Jonathan Shock – Page 23

About Jonathan Shock

I'm a lecturer at the University of Cape Town in the department of Mathematics and Applied Mathematics. I teach mathematics both at undergraduate and at honours levels and my research interests lie in the intersection of applied mathematics and many other areas of science, from biology and neuroscience to fundamental particle physics and psychology.

Some animations of Taylor approximations

Here are some simple animations which might help to understand what we were doing in class today. In each case the function which we are approximating is in red, and the polynomial approximation is in blue.

The following are the different approximations to the function $(1+x)^5$ starting at a constant, then a straight line with non-zero gradient, then a quadratic etc. Each one matching the higher and higher derivatives of the function at $x=0$ . You see that by the time we get to a fifth order polynomial, the match is exact. We know this because we know that there is an exact expansion of this function which is a fifth order polynomial.

This is the same plot, but now with the function $(1+x)^{5.2}$ . This time the polynomial approximation doesn’t stop at the fifth order, but can keep going to any power of $x$ .

This is the same function approximated but now approximated about the point $x=0.25$ .…

By Jonathan Shock| August 17th, 2015|Uncategorized|0 Comments

UCT MAM1000 lecture notes part 23 – Thursday August 20th

Complex numbers

These will be an addition to the notes already on Vula on complex numbers. Please refer to that document as well as I will be taking a slightly alternative approach on occasion.

A philosophical detour

Before we get on to talking about imaginary numbers and complex numbers, let’s try and break down our preconceptions about numbers in general. We look at the world around us and see many things which we categorise. We see a computer, a piece of paper, we see other people, we see our hands. These are labels that we use to categorise the world around us, but these objects seem very physical and very real. We rarely question their existence, though if one wants to take the Cartesian view, we should also question the reality we are in. We are not going to go that far, but let’s try and ask about the existence of numbers.…

By Jonathan Shock| August 16th, 2015|Courses, English, First year, Level: intermediate, MAM1000, Uncategorized, Undergraduate|2 Comments

UCT MAM1000 lecture notes part 22 – Wednesday 19th of August

So, last time we looked at the Maclaurin expression for $e^x$ . The exponential function was particular easy because its derivative is equal to the function itself every time. Let’s look at a slightly more involved example where this is not true: $f(x)=\sin x$ about $x=0$ . Again, we start with the table of derivatives:

$\left( \begin{array}{ccc} \text{ i} & f^{(i)}\text{(x)} & f^{(i)}\text{(0)} \\ 0 & \sin (x) & 0 \\ 1 & \cos (x) & 1 \\ 2 & -\sin (x) & 0 \\ 3 & -\cos (x) & -1 \\ 4 & \sin (x) & 0 \\ 5 & \cos (x) & 1 \\ \end{array} \right)$

Now the values of the derivatives are not always the same. They are zero every other term, and they change in sign when they are not zero. This leads to a very elegant expression for the $\sin$ function expanded around $x=0$ :

$\sin x\approx\sum_{i=0}^n \frac{(-1)^ix^{2i+1}}{(2i+1)!}$

An important point is that here the terms get smaller and smaller as you take more and more of them, so if, for instance, you want to know the value of $\sin 2.4$ you can plug it into the right hand side, take a finite number of terms and you will get an approximation:

$\sum_{i=0}^n \frac{(-1)^i2.4^{2i+1}}{(2i+1)!}$

The higher value you choose for $n$ the more accurate will be your answer, but we can see that we can now, in theory calculate the $\sin$ of any number with pen and paper, so long as you have enough patience and will-power.…

By Jonathan Shock| August 16th, 2015|Courses, English, First year, Level: intermediate, MAM1000, Undergraduate|3 Comments

UCT MAM1000 lecture notes part 21 – Tuesday 18th August

So we’ve now looked at a couple of different functions and found polynomials which approximate the functions to different levels of accuracy. Let’s try and come up with a general method of formulating this. Let’s say that we have some function $f(x)$ and we want to approximate it close to $x=a$ . We will then assume that we can write the polynomial approximation as:

$f(x)\approx \sum_{i=0}^n c_i (x-a)^i$

Note that previously we wrote $a_i$ but it’s good to get used to slightly changeable notation. The context is what should tell you the meaning.

We will first ask that the value of the polynomial is equal to the value of the function at $x=a$ . We do this by setting $x=a$ in both sides of the above. Note that we are being slightly ambiguous in what we mean by the approximation here because in a moment we will go from a $\approx$ sign to an $=$ sign. This is because while the polynomial is only an approximation, we want that certain properties between the two hold exactly at $x=a$ .…

By Jonathan Shock| August 16th, 2015|Courses, English, First year, Level: intermediate, MAM1000, Uncategorized, Undergraduate|1 Comment

UCT MAM1000 lecture notes part 20 – part iii – Monday 17th August

Now we’re going to take a more complicated expression and approximate it by a polynomial function. The function we’re going to look at is $f(x)=2 \sin x+\frac{\cos 3x}{2}$ , but we could choose any function which is well behaved close to where we want to approximate it (there is a much more precise way to phrase this, but for the current discussion, this is enough).

This function looks like:

OK, so how are we going to go about approximating this function? Well, let’s ask about approximating it close to the point $x=2.5$ (this value is arbitrary and we could have asked for any value). What would be the most naive approximation we could make? Well, if we have a function which is a constant, and equal to the original function at $x=2.5$ then that’s a start. At least it matches the value of the function at that point, if nothing else. What is the value of this function at $x=2.5$ ?…

By Jonathan Shock| August 16th, 2015|Courses, English, First year, Level: intermediate, MAM1000, Undergraduate|1 Comment

UCT MAM1000 lecture notes part 20 – part ii – Monday 17th August

So, we saw in the last section that we could work out the polynomial expression for $(1+x)^5$ both using combinatorics as well as using calculus. We had also found previously that for small $x$ we could just take the first couple of terms of the polynomial and it was a good approximation of the function itself, depending on how accurate we needed the answer. For instance, for small $x$ , $1+5x$ is a reasonable approximation. One thing to note is that the value of these two functions is exactly the same at $x=0$ and the derivative of both functions is the same at $x=0$ . The second derivatives are not the same, but had we taken the next term in the polynomial, the second derivatives would have matched as well.

OK, let’s take this one step further and repeat everything exactly as before, but now with the power, not as 5 but as 5.2 and see what happens.…

By Jonathan Shock| August 16th, 2015|Courses, English, First year, Level: intermediate, MAM1000, Uncategorized, Undergraduate|1 Comment

UCT MAM1000 lecture notes part 20 – part i – Monday 17th August

The beginnings of Taylor polynomials and approximations

We’ll start off with some of the discussion we had in class today about treating the binomial expansion as a way to approximate a function.

We know that $(x+y)^5=\sum_{r=0}^5{_5C_r} x^r y^{5-r}$ . We can then see that if we have a function like $(1+x)^5$ we can simply replace the $y$ in the above expression by 1 and we get (writing out the sum explicitly):

$(1+x)^5=1+5x+10 x^2+10x^3+5 x^4+x^5$

If you can’t remember these binomial coefficients offhand, just write out the first 6 lines of Pascal’s triangle and read them off.

ok, so what if I asked you to give me an approximate value for $(1+0.01)^5$ . This isn’t so easy because you’ll have to multiply 1.01 by itself five times. How about if we use the polynomial form from the binomial expansion? Well, we can see that this is:

$(1+0.01)^5=1+5(0.01)+10 (0.01)^2+10(0.01)^3+5 (0.01)^4+(0.01)^5$

But we can also see that each term gets smaller and smaller.…

By Jonathan Shock| August 14th, 2015|Courses, English, First year, Level: intermediate, MAM1000, Uncategorized, Undergraduate|1 Comment

Pascal’s triangle, fractals and the Sierpinski triangle

I mentioned today in class that something rather special happens if you colour the even and odd numbers in Pascal’s triangle differently.

A quick reminder that Pascal’s triangle looks like:

For the first 9 layers.

In the animation below I have simply put an “o” for an odd number and put it in red, and “e” for an even number and put it in brown. Here I am starting with the Pascal triangle with a single entry, and building up in the animation two lines at a time, up to 100 lines. You see that a rather interesting pattern emerges.

The Mathematica code for this is:

tt = Table[With[{n = nn, p = nn},
Graphics[Table[Text[Style[With[{x = Binomial[n – j, n – i]}, If[OddQ[x], Style[o, Red], e]],Round[200/Sqrt[p]]], {Sqrt[3] (i – j/2), 3 j/2}], {i, n}, {j, i}], ImageSize -> 1000]], {nn, 1, 100, 2}]

and then exported as a .gif. (see here for a simplified version and some more Pascal’s triangle code).…

By Jonathan Shock| August 13th, 2015|Uncategorized|2 Comments

Aesthetics and Mathematics

Today I gave a combinatoric argument for the binomial theorem. For me this is the most elegant and clear way to understand it, but there are other proofs which rely on very different techniques.

There is a nice article here, discussing the contrast between three different proofs of this same theorem (using combinatorics, induction and calculus) and arguing why different measures of a beautiful proof would say that each of the three proofs has benefits over the others.

Which proof have you so far seen which you have found most appealing? Can you find a proof that we have already done and write it in a more aesthetically satisfying way?

How clear is this post?

…

By Jonathan Shock| August 13th, 2015|English, Level: intermediate|2 Comments

UCT MAM1000 lecture notes part 19

The Binomial expansion

So, we’re going to start using some of the combinatorics we’ve just learned to answer some questions which without these techniques would seem completely infeasible.

What is $(x+y)^{34}$ ? This seems like an almost impossible task; you’d have to write out $(x+y)(x+y)(x+y)...$ , 34 times, then multiply them all out and it would become incredibly messy. However, everything we’ve done in the last section will allow us to see precisely what this, and any other expression of this form, is given by. Let’s start with a simpler example:

How about $(x+y)^4$ ? This is a simpler example and one that we could think of doing by hand, but we will show that there is a very general way to get any expression of this form. We start by writing this out in long form:

$(x+y)^4=(x+y)(x+y)(x+y)(x+y)$

We are now going to label the $x$ ‘s and $y$ ‘s, though the labels (which will be indices) will just be dummy labels and we will remove them in the end.…

By Jonathan Shock| August 10th, 2015|Courses, First year, MAM1000, Uncategorized, Undergraduate|1 Comment

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