MAC 2312 Calculus II Sanchez

MAP 2302 Differential Equations Sanchez Summary-Power Series Solutions

Power Series: a series of the form  n 2 3 n an(x  a)  a0  a1(x  a)  a2(x  a)  a3(x  a)    an (x  a)     n0 is called a power series centered at a. The number “a” is called the center of the series.

Theorem: for any power series, exactly one of the following is true: a) the series converges only for x=a b) the series converges absolutely for all x c) the series converges absolutely for all x in some finite open interval (a-r, a+r), and diverges if xa+r. At the points x=a-r and x=a+r the series may converge absolutely, converge conditionally, or diverge, depending on the particular series. r is called the radius of convergence of the series. If the series converges only for x=a, r=0. If the series converges for all x, then r  

Absolute convergence. Within the interval of convergence a power series converges absolutely.  n In other words, for a r  x  a  r,  an x  a converges. n0

Definition of Analytic functions at a point. A function f is analytic at a point x=a if it can be represented by a power series in x-a with a positive radius of convergence. Example. The following functions are analytic :sin x, cos x, ex , ln(x  1)

Definition: the set of values of x for which a power series converges is called the interval of convergence.

How to find the interval of convergence and the radius of convergence of a power series a Step 1. Solve the inequality lim n1 1 for x. n an Step 2. Test the end point of the interval for convergence

Techniques for finding Maclaurin and Taylor series for functions: 1. Taylor’s theorem 2. Substitution in a known series 3. Differentiation of a known series does not change the radius of convergence, except at the end-points. 4. Integration of a known series does not change the radius of convergence, except at the end- points. 5. Long division of series. 6. Multiplication of series. 7. The binomial theorem expansion. 8. Synthetic division.

-1- Problem 1. Find the radius of convergence and the interval of convergence of the following power series.  xn a)  n1 n xn1 n1 a n  1 nx nx n lim n1  lim  lim  lim  lim x  x  1 1 x  1 n an n xn n n  1xn n n  1 n n  1 n   1n at x  1,  converges by alternating series test n1 n  1 at x  1,  diverges by p  series test, p  1/ 2  1 n1 n The int erval of convergence is  1 x  1, lenght  2 and radius  1

Problem 2. Find the radius of convergence and the interval of convergence of the following  1n1xn power series:  3 n1 n

xn1 a (n 1)3 n3xn1 n3 lim n1  lim  lim  lim x  x 1 1 x 1 n an n xn n (n  1)3 xn n (n 1)3 n3  1n1(1)n  1 at x  1,  converges absolutely by p  series, p  3,therefore it converges  3  3 n1 n n1 n  1n1 at x 1, converges, because it converges absolutely  3 n1 n The int erval of convergence is 1 x 1, lenght  2 and radius 1

Problem 3. Find the radius of convergence and the interval of convergence of the following  xn power series:  n! n1 xn1 a (n  1)! n!xn1 n! 1 lim n1  lim  lim  lim x  lim x  0  1 for all x n an n xn n (n  1)!xn n (n  1)n! n n  1 n!  The int erval of convergence is    x   lenght   and radius  

-2- Problem 4. Find the radius of convergence and the interval of convergence of the following power series.  a)  n3(x  5)n n1 3 a (n 1)3 (x  5)n1  n 1 lim n1  lim  lim   x  5  x  5 1 1 x  5 1 4  x  6 n an n (n)3 (x  5)n n n   at x  4,  n3(1)2 diverges by n  th term test n1  at x  5,  n3 diverges by n  nth term test n1 The int erval of convergence is 4  x  6, lenght  2 and radius 1

Problem 5. Find the radius of convergence and the interval of convergence of the following  (3x  2)n power series:  n n1 n  3 (3x  2)n1 a (n 1)  3n1  n  3n (3x  2)n1   n  (3x  2)  1 lim n1  lim  lim    lim    3x  2 1 a n  n1 n   n  1  3  3 n n n (3x  2) n n  1 3 (3x  2)  n    (n)  3n 1 5  3x  2  3  3  3x  2  3  1 3x  5    x  3 3 1  (3x  2)n  (3)n  (1)n at x   ,   converges by alternating series test 3  n  n  (n) n1 (n)  3 n1(n)  3 n1 5  (3x  2)n  (3)n  1 at x  ,   divergess by p  series test, p 1 3  n  n  (n) n1 (n)  3 n1(n)  3 n1 1 1 The int erval of convergence is   x   , lenght  2 and radius 1 3 3

Theorem: Taylor’s series expansion. If a function f has a power-series expansion in powers of x-a and all derivatives of f exist in some interval containing a and if term-by-term differentiation is valid, then  ( n ) f ( a ) n f ( x )   ( x  a ) for all x in the interval. n  0 n ! The series is called the Taylor’s series for f about a. A Taylor’s series expansion about a=0 is called a Maclaurin’s series expansion.

-3- n ( k ) f ( a ) k Definition: the polynomial P n ( x )   ( x  a ) is called the Taylor polynomial of k  0 k ! degree n for f(x) about x=a. If a=0, it is called the Maclaurin polynomial

Some important Maclaurin Series Interval of convergence -1

Theorem : eix  cos x  i sin x where i  1  xn x x2 x3 ix i2 x2 i3x3 i4 x4 i5x5 i6 x6 Pr oof : e x  1     ...  eix 1      ...  n! 1! 2! 3! 1! 2! 3! 4! 5! 6! n0 ix  x2  ix3 x4 ix5 1x6  eix 1       1! 2! 3! 4! 5! 6! x2 x4 x6  x x3 x5   eix 1     ...  i    ...  eix  cos x  i sin x 2! 4! 6! 1! 3! 5!   

Theorem- Euler’s formula: ei  1 Proof: eix  cos x  i sin x  ei  cos  i sin  ei  1

-4- eix  eix eix  eix e x  ex e x  ex Theorem: sin x  and sin x  Note :sinh x  and sin x  2i 2 2 2  eix  cos x  i sin x  eix  cos x  i sin x  adding  2cos x  eix  eix     ix  ix  ix ix e  cos x  i sin x e  cos x  i sin x subtracting  2i sin x  e  e  eix  eix cos x   2   eix  eix sin x   2i

Problem 6. Use power series to solve the IVP DE y  xy - 2y  0, y(0)  1, y(0)  0  n 2 3 Let y   a n x  a0  a1x  a2x  a3x  ... n0   Nn1   n1 n1 N n y   annx   annx   aN1N  1x   a n 1(n  1)x n0 n1 N0 n0   Nn 2  n 2 n 2 N y   ann(n  1)x   ann(n  1)x   aN 2 N  2(N  1)x n1 n 2 N0  n   an 2 n  2(n  1)x n0    n n1 n y  xy - 2y  0   an 2 n  2(n  1)x  x  annx  2  a n x  0 n0 n1 n0    n n n   an 2 n  2(n  1)x   annx  2  a n x  0 n0 n0 n0

 n   an 2 n  2(n  1)  nan  2an x n0

 n   an 2 n  2(n  3)  (n  2)an x  0 n0  an 2 n  2(n  1)  (n  2)an  0 n  2  a   a n 2 (n  2)(n  1) n but a0  1, a1  0 0  2  For n  0, a2   a0  1(1)  1 (0  2)(0  1) 1  2 2  2  n  1 a3   a1  0; n  2  a4   a2  0 (1  2)(1  1) (2  2)(2  1)  an  0 for n  3 2 3 2 Therefore : y  a0  a1x  a2x  a3x  ...  1 x

-5- Problem 7. Solve the IVP 1- x2 y  2xy  2y  0, y(0)  0, y(0)  1  n 2 3 Let y   a n x  a0  a1x  a2x  a3x  ... n0   Nn1   n1 n1 N n y   annx   annx   aN1N  1x   a n 1(n  1)x n0 n1 N0 n0

  Nn 2  n 2 n 2 N y   ann(n  1)x   ann(n  1)x   aN 2 N  2(N  1)x n0 n 2 N0  n   an 2 n  2(n  1)x n0 1- x2 y  2xy  2y  0  y  x2y  2xy  2y  0     n 2 n 2 n1 n   an 2 n  2(n  1)x  x  ann(n  1)x  2x  annx  2  a n x  0 n0 n0 n0 n0

    n n n n   an 2 n  2(n  1)x   ann(n  1)x  2  annx  2  a n x  0 n0 n0 n0 n0

 n   an 2 n  2(n  1)  ann(n  1)  2ann  2an x  0 n0  an 2 n  2(n  1)  ann(n  1)  2ann  2an  0 n(n  1)  2n  2 n2  n  2 (n  2)(n  1)  an 2  an  an 2  an  an 2  an (n  2)(n  1) (n  2)(n  1) (n  2)(n  1)

n  1  an 2  an n  1 0  1 1 1 2  1 y(0)  0, y(0)  1 a  0 and a  1, a  a  0, a  a  0,a  a  0 0 1 2 0  1 0 3 1 1 1 4 2  1 2

Therefore an  0 for n  2 2 3 Solution : y  a0  a1x  a2x  a3x  ...  0  1x  0  0  ... y  x

-6- Problem 8. Solve the Hermite DE y - 2xy  2y  0, y(0)  0 and y(0)  2  n 2 3 Let y   a nx  a0  a1x  a2x  a3x  ... n 0

  Nn1   n1 n1 N n y   annx   annx   aN1N  1x   a n 1(n  1)x n 0 n1 N 0 n 0

  Nn 2  n 2 n 2 N y   ann(n  1)x   ann(n  1)x   aN 2 N  2(N  1)x n 0 n 2 N 0

 n   an 2 n  2(n  1)x n0 y - 2xy  2y  0    n n1 n   an 2 n  2(n  1)x  2x  annx  2  a nx  0 n0 n0 n 0

   n n n   an 2 n  2(n  1)x  2  annx  2  a nx  0 n0 n 0 n 0

 n   an 2 n  2(n  1)  2nan  2an x  0 n0

 an 2 n  2(n  1)  2nan  2an  0 2n  2 2n  1  a  a  a  a n 2 n  2(n  1) n n 2 n  2(n  1) n

2(0  1) y(0)  0 and y(0)  2  a  0, a  2, a  a  0 0 1 2 (0  2)(0  1) 0

2(1  1) 2(2  1) a  a  0, a  a  0 3 (1  2)(1  1) 1 4 (2  2)(2  1) 2 2 3 Answer : y  a0  a1x  a2x  a3x  ...  0  2x  0  0  0  ... y  2x

-7- Problem 9. Ue Power series to find the IVP solution of y  2(x  2)y  4y  0, y(2)  1, y(2)  0  n 2 3 Let y   a n x  2  a0  a1x  2 a2 x  2  a3 x  2  . . . n0

  Nn1   n1 n1 N n y   annx  2   annx  2   aN1N  1x  2   a n1(n 1)x  2 n0 n1 N0 n0

  Nn2  n2 n2 N y   ann(n 1)x  2   ann(n 1)x  2   aN2 N  2(N 1)x  2 n0 n2 N0

 n   an2 n  2(n  1)x  2 n0 y  2(x  2)y  4y  0     n n1 n   an2 n  2(n 1)x  2  2(x  2)  annx  2  4  a n x  2  0 n0 n0 n0

   n n n   an2 n  2(n 1)x  2  2  annx  2  4  a n x  2  0 n0 n0 n0

 n   an2 n  2(n  1)  2nan  4an (x  2)  0 n0  an2 n  2(n  1)  2nan  4an  0 2n  4 2n  2  an2  an  an2  an (n  2)(n 1) (n  2)(n  1) y(2)  1, y(2)  0  a0  1, a1  0 20  2 21 2 n  0  a2  a0  2(1)  2; n  1  a3  a1  0 (0  2)(0 1) (1 2)(1 1) 22  2 n  2  a4  a2  0 (2  2)(2 1)  an  0 for n  3 2 3 y  a0  a1x  2 a2 x  2  a3 x  2  . . .  y  1 0(x  2)  2(x  2)2  y  1 2x2  8x  8  y  2x2  8x  7

-8- Problem 7. In problem 6 use the reduction of order technique to find the general solution of y  xy - 2y  0 Solution : 2 y1  1  x is a particular solution of the differential equation. 2 2 2 2 y  vy1  v(1  x )  v  x v  y  v  2xv  x v and y  v  2xv  2v  2xv  x v  x2v  v  2v  4xv y  xy - 2y  0  x2v  v  2v  4xv  x(v  2xv  x2v)  2v  x2v  vx2  1 v(x3  5x)  v(0)  1  x2 v  x3  5xv  0 dw x3  5x w  v  1  x2 w  x3  5xw  0   dx  0 w x2  1 2 dw  4x  x 2   x  dx  0  ln w   2ln(x  1)  c w  x2  1 2 x 2 x 2  2   2 2  x 2 2 Ce 2  lnwx  1     c  wx  1  Ce 2  w  2 2   x2  1 x 2  x 2      y Ce 2 2 Ce 2  v    dx  y  (1  x )  dx 1 x2 2 2  2 2  x  1  x  1     x 2    2  2 2  Ce  The general solution is y  c11  x  c2 (1  x ) dx  2 2   x  1   