Riemann sums to definite integral conversion

In the most recent tutorial there is a question about converting a Riemann sum to a definite integral, and it seems to be tripping up quite a few students. I wanted to run through one of the calculations in detail so you can see how to answer such a question.

 

Let’s look at the example:

 

\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(9\left(4+(i-1)\frac{6}{n}\right)^2-8\left(4+(i-1)\frac{6}{n}\right)+7\right)\frac{13}{n}

 

There are many ways to tackle such a question but let’s take one particular path. Let’s start by the fact that when the limit is defined, the limit of a sum is the sum of the limits. We can split up our expression into 3, which looks like:

 

\lim_{n\rightarrow\infty}\sum_{i=1}^n9\left(4+(i-1)\frac{6}{n}\right)^2\frac{13}{n}-\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(8\left(4+(i-1)\frac{6}{n}\right)\right)\frac{13}{n}+\lim_{n\rightarrow\infty}\sum_{i=1}^n7\frac{13}{n}

 

Let’s tackle each of these separately. Let’s look at the first term:

 

\lim_{n\rightarrow\infty}\sum_{i=1}^n9\left(4+(i-1)\frac{6}{n}\right)^2\frac{13}{n}

 

Well, we can take the factor of 13 outside the front of the whole thing to start with, along with the factor of 9, and this will give

 

13\times 9\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(4+(i-1)\frac{6}{n}\right)^2\frac{1}{n}

 

We see here that we have a sum of terms, and a factor which looks like \frac{1}{n} in each term.…

By | August 23rd, 2017|Courses, First year, MAM1000, Uncategorized, Undergraduate|7 Comments

Some sum identities

During tutorials last week, a number of students asked how to understand identities that are used in the calculation of various Riemann sums and their limits.

These identities are:

 

\sum_{i=1}^n 1=n

\sum_{i=1}^n i=\frac{n(n+1)}{2}

\sum_{i=1}^n i^2=\frac{n(n+1)(2n+1)}{6}

\sum_{i=1}^n i^3=\left(\frac{n(n+1)}{2}\right)^2

 

Let’s go through these one by one. We must first remember what the sigma notation means. If we have:

 

\sum_{i=1}^n f(i)

 

It means the sum of terms of the forms f(i) for i starting with 1 and going up to i=n. Sometimes n will actually be an integer, and sometimes it will be left arbitrary. So, the above sum can be written as:

 

\sum_{i=1}^n f(i)=f(1)+f(2)+f(3)+f(4)+....+f(n-2)+f(n-1)+f(n)

 

We haven’t specified what f is, but that’s because this statement is general and applies for any time of function of i. In the first of the identities above, the function is simply f(i)=1, which isn’t a very interesting function, but it still is one. It says, whatever i we put in, output 1. So this sum can be written as:

 

\sum_{i=1}^n 1=1+1+1+1+....+1

 

Where there are n terms.…

By | August 20th, 2017|Courses, First year, MAM1000, Uncategorized, Undergraduate|1 Comment

MAM1000W 2017 semester 2, lecture 1 (part ii)

The distance problem

If I want to know how far I walked during an hour, I can ask how far I walked in the first five minutes, and how far I walked in the second five minutes, and how far I walked in the third five minutes, etc. and add them all together. ie. I could write:

 

d=d_1+d_2+d_3+d_4+...d_{12}

 

Where d_i is the distance walked in the i^{th} five minutes. To calculate a distance, we need to know how fast we are going, and for how long. In fact:

 

distance=velocity \times time

 

where you can think of velocity as the same thing as speed (though there are subtle differences which you will find out about later). This formula works if the velocity is constant, but what if it is changing. Well, if we have a graph of velocity against time, then we can think about splitting the graph into intervals (like the five minute intervals above), and approximating that during a small interval of time, the velocity is roughly constant.…