The polar coordinates

Transcription

1 The polar coordinates 1

2 2

3 3

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5 Graphing in polar coordinates 5

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9 Area and length in polar coordinates 9

10 10

11 11

12 Partial deravitive 12

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14 14

15 15

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18 18

19 19

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21 Double Integral 21

22 22

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24 24

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26 26

27 27

28 Triple Integrals in Rectangular Coordinates 28

29 29

30 30

31 Triple Integrals in Cylindrical and Spherical Coordinates 31

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39 39

40 The series: is an infinite sequence of numbers and it is a function whose domain is the set of positive integers. The function associated to the sequence 2, 4, 6, 8 2n, Sends 1 to a 1 =2, 2 to a 2 =4, and so on. The general behavior of this sequence is described by the formula a n =2n Example: (a) The sequence of natural numbers 1, 2, 3, 4 n (b) The sequence 1 2, 2 3, 3 4,.., n n+1 (c) The sequence 1, -1, 1, -1, 1, -1 (-1) n Example: find the series of the sequence: (a) a n = n 1 sin n n = 0,1,2,3,. (b) a n+1 n = n = 1,2,3,.. n (b) a n = n! n = 1,2,3, (d) a n 2 n = 1 + ( 1) n n = 0,1,2,3, The sum of series: 1 a n = a 1 + a 2 + a 3 + a 4 + Example: Find the sum series of: 5 ln (n) n 1, 1, e 2n 1 0, x n 2 n x+n 40

41 Series test (1- The Ratio Test: Let a n be a series with positive terms and suppose that a n+1 lim = ρ n a n Then (a) The series converges if ρ < 1. (b) The series diverges if ρ > 1. (c) The test is inconclusive if ρ = 1. (2- The root test: Let a n be a series with positive terms and suppose that lim (a n) 1 n = ρ n Then (a) The series converges if ρ < 1. (b) The series diverges if ρ > 1. (c)the test is inconclusive if ρ = 1. (3- The second Ratio Test: Let a n be a series with positive terms and suppose that lim n(1 a n+1 ) = ρ n a n Then (a) The series converges if ρ > 1. (b) The series diverges if ρ < 1. (c)the test is inconclusive if ρ = 1. Example: find the sequence of the series, and test it: (a) (b) (1 3 )2 + ( 1 4 )3 + ( 1 5 )4 + 41

42 (c) (d) x + x 2 2! + x 3 3! + x 4 4! + Test the convergence or divergence of the series: a) b) c) n=0 n=1 n=1 ( 1) n +1 2 n ( 1) n n 2 +2n+2 ( 1) n +1 n 3n 1 d) 1 n sin 1 1 n=1 n e) ( 1) n n n=2 ln (n) Answers: a) conv. b) conv. c) div. d) conv. e) div. Power Series A power series about x=0 is a series of the form S = n=0 c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + A power series about x=a is a series of the form S = n=0 c n (x a) n = c 0 + c 1 (x a) + c 2 (x a) 2 + c 3 (x a) 3 + in which the center a and the coefficients c 0, c 1, c 2,, c n are constants. The term by term differentiation: If n=0 c n (x a) n converges for a R < x < a + R for some R > 0, it defines a function f: f x = c n x a n n=0 a R < x < a + R Such a function ƒ has derivatives of all orders inside the interval of convergence. We can obtain the derivatives by differentiating the original series term by term: 42

43 f x = f x = n=0 n c n x a n 1 n(n 1) c n x a n 2 n=0 and so on. Each of these derived series converges at every interior point of the interval of convergence of the original series. The Term-by-Term Integration Theorem If f(x) converges for a R < x < a + R for some R > 0 f x = n=0 c n x a n a R < x < a + R Such a function ƒ has integrations of all orders inside the interval of convergence. We can obtain the integrations by integral the original series term by term: f(x) dx = n=0 c n x a n+1 + c n + 1 Series Representations within its interval of convergence the sum of a power series is a continuous function with derivatives of all orders. If a function ƒ(x) has derivatives of all orders on an interval, it can be expressed as a power series on the interval: f x = Or n=0 c n x a n f x = c 0 + c 1 x a + c 2 (x a) 2 + c 3 (x a) 3 + c 4 (x a) 4 + f x = c 1 + 2c 2 (x a) + 3c 3 (x a) 2 + 4c 4 (x a) 3 + f x = 2c c 3 (x a) + 4.3c 4 (x a) 2 + f x = 3.2.1c c 4 (x a)

44 Then, at x=a all derivatives becomes: f a = c 0 f a = c 1 f a = 2c 2 or c 0 = f(a) 0! or c 1 = f a 1! or c 2 = f (a) 2! f a = 3.2.1c 3 or c 3 = f (a) 3! DEFINITIONS: Taylor Series, Maclaurin Series Let ƒ be a function with derivatives of all orders throughout some interval containing a as an interior point, Then the Taylor series generated by ƒ at x=a is: f x = f a + (x a) 1! f a + (x a)2 2! f a + (x a)3 3! f a + = (x a) n n=0 f n (a) Taylor Series n! For Maclaurin series a must equal to zero. Examples: Find Maclaurin series (x=0): e x, e x, sin x, cos x, (x + 1) 3, ln (x + 1) Example: find Taylor series for: ln x at x = 1 1 x 1 at x = 2 x 1 at x = 2 (x + a) n at x = 0 Solutions: The series of e x : f x = f a + (x a) 1! f a + (x a)2 2! f a + (x a)3 3! f a + Where f(x) =e x and a=0 Then: 44

45 f x = e x, f 0 = 1 f x = e x, f 0 = 1 f x = e x, f 0 = 1 f x = e x, f 0 = 1 f x = e x, f 0 = 1 Then: f x = e x = 1 + x 1! + x2 2! + x3 3! + x4 4! + = x n n! - the series of sin(x): n=0 f x = f a + (x a) 1! f a + (x a)2 2! f a + (x a)3 3! f a + f(x)=sin(x), a=0 f x = sin x, f 0 = 0 f x = cos x, f 0 = 1 f x = sin (x), f 0 = 0 f x = cos x, f 0 = 1 f x = sin (x), f 0 = 0 f (5) x = cos x, f (5) 0 = 1 f (6) x = sin x, f (6) 0 = 0 Then: f x = sin x = x x 3 + x 5 x 7 + = 1! 3! 5! 7! ( 1)n x 2n +1 n=0 (2n+1)! Power Series Solutions of Differential Equations and Initial Value Problems When we cannot find a relatively simple expression for the solution of an initial value problem or differential equation, we try to get information about the solution in other ways. One way is to try to find a power series representation for the solution. If we can do so, we immediately have a source of polynomial approximations of the solution, which may be all that we really need. 45

46 The procedures can be concluding: 1) represent the power series 2) Substitute the series at the differential equation. 3) Find the relations between the polynomial for every power of (x) 4) Put all polynomial relative to the first and second polynomials. 5) Find the final solution Example: find the solution of y y = x for y 0 = 1 Solution: Let y = n=0 a n x n = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 + a 5 x 5 + Then: y = n=0 na n x n 1 = a 1 + 2a 2 x + 3a 3 x 2 + 4a 4 x 3 Subs. In the differential equation: + (a 1 + 2a 2 x + 3a 3 x 2 + 4a 4 x 3 + ) a 0 + a 1 x + a 2 x 2 + a 3 x 3 + = x Coefficient of x 0 : a 1 a 0 = 0, a 1 = a 0 Coefficient of x 1 : 2a 2 a 1 = 1, a 2 = (a 1+1) 2 = (a 0+1) 2 = (a 0+1) 2! Coefficient of x 2 : 3a 3 a 2 = 0, a 3 = a 2 = (a 0+1) = (a 0+1) ! Coefficient of x 3 : 4a 4 a 3 = 0, a 4 = a 3 = (a 0+1) = (a 0+1) ! But a 0 =1 due to boundary condition, then: a 1 = 1, a 2 = 2 2!, a 3 = 2 3!, a 4 = 2 4!, a 5 = 2 5!, a n = 2 n! So the solution is: y = x + 2 2! x ! x ! x4 + 46