MA-207 Differential Equations II Ronnie Sebastian Department of - PowerPoint PPT Presentation

Theorem For any power series, ∞ � a n ( x − x 0 ) n n =0 exactly one of these statements is true. 1 The power series converges only for x = x 0 . 2 The power series converges for all values of x . 3 There is a positive number 0 < R < ∞ such that the power series converges if | x − x 0 | < R and diverges if | x − x 0 | > R . R is called the radius of convergence of the power series. 11 / 47

Theorem For any power series, ∞ � a n ( x − x 0 ) n n =0 exactly one of these statements is true. 1 The power series converges only for x = x 0 . 2 The power series converges for all values of x . 3 There is a positive number 0 < R < ∞ such that the power series converges if | x − x 0 | < R and diverges if | x − x 0 | > R . R is called the radius of convergence of the power series. We define R = 0 in case (i) and R = ∞ in case (ii). 11 / 47

Theorem For any power series, ∞ � a n ( x − x 0 ) n n =0 exactly one of these statements is true. 1 The power series converges only for x = x 0 . 2 The power series converges for all values of x . 3 There is a positive number 0 < R < ∞ such that the power series converges if | x − x 0 | < R and diverges if | x − x 0 | > R . R is called the radius of convergence of the power series. We define R = 0 in case (i) and R = ∞ in case (ii). Question. How to compute the radius of convergence? 11 / 47

Theorem (Ratio test) If a n � = 0 for all n and � � a n +1 � � lim � = L � � a n n →∞ � 12 / 47

Theorem (Ratio test) If a n � = 0 for all n and � � a n +1 � � lim � = L � � a n n →∞ � n →∞ | a n | 1 /n = L (Root test) lim sup 12 / 47

Theorem (Ratio test) If a n � = 0 for all n and � � a n +1 � � lim � = L � � a n n →∞ � n →∞ | a n | 1 /n = L (Root test) lim sup ∞ a n ( x − x 0 ) n is � Then radius of convergence of the power series n =0 R = 1 /L . 12 / 47

Theorem (Ratio test) If a n � = 0 for all n and � � a n +1 � � lim � = L � � a n n →∞ � n →∞ | a n | 1 /n = L (Root test) lim sup ∞ a n ( x − x 0 ) n is � Then radius of convergence of the power series n =0 R = 1 /L . For L = 0 , we get R = ∞ and for L = ∞ , we get R = 0 . 12 / 47

Theorem Let R > 0 be the radius of convergence of the power series ∞ � a n ( x − x 0 ) n n =0 Then the power series converges (absolutely) for all x ∈ ( x 0 − R, x 0 + R ) . 13 / 47

Theorem Let R > 0 be the radius of convergence of the power series ∞ � a n ( x − x 0 ) n n =0 Then the power series converges (absolutely) for all x ∈ ( x 0 − R, x 0 + R ) . For R = ∞ , we write ( x 0 − R, x 0 + R ) = ( −∞ , ∞ ) = R . 13 / 47

Theorem Let R > 0 be the radius of convergence of the power series ∞ � a n ( x − x 0 ) n n =0 Then the power series converges (absolutely) for all x ∈ ( x 0 − R, x 0 + R ) . For R = ∞ , we write ( x 0 − R, x 0 + R ) = ( −∞ , ∞ ) = R . The open interval ( x 0 − R, x 0 + R ) is called the interval of convergence of the power series. 13 / 47

Example Find the radius of convergence and interval of convergence (if R > 0 ) of the following three series ∞ ∞ ∞ ( − 1) n x n n ! x n ( ii ) � � � 2 n n 3 ( x − 1) n ( i ) ( iii ) n n 0 10 0 14 / 47

Example Find the radius of convergence and interval of convergence (if R > 0 ) of the following three series ∞ ∞ ∞ ( − 1) n x n n ! x n ( ii ) � � � 2 n n 3 ( x − 1) n ( i ) ( iii ) n n 0 10 0 � a n +1 � � ( n + 1)! � � � � � ( i ) lim � = lim � = lim n →∞ ( n + 1) = ∞ � � � � a n n ! n →∞ n →∞ � � So R = 0 in case (i). 14 / 47

Example Find the radius of convergence and interval of convergence (if R > 0 ) of the following three series ∞ ∞ ∞ ( − 1) n x n n ! x n ( ii ) � � � 2 n n 3 ( x − 1) n ( i ) ( iii ) n n 0 10 0 � a n +1 � � ( n + 1)! � � � � � ( i ) lim � = lim � = lim n →∞ ( n + 1) = ∞ � � � � a n n ! n →∞ n →∞ � � So R = 0 in case (i). Similarly, in case (ii) R = ∞ and in case (iii) R = 1 / 2 . 14 / 47

Example Find the radius of convergence and interval of convergence (if R > 0 ) of the following three series ∞ ∞ ∞ ( − 1) n x n n ! x n ( ii ) � � � 2 n n 3 ( x − 1) n ( i ) ( iii ) n n 0 10 0 � a n +1 � � ( n + 1)! � � � � � ( i ) lim � = lim � = lim n →∞ ( n + 1) = ∞ � � � � a n n ! n →∞ n →∞ � � So R = 0 in case (i). Similarly, in case (ii) R = ∞ and in case (iii) R = 1 / 2 . Interval of convergence : in case (ii) ( −∞ , ∞ ) and in case (iii) (1 / 2 , 3 / 2) 14 / 47

Theorem Let R be the radius of convergence of the power series ∞ � a n ( x − x 0 ) n . We assume R > 0 n =0 15 / 47

Theorem Let R be the radius of convergence of the power series ∞ � a n ( x − x 0 ) n . We assume R > 0 n =0 • We can define a function f : ( x 0 − R, x 0 + R ) → R by ∞ � a n ( x − x 0 ) n f ( x ) = n =0 15 / 47

Theorem Let R be the radius of convergence of the power series ∞ � a n ( x − x 0 ) n . We assume R > 0 n =0 • We can define a function f : ( x 0 − R, x 0 + R ) → R by ∞ � a n ( x − x 0 ) n f ( x ) = n =0 • f is infinitely differentiable ∀ x ∈ ( x 0 − R, x 0 + R ) . 15 / 47

Theorem Let R be the radius of convergence of the power series ∞ � a n ( x − x 0 ) n . We assume R > 0 n =0 • We can define a function f : ( x 0 − R, x 0 + R ) → R by ∞ � a n ( x − x 0 ) n f ( x ) = n =0 • f is infinitely differentiable ∀ x ∈ ( x 0 − R, x 0 + R ) . • The successive derivatives of f can be computed by differentiating the power series term-by-term, that is ∞ � f ′ ( x ) = na n ( x − x 0 ) n − 1 . . . n =0 ∞ f ( k ) ( x ) = � n ( n − 1) . . . ( n − k + 1) a n ( x − x 0 ) n − k n =0 15 / 47

Theorem (continued . . . ) • The power series representing the derivatives f ( n ) ( x ) have same radius of convergence R . 16 / 47

Theorem (continued . . . ) • The power series representing the derivatives f ( n ) ( x ) have same radius of convergence R . • We can determine the coefficients a n (in terms of derivatives of f at x 0 ) as f ′ ( x 0 ) = a 1 , f ′′ ( x 0 ) = 2 a 2 , . . . f ( x 0 ) = a 0 , In general, a n = f ( n ) ( x 0 ) n ! 16 / 47

Theorem (continued . . . ) • The power series representing the derivatives f ( n ) ( x ) have same radius of convergence R . • We can determine the coefficients a n (in terms of derivatives of f at x 0 ) as f ′ ( x 0 ) = a 1 , f ′′ ( x 0 ) = 2 a 2 , . . . f ( x 0 ) = a 0 , In general, a n = f ( n ) ( x 0 ) n ! ∞ � a n ( x − x 0 ) n • We can also integrate the function f ( x ) = 0 term-wise that is if [ a, b ] ⊂ ( x 0 − R, x 0 + R ) , then � b � b ∞ ∞ a n ( x − x 0 ) n dx = � � n + 1( x − x 0 ) n +1 f ( x ) dx = a n a a n =0 0 16 / 47

Example (Power series representation of elementary functions) ∞ x n (i) e x = � − ∞ < x < ∞ n ! 0 17 / 47

Example (Power series representation of elementary functions) ∞ x n (i) e x = � − ∞ < x < ∞ n ! 0 ∞ x 2 n +1 � ( − 1) n (ii) sin x = − ∞ < x < ∞ (2 n + 1)! 0 17 / 47

Example (Power series representation of elementary functions) ∞ x n (i) e x = � − ∞ < x < ∞ n ! 0 ∞ x 2 n +1 � ( − 1) n (ii) sin x = − ∞ < x < ∞ (2 n + 1)! 0 ∞ 1 � x n (iii) 1 − x = − 1 < x < 1 0 17 / 47

Example (Power series representation of elementary functions) ∞ x n (i) e x = � − ∞ < x < ∞ n ! 0 ∞ x 2 n +1 � ( − 1) n (ii) sin x = − ∞ < x < ∞ (2 n + 1)! 0 ∞ 1 � x n (iii) 1 − x = − 1 < x < 1 0 ∞ x 2 n +1 d ( − 1) n d � � � (iv) dx (sin x ) = dx (2 n + 1)! 0 17 / 47

Example (Power series representation of elementary functions) ∞ x n (i) e x = � − ∞ < x < ∞ n ! 0 ∞ x 2 n +1 � ( − 1) n (ii) sin x = − ∞ < x < ∞ (2 n + 1)! 0 ∞ 1 � x n (iii) 1 − x = − 1 < x < 1 0 ∞ x 2 n +1 d ( − 1) n d � � � (iv) dx (sin x ) = dx (2 n + 1)! 0 ∞ ( − 1) n x 2 n � = (2 n )! = cos x 0 17 / 47

Theorem (i) Power series representation of f in an open interval I containing x 0 is unique, that is, if ∞ ∞ a n ( x − x 0 ) n = � � b n ( x − x 0 ) n f ( x ) = 0 0 for all x ∈ I , then a n = b n ∀ n . 18 / 47

Theorem (i) Power series representation of f in an open interval I containing x 0 is unique, that is, if ∞ ∞ a n ( x − x 0 ) n = � � b n ( x − x 0 ) n f ( x ) = 0 0 for all x ∈ I , then a n = b n ∀ n . (ii) If ∞ a n ( x − x 0 ) n = 0 � 0 for all x ∈ I , then a n = 0 for all n . 18 / 47

Theorem (i) Power series representation of f in an open interval I containing x 0 is unique, that is, if ∞ ∞ a n ( x − x 0 ) n = � � b n ( x − x 0 ) n f ( x ) = 0 0 for all x ∈ I , then a n = b n ∀ n . (ii) If ∞ a n ( x − x 0 ) n = 0 � 0 for all x ∈ I , then a n = 0 for all n . Proof. (i) a n = f ( n ) ( x 0 ) = b n for all n. n ! It is clear that (ii) follows from (i). 18 / 47

Algebraic operations on power series Definition ∞ ∞ � � a n ( x − x 0 ) n b n ( x − x 0 ) n If f ( x ) = g ( x ) = 0 0 have radius of convergence R 1 and R 2 respectively, then ∞ � ( c 1 a n + c 2 b n )( x − x 0 ) n c 1 f ( x ) + c 2 g ( x ) := 0 has radius of convergence R ≥ min { R 1 , R 2 } for c 1 , c 2 ∈ R . 19 / 47

Algebraic operations on power series Definition ∞ ∞ � � a n ( x − x 0 ) n b n ( x − x 0 ) n If f ( x ) = g ( x ) = 0 0 have radius of convergence R 1 and R 2 respectively, then ∞ � ( c 1 a n + c 2 b n )( x − x 0 ) n c 1 f ( x ) + c 2 g ( x ) := 0 has radius of convergence R ≥ min { R 1 , R 2 } for c 1 , c 2 ∈ R . Further, we can multiply the series as if they were polynomials, that is ∞ � c n ( x − x 0 ) n ; f ( x ) g ( x ) = c n = a 0 b n + a 1 b n − 1 + . . . + a n b 0 0 It also has radius of convergence R ≥ min { R 1 , R 2 } . 19 / 47

Example Find the power series expansion for cosh x in terms of powers of x n . 20 / 47

Example Find the power series expansion for cosh x in terms of powers of x n . cosh x = 1 2 e x + 1 2 e − x 20 / 47

Example Find the power series expansion for cosh x in terms of powers of x n . cosh x = 1 2 e x + 1 2 e − x ∞ ∞ x n ( − 1) n x n = 1 n ! + 1 � � 2 2 n ! n =0 n =0 20 / 47

Example Find the power series expansion for cosh x in terms of powers of x n . cosh x = 1 2 e x + 1 2 e − x ∞ ∞ x n ( − 1) n x n = 1 n ! + 1 � � 2 2 n ! n =0 n =0 ∞ 2 [1 + ( − 1) n ] x n 1 � = n ! n =0 20 / 47

Example Find the power series expansion for cosh x in terms of powers of x n . cosh x = 1 2 e x + 1 2 e − x ∞ ∞ x n ( − 1) n x n = 1 n ! + 1 � � 2 2 n ! n =0 n =0 ∞ 2 [1 + ( − 1) n ] x n 1 � = n ! n =0 ∞ x 2 n � = (2 n )! n =0 20 / 47

Example Find the power series expansion for cosh x in terms of powers of x n . cosh x = 1 2 e x + 1 2 e − x ∞ ∞ x n ( − 1) n x n = 1 n ! + 1 � � 2 2 n ! n =0 n =0 ∞ 2 [1 + ( − 1) n ] x n 1 � = n ! n =0 ∞ x 2 n � = (2 n )! n =0 Since radius of convergence for Taylor series of e x and e − x are ∞ , the power series expansion of cosh x is valid on R . 20 / 47

Shifting the summation index ∞ ∞ a n ( x − x 0 ) n = � � na n ( x − x 0 ) n − 1 ⇒ f ′ ( x ) = If f ( x ) = n =0 n =1 21 / 47

Shifting the summation index ∞ ∞ a n ( x − x 0 ) n = � � na n ( x − x 0 ) n − 1 ⇒ f ′ ( x ) = If f ( x ) = n =0 n =1 Let us rewrite the series for f ′ ( x ) in powers of ( x − x 0 ) n . Put r = n − 1 , we get ∞ � ( r + 1) a r +1 ( x − x 0 ) r f ′ ( x ) = r =0 21 / 47

Shifting the summation index ∞ ∞ a n ( x − x 0 ) n = � � na n ( x − x 0 ) n − 1 ⇒ f ′ ( x ) = If f ( x ) = n =0 n =1 Let us rewrite the series for f ′ ( x ) in powers of ( x − x 0 ) n . Put r = n − 1 , we get ∞ � ( r + 1) a r +1 ( x − x 0 ) r f ′ ( x ) = r =0 Similarly, ∞ f ( k ) ( x ) = � n ( n − 1) . . . ( n − k + 1) a n ( x − x 0 ) n − k n = k 21 / 47

Shifting the summation index ∞ ∞ a n ( x − x 0 ) n = � � na n ( x − x 0 ) n − 1 ⇒ f ′ ( x ) = If f ( x ) = n =0 n =1 Let us rewrite the series for f ′ ( x ) in powers of ( x − x 0 ) n . Put r = n − 1 , we get ∞ � ( r + 1) a r +1 ( x − x 0 ) r f ′ ( x ) = r =0 Similarly, ∞ f ( k ) ( x ) = � n ( n − 1) . . . ( n − k + 1) a n ( x − x 0 ) n − k n = k ∞ � ( n + k )( n + k − 1) . . . ( n + 1) a n + k ( x − x 0 ) n = n =0 21 / 47

Shifting the summation index ∞ ∞ a n ( x − x 0 ) n = � � na n ( x − x 0 ) n − 1 ⇒ f ′ ( x ) = If f ( x ) = n =0 n =1 Let us rewrite the series for f ′ ( x ) in powers of ( x − x 0 ) n . Put r = n − 1 , we get ∞ � ( r + 1) a r +1 ( x − x 0 ) r f ′ ( x ) = r =0 Similarly, ∞ f ( k ) ( x ) = � n ( n − 1) . . . ( n − k + 1) a n ( x − x 0 ) n − k n = k ∞ � ( n + k )( n + k − 1) . . . ( n + 1) a n + k ( x − x 0 ) n = n =0   ∞ b n ( x − x 0 ) n − k = � � b n + k ( x − x 0 ) n In general,   n = n 0 n = n 0 − k 21 / 47

Example ∞ a n x n . Write ( x − 1) f ′′ as a power series around 0. � Let f ( x ) = n =0 22 / 47

Example ∞ a n x n . Write ( x − 1) f ′′ as a power series around 0. � Let f ( x ) = n =0 ( x − 1) f ′′ = xf ′′ − f ′′ 22 / 47

Example ∞ a n x n . Write ( x − 1) f ′′ as a power series around 0. � Let f ( x ) = n =0 ( x − 1) f ′′ = xf ′′ − f ′′ � ∞ � ∞ � n ( n − 1) a n x n − 2 � n ( n − 1) a n x n − 2 = x − n =2 n =2 22 / 47

Example ∞ a n x n . Write ( x − 1) f ′′ as a power series around 0. � Let f ( x ) = n =0 ( x − 1) f ′′ = xf ′′ − f ′′ � ∞ � ∞ � n ( n − 1) a n x n − 2 � n ( n − 1) a n x n − 2 = x − n =2 n =2 ∞ ∞ n ( n − 1) a n x n − 1 − � � n ( n − 1) a n x n − 2 = n =2 n =2 22 / 47

Example ∞ a n x n . Write ( x − 1) f ′′ as a power series around 0. � Let f ( x ) = n =0 ( x − 1) f ′′ = xf ′′ − f ′′ � ∞ � ∞ � n ( n − 1) a n x n − 2 � n ( n − 1) a n x n − 2 = x − n =2 n =2 ∞ ∞ n ( n − 1) a n x n − 1 − � � n ( n − 1) a n x n − 2 = n =2 n =2 ∞ ∞ ( n + 1) na n +1 x n − � � ( n + 2)( n + 1) a n +2 x n = n =1 n =0 22 / 47

Example ∞ a n x n . Write ( x − 1) f ′′ as a power series around 0. � Let f ( x ) = n =0 ( x − 1) f ′′ = xf ′′ − f ′′ � ∞ � ∞ � n ( n − 1) a n x n − 2 � n ( n − 1) a n x n − 2 = x − n =2 n =2 ∞ ∞ n ( n − 1) a n x n − 1 − � � n ( n − 1) a n x n − 2 = n =2 n =2 ∞ ∞ ( n + 1) na n +1 x n − � � ( n + 2)( n + 1) a n +2 x n = n =1 n =0 ∞ � [( n + 1) na n +1 − ( n + 2)( n + 1) a n +2 ] x n = n =0 22 / 47

Example (Solving ODE) Suppose ∞ � a n ( x − 1) n y ( x ) = n =0 for all x in an open interval I containing x 0 = 1 . 23 / 47

Example (Solving ODE) Suppose ∞ � a n ( x − 1) n y ( x ) = n =0 for all x in an open interval I containing x 0 = 1 . Find the power series of y ′ and y ′′ in terms of x − 1 in the interval I . Use these to express the function (1 + x ) y ′′ + 2( x − 1) 2 y ′ + 3 y as a power series in x − 1 on I . 23 / 47

Example (Solving ODE) Suppose ∞ � a n ( x − 1) n y ( x ) = n =0 for all x in an open interval I containing x 0 = 1 . Find the power series of y ′ and y ′′ in terms of x − 1 in the interval I . Use these to express the function (1 + x ) y ′′ + 2( x − 1) 2 y ′ + 3 y as a power series in x − 1 on I . Find necessary and sufficient conditions on the coefficients a n ’s, so that y ( x ) is a solution of the ODE (1 + x ) y ′′ + 2( x − 1) 2 y ′ + 3 y = 0 23 / 47

Example (Continue . . . ) Solution. Write the ODE in ( x − 1) , that is (1 + x ) y ′′ + 2( x − 1) 2 y ′ + 3 y = ( x − 1) y ′′ + 2 y ′′ + 2( x − 1) 2 y ′ + 3 y 24 / 47

Example (Continue . . . ) Solution. Write the ODE in ( x − 1) , that is (1 + x ) y ′′ + 2( x − 1) 2 y ′ + 3 y = ( x − 1) y ′′ + 2 y ′′ + 2( x − 1) 2 y ′ + 3 y Express each of ( x − 1) y ′′ , 2 y ′′ , 2( x − 1) 2 y ′ and 3 y as a power series in powers of ( x − 1) and add them. 24 / 47

Example (Continue . . . ) Solution. Write the ODE in ( x − 1) , that is (1 + x ) y ′′ + 2( x − 1) 2 y ′ + 3 y = ( x − 1) y ′′ + 2 y ′′ + 2( x − 1) 2 y ′ + 3 y Express each of ( x − 1) y ′′ , 2 y ′′ , 2( x − 1) 2 y ′ and 3 y as a power series in powers of ( x − 1) and add them. ∞ ( x − 1) y ′′ = ( x − 1) � n ( n − 1) a n ( x − 1) n − 2 n =2 24 / 47

Example (Continue . . . ) Solution. Write the ODE in ( x − 1) , that is (1 + x ) y ′′ + 2( x − 1) 2 y ′ + 3 y = ( x − 1) y ′′ + 2 y ′′ + 2( x − 1) 2 y ′ + 3 y Express each of ( x − 1) y ′′ , 2 y ′′ , 2( x − 1) 2 y ′ and 3 y as a power series in powers of ( x − 1) and add them. ∞ ( x − 1) y ′′ = ( x − 1) � n ( n − 1) a n ( x − 1) n − 2 n =2 ∞ � n ( n − 1) a n ( x − 1) n − 1 = n =2 24 / 47

Example (Continue . . . ) Solution. Write the ODE in ( x − 1) , that is (1 + x ) y ′′ + 2( x − 1) 2 y ′ + 3 y = ( x − 1) y ′′ + 2 y ′′ + 2( x − 1) 2 y ′ + 3 y Express each of ( x − 1) y ′′ , 2 y ′′ , 2( x − 1) 2 y ′ and 3 y as a power series in powers of ( x − 1) and add them. ∞ ( x − 1) y ′′ = ( x − 1) � n ( n − 1) a n ( x − 1) n − 2 n =2 ∞ � n ( n − 1) a n ( x − 1) n − 1 = n =2 ∞ � ( n + 1) na n +1 ( x − 1) n = n =1 24 / 47

Example (Continue . . . ) Solution. Write the ODE in ( x − 1) , that is (1 + x ) y ′′ + 2( x − 1) 2 y ′ + 3 y = ( x − 1) y ′′ + 2 y ′′ + 2( x − 1) 2 y ′ + 3 y Express each of ( x − 1) y ′′ , 2 y ′′ , 2( x − 1) 2 y ′ and 3 y as a power series in powers of ( x − 1) and add them. ∞ ( x − 1) y ′′ = ( x − 1) � n ( n − 1) a n ( x − 1) n − 2 n =2 ∞ � n ( n − 1) a n ( x − 1) n − 1 = n =2 ∞ � ( n + 1) na n +1 ( x − 1) n = n =1 ∞ � ( n + 1) na n +1 ( x − 1) n = n =0 24 / 47

Example (Continue . . . ) ∞ 2 y ′′ = � 2 n ( n − 1) a n ( x − 1) n − 2 n =2 25 / 47

Example (Continue . . . ) ∞ 2 y ′′ = � 2 n ( n − 1) a n ( x − 1) n − 2 n =2 ∞ � 2( n + 2)( n + 1) a n +2 ( x − 1) n = n =0 25 / 47

Example (Continue . . . ) ∞ 2 y ′′ = � 2 n ( n − 1) a n ( x − 1) n − 2 n =2 ∞ � 2( n + 2)( n + 1) a n +2 ( x − 1) n = n =0 ∞ 2( x − 1) 2 y ′ = 2( x − 1) 2 � na n ( x − 1) n − 1 n =1 25 / 47

Example (Continue . . . ) ∞ 2 y ′′ = � 2 n ( n − 1) a n ( x − 1) n − 2 n =2 ∞ � 2( n + 2)( n + 1) a n +2 ( x − 1) n = n =0 ∞ 2( x − 1) 2 y ′ = 2( x − 1) 2 � na n ( x − 1) n − 1 n =1 ∞ � 2 na n ( x − 1) n +1 = n =1 25 / 47

Example (Continue . . . ) ∞ 2 y ′′ = � 2 n ( n − 1) a n ( x − 1) n − 2 n =2 ∞ � 2( n + 2)( n + 1) a n +2 ( x − 1) n = n =0 ∞ 2( x − 1) 2 y ′ = 2( x − 1) 2 � na n ( x − 1) n − 1 n =1 ∞ � 2 na n ( x − 1) n +1 = n =1 ∞ � 2( n − 1) a n − 1 ( x − 1) n = n =2 25 / 47

Example (Continue . . . ) ∞ 2 y ′′ = � 2 n ( n − 1) a n ( x − 1) n − 2 n =2 ∞ � 2( n + 2)( n + 1) a n +2 ( x − 1) n = n =0 ∞ 2( x − 1) 2 y ′ = 2( x − 1) 2 � na n ( x − 1) n − 1 n =1 ∞ � 2 na n ( x − 1) n +1 = n =1 ∞ � 2( n − 1) a n − 1 ( x − 1) n = n =2 ∞ � 2( n − 1) a n − 1 ( x − 1) n = ( a − 1 = 0) n =0 25 / 47

Example (Continue . . . ) We have 26 / 47

Example (Continue . . . ) We have ∞ ( x − 1) y ′′ = � ( n + 1) na n +1 ( x − 1) n n =0 26 / 47

Example (Continue . . . ) We have ∞ ( x − 1) y ′′ = � ( n + 1) na n +1 ( x − 1) n n =0 ∞ 2 y ′′ = � 2( n + 2)( n + 1) a n +2 ( x − 1) n n =0 26 / 47

Example (Continue . . . ) We have ∞ ( x − 1) y ′′ = � ( n + 1) na n +1 ( x − 1) n n =0 ∞ 2 y ′′ = � 2( n + 2)( n + 1) a n +2 ( x − 1) n n =0 ∞ 2( x − 1) 2 y ′ = � 2( n − 1) a n − 1 ( x − 1) n ( a − 1 = 0) n =0 Now we get 26 / 47

Example (Continue . . . ) We have ∞ ( x − 1) y ′′ = � ( n + 1) na n +1 ( x − 1) n n =0 ∞ 2 y ′′ = � 2( n + 2)( n + 1) a n +2 ( x − 1) n n =0 ∞ 2( x − 1) 2 y ′ = � 2( n − 1) a n − 1 ( x − 1) n ( a − 1 = 0) n =0 Now we get ∞ ( x − 1) y ′′ + 2 y ′′ + 2( x − 1) 2 y ′ + 3 y = � b n ( x − 1) n n =0 26 / 47

Example (Continue . . . ) We have ∞ ( x − 1) y ′′ = � ( n + 1) na n +1 ( x − 1) n n =0 ∞ 2 y ′′ = � 2( n + 2)( n + 1) a n +2 ( x − 1) n n =0 ∞ 2( x − 1) 2 y ′ = � 2( n − 1) a n − 1 ( x − 1) n ( a − 1 = 0) n =0 Now we get ∞ ( x − 1) y ′′ + 2 y ′′ + 2( x − 1) 2 y ′ + 3 y = � b n ( x − 1) n n =0 where 26 / 47