√70以上 1+2+3+4+5 to infinity=-1/12 proof 341964-1+2+3+4+5 to infinity=-1/12 proof
Heavy Duty Weather Proof Multi Purpose Enclosure Amphenol Broadband 9 x 9 x 4 (Interior approx 8 x 8 x 31/2) Brand Amphenol Broadband 47 out of 5 stars 4 ratingsLet S = Consider S1= Now, this sum should be 0 or 1 based on number of natural numbers taken If infinite numbers are even, S1=0, if odd S1=1 But, Riemann zeta function gives it a value of ½ Mathematical community too agrees that the sum is ½ How?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 a
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1+2+3+4+5 to infinity=-1/12 proof
1+2+3+4+5 to infinity=-1/12 proof-The nth finite sum is 2 1/2^n This converges to 2 as n goes to infinity, so 2 is the value of the infinite sum Submit Your Own Question Create a Discussion Topic This part of the site maintained by (No Current Maintainers)One example of a convergent series is 1/21/41/81/16 This series converges to the number 1 It's pretty easy to see why after the first term, we're halfway to 1
Correction Does 1 2 3 4 1 12 Absolutely Not I Think
Let S = Consider S1= Now, this sum should be 0 or 1 based on number of natural numbers taken If infinite numbers are even, S1=0, if odd S1=1 But, Riemann zeta function gives it a value of ½ Mathematical community too agrees that the sum is ½ How?For example, it will include 1/1, 2/2, 3/3, each of which is equal to 1 In fact, every rational number will be represented infinitely many times Technically, this violates the conditions of strict onetoone correspondenceA quick and easily digestible video to illustrate the peculiarities of the Riemann's Zeta function, and how it applies to real life physicsTo help you in vi
Step 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 toJohn Baez (September 19, 08) "My Favorite Numbers 24" (PDF) The EulerMaclaurin formula, Bernoulli numbers, the zeta function, and realvariable analytic continuation by Terence Tao4 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 induction
1 2 3 4 5 ⋯ = − 1 12 L Euler and S Ramanujan have already provided sufficient proof long ago There is no doubt that the Equation is true S Ramanujan's mathematical powers were absolutely phenomenal Then again, in the popular imagination, he has become known as the man "who knew infinity"Then we introduced the concept of limits, and said "welp, now infinite sums are sometimes equal to a real number, and 1 1/2 1/4 = 2 Now you can tell jokes about a bar and an infinite number of mathematicians coming in, no need to thank me"Euler wanted to know if you could find an answer to endless sums of numbers like 1 plus 1/2 plus 1/3 plus 1/4 on up to infinity, or the squares of those fractions
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1234 to infinity = 1/12 Binomial expansion first three nonzero terms addition Can someone answer a question on my maths exam?The sum to the infinity of the series 1 2/3 6/3^2 10/3^3 14/3^4 is asked Oct 10, 18 in Mathematics by Samantha ( 3k points) sequences and series級数 1 2 3 4 5 の部分和は順に 1, 3, 6, 10, 15, と続き、第 n 部分和は簡単な公式 ∑ = = () によって与えられる
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P p p p 2 i1 i2 i3 i4 This gives N(x) > x/2 Combining this with our upper bound on N(x) we have x/2 < N(x) < sqrt(x) ⋅ 2 i But this is false for all x larger than 2 2i2, completing the proof It is interesting to note that there is a good bound for the partial sums of the reciprocals of the primes If we let S(x) = the1 = 2 1/1 3/2 = 2 1/2 7/4 = 2 1/4 15/8 = 2 1/8 and so on;Smart News Keeping you current The Great Debate Over Whether 1234 ∞ = 1/12 Can the sum of all positive integers = 1/12?
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The proof featured in a video published by the YouTube channel Numberphile ostentatiously titled as ASTOUNDING = 1/12 Physicist Edmund Copeland, the one proving it in the video, after accomplishing the feat, ecstatically admits that "it looks like mathematics hocus pocus"Yes it is possible according to one stupid proof which is absolutely wrong but we can get convinced that it's correct Don't apply that great ramanujan has given this proof for what we are thinking He has given it for a different purpose not forCompute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history
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3n5 2 = 3(2k)5 2 = 25 3k5 2 = 2(24 3k5 3 2)1 = 2(48k5 3 2)1 Since 48k5 3 2 ∈ Z, it follows that 3n5 2 is odd What's wrong with the proof?If you add up every positive number to infinity, you get minus a twelfth this is as true as 11 = 2 How can you know this if it's impossible to add every number to infinity because infinity isIt can, sort of
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To understand it, you have to somehow understand how 1 1 1 1 equals out to 5 and then do some masterful work with 1 2 3 4 in an equation with 1I am currently reading a book called The Music of the Primes which is about prime numbers and in particular the Riemann HypothesisWhy 1 2 3 4 To Infinity 1 12 In 3 Minutes Youtube For more information and source, see on this link https//wwwyoutubecom/watch?v=0yHHadGDj6w
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Correction Does 1 2 3 4 1 12 Absolutely Not I Think
(18) the result can be either 0 or 1 (5) the result does not exist (4) the result is 1 ⁄ 2 (3) the result is 1 (2) the result is infinite (30) no answer The researcher, Giorgio Bagni, interviewed several of the students to determine their reasoning Some 16 of them justified an answer of 0 using logic similar to that of Grandi and RiccatiHillman 43in Black Neoprene Hole Plug Hillman 38in Chrome Plated Steel Hole Plug Hillman 125in Chrome Plated Steel Hole Plug Hillman 2Pack 0375in Black Neoprene Hole Plug Hillman 2Pack 0625in Black Plastic Hole Plug Hillman in Black Neoprene Hole PlugStack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers Visit Stack Exchange
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Some previous threads Straight Dope Message Board – 9 May 07 1234 = 1/12 In what sense?13 MARKOV CHAINS CLASSIFICATION OF STATES 155 2 1 3 4 5 6 06 1 04 03 03 05 05 07 01 03 02 04 07 We observe that 3 can only be reached from 3, therefore 3The proof featured in a video published by the YouTube channel Numberphile ostentatiously titled as ASTOUNDING = 1/12 Physicist Edmund Copeland, the one proving it in the video, after accomplishing the feat, ecstatically admits that "it looks like mathematics hocus pocus"
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Be careful with what you claim to be an integer!!7 1 5 9 13 (4n 3) = 2n2 n Proof For n = 1, the statement reduces to 1 = 2 12 1 and is obviously true Assuming the statement is true for n = k 1 5 9 13 (4k 3) = 2k2 k;Lamb E (14), "Does 123Really Equal –1/12?", Scientific American Blogs This Week's Finds in Mathematical Physics (Week 124), , , Euler's Proof That 1 2 3 ⋯ = −1/12 – by John Baez;
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In mathematics, the harmonic series is the divergent infinite series ∑ = ∞ = ⋯ Its name derives from the concept of overtones, or harmonics in music the wavelengths of the overtones of a vibrating string are 1 / 2, 1 / 3, 1 / 4, etc, of the string's fundamental wavelengthEvery term of the series after the first is the harmonic mean of the neighboring terms;Hence, it suffices to show that 3 2 is a lower bound which we do as follows 2n1 n1 ≥ 3 2 2(2n1) ≥ 3(n1) n ≥ 1 which is true for all n ∈ N Reversing the above argument shows that 3 2 is a lower bound ¤ The central question in this section is "Does every nonempty set of numbers have a sup?" The simple answer is no–the set NAn interwebs firestorm has been raging recently about a Numberphile video that makes the astounding claim that if you add up all the positive whole numbers from one to infinity, the result will be 1/12 To write it out more concisely 1234 = 1/12 , (where the three dots indicate all the rest of the positive numbers up to infinity)
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S2k ‚ 1 1 2 2¢ 1 4 4¢ 1 8 2k¡1 ¢ 1 2k = 1 k 2 for all k Theorem 3 If P1 n=1 j an j converges then P1 n=1 an converges Proof Since P1 n=1 j an j converges the sequence of partial sums of P1 n=1 j an j satisfles the Cauchy criterion Therefore, the sequence of partial sums of P1 n=1 an satisfles the Cauchy criterionThe result is false, but it can be fixed If n is an even integer, then 3n5 2 is even When writing a proof 1 Write a proof so thatThus, you can never provide a list in the form of $\{a_1, a_2, a_3,\cdots\}$ that contains all the elements in, say, $0,1$ This fact can be proved using a socalled diagonal argument, and we omit the proof here as it is not instrumental for the rest of the book
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Below is a visual description of Definition 124 In essence, injective means that unequal elements in A always get sent to unequal elements in B Surjective means that every element of B has an arrow pointing to it, that is, it equals f(a) for some a in the domain of fCorrosion Proof Basket Stainless Steel Mounting Hardware Main Cone Size (Inches) 65" Customers who viewed this item also viewed Page 1 of 1 Start over Page 1 of 1 Previous page Kenwood KFC1653MRW 65" 2Way Marine Speakers Pair (White) 43 out of 5 starsAn interwebs firestorm has been raging recently about a Numberphile video that makes the astounding claim that if you add up all the positive whole numbers from one to infinity, the result will be 1/12 To write it out more concisely 1234 = 1/12 , (where the three dots indicate all the rest of the positive numbers up to infinity)
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4 This follows by induction, using property 2 5 The proof follows from the deflnition F0(s) = d ds Z 1 0 e¡stf(t)dt = Z 1 0 @ @s (e¡st)f(t)dt = Z 1 0 (¡t)e¡stf(t)dt = Lf¡tf(t)g 6 This proof also follows from deflnition Lfeatf(t)g Z 1 0 e ¡steatf(t)dt = Z 1 0 e (s a)tf(t)dt = F(s¡a)Functions 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 x1234 to infinity = 1/12 Binomial expansion first three nonzero terms addition Can someone answer a question on my maths exam?
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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, but1 2 3 4 5 6 7 To Infinity 1 12 For more information and source, see on this link https//dushyantthakur2251wixsitecom/drtech/singlepost//TOWhy 1 2 3 4 To Infinity 1 12 In 3 Minutes Youtube For more information and source, see on this link https//wwwyoutubecom/watch?v=0yHHadGDj6w
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(13) we will prove that the statement must be true for n = k 1 1 5 9 13 4(k 1) 3 = 2(k 1)2 (k 1) (14)Does anyone have a proof of this?Does anyone have a proof of this?
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What you have stumbled upon is the Basel problem, the original Basel problem is to find the following sum math\sum\limits_{n=1}^\infty \frac{1}{n^{2}} = \frac{\pi^{2}}{6}/math The proof of this sum was originally given by Euler, but there exis
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