#1
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Calculus flashback (do my homework)
hey its only 1 problem
I'm sure the answer is obvious it's either 0, 1, 4, 39, or 125 i am so sad |
#2
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Re: Calculus flashback (do my homework)
This is the answer, but probably not what you want to submit
as your homework: Let sigma denote the sum for k=1 to n. The expression is just: sigma [ 3(9k^2 + 12k + 4)/(n^3) ] but sigma (k^2) = n(n+1)(2n+1)/6 = (n^3)/3 + O(n^2) and sigma (k) = n(n+1)/2 = O(n^2) and of course sigma (1) = n = O(n) Thus, the sum is just 27(sigma (k^2))+ 36(sigma(k)) +12(sigma(1))/(n^3) or [27( (n^3)/3 + O(n^2) ) + 36(O(n^2)) + 12((O(n))]/(n^3) = [9(n^3) + O(n^2)]/(n^3) = 9 + O(n^(-1)) Thus, as n approaches infinity, the sum approaches 9. You can also see the above immediately by just looking at the expansion for sigma(k^2) but I included the steps to make it (hopefully) clear. |
#3
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Re: Calculus flashback (do my homework)
I'll add this post just to give you some idea of how to
solve some of these types of problems. (limit n -> infinity) sigma(k=1 to n) (k^r)/(n^(r+1)) = 1/(r+1) ------------------------------------------------------------ The above result can be shown if you already know that the integral of x^r from 0 to 1 is 1/(r+1). (This is simply calculus.) Look at the Riemann sums. Partition the interval from 0 to 1 into the n intervals [0,1/n),[1/n,...),[(n-1)/n,1). For the function f(x) = x^r, the area under the curve f(x) is bounded between the Riemann sums S(lower) = sigma(k=1 to n) {[(k-1)/n]^r}{1/n} and S(upper) = sigma(k=1 to n) {[k/n]^r}{1/n} or rewriting, S(upper) = sigma(k=1 to n) {k^r}/{n^(r+1)} But S(lower) = S(upper) - 1/n {just note that S(upper) has one extra term, an extra [1^r]/n}, so that S(upper)-1/n <= A <= S(upper) where A is the area under f(x)=x^r between x=0 and x=1. Letting n -> infinity and as A = 1/(r+1), S(upper) -> 1/(r+1) which is the result. General summations ------------------ From the above, it can be shown that for s<r, sigma(k=1 to n) {k^s}/{n^(r+1)} -> 0 as n -> infinity. Similarly, for t>r, sigma(k=1 to n) {k^t}/{n^(r+1)} -> infinity as n -> infinity. Thus, the "critical term" of a summation of the form sigma(k=1 to n) p(k)/{n^(r+1)} where p(z) is a polynomial in z is the z^r term. Lower order polynomial terms go to zero as n approaches infinity and any higher order terms cause the summation to approach infinity {or -infinity}. Thus... ------- For the problem in question, the "critical term" is the k^2 term in the polynomial since the denominator is n^3. Using the result that sigma(k=1 to n) {k^2}/{n^3}=1/3, and since there are no terms of order higher than k^2, the summation approaches 9(1/3)(3) = 9. |
#4
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Re: Calculus flashback (do my homework)
Looks like overkill to me. It's much easier to recognize that they're asking for a limit of Riemann sums for a certain integral.
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#5
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Re: Calculus flashback (do my homework)
I take that the expression inside the bracket is
[2 + (3k/n)] instead of [(2+3k)/n], as I thought in my first post. If so, then the answer is 39 since the Riemann sums of the integral of the function f(x)=3[(2+3x)^2] over the interval [0,1) coincide with the summation. f(x)=3(9x^2+12x+4) and so F(x)=3(3x^3+6x^2+4x) is an antiderivative (without constant term) and F(1)-F(0)=3(13)=39. |
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