0
  1. Trang chủ >
  2. Kỹ Thuật - Công Nghệ >
  3. Kĩ thuật Viễn thông >

Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 3 pdf

Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 3 pdf

Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 3 pdf

... x2, x 3 , x 4 } is independent, but not orthogonal in the interval [−1, 1]. Using Gramm-Schmidt orthogo-12 84 12 3 4 56-60 -40 -20Figure 24. 3: log(error in approximation)In Figure 24. 4 we ... =1√πx−1e−x2−1√π∞xt−2e−t2dt=1√πx−1e−x2−1√π−12t 3 e−t2∞x+1√π∞x 3 2t 4 e−t2dt=1√πe−x2x−1−12x 3 +1√π∞x 3 2t 4 e−t2dt=1√πe−x2x−1−12x 3 +1√π− 3 4 t−5e−t2∞x−1√π∞x15 4 t−6e−t2dt=1√πe−x2x−1−12x 3 + 3 4 x−5−1√π∞x15 4 t−6e−t2dt1265The ... 21)π 4 P 4 (x)=1058π 4 [ (31 5 − 30 π2)x 4 + ( 24 2− 270)x2+ (27 − 2π2)]The cosine and this polynomial are plotted in the second graph in Figure 25.1. The le ast squares fit method usesinformation...
  • 40
  • 315
  • 0
Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 5 pdf

Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 5 pdf

... solution and discuss in as much detail as possible what goes wrong. 13 64 Note that this formula is valid for m = 0, 1, 2, . . Similarly, we can multiply by sin(mx) and integrate to solve for bm. ... +n2x2− 2n 3 sin(nx) 137 0 -3 -2-1 12 3 -3 -2-112 3 Figure 28.2: Graph ofˆf(x).would give a better approximation?Least squared error fit. The most common criterion for finding the best ... 21/20x sin(nπx) dx + 211/2(1 − x) sin(nπx) dx= 4 (nπ)2sin(nπ/2)= 4 (nπ)2(−1)(n−1)/2 for odd n0 for even n. 135 6 -3 -2-1 12 3 -1.5-1-0.50.511.5Figure 28.5: Fourier Sine Series.The...
  • 40
  • 223
  • 0
Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 1 pdf

Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 1 pdf

... . . . . . . . . . . . . . . 1 630 34 . 3. 3 Bessel Functions of Non-Integer Order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 633 34 . 3 .4 Recursion Formulas . . . . . . . . . . . ... . . 1910 43 .10Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1911 44 Transform Methods 1918 44 .1 Fourier Transform for Partial Differential ... . . . . . 147 2 30 .3. 2 Singular Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 3 31 The Laplace Transform 147 5 31 .1 The Laplace Transform . . ....
  • 40
  • 619
  • 0
Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 3 pptx

Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 3 pptx

... +∆x 3 6f(x) +∆x 4 24 f(x1),f(x −∆x) = f(x) −∆xf(x) +∆x22f(x) −∆x 3 6f(x) +∆x 4 24 f(x2),where x ≤ x1≤ x + ∆x and x − ∆x ≤ x2≤ x.Hint 3. 20Hint 3. 21a. ... positive for −π < x < 0 and negative for 0 < x < π. Since the sign of ygoes from positive to negative, x = 0 is arelative maxima. See Figure 3. 7.Example 3. 5 .3 Consider y = x 3 and ... Equation 3. 2 is called the forward difference scheme for calculatingthe first derivative. Figure 3. 17 shows a plot of the value of this scheme for the function f(x) = sin x and ∆x = 1 /4. The first...
  • 40
  • 325
  • 0
Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 5 pdf

Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 5 pdf

... =12x 3 e2x− 3 4 x2e2x+ 3 212xe2x−12e2xdxx 3 e2xdx =12x 3 e2x− 3 4 x2e2x+ 3 4 xe2x− 3 8e2x+CSolution 4. 15Expanding the integrand in partial fractions,1x2− 4 =1(x ... +20√x dx=2 3 x 3/ 210+2 3 x 3/ 220=2 3 +2 3 2 3/ 2=2 3 (1 + 2√2)Solution 4. 3 ddxx2xf(ξ) dξ = f(x2)ddx(x2) − f(x)ddx(x)= 2xf(x2) − f(x)15112 3 4 5-101-10112 3 4 5-101Figure ... =12x 3 e2x− 3 212x2e2x−xe2xdxx 3 e2xdx =12x 3 e2x− 3 4 x2e2x+ 3 2xe2xdxLet u = x and dv =e2xdx. Then du = dx and v =12e2x.x 3 e2xdx =12x 3 e2x− 3 4 x2e2x+ 3 212xe2x−12e2xdxx 3 e2xdx...
  • 40
  • 425
  • 0
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 3 ppt

Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 3 ppt

... =ız 4 4−12z2ı1+ı=12+ ıIn this example, the anti-derivative is single-valued.2.Csin2z cos z dz =sin 3 z 3 ıππ=1 3 sin 3 (ıπ) −sin 3 (π)= −ısinh 3 (π) 3 Again ... maxa≤x≤b|f(x)|. 46 6with a, b and c complex-valued constants and d a real constant. Substituting z = x + ıy and expanding productsyields,ax 3 + ı3x2y − 3xy2− ıy 3 + bx2+ ı2xy ... +eıθnıeıθdθ=(ı2+eıθ)n+1n+12π0 for n = −1logı2 +eıθ2π0 for n = −1= 0 3. We parameterize the contour and do the integration.z = reıθ, r = 2 −sin2θ 4 , θ ∈ [0 . . . 4 ) 48 7Solution 10.6Cf(z)...
  • 40
  • 345
  • 0
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 7 pdf

Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 7 pdf

... cosh zz 3 sin z sinh z=1 −z22+z 4 24 − ···1 +z22+z 4 24 + ···z 3 z −z 3 6+z5120− ···z +z 3 6+z5120+ ···=1 −z 4 6+ ···z 3 z2+ z6−1 36 +160+ ... ı /4) (n + 1))zn+ d, for 2 < |z|Solution 12.29The radius of convergence of the series for f(z) isR = limn→∞k 3 /3 k(k + 1) 3 /3 k+1= 3 limn→∞k 3 (k + 1) 3 = ... ···=1z51 −z 4 6+ ···1 −z 4 90+ ···=1z51 −z 4 6+ ···1 +z 4 90+ ···=1z51 −7 45 z 4 + ···=1z5−7 45 1z+ ···Thus we see that the residue is −7 45 . N ow we...
  • 40
  • 357
  • 0
Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 1 docx

Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 1 docx

... polynomial areα1=1 +1 − 3 /4 2= 3 4 , α2=1 −1 − 3 /4 2=1 4 .Thus our two series solutions will be of the formw1= z 3 /4 ∞n=0anzn, w2= z1 /4 ∞n=0bnzn.Substituting ... + 3) For our first solution we have the difference equationa0= 1, a1= 0, a2= 0, a 3 = 0, an +4 = −an(n + 4) (n + 3) . For our second solution,b0= 0, b1= 1, b2= 0, b 3 = 0, bn +4 = ... difference equation for bnthat is of order one less than theequation for an.117812 3 4 56-1-0.50.511.512 3 4 56-1-0.50.511.5Figure 23. 3: The graph of approximations and numerical...
  • 40
  • 249
  • 0
Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 2 pps

Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 2 pps

... relation for the coefficients.(k + 1)kck+1+ b(k + 1)ck+1− kck− ack= 0ck+1=k + a(k + 1)(k + b)ck1 244 23. 6 HintsHint 23. 1Hint 23. 2Hint 23. 3Hint 23 .4 Hint 23. 5Hint 23. 6Hint 23. 7Hint ... =1−1x2dx =x 3 31−1=2 3 1−1P2(x)P2(x) dx =1−11 4 9x 4 − 6x2+ 1dx =1 4 9x55− 2x 3 + x1−1=251−1P 3 (x)P 3 (x) dx =1−11 4 25x6− 30 x 4 + 9x2dx ... 9x2dx =1 4 25x77− 6x5+ 3x 3 1−1=27Solution 23. 3The indicial equation for this problem isα2+ 1 = 0.Since the two roots α1= i and α2= −i are distinct and do not differ...
  • 40
  • 214
  • 0
Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 4 docx

Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 4 docx

... theformula to obtain information about the eigenvalues before we solve a problem.Example 27 .4. 2 Consider the self-adjoint eigenvalue problem−y= λy, y(0) = y(π) = 0. 132 2Example 27 .4. 1 ... eigenfunctionφ. Green’s formula statesφ|L[φ] − L[φ]|φ = 0φ|λφ − λφ|φ = 0(λ − λ)φ|φ = 0Since φ ≡ 0, φ|φ > 0. Thus λ = λ and λ is real. 131 927.7 HintsHint 27.1 132 72. For k = n, φk, ... operatorL[y] = x2y+ 2xy+ 3yhas the adjoint operatorL∗[y] =d2dx2(x2y) −ddx(2xy) + 3y= x2y+ 4xy+ 2y −2xy− 2y + 3y= x2y+ 2xy+ 3y.In Example 27.2.1, the adjoint...
  • 40
  • 366
  • 0

Xem thêm

Từ khóa: advanced mathematical methods for scientists and engineers bender pdfadvanced mathematical methods for scientists and engineers pdf downloadadvanced mathematical methods for scientists and engineers solutions manualadvanced mathematical methods for scientists and engineers djvuadvanced mathematical methods for scientists and engineers downloadadvanced mathematical methods for scientists and engineersadvanced mathematical methods for scientists and engineers free downloadadvanced mathematical methods for scientists and engineers benderadvanced mathematical methods for scientists and engineers bender orszag downloadadvanced mathematical methods for scientists and engineers bender downloadadvanced mathematical methods for scientists and engineers solutionsadvanced mathematical methods for scientists and engineers i pdfmathematical methods for scientists and engineers mcquarrie pdfmathematical methods for scientists and engineers pdfmathematical methods for scientists and engineers donald a mcquarrie pdfBáo cáo thực tập tại nhà thuốc tại Thành phố Hồ Chí Minh năm 2018chuyên đề điện xoay chiều theo dạngNghiên cứu sự hình thành lớp bảo vệ và khả năng chống ăn mòn của thép bền thời tiết trong điều kiện khí hậu nhiệt đới việt namNghiên cứu tổ chức pha chế, đánh giá chất lượng thuốc tiêm truyền trong điều kiện dã ngoạiMột số giải pháp nâng cao chất lượng streaming thích ứng video trên nền giao thức HTTPNghiên cứu tổ chức chạy tàu hàng cố định theo thời gian trên đường sắt việt namđề thi thử THPTQG 2019 toán THPT chuyên thái bình lần 2 có lời giảiĐỒ ÁN NGHIÊN CỨU CÔNG NGHỆ KẾT NỐI VÔ TUYẾN CỰ LY XA, CÔNG SUẤT THẤP LPWANQuản lý hoạt động học tập của học sinh theo hướng phát triển kỹ năng học tập hợp tác tại các trường phổ thông dân tộc bán trú huyện ba chẽ, tỉnh quảng ninhPhát triển du lịch bền vững trên cơ sở bảo vệ môi trường tự nhiên vịnh hạ longPhát hiện xâm nhập dựa trên thuật toán k meansNghiên cứu tổng hợp các oxit hỗn hợp kích thƣớc nanomet ce 0 75 zr0 25o2 , ce 0 5 zr0 5o2 và khảo sát hoạt tính quang xúc tác của chúngSở hữu ruộng đất và kinh tế nông nghiệp châu ôn (lạng sơn) nửa đầu thế kỷ XIXTổ chức và hoạt động của Phòng Tư pháp từ thực tiễn tỉnh Phú Thọ (Luận văn thạc sĩ)Giáo án Sinh học 11 bài 15: Tiêu hóa ở động vậtGiáo án Sinh học 11 bài 15: Tiêu hóa ở động vậtchuong 1 tong quan quan tri rui roGiáo án Sinh học 11 bài 14: Thực hành phát hiện hô hấp ở thực vậtMÔN TRUYỀN THÔNG MARKETING TÍCH HỢPTÁI CHẾ NHỰA VÀ QUẢN LÝ CHẤT THẢI Ở HOA KỲ