# calc help

consider the function f(x)= x^2/4 -6
Rn is the Riemann sum where the sample points are chosen to be the right-hand endpoints of each sub-interval.

Calculate Rn for f(x)= x^2/4 -6 on the interval [0,4] and write your answer as a function of n without any summation signs.
Rn= ?

The k-th term in the summation is:

[(s*k)^2/4 - 6]*s

s is the step between two terms and is 4/n. k ranges from 1 to n. n is the number of parts in which you have divided the interval [0,4]. So, if you take n = 1 then s = 4 and you are then only evaluating the function x = 4, which is the right hand endpoint of the interval.

If you n = 2 then s = 2 and n ranges from 1 to 2. The function is ten evaluated at 1*s and 2*s, i.e. at 2 and 4 and the values are multiplied by 2.

To evaluate the summation we write:

[(k*4/n)^2/4 - 6]*4/n

= k^2 16/n^3 - 24/n

The last term is - 24/n. If we sum this from k = 1 to n then that amounts to mulitplying by n, so that yields -24.

To evaluate te first term we must calculate:

Sum from k=1 to n of k^2.

You can look up the formula for that (it's a third degree polynomial in n), but it is more fun to derive this yourself.

You can use the formula for the geometric series:

Sum from k=0 to n of a^k =

[1-a^(n+1)]/[1-a]

Differentiate both sides w.r.t. a:

Sum from k=1 to n of ka^(k-1) =

- (n+1)a^(n)/(1-a) + [1-a^(n+1)]/[1-a]^2

If you take the limit a-->1 on both sides you get the formula for the sum of k from 1 to n. We want the sum of k^2. If you differentiate again w.r.t. a a factor (k-1) comes down. That's not what we want, we want a factor k. So, you first multiply both sides by a and then you differentiate w.r.t. a. Then you take the limit a -->1 on both sides.

A faster way to calculate the summation is as follows.

Sum from k=0 to n of a^k =

[1-a^(n+1)]/[1-a]

Substitute a = exp[x] in here:

Sum from k=0 to n of exp[kx] =

[exp[(n+1)x] - 1]/[exp(x) - 1]

Expand both sides in powers of x. You can see that the coefficient of x^2 yields 1/2 times the desired summation.

To find the series expansion of the function

[exp[(n+1)x] - 1]/[exp(x) - 1]

you equate it to an unknown series:

c_0 + c_1x + c_2 x^2 + ...

This yields:

[c_0 + c_1x + c_2 x^2 + ...]*

[x + x^2/2 + x^3/6 + ...] =

(n+1)x + (n+1)^2x^2/2 + (n+1)x^3/6 + ...

And you find that:

c_0 = n+1

c_1 = 1/2 n(n+1)

c_2 = 1/6 n(n+1/2)(n+1)

The summation is 2c_2, so:

Sum from k = 1 to n of k^2 =

1/3 n(n+1/2)(n+1)

This means that the Riemann sum of the first term is:

16/3 n(n+1/2)(n+1)/n^3

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