By George A. Anastassiou, Iuliana F. Iatan
ISBN-10: 3642284744
ISBN-13: 9783642284748
ISBN-10: 3642284752
ISBN-13: 9783642284755
Real research is a self-discipline of in depth learn in lots of associations of upper schooling, since it includes important innovations and primary ends up in the examine of arithmetic and physics, of the technical disciplines and geometry. This booklet is the 1st certainly one of its variety that solves mathematical research issues of all 4 comparable major software program Matlab, Mathcad, Mathematica and Maple. in addition to the elemental theoretical notions, the e-book comprises many routines, solved either mathematically and by means of machine, utilizing: Matlab 7.9, Mathcad 14, Mathematica eight or Maple 15 programming languages. The e-book is split into 9 chapters, which illustrate the appliance of the mathematical thoughts utilizing the pc. each one bankruptcy provides the basic strategies and the weather required to unravel the issues contained in that bankruptcy and finishes with a few difficulties left to be solved through the readers. The calculations will be demonstrated through the use of a particular software program similar to Matlab, Mathcad, Mathematica or Maple.
Read Online or Download Intelligent Routines: Solving Mathematical Analysis with Matlab, Mathcad, Mathematica and Maple PDF
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Additional info for Intelligent Routines: Solving Mathematical Analysis with Matlab, Mathcad, Mathematica and Maple
Sample text
D) n≥1 Solutions. a) One obtains lim n→∞ an+1 2 · 7 · 12 · · · · · (5n + 2) 5 · 9 · 13 · · · · · (4n + 1) · = lim n→∞ an 5 · 9 · 13 · · · · · (4n + 5) 2 · 7 · 12 · · · · · (5n − 3) 5 5n + 2 = > 1; = lim n→∞ 4n + 5 4 therefore, using the ratio test it results that the series n≥1 2 · 7 · 12 · · · · · (5n − 3) 5 · 9 · 13 · · · · · (4n + 1) diverges. ]2 (2n)! )2 · (n + 1)2 · (2n)! an+1 · = = 2 an (2n + 2)! )2 (2n + 2) (2n + 1) · (2n)! ) (2n)! is convergent. 4 Tests of Convergence and Divergence of Positive Series 25 c) We can write: π π π π π cos 2n+1 an+1 n + 1 tan 2n+2 n + 1 sin 2n+2 n+2 n+1 · · = = · 2 π · 2 π · ; π π π an n tan 2n+1 n cos 2n+2 sin 2n+1 2n+2 2n+1 therefore an+1 1 = n→∞ an 2 lim and using the ratio test one establishes that n tan n≥1 π 2n+1 is convergent.
F (0) + · · · n! 2! n! 10) namely the particular case of the Taylor series for a = 0 is called the Mac Laurin series. 11. Expand the function f (x) in a series of powers of x: f (x) = 1 1+x ln , x ∈ (−1, 1) 2 1−x b) f (x) = cos3 x, x ∈ R 3x − 5 , x ∈ R\ {1, 3} . c) f (x) = 2 x − 4x − 3 a) f (x) = Solutions. a) One can notice that ⎤ ⎡ f (x) = 1⎢ ⎥ 1 ⎣ln (1 + x) − ln (1 − x)⎦ = [f1 (x) − f2 (x)] , x ∈ (−1, 1) 2 2 f1 (x) f2 (x) and ⎧ ⎧ 1 1 f1 (x) = 1+x f2 (x) = − 1−x ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ f1 (x) = − 1 2 ⎨ f2 (x) = − 1 2 (1+x) (1−x) , 2 2 f1 (x) = (1+x)3 f2 (x) = − (1−x) 3 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ (4) ⎩ (4) 6 6 f1 (x) = − (1+x) f (x) = − .
N tan n≥1 π 2n+1 n · an , a > 0. d) n≥1 Solutions. a) One obtains lim n→∞ an+1 2 · 7 · 12 · · · · · (5n + 2) 5 · 9 · 13 · · · · · (4n + 1) · = lim n→∞ an 5 · 9 · 13 · · · · · (4n + 5) 2 · 7 · 12 · · · · · (5n − 3) 5 5n + 2 = > 1; = lim n→∞ 4n + 5 4 therefore, using the ratio test it results that the series n≥1 2 · 7 · 12 · · · · · (5n − 3) 5 · 9 · 13 · · · · · (4n + 1) diverges. ]2 (2n)! )2 · (n + 1)2 · (2n)! an+1 · = = 2 an (2n + 2)! )2 (2n + 2) (2n + 1) · (2n)! ) (2n)! is convergent. 4 Tests of Convergence and Divergence of Positive Series 25 c) We can write: π π π π π cos 2n+1 an+1 n + 1 tan 2n+2 n + 1 sin 2n+2 n+2 n+1 · · = = · 2 π · 2 π · ; π π π an n tan 2n+1 n cos 2n+2 sin 2n+1 2n+2 2n+1 therefore an+1 1 = n→∞ an 2 lim and using the ratio test one establishes that n tan n≥1 π 2n+1 is convergent.
Intelligent Routines: Solving Mathematical Analysis with Matlab, Mathcad, Mathematica and Maple by George A. Anastassiou, Iuliana F. Iatan
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