48H Problem Set Selected Solutions Chpter 9 # () Tke f(x) = x n, g (x) = e x, nd use integrtion by prts; this gives reduction formul: x n e x dx = x n e x n x n e x dx. (b) Tke f(x) = (log x) n, g (x) =, g(x) = x nd use prts: (log x) n dx = x(log x) n n(log x) n x x dx, so the desired reduction formul is: (log x) n dx = x(log x) n n (log x) n dx. Actully, these re essentilly the sme formul: if you substitute x = e u in (b), you get (). # 3 (i) We hve f (x) = (x + ) / x, so the length is + ( (x + ) / x) dx = + x (x + ) dx (iv) We hve f (x) = tn x, so the length is π/6 + tn x dx = π/6 = x + dx = 3 + = 4 3. sec x dx = log(sec x + tn x) π/6 = log( + ) log( + ) 3 3 = log 3. # 39 From the formul for integrtion by prts u (x)v(x)dx = u(x)v(x) u(x)v (x)dx, (with indefinite integrls), we deduce (using the Second Fundmentl Theorem) version for definite integrls: b b b u (x)v(x)dx = u(x)v(x) u(x)v (x)dx,
If we tke the limit s b, we obtin version for improper integrls: u (x)v(x)dx = u(x)v(x) u(x)v (x)dx, # 4() The integrl e t t x dt is improper t the upper limit nd is improper t the lower limit if x <. We divide the integrl into two prts nd investigte convergence of the two prts seprtely: where e t t x dt = e t t x dt = e t t x dt = lim e t t x dt + e t t x dt, M lim e t t x dt M ɛ ɛ e t t x dt Since the integrnd is positive, we cn show tht the first integrl converges if we cn find function f(t) such tht e t t x f(t) (t lest for t lrge) nd such tht f(t) dt converges. Tke f(t) = e t/ : then e t t x /f(t) = t x /e t/ which pproches s t (the exponentil grows fster thn ny power of t) so e t t x f(t) for t lrge enough. Finlly, since f(t) dt = lim M M e t/ dt = M lim M e t/ = e /, the integrl f(t) dt is convergent, nd so e t t x dt is convergent, too. For the other prt of the integrl, we use the estimte e t for t, so e t t x t x for t. Now lim t x t x dt = lim ɛ ɛ x = x, ɛ for ny x >, so tx dt converges, nd so e t t x dt converges s well. Hence e t t x dt converges for ny x >. (b) Use integrtion by prts: Γ(x + ) = e t t x dt = t x ( e t ) ( e t )xt x dt = x Here x is ny fixed positive number. e t t x dt = xγ(x) (c) If we tke x = we get n integrl we cn do: e t dt = e t =. ɛ
3 Hence Γ() = =!. If Γ(n) = (n )!, then, by (b), Γ(n + ) = nγ(n) = n (n )! = n!. So Γ(n) = (n )! for ll integers n, by induction. # 4() This follows from the reduction formul on pg. 377: note tht n is n integer, so tht sin n (x) = when x = or x = π/. (b) This is immedite from prt (): if we let I n = π/ sin n x dx, then formul in () cn be written thus for ny n we hve I n+ = n + n + I n, for ny integer n ; I n+ = n n + I n = n n + n n I n 3. = n n + n n 3 I nd I = π/ sin x dx = cos x π/ =. Similrly, for ny n, nd I = π/ dx = π/. Hence so I n I n+ = (c) From the inequlites I n = n n = n n n n. = n n n 3 n n n+ n n 3 = n + n n n I n n 3 n I n 4 n 3 n I π n n n 3 n 3 π π = 5 n n n n + sin n+ x sin n x sin n x I n I n+ (which hold for ll x [, π/] nd ny n, since sin x for ll x [, π/]), we obtin by integrtion I n+ I n I n.
4 Moreover, ll the integrls I n re positive numbers, by prt (b). Divide by I n+ : then we get I n I n+ I n I n+ = I n n n+ I = n + = + n n n. Thus I n /I n+ pproches s n, nd hence lso π = 5 6 5 6 7 in the sense tht the more terms we tke the closer the product comes to π/. Chpter # 4(ii) We cn use the estimte mde in clss: from Tylor s Theorem we hve e x = + x + x + + xn n! + R n,,exp(x), tke x =, nd use Lgrnge s form of the reminder: e = + + + + n! + for some c between nd. Since e c 3, we see tht e c (n + )! e ( + + + + ) n! provided tht we choose n so tht 3/(n + )!, i.e. if (n + )! 3. So how lrge do we hve to tke n? We might estimte (n + )! roughly s follows: (n + )! = 9! (n + ) 4 n+ 9 since 9! > 4 nd (n + ) contins n + 9 fctors, ech. So if n + 9, i.e. if n is t lest 8, then < R n,,exp (), so e is given by the sum with n error less thn. # 6() Problem 5.9 is the identity 8 k= k! rctn x + rctn y = rctn x + y xy. provided tht xy < (see pge 644 for the proof). We hve rctn + rctn 3 = rctn + 3 3 = rctn = π 4.
5 For the second identity we hve nd so nd so rctn 5 = rctn 5 + 5 5 5 = rctn 5, 4 rctn 5 = rctn 5 5 = rctn + 5 5 5 = rctn 9 4 rctn 5 rctn ( = rctn 39 9 + rctn ) 39 = rctn 9 39 + 9 39 = rctn = π 4. (b) We will pproximte π to within 6, which should tell us π to 6 plces with n mbiguity of in the 6 th digit; this should llow us to be certin of the first five plces of π (in fct it will llow us to be certin in this cse; but there is quirk to the deciml system tht could hve gotten in the wy: e.g. suppose we know tht x =.399999999 with n error of t most ; then we still cn t sy for sure whether the nd plce of x is 3 or 4). We use π = 6 rctn 5 4 rctn 39 from prt (). We use the estimte of the error term for the rctngent from pge 4: rctn x x x3 3 + x5 xn+ + ( )n 5 n + with error less thn x n+3 n+3. Thus for 6 rctn 5 or which holds for n = 4. So 6 <.5 6 (n + 3)5n+3 3 6 < (n + 3)5 n+3 we wnt to choose n so tht 6 rctn 5 = 6( 5 3 5 3 + 5 5 5 7 5 7 + 9 5 9 ) + ɛ = 3.5838986... + ɛ where ɛ <.5 6 =.5. Similrly, for 4 rctn tht 4 <.5 6 (n + 3)39n+3 or 8 6 < (n + 3)39 n+3 39 we wnt to tke n so
6 which holds for n =. So 4 rctn 39 = 4( 39 ) + ɛ =.67364... + ɛ where ɛ <.5 6 =.5. (Oky, I confess I hve been using clcultor to do the rithmetic; but I hven t used the rctn key, so I could hve used vintge 6 s 4-function clcultor... or even done it by hnd... ) Therefore we hve proved tht π = 3.5838986....67364... + ɛ = 3.45958... + ɛ where ɛ <.; in other words 3.459358... < π < 3.45958... Therefore π = 3.459..., to 5 plces. The next digit could be,, or 3. # 8() The Tylor polynomil of sin of degree n + t is P (x) = x x3 3! + x5 xn+ + ( )n 5! (n + )!, nd stisfies (sin x P (x))/x n+ ) s x. Hence sin(x ) P (x ) lim x x 4n+ = s well. Therefore P (x ) is the Tylor polynomil of f(x) = sin(x ) of degree 4n+ t : nd P (x ) equls x x6 3! + x 5! x4n+ + ( ) n (n + )!. (b) Since P (x ) is the Tylor polynomil of f(x) = sin(x ) of degree 4n + t, we hve x x6 3! + x 5! x4n+ + ( ) n (n + )! = f() + f ()x + f () x + + f 4n+ () (4n + )! x4n+. for ny n. Hence { f (k) if k, 6,,... () = ( ) n (4n + )!/(n + )! if k = 4n + for some integer n