
#1
Apr1108, 03:39 PM

P: 4

1. The problem statement, all variables and given/known data
Find the asymptotic formula for [tex]\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{3}+...+\sqrt{n}[/tex] in the form of [tex]c \ast n^{\alpha}[/tex] Identify c and alpha. (Do NOT use the fundamental theorem of calculus) 2. Relevant equations Area under curve [tex]y = \sqrt{x}[/tex] in [0, 1] is [tex]\frac{1}{n^{\frac{3}{2}}}\sum^{n}_{i=1}\sqrt{i}[/tex] 3. The attempt at a solution ...I'm really not sure from where to start... I know [tex]c = \frac{2}{3}[/tex] and [tex]\alpha = \frac{3}{2}[/tex], but do not know how to prove it. 



#2
Apr1108, 03:47 PM

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Go with the area under the curve idea. The sum from 1 to n of sqrt(n) is approximated by the area under the curve sqrt(x) from x=0 to n. Set that up as an integral and evaluate it.




#3
Apr1108, 04:10 PM

P: 4

Sorry I forget to say that the fundamental theorem of calculus cannot be used, otherwise it is pretty straight forward.
Thanks for the quick reply. 



#4
Apr1108, 04:35 PM

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Find formula for sum of square roots
Ok, then. If c is the area under the curve sqrt(x) on [0,1] and I(n) is your sum of square roots, then you have c=limit n>infinity (1/n^(3/2))*I(n) from your Riemann integral type expression (were you just given that?). So for large n, I(n)~c*n^(3/2). There's the 3/2. The easy way to get c is to integrate to get the area. But not allowed? Ok, then look at I(n)I(n1)~c*(n^(3/2)(n1)^(3/2)). Separate the n's from the c and take the limit as n > infinity. Is that allowed? (All of this is really just doing calculus the HARD WAY).




#5
Apr1108, 05:01 PM

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#6
Apr1108, 05:45 PM

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Hint: when n > ∞, cn^α + √n > c(n+1)^α. 



#7
Apr1208, 03:07 AM

P: 4

I(n)I(n1)~c*(n^(3/2)(n1)^(3/2)) will work when alpha is found. So, c = [tex]\frac{\sqrt{n}}{n^\frac{3}{2}(n1)^\frac{3}{2}}[/tex] ... = [tex]\frac{1+(11/n)^\frac{3}{2}}{33/n3/n^2} [/tex] lim c = 2/3  then for the first half of question lim Area = lim 1/(n^3/2) * ∑√i ( lim n > infinity) > Area * n^3/2 = ∑√i = c * n^a > alpha = 3/2.. Is this precise enough to conclude the value of alpha? 



#8
Apr1208, 10:50 AM

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Sure, if you presume that the area is a well defined constant, the riemann sum immediately gives you the n^(3/2) dependence. You just have to work a little harder to get the c.



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