Calculus flashback (do my homework)

[LittleWing]

hey its only 1 problem

I'm sure the answer is obvious

http://img20.imageshack.us/img20/3461/math1tp8.jpg


it's either 0, 1, 4, 39, or 125

i am so sad

[bigpooch]

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.

[bigpooch]

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.

[astvald]

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.

[bigpooch]

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.