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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;

The Ramanujan Summation 1 2 3 1 12 By Mark Dodds Cantor S Paradise Medium

1+2+3+4+5 to infinity=-1/12 proof pdf

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

Countertops 1/4", 1/2", 5/8" thickness Minimum 3/4" APA spanrated plywood or OSB with an Exposure 1 classification or better Cover base with USG Durock™ Brand Tile Membrane, 15 lb felt or 4mil polyethylene and attach with 1/4" galvanized staples Walls and Ceilings 1/2" andExercise 26Use the following theorem to provide another proof of Exercise 24 Theorem 21 For any realvalued sequence, s n s n!0 ()js nj!0 s n!0 Proof Every implications follows because js nj= jjs njj= j s nj Theorem 22 If lim n!1 a n= 0, then the sequence, a n, is bounded That is, there exists a real number, M>0 such that jaBut if we decide that xwill be within 1=2 of 1, then jxjis at most 3=2 So jx2 x 1j jxj2 jxj j1j 9=4 3=2 1 = 19=4

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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 = ?

Infinite Series

The Great Debate Over Whether 1 2 3 4 1 12 Smart News Smithsonian Magazine

= (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;

Q How Does 1 2 3 4 5 1 12 Make Any Sense Ask A Mathematician Ask A Physicist

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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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