Giáo trình giải tích 1 part 7 docx

... ×···[an,bn].A v(A)=(b 1 − a 1 ) ···(bn− an)P A [ai,bi],i =1, ··· ,nai= ci0<ci1< ···<cimi= bim 1 m2···mnAS =[c1i 1 ,c1i 1+ 1] ×···×[cnin,cnin +1 ].S ∈ Pf : A ... f : A =[a 1 ,b 1 ] ×[an,bn] → RPN[ai,bi],i =1, ··· ,n N +1 cik= ai+k(bi− ai)N,k=0, ··· ,N N →∞(b 1 − a 1 ) ···(bn− an)NnNk 1 ,···,kn =1 f(c1k 1 , ··· ,cnkn) ... =A 1 f +A2f −A 1 ∩A2f.f ≤ g AAf ≤Ag.|f| A |Af|≤A|f|.f A c ∈ AAf = f(c)v(A).u, v x, y x =1, y = 1, u =1, v = 1 x =0,y =1, u=0,v=0∂u∂xx =1, y = 1 x =0,y =1 f(x,...

... NK 1 × K2[a, b]K Rnf ∈ C[0, 1] nk fBk(x 1 , ··· ,xn)=0≤p 1 ,···,pn≤kCp 1 k···Cpnkf(p 1 k, ··· ,pnk)xp 1 1···xpnn (1 x 1 )k−p 1 ··· (1 xn)k−pn.(Bk) fA = ... TPk(x)=ak,0+Nkp =1 (ak,psin(2πpxT)+bk,pcos(2πpxT)).R T>0C[0,T]RnnK 1 ⊂ Rn 1 K2⊂ Rn2A 1 A2K 1 ,K2A 1 A2f ∈ C(K 1 × K2)ki =1 gi(x)hi(y)gi∈A 1 ,hi∈A2,k ∈ NK 1 × ... ,fm(x 1 , ··· ,xn))g(y)=(g 1 (y 1 , ··· ,ym), ··· ,gp(y 1 , ··· ,ym))y = f(x)h(x)=g ◦f(x)=(h 1 (x 1 , ··· ,xn), ··· ,hp(x 1 , ··· ,xn))∂h 1 ∂x 1 ···∂h 1 ∂xn···...

... 1 a 1 + a2+ a3+ ··· S a2+ a3+ ···S − a 1 1 1. 2+ 1 2.3+ 1 3.4+ 1 4.5+ 1 5.6+ ··· 1 1.4+ 1 4 .7 + 1 7 .10 + 1 10 .13 + ··· 1 1.3+ 1 4.6+ 1 7. 9+ 1 10 .12 + 1 13 .15 + ··· 1 2− 1 4+ 1 8− 1 16+ 1 32+ ... bk)2,∞k=0√akk 1 − 1 2+ 1 3− 1 4+ 1 5− 1 6+ ··· =1+ ( 1 2− 1) + 1 3+( 1 4− 1 2)+ 1 5+( 1 6− 1 3)+···= (1+ 1 2+ 1 3+ 1 4+ ···) − 1 − 1 2− 1 3− 1 4−···= (1+ 1 2+ 1 3+ 1 4+ ···) − (1 + 1 2+ 1 3+ 1 4+ ... + 1 2+ 1 3+ 1 4+ ···)=0. 1+ x2+ x + x4+ x6+ x3+ x8+ x 10 + x5+ ···= 1 1 − x|x| < 1 0, 611 11 ··· 1, 33333 ··· −2, 343434 ··· e π ln 2S =1+ 2+4+8 +16 +··· 2S =2+4+8+···= S − 1 S = 1 a 1 +...

... limn→∞sinnπ2=0a0 =1, an=√ 1+ an 1 (an)ϕ = limn→∞antn= 1+ 1 nn +1 1n +1 < ln 1+ 1 n< 1 nan =1+ 1 2+ ···+ 1 n−ln n (an)γ = limn→∞an=0, 577 215 6649 ···(an) an +1 − an≤ 1 n(an) 1+ x<ex(x ... FR ∈{ 1 10, 1 100, ··· , 1 10n} Nnn +1 − 1 <, ∀n ≥ N Nlimn→∞nn +1 =1 N 1 √n +1 < 0, 03, ∀n ≥ N limn→∞ 1 √n +1 =0an= 1 2nan=sinnπ2an =10 nan= ... n√9n4 +1 limn→+∞(n2+5−n2+3) limn→+∞√n(√n +1 √n +2)limn→+∞ 1 1.2+ 1 2.3+ ···+ 1 n(n +1) limn→+∞ (1 − 1 22) (1 − 1 32) ··· (1 − 1 n2)limn→+∞ 1+ a + ···+ an 1+ b +...

... =2kπ.∞k =1 1k2 =1+ 1 22+ 1 32+ ···Sn =1+ 1 22+ 1 32+ ···+ 1 n2≤ 1+ 1 1.2+ 1 2.3+ ···+ 1 (n 1) n≤ 1+ 1 1− 1 2+ 1 2− 1 3+ ···+ 1 (n 1) − 1 n< 2 − 1 n∞k=0( 1) k =1 1+ 1− 1+ ··· ... = 1 2− 1 4+ 1 6− 1 8+ 1 10− 1 12+ ···=0+ 1 2− 0 − 1 4+0+ 1 6− 0 − 1 8+ ···ln 2 + 1 2ln 2 = (1 + 0) + (− 1 2+ 1 2)+( 1 3− 0) + ( 1 4− 1 4)+( 1 5+0)+(− 1 6+ 1 6)+··· =1+ 1 3− 1 2+ 1 5+ 1 7 − 1 4+ 1 9+ 1 11 − 1 6+ ... 1+ 1 1+ ··· (1 1) + (1 1) + ···=0 1+ ( 1+ 1)+( 1+ 1)+··· =1 ∞k=0akσ : N → N∞k=0aσ(k)kakSkaσ(k)S∞k =1 ( 1) k +1 kln 2ln 2 = 1 − 1 2+ 1 3− 1 4+ 1 5− 1 6+ 1 7 − 1 8+ ··· 1 2ln...

... =tanx2dx 1+  cos x(0 << ;1) t =tanx2x = 2 arctan t, dx =2dt 1+ t2, cos x = 1 − t2 1+ t2dx 1+  cos x=2dt (1 − )t2 +1+ =2 1 − dtt2+ 1+  1 =2 1 −  1+  1 − arctan ... x2x =a(t2− 1) t2 +1 dx =4atdt(t2 +1) 2 1 2a22t2+2t4 +1 dt = 1 2a2 1 t2+√2t +1 + 1 t2−√2t +1 dt= 1 a2√2(arctan(√2t + 1) + arctan(√2t − 1) ) + Ct =a ... (x)cosaxdxPIn=xnln xdxn = 1 u =lnx ⇒ du =dxxdv = xndx v =xn +1 n +1 In=xn +1 n +1 ln x − 1 n +1 xndx =xn +1 n +1 ln x −xn +1 (n +1) 2+ Cn = 1 I 1 =ln xxdx =ln xd(ln...

... [ 1, 1] max, minx f(x)=0f( 1) ,f (1) f(x)= 1 − 2x2√ 1 − x2=0 ⇔ x = ± 1 √2f( 1 √2)= 1 2,f(− 1 √2)=− 1 2,f( 1) = 0,f( +1) = 0fmax= f = f( 1 √2)= 1 2,fmin= f(− 1 √2)=− 1 2Sr ... t)f(x2)fx 1 ,x2∈ I 0 <t< ;1 f(tx 1 + (1 t)x2) ≥ tf(x 1 )+ (1 t)f(x2)f ff f|Rn| = |eθ(n +1) !|≤3(n +1) ! =10 −3n =6 =10 −6n =9limx→+∞(x − x2ln (1 + 1 x))ln (1 + 1 x)) = 1 x− 1 2x2+ ... ln x= e0 =1 limx→0 (1 + x2) 1 ex−x 1 1∞y = (1+ x2) 1 ex−x 1 ln y =ln (1 + x2)ex− x − 1 00limx→0ln y = limx→02x 1+ x2ex− 1 = limx→0 1 1+x2limx→02xex− 1 = limx→02ex=2limx→0(1...

... ···+( 1) n 1 xnn+( 1) nxn +1 (n + 1) (1 + θx)n +1 (1 + x)α =1+ αx + ···+α(α − 1) ···(α −n +1) n!xn+α(α − 1) ···(α −n) (1 + θx)α−n 1 (n +1) !xn +1 6 0e−x2 =1+ (−x2)+ 1 2!(−x2)2+ 1 3!(−x2)3+ ... + ···+ 1 n!f(n)(x0)∆xn|Rn(∆x)| =|f(n +1) (x0+ θ∆x)|(n +1) !|∆x|n +1 = o(∆xn)n√ 1+ x xn√ 1+ x x0 =1 n√ 1+ x ≈n√ 1+ (n√ 1+ x)|x =1 x =1+ 1 nxe <e ≈ 1+ 1 1!+ 1 2!+ ... ···+xnn!+eθx(n +1) !xn +1 sin x = x −x33!+ ···+( 1) n 1 x2n 1 (2n − 1) !+( 1) ncos θx(2n +1) !x2n +1 cos x =1 x22!+ ···+( 1) nx2n(2n)!+( 1) n +1 cos θx(2n +2)!x2n+2ln (1 + x)=x −x22+...