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62 Solution of initial value problems (4) Topics † Properties of Laplace transform, with proofs and examples † Inverse Laplace transform, with examples, review of partial fraction, † Solution of initial value problems, with examples covering various cases Properties of Laplace transform 1 Linearity Lfc1f(t)c2g(t)g = c1Lff(t)gc2Lfg(t)g 2 First derivative Lff0(t)g = sLff(t)g¡f(0)Math131 Calculus I The Limit Laws Notes 23 I The Limit Laws Assumptions c is a constant and f x lim ( ) →x a and g x lim ( ) →x a exist Direct Substitution Property If f is a polynomial or rational function and a is in the domain of f, then = f x lim ( ) x aUnit 3 Algebra 1;

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1+2+3+4+5 to infinity=-1/12 proof pdf-1 2 ln 1x 1− x Q 1(x)=P 1(x)Q 0(x)− 1 Q 2(x)=P 2(x)Q 0(x)− 3 2 x Q 3(x)=P 3(x)Q 0(x)− 5 2 x2 2 3 showing the even order functions to be odd in x and conversely The higher order polynomials Q n(x) can be obtained by means of recurrence formulas exactly analogous to those for P n(x) Numerous relations involving the Legendre functionsLet us try adding up the first few terms and see what happens If we add up the first two terms we get 1 2 1 4 = 3 4 The sum of the first three terms is 1 2 1 4 1 8 = 7 8 The sum of the first four terms is 1 2 1 4 1 8 1 16 = 15 16 And the sum of the first five terms is 1 2 1 4 1 8 1

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;Z nare sometimes called \Normalized Spacings" Imagine the 'window' in time extending from t= 0 until t= T (1) The total amount of lifetime observed during this window is nT (1), since all n subjects are alive througout this time period This is just Z 1 Next consider the 'window23 Proof by Mathematical Induction To demonstrate P )Q by induction we require that the truth of P and Q be expressed as a function of some ordered set S 1 (Basis) Show that P )Q is valid for a speci c element k in S 2 (Inductive Hypothesis) Assume that P )Q for some element n in S 3 Demonstrate that P )Q for the element n 1 in SNumber Sense Workbook and Video Lessons Volume 1;

Let us try adding up the first few terms and see what happens If we add up the first two terms we get 1 2 1 4 = 3 4 The sum of the first three terms is 1 2 1 4 1 8 = 7 8 The sum of the first four terms is 1 2 1 4 1 8 1 16 = 15 16 And the sum of the first five terms is 1 2 1 4 1 8 1Unit 8 Algebra 1;So putting #x=1# we find #3c = = 1/(11)^2 = 1/4# Then dividing both ends by #3# we get #c = 1/12# Note that it is not really valid to manipulate divergent infinite series in these ways The calculations above are a shadow of the real derivation of the Ramanujan Sum of the series #1234#, which is more properly presented using the Riemann Zeta function and analytic continuation The useful thing about the above nonrigorous derivation is that it gives a very

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2 − π 4 cosx 12 cos3x 32 cos5x 52 cos7x 72 ··· (15) The constant of integration is a 0 Those coefficients a k drop off like 1/k2Theycouldbe computed directly from formula (13) using xcoskxdx, but this requires an integration by parts (or a table of integrals or an appeal to Mathematica or Maple) It was much2 = (n 1)T (2) T (1) Z n = T (n) T (n 1) Z 1;Z 2;42 The Cauchy condition 63 Example 410 The geometric series P an converges if jaj

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Math Team Academic Team Travel Information;John Baez (September 19, 08) "My Favorite Numbers 24" (PDF) The EulerMaclaurin formula, Bernoulli numbers, the zeta function, and realvariable analytic continuation by Terence TaoK=1 1 2k = 1 2 1 4 1 8 = ?

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= (1–23–45–6⋯)(⋯) Because math is still awesome, we are going to rearrange the order of some of the numbers in here so we get something that looks familiar, butINFINITE SERIES 415 n 1 5 10 2 3 n We observe that as n becomes larger and larger, 2 3 n becomes closer and closer to zeroUnit 10 Algebra 1;

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Case 2 Suppose n = 3m2 Then n 2= (3m2)2 = 9m 12m4 = 9m2 12m31 = 3(3m2 4m 1) 1 So we can write n2 = 3k 1 for k = 3m2 4m 1 Since we have proven the statement for both cases, and since Case 1 and Case 2 re ect all possible possibilities, the theorem is true 12 Proof by induction We can use induction when we want to show aThe proof of Theorem 2 is divided into two parts rst, a proof that A is the union of the equivalence There are ve distinct equivalence classes, modulo 5 0;1;2;3, and 4 The last examples above illustrate a very important property of equivalence classes, namely that an equivalence class may have many di erent names In the aboveStep one of the "proof" tries to persuade you of something rather silly, namely that the infinite sum (easiest example is remove all even numbers from 1,2,3,4,5 and you are left with an infinite set, compared to match up 12, 24, 36 etc and show that none are left behind) Exposure to such conflicts is a great and fun way to

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INFINITE SERIES 415 n 1 5 10 2 3 n We observe that as n becomes larger and larger, 2 3 n becomes closer and closer to zeroSetswithEqualCardinalities 219 N because Z has all the negative integers as well as the positive ones Definition131settlestheissue Becausethebijection f N!Z matches up Nwith Z,itfollowsthat jj˘jWesummarizethiswithatheorem Theorem131 Thereexistsabijection f N!ZTherefore jNj˘jZ The fact that N and Z have the same cardinality might prompt usSince there are five 3's and one six we expect roughly 5/6 of the rolls will give 3 and 1/6 will give 6 Assuming this to be exactly true, we have the following table of values and counts value 3 6 expected counts 5000 1000 The average of these 6000 values is then 5000 · 3 1000 · 6 5 1 = · 3 · 6 = 35 6000 6 6

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52 Closed sets 93 Example 519 To verify that the closed interval 0;1 is closed from Proposition 518, suppose that (x n) is a convergent sequence in 0;1Then 0 x n 1 for all n2N, and since limits preserve (nonstrict) inequalities, we haveWith both {1,2,3}and {1,2,3,4}(say)?Dot product and vector projections (Sect 123) I Two definitions for the dot product I Geometric definition of dot product I Orthogonal vectors I Dot product and orthogonal projections I Properties of the dot product I Dot product in vector components I Scalar and vector projection formulas There are two main ways to introduce the dot product Geometrical

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Unit 1 Algebra 1;4 TYLER CLANCY which we can see holds true to the formula The equation for m = 2 also proves true for our formula, as un2 = un1 un = un 1 un un = un 1 2un = un 1u2 unu3 Thus, we have now proved the basis of our inductionWith mean 3/2 Thus, the distribution of X¯ is nearly symmetric around 3/2 From the figure, we can see that the interval from 14 to 15 under the function g maps into a longer interval above =3than the interval from 15 to 16 maps below =3 Thus, the function g spreads the values of X¯ above =3more than below Consequently, we

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1 = T 2 = T 3 = 1 and T n = T n 1 T n 2 T n 3 for n 4 Prove that T n < 2n for all n 2Z Proof We will prove by strong induction that, for all n 2Z , T n < 2n Base case We will need to check directly for n = 1;2;3 since the induction step (below) is only valid when k 3 For n = 1;2;3, T n is equal to 1, whereas the righthand side of is21 Sequences and Their Limits 25 In this case, we call thenumber a a limit of thesequence {a n}Wesay that thesequence{a n}converges (or is convergent or has limit) if itconverges to some numbera A sequencediverges (or is divergent) if it does not convergeUnit 6 Algebra 1;

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Infinity2 5% NFPA 701 TM#1 (small scale), NFPA 101 (Class A Rating), CAN/ULCS 109 (large scale), IBC Section (Class A Rating), BS 5867 Part 2 Type B Performance, CAN/CGSB M80 Bacteria and Fungal Resistance0 1 2 3 4 5 6 O O O O O O O O In this picture, observation begins at time t = 0 and time passing is denoted by moving to the right on the number lineUnit 4 Algebra 1;

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A première vue oui, mais quand on y regarde de plus près, on découvre l'une des plus bHere's a fun little brain wrinkle pinch for all you nonmath people out there (that should be everyone in the world*) the sum of all natural numbers, from one to infinity, is not a= 1 , directly from definition 31 Solution According to definition 31, we must show (2) given ǫ > 0, n−1 n1 ≈ ǫ 1 for n ≫ 1 We begin by examining the size of the difference, and simplifying it ¯ ¯ ¯ ¯ n−1 n1 − 1 ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ −2 n1 ¯ ¯ ¯ ¯ = 2 n1 We want to show this difference is small if n

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De nining property of r, we have 1 = f2 = ar2 = a(r 1) = ar a Thus we have 1 = ar 1 = ar a which cannot be satisfed by any value of a But there's one more trick we can still deploy x2 = x1 is a quadratic equation, and it has two solutions r = 1 p 5 2 and s = 1 p 5 2 Any statement P(n) of the form fn = arn bsnL'addition de tous les nombres entiers positifs donne 1/12 Absurde ?2 − π 4 cosx 12 cos3x 32 cos5x 52 cos7x 72 ··· (15) The constant of integration is a 0 Those coefficients a k drop off like 1/k2Theycouldbe computed directly from formula (13) using xcoskxdx, but this requires an integration by parts (or a table of integrals or an appeal to Mathematica or Maple) It was much

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1 = 10 Hz and F 2 = 12 Hz sampled with the sampling interval T = 002 seconds Consider various data lengths N = 10,15,30,100 with zero padding to 512 points DFT with N = 10 and zero padding to 512 points Not resolved F 2 −F 1 = 2 Hz < 1/(NT) = 5 Hz EE 524, Fall 04, # 5 11Archimedes was fascinated with calculating the areas of various shapes—in other words, the amount of space enclosed by the shape He used a process that has come to be known as the method of exhaustion, which used smaller and smaller shapes, the areas of which could be calculated exactly, to fill an irregular region and thereby obtain closer and closer approximations to the total areaRead this too http//wwwbradyharanblogcom/blog/15/1/11/thisblogprobablywonthelpMore links & stuff in full description below ↓↓↓EXTRA ARTICLE BY TONY

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K=1 1 2k = 1 2 1 4 1 8 = ?(Section 21 An Introduction to Limits) 213 lim x 1 fx()= lim x 1 3x2 x 1 WARNING 3 Use grouping symbols when taking the limit of an expression consisting of more than one term = 31() 2 ()1 1 WARNING 4 Do not omit the limit operator lim x 1 until this substitution phase WARNING 5 When performing substitutions, be prepared to useSolution 0,1 −1 1,1 1 2,1 −1 3,1 1 4,··· This is 0, 3 2, 2 3, 5 4,··· Picture drawn in class on the real number line of dots alternating between being to the left and to the right of 1, but always getting closer and closer to 1 These dots each get a number above them 1 above 0, write 2 above 3 2, write 3 above 2 3, and so on

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ORDERS OF ELEMENTS IN A GROUP 3 When gn = e, nmight not be as small as possible, so the repetition in the powers of g may really occur more often than every nturns For example, ( 1)4 = 1, so Theorem31 says the only powers of 1 are ( 1)k for k2f0;1;2;3g, but we know that in fact a more economical list is ( 1)k for k2f0;1g This is connected with the fact that ( 1)2 = 1P197 The main pointin the section is to define vector spaces and talk about examples The following definition is an abstruction of theorems 412 and theorem 414 Definition 421 Let V be a set on which two operations (vector addition and scalar multiplication) are definedFunctions like 1/x approach 0 as x approaches infinity This is also true for 1/x 2 etc A function such as x will approach infinity, as well as 2x, or x/9 and so on Likewise functions with x 2 or x 3 etc will also approach infinity But be careful, a function like "−x" will approach "−infinity", so we have to look at the signs of x

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Unit 5 Algebra 1;3 and tanθ= sinθ cosθ = b a 21 Example Find argument and absolute value of z = 2 i Solution z = √ 22 12 = 5 zlies in the first quadrant so its argument θis an angle between 0 and π/2 From tanθ= 1 2 we then conclude arg(2 i) = θ= arctan 1 2 3 Geometry of Arithmetic Since we can picture complex numbers as points in theI thank the following people for their help in note taking and proof reading Mark Gockenbach, Ryan McNamara, Kaylee Walsh, Tjeerd Woelinga iii iv Contents Ackowledgementsiii 5 g 2 g 4 g 1 g 3 e The above latin square is not the multiplication table of a group, because for this square (g 1 g 2) g 3 = g 3 g 3 = e but g 1 (g 2 g 3) = g 1

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Here's a fun little brain wrinkle pinch for all you nonmath people out there (that should be everyone in the world*) the sum of all natural numbers, from one to infinity, is not aOfcourse, everydayexperience saysthat this is impossible However, mathematicians always take the point of view that if something is really obvious, then it ought to be easy to justify 2 π Here's the proof that g and g−1 are inverses gUnit 7 Algebra 1;

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