Maclaurin series for e^x and Riemann sum

Question

Find the Maclaurin series using e¹ for the lim n->∞ nsin(2πen!):
The Maclaurin series for e×=1+x+x²/2!+x³/3!+...=∑x^k/k! as k goes from 0 to ∞. Time=Distance/Rate.
Is the Maclaurin series for e× modified to remove the first two terms in the series, 1+x?

A>-1 and b>-1, use Riemann sum to compute lim n->∞ n^b-a (1^a+2^a+3^a...n^a/1^b+2^b+3^b...n^b). A log with base 4 is y=log(8), 4^y=8, 2^2y=2³, 2y=3, y=3/2. In mathematics, a Riemann sum is an approximation that takes the form ∑ f(x) Δx. It is named after German mathematician Bernhard Riemann. One very common application is approximating the area of functions or lines on a graph, but also the length of curves and other approximations...
Let f : D → R be a function defined on a subset, D, of the real line, R. Let I = [a, b] be a closed interval contained in D, and the partition of I where a=b is defined.
  

Answer 1

Anthony,

 

What is the argument of sin(2nen!)? sin( 2*n*e*n!)?

 

The Riemann sum needs a definition of a function f(x). Is your function [n^(b-a)][1^a + 2^a + ... + n^a]/[1^b + ... + n^b]?

 

Do you need to calculate the area under this function?

 

What does "A log with base 4..." have to do with this problem?

 

Michael.

 

 

 

You are right about most of the pieces, Anthony. My apologies.

These are definitely two unusual problems.

I think the way to handle problem 3 is to recognize it as a ratio of Riemann Sums that approximate a definite integral. There is no need to identify the function f(xi) and the width of the partition Δxi

limn→∞ Σ f(xi) Δxi = ∫lu f(x)dx

because the Riemann Sum(s) are already given. The problems is half done. You are "simply" required to evaluate the Riemann Sums.

Notice that the Example for Riemann Sums in Wikipedia for a = b =2 are similar to what is here:

the numerator (1^a + 2^a + ... + n^a) and the denominator (1^b + 2^b + ... + n^b)

These sums can be evaluated with Bernoulli numbers Bm and binomial combinations c+1Cm (see http://www.maa.org/press/periodicals/convergence/sums-of-powers-of-positive-integers-conclusion):

∑(k=1,n) k^c =1/(c+1) ∑(m=0,c) c+1Cm Bm n^(c+1−m)

We can handle the numerator and denominator as independent limits.

Multiplying the numerator and the denominator through by the numerator and denominator of n^(b-a) = n^(-a) / n^(-b), that is, by c =a and c=b, respectively, converts each n^(c+1-m) into n^(1-m).

Factoring out n^1 from n^(1-m) in the numerator and the denominator cancels the respective factors n^1 and leaves n^(-m) = (1/n)^m.

In the limit, all the powers of 1/n in the sum over m go to zero--except for m=0.

We are finally left with the ratio: 1/(a+1) a+1C0 B0 / [1/(b+1) b+1C0 B0] = (b+1) / (a+1) since the combinations are each 1 and the B0 factors cancel.

Note that a > 1 or a+1 > 0 and b >1 and b+1 > 0 so that the combinations have parameters with appropriate values.

That's it!


I think the way to handle Problem 6 is to use the Maclaurin series for sin(x) and then expand e in the argument x = 2 pi e n!. This produces a double sum that might be able to be inverted or perhaps multiplying through by the n or the n! will simplify some of the terms so that in the limit they are finite.

I don't see the answer immediately, so I'm going to let you or someone else work it out.

Good Luck!

Michael.