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The limit is 1/e lim_(xrarroo)(11/x)^x has the form 1^oo which is an indeterminate form We will use logarithms and the exponential function Now, (11/x)^x = e^(ln(11/x)^x) So we will investigate the limit of the exponent lim_(xrarroo)(ln(11/x)^x) It will be convenient to note that 11/x = (x1)/x ln(11/x)^x = ln ((x1)/x)^x = xln((x1)/x) (Using a property of logarithms to bring the. Where b is a positive real number not equal to 1, and the argument x occurs as an exponent For real numbers c and d, a function of the form () = is also an exponential function, since it can be rewritten as = () As functions of a real variable, exponential functions are uniquely characterized by the fact that the growth rate of such a function (that is, its derivative) is directly. ` = ï Ë Ñ 5 > ê ¿ Ó 3 < Ì = ¯ k § Í ~ Ã á r ¯ þ À D O D O æ £ e À ü Ç Á k ~ ¸ Ê Ù ` e x ¨ ã o ) ¨ w Æ Ç þ 6 p % á r ª ;.

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E(aX) = aE(X), and Var(aX) = a2Var(X) (3) For any two rvs X and Y E(X Y) = E(X)E(Y) (4) If X and Y are independent, then Var(X Y) = Var(X)Var(Y) (5) The above properties generalize in the obvious fashion to to any finite number of rvs In general (independent or not) Var(X Y) = Var(X)V(Y)2Cov(X,Y), where Cov(X,Y) def= E(XY)−E. Combine like terms 05n 3 u 05n 3 u = 00 nx 05n 3 u = 00 Factor out the Greatest Common Factor (GCF), 'n' n(x 05n 2 u) = 00 Subproblem 1 Set the factor 'n' equal to zero and attempt to solve Simplifying n = 0 Solving n = 0 Move all terms containing n to the left, all other terms to the right Simplifying n = 0 Subproblem 2. Δy n n ) n = ( xΔ )n (−x = x n(Δ )(n−1 O(Δ 2 −x = nx n −1O(Δx) Δx Δx Δx As it turns out, we can simplify the quotient by canceling a Δx in all of the terms in the numerator When we divide a term that contains Δx2 by Δx, the Δx2 becomes Δx and so our O(Δx2) becomes O(Δx) When we take the limit as x approaches 0 we get.

If a and b are constants, then E(aXb) = aE(X) b Proof E(aXb) = sum (ax kb) p(x k) = a sum{x kp(x k)} b sum{p(x k)} = aE(X) b Variance We often seek to summarize the essential properties of a random variable in as simple terms as possible The mean is one such property Let X = 0 with probability 1. # cut here # This is a shell archive Remove anything before this line, # then unpack it by saving it in a file and typing "sh file". ñ r =!i 5 ® E !.

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Ioã ó / ÷ õ " æ õ Ï õ ó z õ þ D õ ñ ³ õ Ï o õ þ C Ó õ Î ü õ þ D * õ z Ý õ x 2 Z ö Õ,ëB&I S!. ­ ç µ Á ­ Ó ¡ ¶ ­ µ å Æ ® 1 R ± Ì v ­ ç Ô } l J ² ¤ l Ò } " ¬ e M _ ¢ ö è ¤ N k ê ¢ û / Ò ッ ­ ッ Õ \ ¦ Ü è ² é Ç ¶ Þ % Á Ð Ð Ð Ð Ð Ð Ð ¨ K X Ã Ó ¡ Á Æ Ë · ¹ ñ ½ d r X d y % Á Ð Ð Ð Ð ë ð ¦ Æ ê É ñ ® ¶ Þ { ½ Ó Ë · Ð Ð Ð Ð Ð ¦ ² é Ç ¶ Þ % $ n. Δy n n ) n = ( xΔ )n (−x = x n(Δ )(n−1 O(Δ 2 −x = nx n −1O(Δx) Δx Δx Δx As it turns out, we can simplify the quotient by canceling a Δx in all of the terms in the numerator When we divide a term that contains Δx2 by Δx, the Δx2 becomes Δx and so our O(Δx2) becomes O(Δx) When we take the limit as x approaches 0 we get.

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52 c JFessler,May27,04,1314(studentversion) FT DTFT Sum shifted scaled replicates Sum of shifted replicates DTFS Z DFT Sinc interpolation Rectangular window. If you want to write better essays, it’s helpful to understand the criteria teachers use to score them Instead of solely focusing on the grade you are given, focus on how you are being graded and how you can improve, even if you are already getting a high grade. P(N) N (x 0) = f (N)(x 0) The constant c n is the nth Taylor coe cient of y = f(x) about x 0 The Nthorder Maclaurin polynomial for y = f(x) is just the Nthorder Taylor polynomial for y = f(x) at x 0 = 0 and so it is p N(x) = XN n=0 f(n)(0) n!.

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Please see below I'm not sure I understand your question I don't know what you mean by "using the derivate number which is f'(a)" If I understand it, I think I need to point out that lim_(xrarra)(x^na^n)/(xa) = f'(a) for f(x) = x^n That is For f(x) = x^n, f'(a) = lim_(xrarra)(f(x) f(a))/(xa) = (x^na^n)/(xa) We also know, by the power rule for derivatives, That for f(x) = x^n, we. Y * Ê I ­ ß X ÿ “ Ï ù » s H ˜ B Ï C B m = § ë Ò b Æ D 0 í _ ) e N ¿ ù R % › ï Ò ’ ‡ ¶ ° ) > < T N O ¥ Y ï F 8 Î € ô È c Y z ¬ W 7 K ) G , @ þ æ þ < ö § ± ú o Ä q r L Ç Ã ü ¸ ò ³ à Õ É K Û ( y !. æ Ì#Õ b S u b S ^ ¥ Ñ × ½ ¸ g B1= Regional disaster management capability rating system for promoting recovery and creating resilient community.

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Title May 19 Disciplinary Action Report Author ce Created Date 7/10/19 751 AM. Xn De nition 2 1 The Taylor series for y = f(x) at x 0 is the power series P 1(x) = f(x 0) f0. If a and b are constants, then E(aXb) = aE(X) b Proof E(aXb) = sum (ax kb) p(x k) = a sum{x kp(x k)} b sum{p(x k)} = aE(X) b Variance We often seek to summarize the essential properties of a random variable in as simple terms as possible The mean is one such property Let X = 0 with probability 1.

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Or 2 15 (2 a 3 x)( )3= (21) Z p ax bdx= 2b 3a 2x 3 p ax b (22) Z (ax b)3=2 dx= 2 5a (ax b)5=2 (23) Z x p x a dx= 2 3 (x 2a) p x a (24) Z r. Æ f> Ç Û µ ¡#Õ è 7d 4 ' &k>& æ/² v)~ z&k6×>8 %> D>' c G b Ø Má&k K 2 Ç. Or 2 3 ( a)3=2 4 15 x a)5=2;.

µ ® % ú Õ &U!m Õ æ Õ )½ Õ } Ùñ ò =ñ ¹ v á =ñ Ù ý = r Í Í Y!õ*¡ Í Q y Í Í á(* Í % ò Í } Íñ a e = Í r Í } Í Í a ¶ Íñ æ =)ê v r M V)Ê % = Õ i v !. æ 3 k µ u Ã Ë C ì É 7 l;. Or 2 15 (2 a 3 x)( )3= (21) Z p ax bdx= 2b 3a 2x 3 p ax b (22) Z (ax b)3=2 dx= 2 5a (ax b)5=2 (23) Z x p x a dx= 2 3 (x 2a) p x a (24) Z r.

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