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Z ef(y)dy G(x) Finally, dividing by ey gives u(x;y) = e y Z eyf(y)dy e yG(x) = F(y) e yG(x);X 4 k(x,z) = g(x)g(z), for g X !Ie, f Rn!Ris convex if g( ) = f(x 0 v) is convex 8x 0 2dom(f) and 8v2Rn We just proved this happens i g00( 2) = vTrf(x 0 v)v 0;

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G x z x 0 f t dt-For some z2x;y)f(y) f(x) f0(x)(y x) Now to establish (ii) ,(iii) in general dimension, we recall that convexity is equivalent to convexity along all lines;X C o m t s f h D T r a l ­ O c y u h t p / w o c u a i n l f m e d x _ 1 2 7 or serving in other jobs which qualify) The following is an explanation of the various levels of specific vocational preparation Level Time 1 Short demonstration only 2 Anything beyond

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Proposition 17 Let f;f 0 X !Y and g;g Y !Z be continuous maps, and let g f;g 0 f X !Z be the respective composite maps If f 'f0and g 'g0, then g f 'g0 f0 Proof Let F X I !Y be a homotopy between f and f0and G Y I !Z be a homotopy between g and g0 De ne a map H X I !Z by H(x;t) = G(F(x;t);t) Clearly, H is continuousZ ∞ −∞ G(−x y′,t)f(y′)dy′ = Z ∞ −∞ G(x −y′,t)f(y′)dy′ =u(x,t) In the last line, we used that G is an even function of its first arguement Smooth even functions have zero slope at x =0, ie, ux(0,t)=0 So we solve our semiinfinite domain problem by extending the initial data to −∞(e)Write g(t) in terms of f(t) and use the three previousproperties to solve y(t) = f(t) ∗ g(t) in terms of x(t) from part a (f) Solve and then sketch the function z(t) = g(t 2) ∗ g(t) (hint use shifted versions of x(t) from part a)

DOE A to Z The State of NJ site may contain optional links, information, services and/or content from other websites operated by third parties that are provided asThe Indefinite Integral of f(x) is the General Antiderivative of f(x) Z f(x)dx = F(x) C Z 2xdx = x2 C Definition Riemann Sum The Riemann Sum is a sum of the areas of n rectangles formed over n subintervals in a,b Here the subintervals are of equal length, but they need not be The height of the ith rectangle, is3 Find the area bounded between the regions y= 1 2x2 and y

For f(x) and g(x) defined on 0 £ x < ¥, such as in the case of the semiinfinite string, the solution is not welldefined For positive c and t > 0, we have that f(xct) is not defined for xct < 0, or t > x/c This also affects the range of integration over values where g is not defined∂x (x,t)dt Z x 0 ∂u ∂y (s,0)ds Of course if Gisn't a ball we might not be able to integrate along quite this path,butsimilarargumentswork Exercise 12 LetGbeanopensubsetofC DefineG= z z∈G Suppose that f G→C is analytic Show that f?1 k(x,z) = αk1(x,z)βk2(x,z), for α,β 0 2 k(x,z) = k1(x,z)k2(x,z) 3 k(x,z) = k1(f(x),f(z)), where f X !

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(f ∗g)(t) = Z t 0 f(τ)g(t −τ)dτ Remarks I f ∗g is also called the generalized product of f and g I The definition of convolution of two functions also holds in the case that one of the functions is a generalized function, like Dirac's delta Convolution of two functionsProblem 32 Let A,W, and t 0 be real numbers such that A,W > 0, and suppose that g(t) is given by g(t) A t 0 t 0 − W 2 t 0 W 2 Show the Fourier transform of g(t) is equal to AW 2 sinc2(Wω/4) e−jωt0 W using the results of Problem31 and the propertiesof the Fourier transformNow f(t)=tdoes not depend on x, so Z b 0 g(x)dx= Z b 0 f(t) t Z t 0 dx dt And noting that Z t 0 dx= t we have Z b 0 g(x)dx= Z b 0 f(t) t tdt= Z b 0 f(t)dt As fis integrable on 0;b, the integral is nite, so gis also integrable on 0;b 1 Proposition 02 (Exercise 7) Let fbe a

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X(t) = F Y(t) for all t 2 Expected Values The mean or expected value of g(X) is E(g(X)) = Z g(x)dF(x) = Z g(x)dP(x) = (R 1 1 g(x)p(x)dx if Xis continuous P j g(x j)p(x j) if Xis discrete Recall that 1 Linearity of Expectations E P k j=1 c jg j(X) = P k j=1 c jE(g j(X)) 2 If X 1;;X n are independent then E Yn i=1 X i!Example 4 Find the tderivative of z = f (x(t),y(t)), where f(x,y) = x5y6,x(t) = et, and y(t) = √ t Solution Because f(x,y) is a product of powers of x and y, the composite function f (x(t),y(t)) can be rewritten as a function of t We obtain f (x(t),y(t)) = x(t)5y(t)6 = (et)5(t1/2)6 = e5tt3 Then the Product and Chain Rules for oneJun 11, 14 · z is local to the function fThere's also a z declared in the global environment but it doesn't have an impact on the calculation of f, it's just there to try to throw you off;

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For arbitrary functions f and g, thus proving our claim ⁄ Geometric Interpretation The general solution of the wave equation is the sum of two arbitrary functions f and g where f = f(xct) and g = g(x¡ct)In particular, f(xct) is a wave moving to the left with speed c, while g(x¡ct) is a wave moving to the right with speed c 53 Initial Value ProblemRather, in the case where z = f ⁢ (x, y), x = g ⁢ (t) and y = h ⁢ (t), the Chain Rule is extremely powerful when we do not know what f, g and/or h are It may be hard to believe, but often in "the real world" we know rateofchange information (ie, information about derivatives) without explicitly knowing the underlying functionsIntuitively, a function is a process that associates each element of a set X, to a single element of a set Y Formally, a function f from a set X to a set Y is defined by a set G of ordered pairs (x, y) with x ∈ X, y ∈ Y, such that every element of X is the first component of exactly one ordered pair in G In other words, for every x in X, there is exactly one element y such that the

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H t t ps //w w w yo u t u be c o m /w a t c h ?R(t) = f(t)ig(t)jh(t)k = hf(t),g(t),h(t)i for some scalar functions f, g, and h, the component functions or r In 3 dimensions, we get a space curve, and in 2 dimensions, a plane curve Problem (Page 863 #48) Find the intersection of z = p x2 y2 and y 2z = 2 First eliminate z z = p x2 y2 = 1 2 (2y) =) 4(x2 y2) = (2y)2 = y2 4y 4 =) 3y2L H ` L 4 e V = e V c M d @ = h p X c 4 e V 5 e U g L J V d Q U p X Z O l X J k L ;

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Z = f (x, y) where x and y are functions of t, gives z = h(t) = f (x(t), y(t)) z x y t t @z @x dx dt @z @y dy dt z = f (x, y) depends on two variables Use partial derivatives x and y each depend on one variable, t Use ordinary derivative To compute dz dt There are two paths from z at the top to t's at the bottom Along each pathR 5 k(x,z) = f(k1(x,z)), where f is a polynomial with positive coefficients Proof Since each polynomial term is a product of kernels with a positive coefficient, the proof follows by applyingThe form z= f(xat) g(x−at) is a solution of the wave equation ∂2z ∂t2 = a2 ∂2z ∂x2 Solution Let u= xatand v= x−at Then z= f(u) g(v) and the chain rule gives ∂z ∂x = df du ∂u ∂x dg dv ∂v ∂x = df du dg dv, ∂z ∂t = df du ∂u ∂t dg dv ∂v ∂t = a df du −a dg dv Thus ∂2z ∂x2 = ∂ ∂x ∂z ∂x

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If g(x) = Z x a f(t)dt, then g0(x) = f(x) Z b a f(x)dx= F(b) F(a), where Fis any antiderivative of f 2 Give the de nition of the de nite integral Z b a f(x)dx= lim n!1 i=1 f(x i) x iOr you may be more speci c and use right endpoints Z b a f(x)dx= lim n!1 i=1 f a b a n i!Compose (f,g,z) returns f (g (z)) where f = f (x), g = g (y), and x and y are the symbolic variables of f and g as defined by symvar example compose (f,g,x,z) returns f (g (z)) and makes x the independent variable for f That is, if f = cos (x/t), then compose (f,g,x,z) returns cos (g (z)/t) whereas compose (f,g,t,z) returns cos (x/g (z))Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history

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Transcribed image text 3 UTM Z G UTA IM, UTM A periodic function f(x) is defined by f(x) =it for 2 < x < 2 and f(x) = f(x4) i) Sketch the graph of the function for 63 UTMB UNEM < X < 6 Then, determine IMU ether f(x) is even, odd or heither marks) ii) Hence, find its Fourier series= i E(X i) 3 We4 f(z)=g(z), where de ned (ie where g(z) 6= 0) 5 (g f)(z) = g(f(z)), the composition of g(z) and f(z), where de ned 23 Complex derivatives Having discussed some of the basic properties of functions, we ask now what it means for

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V = Q Z 1 U u 2 MR X T Y & l i s t = P L_ C l x R t MU w Z u 3 m C 7 d s c B 9 v R 9 6 E X l 3 m 1 3 h t t ps //w w w yo u t u be c o m /w a t c h ?W(ξ(x,t),η(x,t)), we find that u(x,y) = F(xct) G(x−ct) (62) Conversely, if F and Gare of class C2, then udefined by u(x,t) = F(x ct) G(x−ct) is a classical solution of (61) T The families of lines x−ct= constant and xct= constant, 42G→C defined by f?(z) = f(z) is alsoanalytic Lecture2MöbiusTransformations 2

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F ∗ g(t) = Z t x=0 f (x)g(t − x)dx is well defined and finite for every positive value of t In other words, f ∗ g is a welldefined function on (0,∞), at least whenever f and g are both piecewise continuous on (0,∞) (In fact, it can then even be shown that f ∗ gChain Rules for One or Two Independent Variables Recall that the chain rule for the derivative of a composite of two functions can be written in the form d dx(f(g(x))) = f′ (g(x))g′ (x) In this equation, both f(x) and g(x) are functions of one variable Now suppose that f is a function of two variables and g is a function of one variableWhere we have replaced the arbitrary function e y R eyf(y)dywith another we call F for convenience Exercise 4 Solve the wave equation subject to the initial conditions u(x;0) = xe x2;

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F(Xt,Yt)=g(Zt) (2) where Xt, Yt and Zt are strictly positive variables This equation is clearly also valid at the steady state f(X,Y)=g(Z) (3) To find the loglinearized version of (2), rewrite the variables using the identity Xt = exp(log(Xt))1 andthentakelogsonbothsides log(f(elog(Xt),elog(Yt))) = log(g(elog(Zt))) (4)Background Platinumbased concurrent chemoradiotherapy is the standard of care for patients with locoregionally advanced nasopharyngeal carcinoma Additional gemcitabine and cisplatin induction chemotherapy has shown promising efficacy in phase 2 trials Methods In a parallelgroup, multicenter, randomized, controlled, phase 3 trial, we compared gemcitabine and cisplatinC o nta c t i nfo r ma ti o n i s di ffe re nt fo r a ny o f the a utho r i z e d pro g r a ms, pl e a se c he c k the pro g r a m na me be l o w a nd pro v i de the c o nta c t i nfo r ma ti o n T he Sta te c o nta c t i nfo r ma ti o n i s di ffe re nt fo r the fo l l o w i ng pro g r a ms

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V = N K 8 _ T v u 6 bJ k & t = 5 s Reading/ Writing minutes S t o r y o f t h e W e e k ' O l dC s G J k L 7 j L T e 4 4 Z e ^ V j ` L ` U 4 Z e o L j w ` ;Student Solutions Manual for Stewart's Essential Calculus (2nd Edition) Edit edition Solutions for Chapter 11R Problem 35E Suppose z = f(x, y), where x = g(s, t), y = h(s, t), g(1, 2) = 3, gs(1, 2) = −1, gt(1, 2) = 4, h(1, 2) = 6, hs(1, 2) = −5, ht(1 2) = 10, fx(3, 6) = 7, And fy(3, 6) = 8, Find ∂z/∂s and ∂z/∂t when s = 1 and t

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2 f(z) g(z);E 4 1$9 o N X h w U L p N X s G r L ` L e 7 HConsider the function f (x, y, z) = x ln y − x y z 2 e z, y > 0 (a) Find the partial derivatives f x, f y and f z of f at any point (x, y, z) with y > 0, and the directional derivative of f at the point P = (2, 1, 0) in the direction of the vector v = (− 2, 2, 1) 4 marks (b) Find the direction in which the function f (x, y, z

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8x 0 2dom(f);8v2Rn and 8 st x 0Initially by f(x)att= 0 remains unchanged in form as tincreases, ie, f propagates to the right at the speed c Similarly gpropagates to the left at the speed c The lines x−ct=constant and xct= contant are called the characteristic curves (lines) along which signals propagate Note that another way of writing (12) is φ(x,t)=F(t− x/c)GAll we need to do

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The assignment z < 10 (or any other value) before the call to f will not change the f(3) y is a formal parameter for g which will be evaluated when needed When you compute f(3), you will calculate 3 g(3)The Fundamental Theorem of Calculus, Part 1 If f is continuous on a,b, then the function g defined by g(x) = Z x a f(t)dt a ≤ x ≤ b is continuous on a,b and differentiable onX(k) = x(k 1) t kG t k (x(k 1)) (87) where G tis a generalized gradient of f, G t(x) = x prox t(x trg(x)) t () Key Points Proximal mapping prox t() can be computed analytically for a lot of important hfunctions The mapping prox t() doesn't depend on gat all, only on h This also means that gcan be a complicated function;

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Assuming z = f(x;y), x= g(t), y= h(t), we start with z 2 From z, we draw a branch for each variable zdepends on, xand yin this case, so we draw two branches 3 From each of these variables, we repeat the procedure, that is draw a branch for each variable itCHAPTER 1 CALCULUS OF EUCLIDEAN MAPS 2 Distance Function on IRn d(x;y) = jx 2yj= p (x1 y1)2 (x2 y)2 (xn yn)2 v u u t i=1 (xi yi)2 Open sets in Rn B r(p) = open ball of radius rcentered at p = fx2Rn d(x;p) 0 such that B "(p) ˆU Euclidean Mappings F Rn!IRm These are the types of maps that will ariseSince g(x) 0, mg(x) f(x)g(x) Mg(x) Hence by Theorem 69, m Z b a g(x)dx Z b a f(x)g(x)dx M Z b a g(x)dx If R b a g(x)dx= 0, then R b a f(x)g(x)dx= 0 and so (4) is trivially satis ed by any c Otherwise, set k= R b a f(x)g(x)dx R b a g(x)dx and note that k2m;M By Theorem 46 (IVT), there exists c2(a;b) such that f(c) = k 623 If f2Ra;b

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F x = λg x f y = λg y f z = λg z g(x, y, z) = k This is a system of four equations in the four unknowns x, y, z, and λ, but it is not necessary to find explicit values for λ For functions of two variables the method of Lagrange multipliers is similar to the method just described

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