3 1 sqrt 2 x

• [PDF File]Finding Square Roots Using Newton’s Method

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  gence property, it thus converges to some limit x. I claim that x2 = A. Rewrite (2) as A − x2 k = 2xk(xk+1 − xk) and let k → ∞. Since xk+1 −xk → 0 and xk is bounded, this is obvious. We now know that √ A exists as a real number. then it is simple to use (1) to verify that xk+1 − √ A = 1 2xk (xk − √ A)2. (3) Equation (3 ...

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• [PDF File]Techniques of Integration - Whitman College

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  2 3 cos3 x− 1 5 cos5 x+ C. 203. 204 Chapter 10 Techniques of Integration EXAMPLE 10.1.2 Evaluate Z sin6 xdx. Use sin2 x = (1 − cos(2x))/2 to rewrite the function: Z sin6 xdx = Z (sin2 x)3 dx = Z (1− cos2x)3 8 dx = 1 8 Z 1−3cos2x+3cos2 2x− cos3 2xdx. Now we have four integrals to evaluate: Z 1dx = x

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• [PDF File]QUANTUM MECHANICS Examples of operators

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  KEx = (1/2) m vx 2 = p x 2/(2m) The classical Hamiltonian function is defined as the sum of the kinetic energy (a function of momentum) & the potential energy (a function of cordinates) H = px 2/(2m) + V(x) for a 1-dimensional system Comparison to the Schrödinger Eq. shows that (-h2/2m) d2/dx2 ↔ p x 2/(2m) Some Postulates of Quantum Mechanics:

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• [PDF File]Math 431 - Real Analysis I Solutions to Homework due ...

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  n+1 = p 2 + x n: (a) For n= 1;2;:::;10, compute x n. A calculator may be helpful. (b) Show that x n is a monotone increasing sequence. A proof by induction might be easiest. (c) Show that the sequence x n is bounded below by 1 and above by 2. (d) Use (b) and (c) to conclude that x n converges. Solution 1. (a) n x n 1 1 2 1:41421 3 1:84776 4 1 ...

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• [PDF File]Square Roots via Newton’s Method

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  n+1 = 1 2 x n + a x n : The intuition is very simple: if x n is too big (> p a), then a=x n will be too small (< p a), and so their arithmetic mean x n+1 will be closer to p a. It turns out that this algorithm is very old, dating at least to the ancient Babylonians circa 1000 BCE.1 In modern times, this was seen to

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• [PDF File]Solutions to HW5 Problem 3.1 - IUPUI

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  ECE302 Spring 2006 HW5 Solutions February 21, 2006 3 Problem 3.2.1 • The random variable X has probability density function fX (x) = ˆ cx 0 ≤ x ≤ 2, 0 otherwise.

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• [PDF File]Homework 8 Solutions - Stanford University

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  3 2x 22 + 3 23 j 3 x2 2 >1: This shows that f(x) = x3 is not uniformly continuous on R. 44.5. Let M 1; M 2, and M 3 be metric spaces. Let gbe a uniformly continuous function from M 1 into M 2, and let fbe a uniformly continuous function from M 2 into M 3. Prove that f gis uniformly continuous on M 1. Solution. Let >0. Since fis uniformly ...

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• [PDF File]Volumes by Cylindrical Shells: the Shell Method

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  Ex. Find the volume of the solid generated by revolving the region bounded by y = x2, y = 0, x = −1, and x = 1, about the line x = 2. The axis of rotation, x = 2, is a line parallel to the y-axis, therefore, the

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• [PDF File]Section 1.5. Taylor Series Expansions

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  and set x =a to obtain f00(a)=c 2¢2¢1=) c2= f00(a) 2!; take derivative again on (5) f(3)(x)= X1 n=3 cnn(n¡1)(n¡2)(x¡a)n¡3=c 33¢2¢1+c44¢3¢2(x¡a)+c55¢4¢3(x¡a) 2+::: and insert x =a to obtain f(3)(a)=c 33¢2¢1=) c3= f(3)(a) 3!: In general, we have cn = f(n)(a) n!; n =0;1;2;::: here we adopt the convention that 0!=1: All above process can be carried

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• [PDF File]WA 3: Solutions - Drexel University

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  WA 3: Solutions Problem 1. Prove that: (i) If f = u + iv is analytic and satisfies u2 = v in a domain D, then f is a constant. (ii) If f is a real-valued analytic function in a domain D, then f is a constant.

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• [PDF File]2003 AP Calculus BC Scoring Guidelines - College Board

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  3 4 2 0 5 1 3 ¨ + yydy ¦ = 0.346 or 0.347 3 : 1 : limits 1 : integrand 1 : answer £¦ ¦¦ ¦ ¤ ¦¦ ¦¦ ¦¥ (c) xr= cosR; yr= sinR 22 22 22xy r r = 1cos sin 1º RR= 2 22 1 cos sin r RR = 2 : 22 2 1 : substitutes cos and sin into 1 1 : isolates xr yr x y r R R £¦¦ = ¦¦ ¦¦¤ = = ¦¦ ¦¦ ¦¥ (d) Let C be the angle that segment OP ...

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• [PDF File]Integration by substitution

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  1+x2 dx (3) In this example we make the substitution u = 1+x2, in order to simplify the square-root term. We shall see that the rest of the integrand, 2xdx, will be taken care of automatically in the substitution process, and that this is because 2x is the derivative of that part of the integrand used in the substitution, i.e. 1+x2. As before ...

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• [PDF File]Chapter 3 Solved Problems - HVL

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  >> y=(x+x.*sqrt(x+3)).*(1+2*x.^2)-x.^3 y =-28.0000 -14.9791 -6.2426 -1.8109 0 2.0281 8.0000 22.3759 50.2492 4. For the function , calculate the value of y for the following values of x using element-by-ele ment operations: . Solution >> x=15:10:65 x = 15 25 35 45 55 65 ...

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• [PDF File]PLOTTING AND GRAPHICS OPTIONS IN MATHEMATICA

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  Plot Sqrt 1-x^2 ,-Sqrt 1-x^2 , x,-1, 1 , AspectRatio ÆAutomatic -1.0 -0.5 0.5 1.0-1.0-0.5 0.5 1.0 voila. Or, we can use the AspectRatio command to make an even more oblate shape (but the figure is still a circle):

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• [PDF File]Does it converge or diverge? If it converges, find its ...

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  +1 = ∞ The sequence diverges. 3. X∞ n=2 n2 +1 n3 −1 The terms of the sum go to zero, since there is an n2 in the numerator, and n3 in the denominator. In fact, it looks like P 1 n, so we compare it to that: lim n→∞ n2−1 n3−1 1 n = lim n→∞ n3 −n n3 −1 = 1 Therefore, the series diverges by the limit comparison test, with P 1 ...

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• [PDF File]Techniques of Integration - Whitman College

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  1 3 u3/2 +C. Then since u = 1− x2: Z x3 p 1− x2 dx = 1 5 (1−x2)− 1 3 (1−x2)3/2 + C. To summarize: if we suspect that a given function is the derivative of another via the chain rule, we let u denote a likely candidate for the inner function, then translate the given function so that it is written entirely in terms of u, with no x ...

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• [PDF File]NUMERICAL INTEGRATION: ANOTHER APPROACH

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  Z 1 0 sqrt(x) dx= 2 3 nI−In Ratio 2 −7.22E −3 4 −1.16E −36.2 8 −1.69E −46.9 16 −2.30E −57.4 32 −3.00E −67.6 64 −3.84E −77.8 The column labeled Ratio is defined by I−I1 2n I−In It is consistent with I−In≈ c n3, which can be proven theoretically. In comparison for the trapezoidal and Simpson rules, I−In≈ c n1.5

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• [PDF File]Lecture 4 : Calculating Limits using Limit Laws

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  x!3 x3+2x2 x+1 x 1 (b) lim x!1 3 p x+ 1 2 (c) Determine the in nite limit (see note 1 above, say if the limit is 1, 1 or D.N.E.) lim x!2 x+1 (x 2). Polynomial and Rational Functions Please review the relevant parts of Lectures 3, 4 and 7 from the Algebra/Precalculus review page. This demonstration will help you visualize some rational functions:

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• [PDF File]Math 121 Homework 7: Notes on Selected Problems

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  — 2 ‡10–. 13.1.2. Show that x3 2x 2 is irreducible over Q and let be a root. Compute —1‡ –—1‡ ‡ 2–and 1‡ 1‡ ‡ 2 in Q— –. Solution. The polynomial x3 2x 2 is irreducible by Eisenstein’s cri-terion with the prime 2. (Alternatively, by the rational roots test, the only possible rational roots of x3 2x 2 are 1; 2 ...

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• [PDF File]TRIGONOMETRY LAWS AND IDENTITIES

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  x x y ⇡ 2 ⇡ 3⇡ 2 2⇡ 1 1 y =sin(x) x y ⇡ 2 ⇡ 3⇡ 2 2⇡ 1 1 y = cos(x) x y ⇡ 2 ⇡ 3⇡ 2 2⇡ 1 1 y = tan(x) x y 0 30 60 90 120 150 180 210 240 270 300 330 360 135 45 225 315 ⇡ 6 ⇡ 4 ⇡ 3 ⇡ 2 2 3 3 5 ⇡ 7⇡ 6 5⇡ 4 4⇡ 3 3⇡ 2 5⇡ 3 7⇡ 4 11⇡ 6 2⇡ ⇣p 3 2, 1 ⌘ ⇣p 2 2, p 2 ⌘ ⇣ 1 2, p 3 2 ⌘ ⇣ p 3 1 ...

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