Wikipedia:Reference desk/Archives/Mathematics/2010 February 14

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February 14

Integration

Can anyone tell me what ∫x dx({} refers to the "fractional part" function) will be in the limit of 2 to 20. Actually, it'll be better if you gave me the reasons rather than the actual steps. I mean, from 2 to 20, there are an awful number of numbers having non-zero fractional parts (apart from the integers, obviously). So how on earth can anyone integrate it? It can't be integrated piece-wise either. I tried to. Can anyone help?117.194.225.72 (talk) 12:10, 14 February 2010 (UTC)

Did you mean to put the curly brackets round x in the integration? Not that I recognize that notation anyway. It sounds like you are simply supposed to integrate the sawtooth wave function. Dmcq (talk) 12:32, 14 February 2010 (UTC)

(edit conflict) You want to integrate the fractional part of x from x = 2 to x = 20 ? Remember that ∫ f(x) dx from a to b is just the (signed) area enclosed between the graph of f(x) and the x-axis between x = a and x = b. This is quite straightforward to determine geometrically when f(x) is the fractional part of x. Note that this is the same as integrating piece-wise - between n and n+1 the fractional part of x is x-n, so you can integrate these pieces and then add them together. Gandalf61 (talk) 12:39, 14 February 2010 (UTC)

The fractional-part function is periodic, so if you integrate it from 0 to 1, you get the same thing as if you integrate it from 2 to 3, or from 3 to 4, or from 4 to 5, etc. There are 18 such intervals between 2 and 20, so the integral from 2 to 20 is 18 times that from 0 to 1. For x between 0 and 1, the fractional part of x is just x itself. So

$\int _{2}^{20}\{x\}\,dx=18\int _{0}^{1}x\,dx.$

Michael Hardy (talk) 14:09, 14 February 2010 (UTC)

What about the fractional part of numbers between 0 to 1 (having non-zero fractional part)? 0.556 ,for example? They should be taken into account too, right?In that case, integrating it should give an answer other than the value of x itself.... Correct me if I'm wrong. 117.194.231.5 (talk) 15:46, 14 February 2010 (UTC)

If x = 0.556, then the fractional part of x is also 0.556, i.e. the fractional part of x is x itself, just as stated above. Michael Hardy (talk) 22:40, 14 February 2010 (UTC)

Consider what the difference is between the fractional part of x and x itself between 0 and 1. Also, you should revisit your original impression that it can't be integrated piece-wise. Why did you say that? Finally, the suggestion above to consider its graph is a great idea. Have you sketched out a quick graph of the function and shaded in the area of integration yet? 58.147.58.28 (talk) 16:06, 14 February 2010 (UTC)

Thank you everyone. I just realised where I went wrong. I'm new to this stuff, so, while I knew that ∫f(x)dx is actually the area bounded by the function's graph and the x-axis, while trying to solve the above problem, I was constantly thinking of the summation of all the values of {x}. That would have made a different problem, something that went like: 0+0.001+0.002+0.003 and so on.. Thanks a bunch. Sketching the saw-tooth graph helped a lot. As simple as calculating the area of a triangle and multiplying it by 18, right? 117.194.227.81 (talk) 05:32, 15 February 2010 (UTC)

Correct. Michael Hardy (talk) 06:05, 15 February 2010 (UTC)

Sum of all combinations of n

What is the equation for the sum of all unique combinations of n objects. Eg, if n=2 (let's call the objects, A and B), the sum is 3 (combinations are A, B, and AB); for n=3, sum is 7 (combinations are A, B, C, AB, AC, BC, ABC). I know this seems similar to triangular numbers which is the sum of unique interactions between 2 objects among n numbers. Ephilei (talk) 19:06, 14 February 2010 (UTC)

Never mind. I found it. It's 2^n-1, one less than n's power set.Ephilei (talk) 19:23, 14 February 2010 (UTC)

Actually if you consider the empty subset too, they are exactly 2ⁿ. --84.221.209.11 (talk) 20:47, 14 February 2010 (UTC)

It is not very hard to prove either. We need to know about the binomial theorem and the binomial coefficients. Let us say that you have n objects. How many ways to pick none? There are nC0 (read as n choose none) ways of doing that. How many ways to choose one? There are nC1 (read as n choose one) ways of doing that. How many ways to choose two? There are nC2 ways of doing that. ... How many ways to choose n? There are nCn ways of doing that. So you need to make the following addition: nC0 + nC1 + nC2 + ... + nCn. This is given by applying the binomial theorem to (1+1)^n (which equals 2^n since 1+1=2). But you don't seem interested in the number of ways of choosing none, so we subtract nC0 = 1. That means there are 2^n-1 ways. •• Fly by Night (talk) 22:08, 14 February 2010 (UTC)

Quicker is to consider 3-digit binary numbers, with (say) A corresponding to a "1" in the RH position, B to a "1" in the central position and C to a "1" in the LH position. Then there are 8 possibilities 000, 001, 010, 011, 100, 101, 110 and 111, leaving 7 = 2^3 - 1 if the one containing neither A, B nor C is removed. Obviously extendable to any value for the number of objects—86.132.238.199 (talk) 23:06, 14 February 2010 (UTC)

Interesting! Thanks Ephilei (talk) 16:06, 15 February 2010 (UTC)