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 one year ago
Can you please explain about all the functions used in maths like greatest integer, lowest integer, modulus , sigma function with examples?
 one year ago
Can you please explain about all the functions used in maths like greatest integer, lowest integer, modulus , sigma function with examples?

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Topi
 one year ago
Best ResponseYou've already chosen the best response.0No I can't. There are so many(1) functions in math that I just cannot describe them all and furthermore the description of some of them needs so much prior knowledge that the cold description doesn't even mean anything. (1) And new ones are coined all the time. But I'll take a shot for those four you mentioned. Greatest integer is actually called ceiling and is the least integer that is greater or equal to a given real. In the same vein lowest integer, which is called floor, is the greatest integer that is lower or equal to the given real. Here are examples: \[\lceil \sqrt{2} \rceil = 2, \lceil \sqrt{2} \rceil = 1\] \[\lfloor \sqrt{2} \rfloor = 1, \lfloor \sqrt{2} \rfloor = 2\] Modulus is not a function, it's a relation. You get modulus by dividing two intgers and check what the remainder is. Examples: 5 = 1 mod 2, 27 = 2 mod 5 and 29 = 1 mod 7. Sigma function can have diffrent meanings. So, just writing 'sigma function' doesn't mean anything. Anyways, all sigma functions I know are easy in concept but not so easy in actualisation. Do a search in wikipedia and you can see yourself. Then there's also sigmanotation which is not a function at all but marks summation of a function over a range: \[\sum_{0}^{n} n^2 = 0^2+1^2+2^2+3^2 ... (n1)^2 + n^2 = \frac{ n(2n+1)(n+1) }{ 6 }\]

Topi
 one year ago
Best ResponseYou've already chosen the best response.0BTW. I can prove that last sum with a very nice calculus related way: Let's consider the following integral:\[\int\limits_{0}^{b}x^2dx\]If we take the Riemann sum with equal spacing of this function we get:\[\Delta x = b/n, \sum_{k=1}^{n} (\frac{ kb }{ n })^2 \Delta x = (\frac{ b }{ n })^3\sum_{k=1}^{n}k^2\]And now when we take the limit as n goes to infinity we get:\[n \rightarrow \infty, b^3 \sum_{k=1}^{n}\frac{ k^2 }{ n^3 }\]So as we know that the given integral has a finite answer then the term:\[\frac{ 1 }{ n^3}\sum_{k=1}^{n} k^2\]has to be finite. So we try to find a polynomial answer to the sum. Noticing that P4(n)/n^3 goes to infinity when n goes to infinity and P2(n)/n^3 goes to zero when n goes to infinity we only need to try a polynomial P3(n):\[an^3+bn^2+cn+d\]We also notice that we want to have:\[\sum_{k=0}^{0} k^2 = 0\]so d=0. For n=1, n=2 and n=3 we get the following three equations:\[a+b+c = 1\]\[8a+4b+2c = 5\]\[27a+9b+3c = 14\]So by subtracting the first multipied by 2 from the second and subtracting the first multiplied by 3 from the third we get:\[6a+2b = 3\]\[24a+6b =11\]Now we subtract the first multiplied by 3 from the second and we get:\[6a = 2 \rightarrow a=\frac{ 1 }{ 3 }\]Now substituting back we get:\[6\times \frac{ 1 }{ 3 }+2b = 3 \rightarrow 2b = 32 \rightarrow b = \frac{ 1 }{ 2 }\]And substituting both of those values we get:\[\frac{ 1 }{ 3 }+\frac{ 1 }{ 2 }+c = 1 \rightarrow c = \frac{ 1 }{ 6 }\]So the final answer is:\[\sum_{k=0}^{n}k^2 = \frac{ 1 }{ 3 }n^3 + \frac{ 1 }{ 2 }n^2+\frac{ 1 }{ 6 }n = \frac{ 2n^3+3n^2+n }{ 6 }=\frac{ n(2n^2+3n+1) }{ 6 }= \frac{ n(2n+1)(n+1) }{ 6 }\]Which is what I claimed in the previous post. To prove that this works for all natural numbers we must calculate the following:\[\sum_{k=0}^{n+1}k^2 = (n+1)^2+\sum_{k=0}^{n}k^2= (n+1)^2 + \frac{ n(2n+1)(n+1) }{ 6 }= \frac{ 6(n+1)^2+n(2n+1)(n+1) }{ 6 }\]\[=\frac{ (n+1)(6n+6 +2n^2+n) }{ 6 }=\frac{ (n+1)(2n^2+7n+6) }{ 6 }= \frac{ (n+1)(2n+3)(n+2) }{ 6 }\]\[=\frac{ (n+1)(2(n+1)+1)((n+1)+1) }{ 6 }\]So we've proven that:\[\sum_{k=0}^{n}k^2=\frac{ n(2n+1)(n+1) }{ 6 }\]Now to get back to the Riemann sum we have:\[x = b/n, \sum_{k=1}^{n} (\frac{ kb }{ n })^2 \Delta x = (\frac{ b }{ n })^3\sum_{k=1}^{n}k^2=b^3(\frac{ 1 }{ n^3}(\frac{ 1 }{ 3 }n^3+\frac{ 1 }{ 2 }n^2+\frac{ 1 }{ 6 }n))\]Which, when n goes to infinity, convergenses to:\[\frac{ b^3 }{ 3 }\] So here was a analytical proof that:\[\int\limits_{0}^{b}x^2dx = \frac{ b^3 }{ 3 }\]

Topi
 one year ago
Best ResponseYou've already chosen the best response.0Sorry that you cannot see all the equations correctly. But that is a problem of the presentation interface. The first equation that goes out of bounds is just the same as what is in the conclusion. The second is an intermediate one where I'm calculating the induction proof from n to n+1. The missing part is just the multiplicant (n+1), which you can see in the previous equation.
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