1 $10.2 - Taylor series (p. 551, Q32) Use the fact that the Taylor series of g(x) = sin(x2) is -10 -14 x 2 _ + 3! 5! 7! to find g"(0), g"(0) and g(10)(0). (There is an easy way and a hard way to do this - choose wisely!) 2 $10.3 - Using Taylor series (p. 558, Q40) One of Einstein's most amazing predictions was that light traveling from distant stars would bend around the sun on the way to earth. His calculations involved solving for in the equation sing + b(1 + cos $ + cos $) = 0, where b is a very small positive constant. (a) Explain why the equation could have a solution for o that is near 0. (b) Expand the left-hand side of the equation in Taylor series about o = 0, disregard- ing terms of order q and higher. Solve for $. (Your answer will involve b.)9 ( x ) = sincu ) = x x 6 y + 3 1 5 ! 7 1 g ' ( x ) = 2x - 62, + 102 13 14x 3 ! 4 - - - 5 ! 7 ! 9"( x) = 2 - 30x + 9 0 x 19 182x 3 ? 4 - - 5 1 7 1 9"1 (0) = 2 3 gill ( x ) = 0 - 120x + 720x 2184x 3 1 7 ! 9lll lo] = 6 9lo (o ) = 10 ! 51 = 30240Step 1! 2 Explanation: first we substitute (j) directly, Next using the taylor's series of sine, cosine functions we nd answer Solution: given equation sin (in + b(1+ cosgq + cos (in) = 0 a} if qb ) 0 substitute in equation LHS=5iJJO + b(1 + (3052(0) + cos [0} =U+b[1+l+1] =3b implies 3km: but b is small positive constant which is satisfies the equation there fore, the equation have a solution for gift that is near 0 Step 2/2 b) given equation sin d + b(1 + cos'd + coso) = 0 now, we write the Taylor series for sine, cosine about $ = 0 Taylor series at a=0 is f(x)=f(0)+ - W x _ _(0) f"(0) 2! X + - 3! X + . . . . . . here a=0 given of and higher powers neglect sin 0 = 0 - - $5 3! + + . ... then sino ~ o cos 0 = 1 - + T. . then cos 0 1 cos'd = cos . cos d = (1 + 4 ! Fil + . . . . . ). 1 2! + 4 ! - 6! + . .... cos'd = 1.1 ~1 substitute in equation o+b(1+1+1)=0 then solution of the equation for d is d ~ -3bFinal answer a} the equation have a solution for cf: that is near 0 b] solution of the equation for q'a is d) 2:: 3b