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FAQ

How do we know the eligibility to fill out Form 12 BB?
Every year as a salaried employee many of you must have fill Form 12BB, but did you ever bothered to know its purpose. Don’t know??It is indispensable for both, you and your employer. With the help of Form 12BB, you will be able to figure out how much income tax is to be deducted from your monthly pay. Further, with the help of Form 12BB, you will be in relief at the time of filing returns as at that time you will not have to pay anything due to correct TDS deduction.So, before filing such important form keep the below listed things in your mind so that you may live a tax hassle free life.For More Information:- 7 key points which must be known before filling Form 12BB
How do I fill out the class 12 performance check that is required for the JoSAA?
Thanks for A2A.It's a simple job.Just take printouts ofPage 51 ( for iit's).Page 52(for nit's , iiit's and other gfti's).Page 53 ‡ medical certificatePage 54- undertaking by candidate.Page 55 ‡ switch over between slide , float and freezeComing to the performance check,U have to fill them by yourself.Category cut off marks are uploaded in the official site of josaa.So, have a visit at it and note down the category cut off for top 20 percentile.
Is it true that the sum of all natural numbers is -1/12?
And so we return to the infamous sum of natural numbers, probably the most popular and most confused piece of mathematics circulated through the media and the general public. Thousands have attempted to portray the strange beauty that lies within it, though very few actually hit the nail on the head.As one angry Quoran noted on a different answer to this question, there are countless Quora answers devoted to the confusing yet elegant statement often written as[math]\displaystyle \sum_{n=1}^{\infty}{n} = -\frac{1}{12}[/math]Being not formally trained in complex analysis and probably way less qualified to talk about the topic than other Quorans, my answer is a summary of my own understanding of this strange result, and probably lacks in some amount of formal definition and jargon that would otherwise characterize a true mathematical evaluation of it. Do not take this as any official statement on the matter. But, nonetheless, I write, and if you feel so compelled as to continue reading, I thank you.Let us begin.Adding things is easy. Whether adding [math]2+2[/math] or [math]f(x)+g(x)[/math], addition is a rather simple and natural mathematical concept to us. It is a well-defined[1]operation, setting the basis for nearly every other operation among the natural numbers, integers, rationals, complex numbers, and the plethora of rings[2] that can be obtained from them.Unsurprisingly, mathematicians developed a symbol to represent adding a whole bunch of things, to avoid the hassle of writing all the terms out. Take the sum [math]1+3+5+7=16[/math], the sum of the first four odd numbers. Using sigma notation[3], this can be expressed as[math]\displaystyle \sum_{n=1}^{4}{(2n-1)} = 16[/math]where [math]n[/math] takes on the values 1 to 4 and is substituted into the expression [math]2n-1[/math] for each one. Add them up, and you get, in our case, 16.Sigma notation allows for useful generalizations of summing. For example, if we let the upper bound of our sum be some variable [math]m[/math], we can write[math]\displaystyle \sum_{n=1}^{m}{(2n-1)} = m^2[/math]which is a succinct way of expressing the fact that the sum of the first [math]m[/math] odd numbers is [math]m^2[/math].(The formal proof of this is a fun exercise, or you can find it elsewhere on Quora.)The past four centuries have seen ample study of interesting sums, results such as Faulhaber’s formula[4] and even integral calculus itself is all about studying sums and their properties.Calculus specifically looks into what happens when you add an infinite number of things. Now think about that for a second…what does it even mean to add infinitely many things? If I add [math]1[/math] to itself over and over again, what would I get? Is it infinity? Undefined? Some strange Greek letter?[math]\displaystyle \sum_{n=1}^{\infty}{1} = 1+1+1+\dots =?[/math]Mathematicians of old pondered such a question, and decided that there is indeed a way to deal with some of these strange infinite sums. Specifically, consider the sum[math]\displaystyle \sum_{n=1}^{\infty}{\frac{1}{2^n}} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots[/math]The terms of the sum are getting smaller and smaller, and so pretty quickly. More formally, this series does some important things. Let the [math]n[/math]th term be denoted as [math]a_n[/math]. Then we can observe thatThe terms get closer and closer to zero: [math]\lim_{n \rightarrow \infty}{a_n}=0[/math]The partial sums of the terms are bounded above by some value: [math]\displaystyle \sum_{n=1}^{m}{a_n} L[/math]These facts allows us to assign a value to this infinite sum. Given that the partial sums are bounded by [math]L=1[/math], we say that[math]\displaystyle \sum_{n=1}^{\infty}{\frac{1}{2^n}} = 1[/math]It is a little less intuitive, but not impossible to understand.Modern mathematicians are quite satisfied with the rules they developed for these sums, as they are consistent with other mathematical operations. For example if[math]\displaystyle \sum_{n=0}^{\infty}{a_n} = A[/math]and[math]\displaystyle \sum_{n=0}^{\infty}{b_n} = B[/math]then[math]\displaystyle \sum_{n=0}^{\infty}{a_n} + \sum_{n=0}^{\infty}{b_n} = A+B[/math]exactly as we’d expect with regular summing. Mathematicians also found ways to determine the values of these sums, like Euler’s solution to the Basel problem[5][math]\displaystyle \sum_{n=1}^{\infty}{\frac{1}{n^2}} = \frac{\pi^2}{6}[/math](The proof of this is quite the beast, though not impossible if you know some calculus.)Great! We can add infinitely many things in a way that makes sense…sometimes. Mathematicians like Ramanujan[6] were not satisfied with sometimes. They wanted to extend the idea of summing even further, to evaluate sums nonsensical in every way in a way that was completely, er, sensical.Enter Bernhard Riemann[7], a German mathematicians very interested in the following sum[math]\displaystyle \sum_{n=1}^{\infty}{\frac{1}{n^s}}[/math]for various [math]s[/math]. As we discussed earlier, a positive value of [math]s[/math] makes sense if this sum is to converge. Denote this sum as [math]\zeta(s)[/math], for example[math]\zeta(2) = \dfrac{\pi^2}{6}[/math] (Euler’s result)[math]\zeta(3) \approx 1.2021[/math] [8][math]\zeta(4) = \dfrac{\pi^4}{90}[/math] (also due to Euler)and so on[9]. Now, what if [math]s=1[/math]? We run into a problem: Though the terms of [math]\zeta(1)[/math] do converge to zero, their partial sums are not bounded, and thus the value is undefined[10]. This series is called the harmonic series, and there really is no way to tackle it.Ignoring that pesky value at 1, Riemann found that it was possible to define his sum for real and complex [math]s[/math], provided that [math]\Re(s) 1[/math]. He also described a way whereby one could defined [math]\zeta(s)[/math] for all complex [math]s[/math] save [math]s=1[/math]. The process he used is called analytic continuation[11] , not being an expert on it myself, I implore you to watch this video:Visualizing the Riemann zeta function and analytic continuationwhich gives an amazingly intuitive way to understand Riemann’s proof. Essentially, however, Riemann extended the definition of his sum to other values in a way that was consistent, that could withstand any sort of mathematical twists or turns thrown at it.With this, the Riemann Zeta Function[12] was born! Riemann found there were also sorts of cool properties of this function, such as a connection to prime numbers[13] (though we won’t get into that now). One consequence of Riemann’s analytic continuation was a value for [math]\zeta(-1)[/math], specifically[math]\zeta(-1) = -\dfrac{1}{12}[/math]Now, because [math]\zeta(s)[/math] is not normally defined for [math]s=-1[/math], it is dangerous to consider this value as a ‘sum’, despite the fact that the original definition of the Zeta function is based on an infinite sum. If we were to interpret this result as a sum, we obtain[math]\displaystyle \sum_{n=1}^{\infty}{\frac{1}{n^{-1}}} = \sum_{n=1}^{\infty}{n} = -\frac{1}{12}[/math]and all the confusion suddenly makes sense. Many unskilled observers (or skilled observers attempting to make the result more accessible to the public) will take the summation definition of the Zeta function as truth, which is not valid if [math]\Re(s) \leq 1[/math]. To answer your question, NO, there is no truly logical reason that the sum of all natural numbers would converge to [math]-1/12[/math]. It just does not make sense under our normal definition of summation.But, as Riemann showed, there do exist ways in which we can tackle such a sum and obtain some well-defined and well-behaved value, there is no other logical way to define [math]\zeta(-1)[/math] other than [math]-1/12[/math]. Yet we cannot interpret this result the same way we have understood our other, nicer sums. It is a result of deeper mathematical thinking, a process which lies in the abstract and must be treated as such.Perhaps one day we will discover the existence of infinity in our universe. Based on what we know of it mathematically, it would be hella weird, and imply all kinds of crazy results[14] . And perhaps then [math]\zeta(-1)[/math] will also make sense as a sum, and Numberphile[15] will not have deceived after all. Until then, take such infinities with a grain of salt, and be wary of the dangers of trying to tackle that which we cannot perceive.EDIT:Thank you all for 7̶5̶+1̶0̶0̶+ 150 upvotes! This is my most viewed and upvoted answer on Quora to date, so I thank everyone for their support. It seems my explanation was not too bad.Footnotes[1] Peano axioms - Wikipedia[2] Ring (mathematics) - Wikipedia[3] Summation - Wikipedia[4] Faulhaber's formula - Wikipedia[5] Basel problem - Wikipedia[6] Srinivasa Ramanujan - Wikipedia[7] Bernhard Riemann - Wikipedia[8] Apéry's constant - Wikipedia[9] Particular values of the Riemann zeta function - Wikipedia[10] Harmonic series (mathematics) - Wikipedia[11] Analytic continuation - Wikipedia[12] Riemann zeta function - Wikipedia[13] Riemann hypothesis - Wikipedia[14] Banach–Tarski paradox - Wikipedia[15] Riemann Hypothesis - Numberphile
In which class does one have to fill out the NDA for class 11 or 12?
See, you become eligible to apply in the second NDA exam of any year(June notification) once you pass class 11th and probably may have started your class 12th. After that you can apply for as many times provided that your age is in the prescribed limit
What percentage is needed in 12 class to apply for the NDA exam?
According to the latest notification ,there is no minimum percentage required in for joining NDA, but you should be pass in all the subjects , in terms of percentage you can take it as 33% . But aim to score a minimum of 80% so you can apply for other fields in the future also.