggarapu (Dept. of Maths) Review of Power Series September, / 25

8 Differentiation of power series: Suppose that the power series (0.) converges for x < R with R > 0 and denote the sum by f (x): f (x) = a n x n = a 0 + a x + a 2 x 2 + a 3 x 3 +. Then f (x) is automatically is continuous and has the derivatives of all orders for x < R. Also, f (x) = na n x n = a + 2a 2 x + 3a 3 x 2 +, (0.6) f (x) = n= n(n )x n 2 = 2a a 3 x +, (0.7) n=2 and so on, and each of the resulting series converges for x < R. And we can link the coefficient a n to f (x) and its derivative via the following formula a n = f (n) (0) n! (0.8) Pradeep Boggarapu (Dept. of Maths) Review of Power Series September, / 25

9 Algebra of power series: Let f (x) = a nx n and g(x) = b nx n be two power series with radius of convergence at least R > 0, then these power series can be added or subtracted termwise: f (x) ± g(x) = (a n ± b n )x n = (a 0 ± b 0 ) + (a ± b )x +. They can also be multiplies as they were polynomials, in the sense that f (x)g(x) = where c n = a 0 b n + a b n + + a n b 0. c n x n If f (x) = g(x) for x < R if and only if a n = b n for all n i.e. If both series converges to the same function for x < R if and only if they have the same coefficients. Pradeep Boggarapu (Dept. of Maths) Review of Power Series September, / 25

10 Power series representation of a function: Let f be a continuous function defined for x < R or ( R, R) which has derivatives of all orders in the interval ( R, R). Can f (x) be represented by a power series about the point zero? Equivalently will the following hold f (x) = f (n) (0) n! throughout the interval ( R, R)? x n = f (0) + f (0)x + f (0) x 2 + (0.9) 2! This is often true, but unfortunately it is some times false. The above expansion is valid if the error term R n (x) in Taylor s formula: f (x) = n convegres to zero as n tends to infity. f (k) (0) x k + R n (x) k! Pradeep Boggarapu (Dept. of Maths) Review of Power Series September, / 25

11 By means of the procedure explained in the previous slide, it is quite easy to obtaion the following familiar expansions, e x x n = n! = + x! + x 2 2! + x 3 3! + ; (0.0) sin x = cos x = ± x = log ( + x) = tan x = ( ) n x 2n+ (2n + )! = x x 3 3! + x 5 5! ; (0.) 2n x 2n ( ) (2n)! = x 2 2! + x 4 4! ; (0.2) (±) n x n = ± x + x 2 ± x 3 + ; (0.3) ( ) n x n n= n = x x x 3 3 x 4 ( ) n x 2n+ (2n + ) = x x ; (0.4) 3 + x 5 5 ; (0.5) Pradeep Boggarapu (Dept. of Maths) Review of Power Series September, 205 / 25

12 The function f for which the above series expansion (0.9) is valid for some neighbourhood of zero is said to be analytic at x = 0. More generally the analyticity at any point is defined as follows. Analytic at a point: A function f (x) with the property that a power series expansion of the form f (x) = a n (x x 0 ) n valid in some neighbourhood of the point x 0 is said to be analytic at x 0. In this case the coefficients a n s are necessarily given by a n = f (n) (x 0 ), n! and the above series is called the Taylor series of f (x) at x 0. Pradeep Boggarapu (Dept. of Maths) Review of Power Series September, / 25

13 Facts about analytic functions: Polynomials and the functions e x, sin x and cos x are analytic at all points. If f (x) and g(x) are analytic at x 0, then f (x) + g(x), f (x)g(x), and f (x)/g(x) [if g(x 0 ) 0] are also analytic at x 0. If f (x) is analytic at x 0, f (x 0 ) 0 and f (x) is a continuous inverse, then f (x) is analytic at f (x 0 ). If g(x) is analytic at x 0 and f (x) is analytic at g(x 0 ), then f g(x) = f (g(x)) is analytic at x 0. The sum of a power series is analytic at all points inside the interval of convergence. Pradeep Boggarapu (Dept. of Maths) Review of Power Series September, / 25

14 Series solutions of first order equations: We have repeatedly emphasized that many interesting and important differential equations cannot be solved by any of the methods discussed so far. And also note that solutions for equations of this kind can often be found in terms of power series. Our purpose in this section is to discuss that how we use power series representation to solve a differential equation by demonstrating an example with a first order equations that are easy to solve by elementary methods. Problem : Solve the first order differential equation y = y by using power series representation. Solution: We assume that the given equation has a power series solution of the form y = a n x n = a 0 + a x + a 2 x 2 + a 3 x 3 + (0.6) that converges for x < R with R > 0. Pradeep Boggarapu (Dept. of Maths) Review of Power Series September, / 25

15 That is, we are assuming that the differential equation has a solution say y that is analytic at the origin. We know that y = na n x n = a +2a 2 x +3a 3 x 2 + +(n +)a n+ x n +, (0.7) n= in the interval of convergence. Since y = y, we have that na n x n = n= a n x n a + 2a 2 x + 3a 3 x 2 + = a 0 + a x + a 2 x 2 + a 3 x 3 +. The above both series must have the same coefficients: a = a 0, 2a 2 = a, 3a 3 = a 2, (n + )a n+ = a n,. These equations enables us to express each a n in terms of a 0 : a = a 0, a 2 = a 2 = a 0 2, a 3 = a 2 3 = a 0 2 3,, a n = a 0 n!,. Pradeep Boggarapu (Dept. of Maths) Review of Power Series September, / 25

16 If you substitute all these values of coefficients in (0.6), we obtain our power series solution ( y = a 0 + x! + x 2 2! + x 3 3! + + x n ) n! + We can easily recognise that the above series as the power series expansion of e x, so the above can be written as y = a 0 e x., This example suggests a useful method for obtaining the power series expansion of a given function: Find the differential equation with initial condition satisfied by the function, and then solve this equation by power series. We consider an example (to understand this idea). Proplem 2: Find the power series expansion of y = ( + x) p about the origin (where p is a constant) by using the method described in the above. Use the result to show that 2 = Pradeep Boggarapu (Dept. of Maths) Review of Power Series September, / 25

17 Part : First find a differential equation satisfied by y: ( + x)y = py, y(0) = As before we assume that the above equation has a power series solution y = a n x n = a 0 + a x + a 2 x 2 + a 3 x 3 + (0.8) with positive radius R > 0 of convergence. It follows that y = a + 2a 2 x + 3a 3 x (n + )a n+ x n +, xy = a x + 2a 2 x na n x n +, py = pa 0 + pa x + pa 2 x pa n x n +. Since the d.e. is ( + x)y = py, the sum of the first two series must equal the third, so equating the coeffiecients of the successive powers of x gives a = pa 0, 2a 2 + a = pa, 3a 3 + 2a 2 = pa 2,..., (n + )a n+ + na n = pa n,... Pradeep Boggarapu (Dept. of Maths) Review of Power Series September, / 25

18 The initial condition y(0) = implies that a 0 =, so a = p, a 2 = a (p ) 2 a 3 = a 2(p 2) 3 = = p(p ), 2 p(p )(p 2),..., 2 3 p(p )(p 2) (p n + ) a n =,.... n! With these coefficints, the solution (0.8) becomes p(p ) y = + px + x 2 p(p )(p 2) + x 3 + 2! 3! p(p )(p 2) (p n + ) + x n +. (0.9) n! To conclude that (0.9) actually is the desired solution, it suffices to observe that the series converges for x < R for some R > 0. Pradeep Boggarapu (Dept. of Maths) Review of Power Series September, / 25

19 We can show that R = by using the ratio tests or by using first of the formulas given previously R = lim n a n n + = lim =. a n+ n 0 p n On comparing the two solutions y = ( + x) p and (0.9), and using the fact that the initial value problem has only one solution, we have ( + x) p p(p ) = + px + x 2 p(p )(p 2) + x 3 + 2! 3! p(p )(p 2) (p n + ) + x n + (0.20) n! for x <. This expansion is called the binomial series and this formula generalises the binomial theorem to the case of an arbitrary exponent. Pradeep Boggarapu (Dept. of Maths) Review of Power Series September, / 25

20 Part2: Now we substitute x = 2 and p = 2 in the power series (0.20) to get the required indentity. ( ) 2 2 ( = + ) ( 2 ) + 2 ( 2 + ( 2 )( 3 2 3! )( ) 3 2 2! )( 5 2 After simplying the above you get the required identity ) ( 2 = ) 2 2 ( ) Pradeep Boggarapu (Dept. of Maths) Review of Power Series September, / 25

21 Problem 3: Express sin x in the form of a power series a n x n by solving y = ( x 2 ) /2 in two ways. (Hint: Remember the binomial series.) Use this result to obtain the formula π 6 = Solution: Part : We first find an intial value problem satisfied by y = sin x: y = ( x 2 ) /2, y(0) = 0 As before we assume that the above equation has a power series solution y = a n x n = a 0 + a x + a 2 x 2 + a 3 x 3 + (0.2) with positive radius R > 0 of convergence. It follows that y = a + 2a 2 x + 3a 3 x 2 + 4a 4 x (n + )a n+ x n + (0.22) Pradeep Boggarapu (Dept. of Maths) Review of Power Series September, / 25

22 We know that ( + x) p p(p ) = + px + x 2 p(p )(p 2) + x 3 + 2! 3! p(p )(p 2) (p n + ) + x n +. n! Now we replace p by /2 and x by x 2 in the above to get ( x 2 ) /2 = + 2 x x x (2n ) (2n) x 2n +. Since the d.e. is y = ( x 2 ) /2, so equating the coeffiecients of the successive powers of x on both series gives 2a 2 = 4a 4 = a 2n = = 0; a = ; 3a 3 = 2 ; 5a 5 = ; 7a 7 = ;... ; (2n + )a 2n+ =, (2n ) (2n) Pradeep Boggarapu (Dept. of Maths) Review of Power Series September, / 25

23 Further simplification implies 2a 2 = 4a 4 = a 2n = = 0; a = ; a 3 = 2 3 ; a 5 = ; a 7 = ;... ; a 2n+ =, (2n ) (2n) 2n + Since y(0) = 0, a 0 = 0, now substitute all the values of coefficients in power series solution (0.2) and the initial value problem has only one solution which means the power series solution is nothing but sin x, therefore sin x = x x x x 7 +. If x = 2 then sin x = π 6 and hence in view of the above we get the required identity π 6 = Pradeep Boggarapu (Dept. of Maths) Review of Power Series September, / 25

24 Problem 4: Find the power series solution of the initial value problem ( + x)y = ; y(0) = 0, and also find solution of the same by using variable separable method to get the follwoing identity log e 2 = n= n 2 n = Problem 5: Find the power series solutions of the each of the following first order differentail equations: () y y = 0 (2) y = e x2 y (3) y xy = 0 (4) ( x)y = y (5) y y = x 2 (6) y + xy = + x (7) ( + x 2 )y = 0. Pradeep Boggarapu (Dept. of Maths) Review of Power Series September, / 25

25 Thank you for your attention Happy Weekend! Pradeep Boggarapu (Dept. of Maths) Review of Power Series September, / 25