e y [cos(x) + i sin(x)] e y [cos(x) i sin(x)] ] sin(x) + ey e y x = nπ for n = 0, ±1, ±2,... cos(nπ) = ey e y 0 = ey e y sin(z) = 0,

Transcription

1 Worked Solutions 83 Chapter 3: Power Series Solutions II: Generalizations Theory 34 a Suppose that e z = 0 for some z = x + iy Then both the real imaginary parts of e z must be zero, e x cos(y) = 0 e x sin(y) = 0, since e x = 0 for every real value x, we must have cos(y) = 0 sin(y) = 0 But this is impossible After all, if sin(y) = 0 for some real value y, then y = nπ for some integer n, which then means that cos(y) = cos(nπ) = ( ) n = 0 So it is impossible for e z = 0 In other words, e z = 0 for every z in the complex plane 34 b We will just verify the claims concerning sin(z) First of all, sin(x + iy) = ei(x+iy) e i(x+iy) i = e i x y i e x+y] i = e y e i x e y i e x] i = e y cos(x) + i sin(x)] e y cos(x) i sin(x)] ] i = i e y + e y] sin(x) e y e y] cos(x) ] i = i i ey + e y sin(x) i ey e y cos(x) = ey + e y sin(x) i i i ey e y cos(x) = ey + e y sin(x) + ey e y cos(x) Next, assume sin(z) = 0 for some z = x + iy Then both the real part imaginary parts of sin(x + i y) must equal 0, which, from above, means e y + e y sin(x) = 0 Note that since e y + e y > 0, But then e y e y cos(x) = 0 e y + e y sin(x) = 0 sin(x) = 0 x = nπ for n = 0, ±, ±, 0 = ey e y cos(x) = ey e y cos(nπ) = ey e y ( ) n, which means that e y e y = 0 Solving this for y yields y = 0 So, if sin(z) = 0,

2 84 Power Series Solutions II: Generalizations Theory then z = x + iy = nπ + i0 = nπ with n = 0, ±, ±, 34 c This is done in much the same way as the previous Start by showing sinh(x + iy) = ex+iy e (x+iy) = e x cos(y) + i sin(y)] e x cos(y) i sin(y)] ] = = ex e x cos(y) + i ex + e x sin(y) Next, assume sinh(z) = 0 for some z = x + iy Then both the real part imaginary parts of sinh(x + i y) must equal 0, which, from above, means e x e x cos(y) = 0 e x + e x sin(y) = 0 Solving this pair of equations then yields z = x + iy = i n + ] π with n = 0, ±, ±, 35 a Both coefficients e x are analytic,, from exercise 34, we know neither is ever zero So, lemma 34 assures us that there are no singular points Hence, also, the radius of analyticity R is 35 c All three coefficients sin(x), x e x are analytic everywhere, (from exercise 34) we know e z = 0 for every z in the complex plane, while sin(z) = 0 z = 0, ±π, ±π, ±3π, Lemma 35 then assures us that the singular points are all given by z s = 0, ±π, ±π, ±3π,, since z s = π is the singular point closest to x 0 =,, R = π = π 35 e The coefficients are all analytic on the entire complex plane, (from exercise 34) we know sinh(z) = 0 z = inπ for n = 0, ±, ±, ±3, ( ) The other coefficients z sin(x) are clearly nonzero at all these points except z = 0 So we need to check the point z = 0 more carefully, by computing x lim x 0 sinh(x) sin(x) lim x 0 sinh(x)

3 Worked Solutions 85 Using L Hôpital s rule, we have lim x 0 sin(x) lim x 0 sinh(x) x sinh(x) = lim x 0 cos(x) = lim x 0 cosh(x) x cosh(x) = 0 = 0 = = Since both of these limits are finite, lemma 35 tells us that the point x = z = 0 is an ordinary point Hence, all the values of z given in line ( ) are singular points except z = 0 So, z s = inπ with n = ±, ±, Since the singular points closest to x 0 = are ±iπ, R = (±iπ) = + π = 4 + π 35 g Multiplying through by e x, we get e x ] y + + e x] y = 0 Both coefficients are analytic on the entire complex plane,, letting z = x + iy, e z = 0 = e z + i0 = e x+iy = e x cos(y) + i sin(y)] = e x cos(y) 0 = e x sin(y) = e x cos(y) y = nπ for n = 0, ±, ±, = e x cos(nπ) = e x ( ) n for n = 0, ±, ±, x = 0 n is an even integer z = x + iy = 0 + ikπ for k = 0, ±, ±, ±3, ( ) Checking, we see that, if z = ikπ for any integer k, + e z = + e ikπ = + cos(kπ) + i sin(kπ) = = 0 Hence, all the points given in line ( ) are singular points, z s = ikπ for k = 0, ±, ±, ±3,, R = x 0 closest singular point = 3 0i = 3

4 86 Power Series Solutions II: Generalizations Theory 35 i All the coefficients are analytic on the entire complex plane, with the first being zero if only if x = z = 0 But we also have e z = 0 if z = 0 So we need to see if is finite Using L Hôpital s rule, we see that e lim x x 0 x e lim x x 0 x e = lim x x 0 =, which is finite, telling us that z = 0 is an ordinary point Hence, there are no singular points, R = 36 a The coefficients of this equation e x are analytic everywhere, the first is simply a nonzero constant So there are no singular points the power series solution will be valid on I = (, ) The solution is k=0 a k x k with a 0 being arbitrary a k k = p k j for k > 0 k where, in this case, p n x n = P(x) = e x = n! xn = n! xn So, p n = n! for n = 0,,, 3,, a k k = p k j k = k k (k j)! = k k (k j)! In particular, a = ( j)! = 0 (0 j)! = a 0 (0 0)! = a 0, a = ( j)! = = ( j)! a 0 ] ( 0)! + a = ( )! a 0 + a 0 ] = a 0,

5 Worked Solutions 87 a 3 = 3 3 (3 j)! = 3 ( j)! = ] a 0 3 ( 0)! + a ( )! + a = ( )! 3 a 4 = 4 4 (4 j)! 3 = 4 = 4 = 4 (3 j)! a 0 a 0 a 0 ] (3 0)! + a (3 )! + a (3 )! + a 3 (3 3)! ( ) ] 6 + a 0 + a a 0 = 5 8 a 0 ] + a 0 + a 0 = 5 6 a 0 Thus, the 4 th -degree partial sum of the general power series solution is 4 a k x k = a 0 + a x + a x + a 3 x 3 + a 4 x 4 k=0 = a 0 + a 0 x + a 0 x a 0x a 0x 4 = a 0 + x + x x x4 ] 36 c The coefficients of this equation cos(x) are analytic everywhere, the first is simply a nonzero constant So there are no singular points the power series solution will be valid on I = (, ) The solution is k=0 a k x k with a 0 being arbitrary a k k = p k j for k > 0 k where, in this case, p n x n = P(x) = cos(x) = ( ) m (m)! xm m=0 = x 0! x + 4! x4 6! x6 +

6 88 Power Series Solutions II: Generalizations Theory = ( ) 0 / x 0 + 0x + ( ) /! x + 0x 3 + ( ) 4 / 4! x4 + 0x 5 + ( ) 6 / 6! x6 + 0x 7 + So, p 0 =, p = 0, p =! =, p 3 = 0, in general, p n = { ( ) n/ if n is even n! 0 if n is odd Applying the above formulas for a k p k, we get: a = p j 0 = p 0 j = a 0 p 0 = a 0 = a 0, a = p j = p j = a 0 p + a p 0 ] = a 0 0 a 0 ] = a 0, a 3 = 3 p 3 j 3 = 3 a 0 p + a p + a p 0 ] = ( ) a 0 a ] 3 a 0 a 4 = 4 p 4 j 4 = 0 = 4 a 0 p 3 + a p + a p + a 3 p 0 ] = ( ) a ( a 0 ) + ] 4 a = 8 a 0

7 Worked Solutions 89 Thus, the 4 th -degree partial sum of the general power series solution is 4 a k x k = a 0 + a x + a x + a 3 x 3 + a 4 x 4 k=0 = a 0 a 0 x + a 0x + 0a 0 x 3 8 a 0a 0 x 4 = a 0 x + x 8 x4] 37 a The equation is in reduced form y + Py + Qy = 0 with P(x) = 0 = Q(x) = e x = 0x n for x < n! xn for x < Since the radii of convergence for each of these two series is, so is the radius of convergence for the power series solution Thus, the power series solution will be valid on I = (, ) Moreover, by the above, p k x k = 0x n q n x n = n! xn, which means In particular, let s note that p n = 0 for n = 0,,, 3, q n = n! = 0,,, 3, q 0 = 0! =, q =! = q =! = The power series solution is given by k=0 a k x k where a 0 a are arbitrary, a k = k ] ( j + )+ p k j + q k j for k > k(k ) In this case, since every p n is 0, the above recursion formula simplifies to a k = k q k j for k > k(k )

8 90 Power Series Solutions II: Generalizations Theory Applying this recursion formula, we get a = q j ( ) = 0 q j = a 0q 0 = a 0( ) = a 0, a 3 = 3 q 3 j 3(3 ) = q j 6 = 6 a 0q + a q 0 ] = 6 a 0( ) + a ( )] = 6 a 0 + a ] a 4 = 4 q 4 j 4(4 ) = q j = a 0q + a q + a q 0 ] = ( ) a 0 + a ( ) + ] a 0( ) = a 0 + a ] Thus, the 4 th -degree partial sum of the general power series solution is 4 a k x k = a 0 + a x + a x + a 3 x 3 + a 4 x 4 k=0 = a 0 + a x + a 0x + 6 a 0 + a ] x 3 + a 0 + a ]x 4 = a 0 + x + 6 x3 + x4] + a x + 6 x3 + x4] 37 c Dividing through by x, we get the equation into reduced form, which is y 3y + sin(x) y = 0, x y + Py + Qy = 0 with P(x) = 3 Q(x) = sin(x) x Obviously, P Q are analytic everywhere except, possibly, for a singularity at x = 0 for Q(x) (because of the x in the denominator) But, using L Hôpital s rule, we see that sin(x) lim Q(x) = lim x 0 x 0 x cos(x) = lim x 0 =,

9 Worked Solutions 9 which is finite So Q(x) is analytic at x = 0, as well as everywhere else Hence, there are no singular points, I = ( in f ty, ) The power series for P Q about x 0 = 0 are easily found For P, we trivially have p n x n = P(x) = 3 = 3 + 0x + 0x + 0x 3 + which means p 0 = 3 while p n = 0 for n For Q, we use the well-known series for the sine as follows: q n x n = Q(x) = x sin(x) Hence, In general, = x m=0 ( ) m (m + )! xm+ = x x 3! x3 + 5! x5 ] 7! x7 + = x 0 3! x + 5! x4 7! x6 + = x 0 + 0x 3! x + 0x 3 5! x4 + 0x 5 7! x6 + q 0 =, q = 0, q = 3! = 6 ( ) n / if n is even q n (x) = n + 0 if n is odd Using this with recursion formula (34), we get a = ( ) ] ( j + )+ p j + q j = 0 ] ( j + )+ p 0 j + q 0 j = ] a0+ p a 0 q 0 0 = a p 0 + a 0 q 0 ] = a ( 3) + a 0 ] = 3a a 0 ], a 3 = 3(3 ) 3 ] ( j + )+ p 3 j + q 3 j = ] ( j + )+ p j + q j 6, q 3 = 0,

10 9 Power Series Solutions II: Generalizations Theory = 6 = ( 6 = 6 ( (0 + )a0+ p 0 + a 0 q 0 ] + ( + )a+ p + a q ] ) ) a p + a 0 q ] + a p 0 + a q 0 ] ( a 0 + a 0 0] + 3a a 0 ] ( 3) + a ]) = a a a 4 = 4(4 ) 4 ] ( j + )+ p 4 j + q 4 j = ] ( j + )+ p j + q j = ( ) a p + a 0 q ] + a p + a q ] + 3a 3 p 0 + a q 0 ] = ( ( )] a 0 + a 0 + a 0 + a 0] 6 ( + 3 a ) 3 a ( 3) + ] ) 3a a 0 ] = 3 7 a a Thus, the 4 th -degree partial sum of the general power series solution is 4 a k x k = a 0 + a x + a x + a 3 x 3 + a 4 x 4 k=0 = a 0 + a x + 3a a 0 ] x a ] 8 a x 4 a ] 3 a x 3 = a 0 x x3 3 7 x4] + a x + 3 x x x4] 37 e The first coefficient is 0 if only if x = 0, while the coefficient on y is never 0 From that it follows that z s = 0 is the only singular point, R = z s x 0 = 0 = I = (x 0 R, x 0 + R) = (0, ) Dividing the equation by the first coefficient letting Y(X) = y(x) with X = x converts the differential equation to Y + X + Y = 0 with X 0 = 0, which is Y + PY + QY = 0 with P(X) = 0 Q(X) = (X + ) /

11 Worked Solutions 93 Obviously, P(X) = p n X n with p n = 0 for all n Thus, the recursion formula for the power series of Y about X 0 reduces a bit, a k = k(k ) k ] ( j + )+ p k j + q k j = k q k j k(k ) For Q, we use the Taylor series, q n X n = Q(x) = Q (n) (0) X n n! So, q 0 = Q(0) (0) 0! q = Q (0)! q = Q (0)! q 3 = Q (0) 3! = Q(0) = (0 + ) / =, = Q (0) = (0 + ) 3 / = =! Q (0) =! 3 (0 + ) 5 / = 3 8 = 3! Q(3) (0) = 3! (0 + ) 7 / 5 = 6 Plugging the above into our recursion formula, we get,,, a = q j = ( ) a 0q 0 = a 0 = a 0, a 3 = 3 q 3 j 3(3 ) = 6 a 0q + a q 0 ] = 6 a 4 = 4 q 4 j 4(4 ) ( a 0 ) ] + a = a 0 6 a, = a 0q + a q + a q 0 ] = ( 3 ) ( a 0 + a ) ] 8 a 0 = 96 a a

12 94 Power Series Solutions II: Generalizations Theory Thus, the 4 th -degree partial sum of the general power series for Y(X) about X 0 = 0 is 4 a k X k = a 0 + a X + a X + a 3 X 3 + a 4 X 4 k=0 = a 0 + a X a 0 X + a 0 ] 6 a X a 0 + ] 4 a X 4 = a 0 X + X X 4] + a X 6 X X 4] Replacing X with x then gives us the 4 th -degree partial sum of the general power series solution about x 0 = to the original differential equation: 4 k=0 a k (x ) k = a 0 (x ) + (x )3 + (x )4] 96 + a (x ) 6 (x )3 + 4 (x )4]