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PostPosted: Wed, 24 Sep 2008 12:16:50 UTC 
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As commutative said, this is a special case of a general result. Move the $\frac{n-k}{n} inside the integral.
Also, read this thread particularly to commutative's post at the bottom http://www.sosmath.com/CBB/viewtopic.php?t=35977&highlight=

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 Post subject: Re: Calculus (13)
PostPosted: Thu, 25 Sep 2008 00:46:34 UTC 
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I think I have the actual answer now:
Spoiler:
$ \phi = \lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}(n-k)\int_{\frac{k}{n}}^{\frac{k+1}{n}}\frac{dx}{x^4 + 1}.
Let y_{k}=\frac{k}{n}. The kth term of the sum is (n-k) times the area under the curve from y_{k} to y_{k+1}. Summing this over k is equivalent to summing the value of the integral from 0 to y_{k}.
Or: $\sum_{k=0}^{n-1}(n-k)\int_{\frac{k}{n}}^{\frac{k+1}{n}}\frac{dx}{x^4 + 1} = \sum_{k=0}^{n} \int_{0}^{\frac{k}{n}} \frac{dx}{1+x^{4}}

Plugging this in gives $\phi = \lim_{n\to\infty} \sum_{k=0}^{n} \frac{1}{n} \int_{0}^{\frac{k}{n}} \frac{dx}{1+x^{4}}
Letting f(y) = \int_{0}^{y} \frac{dx}{1+x^{4}} and \Delta y = \frac{1}{n} we see that this is really just the riemann sum for $ \int_{0}^{1} \int_{0}^{y} \frac{1}{1+x^{4}} dx \: dy

The answer is somewhere around .47
If I'm correct, the general case of this is taking the f(y) to be anything of the form \int_{0}^{y} g(x) dx

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Last edited by simplethinker on Fri, 26 Sep 2008 15:30:21 UTC, edited 1 time in total.

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 Post subject: Re: Calculus (13)
PostPosted: Thu, 25 Sep 2008 18:24:00 UTC 
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simplethinker wrote:
Or: $\sum_{k=0}^{n-1}(n-k)\int_{\frac{k}{n}}^{\frac{k+1}{n}}\frac{dx}{x^4 + 1} = \sum_{k=0}^{n} \int_{0}^{\frac{k}{n}} \frac{dx}{1+x^{4}}


Hi simplethinker, may you please explain this equality again. This is the only line that I do not understand from your solution.

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 Post subject: Re: Calculus (13)
PostPosted: Thu, 25 Sep 2008 21:57:47 UTC 
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bidentate wrote:
simplethinker wrote:
Or: $\sum_{k=0}^{n-1}(n-k)\int_{\frac{k}{n}}^{\frac{k+1}{n}}\frac{dx}{x^4 + 1} = \sum_{k=0}^{n} \int_{0}^{\frac{k}{n}} \frac{dx}{1+x^{4}}


Hi simplethinker, may you please explain this equality again. This is the only line that I do not understand from your solution.


Set $ g(k)= \int_{0}^{\frac{k}{n}} \frac{dx}{1+x^{4}} so $\int_{\frac{k}{n}}^{\frac{k+1}{n}} \frac{dx}{1+x^{4}} = g(k+1) - g(k)
If we set n=4, then the sum is: (4-0)*(g(1) - g(0)) + (4-1)*(g(2) - g(1)) + (4-2)*(g(3) - g(2)) + (4-3)*(g(4) - g(3)) = 4*g(1) - 4*g(0) + 3*g(2) - 3*g(1) + 2*g(3) - 2*g(2) + 1*g(4) - 1*g(3) = (4-3)*g(1) + (3-2)*g(2) + (2-1)*g(3) + (1)*g(4) - 4*g(0) = (g(1)-g(0)) + (g(2)-g(0)) + (g(3)-g(0)) + (g(4)-g(0))
[ for this case g(0)=0, so the above could be simplified by removing the g(0)'s, but I don't think it would generalize well for other functions such that g(0)=0]
Since $ g(k)-g(0) = \int_{0}^{\frac{k}{4}} \frac{dx}{1+x^{4}}, we have $\sum_{k=0}^{4-1}(4-k)\int_{\frac{k}{4}}^{\frac{k+1}{4}}\frac{dx}{x^4 + 1} = \sum_{k=0}^{4} \int_{0}^{\frac{k}{4}} \frac{dx}{1+x^{4}}
And this can be generalized to arbitrary n.

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PostPosted: Fri, 26 Sep 2008 14:12:10 UTC 
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Very nice, good job!

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PostPosted: Fri, 26 Sep 2008 15:29:28 UTC 
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Thank you :)
I also just noticed that there was an error in my first explanation of the sum. It should be (n-k) times the area under the curve from y_k to y_k+1.

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Make It Fun And Complicate It Further, or MIFACIF. Ever try and find the volume of a cube in spherical coordinates?


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PostPosted: Fri, 19 Dec 2008 23:06:13 UTC 
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let's see your solutions to the generalized version of the problem, i.e. prove that: $ \lim_{n\to\infty} \frac{1}{n}\sum_{k=0}^n (n - k)\int_{\frac{k}{n}}^{\frac{k+1}{n}} f(x)dx=\int_0^1(1-x)f(x)dx.


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PostPosted: Sat, 20 Dec 2008 06:05:51 UTC 
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I think my solution just needs a bit of tweaking (and I finally figured out how to switch the order of integration):
Spoiler:
$\lim_{n\to\infty} \frac{1}{n} \sum_{k=0}^{n-1}(n-k)\int_{\frac{k}{n}}^{\frac{k+1}{n}} f(x) dx = \lim_{n\to\infty} \frac{1}{n} \sum_{k=0}^{n} \int_{0}^{\frac{k}{n}} f(x) dx (for the reason see my above post). Set $g(y) = \int_{0}^{y} f(x) dx, \; y_k = \frac{k}{n}, \; \Delta y_k = \frac{1}{n}, so the original expression becomes $ \lim_{n\to\infty} \sum_{k=0}^{n}g(y_k) \Delta y_k = \int_{0}^{1} g(y) dy, and remembering the definition of g(y), we have $ \int_{0}^{1} \int_{0}^{y} f(x) dx.
This integral is over the region 0 \leq x \leq y, \; 0 \leq y \leq 1, which can also be expressed as x \leq y \leq 1, \; 0 \leq x \leq 1, so finally the expression becomes $ \int_{0}^{1} \int_{x}^{1} f(x) dy dx = \int_{0}^{1} (1-x) f(x) dx


Another correction for my solution a couple posts up: I said I didn't think it would generalize well for functions such that g(0)\neq0, but by my definition g(0) has to be identically zero.

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Some people live their lives by the KISS principle (Keep It Simple Stupid), but I prefer a slightly different one:
Make It Fun And Complicate It Further, or MIFACIF. Ever try and find the volume of a cube in spherical coordinates?


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