EE1. Solutions of Problems 7

Transcription

1 EE1 Soution of Probem 7 1. a) Appy the Ratio Tet. Puta n = n th term. a n = x n. Fix x 0. = x n+1 x n = x x a n. By the Ratio Tet the erie x n converge if the at imit x i < 1 1) n=0 and diverge if the at imit x i > 1. ) Hence the radiu of convergence i 1, and the interva of convergence i 1 < x < 1. [ In the cae x = 0, x n = 0 for a n 1 o that the erie n=0 xn mut converge. Simiar happen in cae b) for x = 0, in cae b) for x = 100 and in cae c) forx =.] b) Same method a a). c) Appy the Ratio Tet. Put a n = n th term. a n = x 100)n n. Fix x 100. = x 100) n+1 n n+1) x 100) n = n = x 100 n+1) x 100 a n. Hence x 100) n n=1 n converge if the at imit x 100 i < 1 and diverge if the at imit x 100 i > 1. So the radiu of convergence i 1, and the interva of convergence i 99 < x < 101. d) Appy the Ratio Tet. Put a n = n th term. a n = x+)n. Fix x. = x+) n+1 n+1)! x+) n = x+ 0 a n. n+1 im a n+1 n a n = 0. So the erie converge for a vaue of x, giving radiu of convergence, with the interva of convergence coniting of ALL vaue of x.. fx) = n, f x) = 1 1+x, f x) = 1, f x) =, 3 f 4) x) = 3. 4 fx) = f0) + f 0)x + f 0) x + f 0) x3 + Ox 4 )

2 i.e. n = x x 3 + Ox4 ). where the remainder termr 4 = f 4) ξ) x4 for omeξ between0andx. 4! The given approximation i obtained by putting x = 1/ in the above and ignoring term of Ox 4 ). Then the error i wort cae 1/64 from f 4) ξ) = 6/1+ξ) 4 evauated at ξ = 0 wort cae). 3. Cacuating derivative ofyx) = inx up toy 5) = cox and ubtituting into yx) = y0) + y 0)x y 4) 0) x4 + Ox 5 ) 4! give inx x dx 1 0 Hence 1 0 i e than inx = x x3 + Ox 5 ). ) 1 x dx = = The error on thi 4. i) Find the firt and econd derivative, or expand uing the Binomia Serie. ii) Put y = tan 1 x. y = 1 + x ) 1. Serie, y = 1 x +x Integrate to obtain Then, uing the Binomia y = c+x x3 3 +x5..., where the contant termci0incey0) = 0. 5 iii) h = ecx = cox) 1, h = inxcox). h = coxcox) +in xcox) 3 = cox) 1 +1 co x)cox) 3, o h = cox) 3 cox) 1. h = 6inxcox) 4 inxcox). h 4) = 6coxcox) 4 +4in xcox) 5 coxcox) +in xcox) 3. At x = 0, h = 1. h = 0, h = 1, h 0, h 4) = 5, giving Macaurin erie h0)+h 0)x+ h 0) x +... = 1 + x x OR Mutipy the unknown erie forecx into the known erie forcox and equate coefficient of power ofx. 5. Firt part Step 1) : the cae n = 1: Dinx = d dx inx = cox, and inx+π/) = inx coπ/ + cox inπ/ = cox. So true forn = 1. Step ) : now aume true for ome integer n 1 :

3 D n+1 inx = DD n inx) = D inx+nπ/), uing at aumption, = cox+nπ/), = inbx+n+1)π/), ince inbx+n+1)π/) = inbx+nπ/) + π/) = inbx+nπ/) coπ/ + cobx+nπ/) inπ/ = cobx+nπ/). Hence if the formua i true forn, then it i true forn+1. From tep 1) and ), the Principe of Induction how that D n inx = inx+nπ/) i true for a integer n 1. Now putting fx) = inx, Tayor Theorem give fx) = fa) + f a)x a) + f a) x a) f n 1 a) x a)n 1 n 1)! + R n i.e. inx = ina + in ) a+ ) π x a) + in a+ π x a) in ) a+n 1) π x a) n 1 + R n n 1)! where R n = f n) ξ) xn = in ) ξ +n π x a) n for ome ξ between a and x, o that R n = in ) ξ +n π x a) n x a n ince in ξ +n) π 1. Now putx = 6 o anda = 30 o, o thatx a = 4 o i ma. In radian, a = π/6 and x a = π/45. Then x a n = ) π n/ 45 which, by cacuator, one find < forn = 4. Hence, ignoring R n with n = 4, give in6 o = in π 6. n ) = n) n) = n+n ) 1+ ) n n 1 But, uing the ogarithmic erie n = x x 1 < x < 1, and temporariy putting x = give n ) = = x x x+ x3 3 + x ) ) 6 π 45) , for x x x ) o that a = n ) ) = ) + 1 ) 3 ) ) +...

4 = = 1 3 ) ) 15 ) ) ) )... ) 4 ) +..., and the reut foow. 7. y = tanx. y = ec x = 1+tan x. Hence y = 1+y, y = yy, y = y +yy,y 4) = 6y y +yy,y 5) = 6y +8y y +yy 4). y0) = 0, y 0) = 1, y = yy y 0) =, y 4) 0) = 0, y 5) 0) = 16. So the Macaurin erie i y0)+y 0)x+y 0) x +y 0) x = x+ x3 3 + x y = inkin 1 x) y = cokin 1 x) d dx kin 1 x = cokin 1 x) k 1 x. Hence y 1 x ) 1/ = k cokin 1 x). Differentiating the at, y 1 x ) 1/ y x1 x ) 1/ = k 1 x ) 1/ inkin 1 x) giving 1 x )y xy +k y = 0. = k 1 x ) 1/ y D n{ 1 x )y } D n {xy } + D n k y = D n 0 = 0. So uing Leibnitz Formua, 1 x )D n y + n C 1 D1 x )D n 1 y + n C D 1 x )D n y + 0 xd n y + n C 1 DxD n 1 y + 0) + k D n y = 0. So on putting x = 0, y n+) 0) = n k )y n) 0). y0) = 0 Uing the expreion for y and the differentia equation, y 0) = k, y 0) = 0. Puttingn = 1,,... in the recurrence reation, y 0) = 1 k )y 0) = 1 k )k, and imiary y 4) 0) = 0, y 5) 0) = 3 k )1 k )k. The Macaurin Serie i y0) +y 0)x +y 0) x +... = kx + 1 k )k x3 + 3 k )1 k )k x5 5! Since y n) = 0 forneven, the generan th non-zero term i yn 1) 0) n 1)! xn 1.

5 Appy the Ratio Tet for convergence, fixingx 0. n+1) th term n th term = y n+1) 0) x n+1 n 1)! n+1)! y n 1) 0) x n 1 = y n+1) 0) x y n 1) 0) n+1)n = n 1) k x, uing the recurrence reation with n THERE repaced n+1)n byn 1, x a n, howing the Macaurin Serie converge for 1 < x < Thi i a homogeneou difference equation, o anatz a n = Aλ n, with λ = λ+1, oλ ± = 1± 5)/. Froma 0 = 0 it foowa n = Aλ n λ n ) and from a 1 = 1 we have A = 1/ 5, o a n = λ n + λ n )/ 5. To take the imit, we note that λ + > λ and thu q = λ /λ + ha moduu e than unity in fact, λ < 1 aready, but for the ake of the argument we continue withq). Writing a n = λ n +1 q n )/ 5 the imit i which i the goden ratio. a n+1 im = λ + = 1 n a n 1+ 5) 10. Write H n x) = n i=0 a n,ix i and find term by term i + )i + 1)a n,i+ ia n,i = na n,i and therefore i n a n,i+ = i+)i+1) a n,i. Note: If a n,i vanihe for ome i, a n,i+k vanihe for a non-negative, integer k. Moreover a n,i+ vanihe for i = n regarde of a n,i. Note ao that the equence for even and odd i remain eparate: If n i even odd), then a n,i vanihe for either a or none of the odd even) i. To have a finite number of term,a n,0 = 0 for oddnanda n,1 = 0 for evenn. Choing a n,1 = 1 for oddnanda n,0 = 1 for odd n one find: H 0 x) = 1 H 1 x) = x H x) = 1 x H 3 x) = x 3 x3 H 4 x) = 1 4x x4 which are mutipe of Hermite poynomia. For fun: Show that H i H j exp x ) vanihe if i j.