## 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.…

## 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.…

## 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.…

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## MAM1000W 2017 semester 2, lecture 1 (part i)

I wanted to put up a little summary of some of the most important things to remember from the end of last semester. There was a sudden input of new concepts, so let’s put some of them down here to get a clear reminder of what we need to know. A few things in this post:

• The antiderivative
• Sigma notation
• Areas under curves

Antiderivatives

An antiderivative of a function $f$ on an open interval $I$ is a function $F$ such that: $F'(x)=f(x)$ for every $x\in I$

Note that we say an antiderivative, not the antiderivative. There can be many functions whose derivatives give the same thing. While we know that: $\frac{d}{dx}\sin x=\cos x$

and therefore $\sin x$ is an antiderivative of $\cos x$, we can also say that: $\frac{d}{dx}(\sin x+3)=\cos x$

So $\sin x+3$ is also an antiderivative of $\cos x$. In fact for any constant $c$ it is true that $\sin x+c$ is an antiderivative of $\cos x$. We will come up with some clever notation for the antiderivative soon.…

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## Checking direction fields

I was recently asked about how to spot which direction field corresponds to which differential equation. I hope that by working through a few examples here we will get a reasonable intuition as to how to do this.

Remember that a direction field is a method for getting the general behaviour of a first order differential equation. Given an equation of the form: $\frac{dy}{dx}=f(x,y)$

For any function of x and y, the solution to this differential equation must be some function (or indeed family of functions) where the gradient of the function satisfies the above relationship.

The first such equation that we looked at was the equation: $\frac{dy(x)}{dx}=x+y(x)$.

We are trying to find some function, or indeed family of functions y(x) which satisfy this equation. We need to find a function whose derivative (y'(x)) at each point x is equal to the value of the function (ie. y(x)), plus that value of x.…

## Radius of convergence of a series, and approximating polynomials

I hinted today that there were sometimes issues when you did a polynomial approximation, that if you tried to find the value of a function a long way from the region about which you’re approximating, that sometimes you wouldn’t be able to do it. This is related to an idea called the radius of convergence of a series. In the following we are just plotting polynomials, but you can see that whereas in the polynomial approximation for sin(x) (on the right), as we get more and more terms, we approximate the function better and better far away from the point x=1 (which is the point about which we are approximating the function). However, for the function $\sqrt{1+x}$, after x=3, the approximations are nowhere near the function itself. This is because that function has a radius of convergence of 2, when expanded about x=1. This is due to the behaviour of the function at x=-1, which is a distance 2 away.…

## Integration by Parts – Lightbulb Education

Integration by Parts

 How clear is this post?

## Integration By Substitution – Lightbulb Education

Integration By Substitution

 How clear is this post?
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## Does knowledge of past years affect the Keynesian Beauty Contest result?

Last year I played the 2/3 numbers game, also called the Keynesian Beauty Contest with my first year maths class. The discussion can be found here: http://www.mathemafrica.org/?p=11143 I wanted to know if, telling my class the results from last year (including sketching for them the histogram of results), would change how they chose their numbers this year. Of course I can’t tell if it changed them, but what is fascinating is that either:

1. Their guesses (if I didn’t tell them about the results from last year) would have been very different from those last year, or:
2. They were completely unaffected by knowing what people did last year, which really means that they believed that the rest of the class would have been unaffected.

I plot here the results from the last three years and you can see how similar the results are, year on year. You can see that the distributions are relatively similar, and the means are extremely close.…

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## Picturing volumes of revolution

One of the homework questions this week was the following:

Let $R=\{(x,y)\in \mathbb{R}^2:y\ge 0, \cos x\le y\le \sin x\,\,and\,\,0\le x\le\pi\}.$

a) Sketch the region R and find its area.
b) Find the volume of the solid obtained by rotating the region R around the y-axis.

The first thing to do is to sketch the graphs of $y=\cos x$ and $y=\sin x$. Once you’ve done that, the other parts of the inequalities should be clear. It should look like the red region in the following plot: Now we have to imagine bringing out a third axis, perpendicular to the picture above, ie. coming out towards us. We then want to rotate the red form here about the vertical axis. This we can imagine doing in the following animation: Given this form we can then think about either taking horizontal cross-sections through it, which will give us thin annuli, or we can take vertical, circular slices to give us thin shells. Adding these together and integrating should give us the same answer whichever way we choose to slice it, but one way will be considerably easier.…