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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

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