3 1 sqrt 2 x 2

• [PDF File]Math 104: Introduction to Analysis SOLUTIONS

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  4 Applying other theorems about behavior of limits under arithmetic operations with sequences, we conclude that lim 1 2 q 1+ 1 4n +2 = 1 2·1+2 = 1 4. 9.5. Let t1 = 1 and tn+1 = (t2 n + 2)/2tn for n ≥ 1. Assume that tn converges and find the limit.

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• [PDF File]Math 104: Improper Integrals (With Solutions)

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  1 (x−1)2/3 dx= 3[1+2 1/3]. RyanBlair (UPenn) Math104: ImproperIntegrals TuesdayMarch12,2013 11/15. ImproperIntegrals Tests for convergence and divergence The gist: 1 If you’re smaller than something that converges, then you converge. 2 If you’re bigger than something that diverges, then you diverge.

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• [PDF File]Python Question and Answers Built-in Functions

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  The functions sqrt and factorial are a part of the math module. The print function is a built-in function which prints a value directly to the system output. 2. What is the output of the expression: ... 2), (1, 3)] d) [(2, 3)] Answer: c EXPLANATION: The built-in function enumerate() accepts an iterable as an argument. The function shown in the ...

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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]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]Double integrals

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  x3 2 # 1 x=0 = 1 Note that Methods 1 and 2 give the same answer. If they don’t it means something is wrong. 0.11 Example Evaluate ZZ D (4x+2)dA where D is the region enclosed by the curves y = x2 and y = 2x. Solution. Again we will carry out the integration both ways, x first then y, and then

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• [PDF File]Math 2260 Exam #3 Practice Problem Solutions

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  Now we check the endpoints. When 2x 5 = 3, the series becomes X1 n=1 3n n23n = X1 n=1 1 n2; which converges. Likewise, when 2x 5 = 3, then series becomes X1 n=1 ( 3) n n23n = X1 n=1 ( n1)n3 n23n = X1 n=1 ( 1) n2; which also converges. Therefore, the series converges for all xso that 3 2x 5 3; which is the interval [1;4]. 2

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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]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]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]Integration by substitution

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  used in the substitution, i.e. 1+x2. As before, du = du dx dx and so with u = 1+x2 and du dx = 2x It follows that du = du dx dx = 2xdx So, substituting u for 1+x2, and with 2xdx = du in Equation (3) we have Z 2x √ 1+x2 dx = Z √ udu = Z u1/2 du = 2 3 u3/2 +c We can revert to an expression involving the original variable x by recalling that u ...

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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]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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• Midterm 2 solutions for MATH 53

  1. Find the volume of the solid that lies under the hyperbolic paraboloid z= 3y2 x2 +2 and above the rectangle R= [ 1;1] [1;2] in the xy-plane. Solution: We set up the volume integral and apply Fubini’s theorem to convert it to an iterated integral: ZZ R 3y 2 2x + 2 dA= Z 1 1 Z 2 1 3y 2x2 + 2 dydx= Z 1 1 [y3 yx + 2y]2 1 dx = Z 1 1 [23 22x+4 ...

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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]z=8−x2−y2 S z=x2+y2 R x2 +y2

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  3 O plot3d({sqrt(9-x^2), sqrt(9-y^2)},x=-3..3,y=-3..3); Figure 2. Part of the region S bounded by x2+z2 = a2 and x2 +y2 = a2 for x ≥ 0 Note that the projection of region S1 on the y − z plane, call it R is a a square 0 ≤ y ≤ a, 0 ≤ z ≤ a. We break

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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]PLOTTING AND GRAPHICS OPTIONS IN MATHEMATICA

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  Now with axes labelled and a plot label : Plot x, x^2, x^3, x^4 , x, 1, 1 , AxesLabel x, y , PlotLabel "Graph of powers of x" -1.0 -0.5 0.5 1.0 x-1.0-0.5 0.5 1.0 y Graph of powers of x Notice that text is put within quotes. Or to really jazz it up (this is an example on the Mathemat-

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• [PDF File]Physics 505 Homework No. 5 Solutions S5-1

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  1 2 1 r (r ·p)+(p·r) 1 r , (1) with the Hermitian operator pr = ¯h i ∂ ∂r + 1 r . (2) Show the operator defined in equation (1) is the same as that in equation (2). Show that pr in equation (2) is Hermitian (consider ψ(r,θ,φ) and ϕ(r,θ,φ)) and that when used in the Hamilton, pr of equation (2) gives the correct Schroedinger ...

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• [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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