Level: intermediate – Page 6

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

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

The mean value theorem for integrals – proof

I left out one line at the beginning, which was pointed out in class.

We start with some function, $f(t)$ which is continuous on $[a,b]$ . We define:

$F(x)=\int_a^x f(t)dt$

By the fundamental theorem of calculus, this is continuous of $[a,b]$ and differentiable on $(a,b)$ .

Now, we use the mean value theorem for derivatives (somewhere in an interval, a function has a gradient which is equal to its average gradient) which say that if $g$ is differentiable on $(a,b)$ and continuous on $[a,b]$ then there exists some $c$ between $a$ and $b$ such that $g'(c)=\frac{g(b)-g(a)}{b-a}$ .

Now simply using our function $F(x)$ in the mean value theorem for derivatives, we have that:

$F'(c)=\frac{F(b)-F(a)}{b-a}$

But we know that $F'(c)=f(c)$ . We also have defined $F(x)$ above, so we can plug in $a$ and $b$ to get:

$f(c)=\frac{\int_a^b f(t) dt-\int_a^a f(t) dt}{b-a}$

The second term on the right hand side is zero as the two limits are the same, so we have:

$f(c)=\frac{\int_a^b f(t) dt}{b-a}$

But we know that $\int_a^b f(t) dt=\int_a^b f(x) dx$ as $t$ and $x$ are just dummy variables, and so we have proved that:

$f(c)=\frac{\int_a^b f(x) dx}{b-a}$

This finishes the proof.…

By Jonathan Shock| August 7th, 2015|English, Level: intermediate|0 Comments

Areas between curves

We know how to calculate the area between a curve and the horizontal ( $x$ ) axis of a graph. We have learnt a number of very sophisticated techniques which allow us to get an analytic answer.

It is a fairly simple extension to study not the area between a curve and the horizontal axis, but between two curves. You can see here the link between the two:

We can see in the figure below how we can approximate the area by a sum of rectangles. In this case we have used the left point approximation. Each rectangle here has $\Delta x=0.02$ and the heights are $f(x_i)-g(x_i)$ where the $x_i$ are the left points of each rectangle, in this case $=0.02(i-1)$ . So $x_1=0$ , $x_2=0.02$ , etc.

We started off studying integration by taking the limit of a Riemann sum. We can do exactly the same thing here, taking the simplest possible Riemann sum, with rectangular regions which stretch between the two curves (see the diagrams on page 448 in Stewart) and then taking the limit to find an expression in terms of integrals.…

By Jonathan Shock| July 30th, 2015|Courses, English, First year, Level: intermediate, MAM1000, Uncategorized, Undergraduate|4 Comments

UCT MAM1000 lecture notes part 8

Last time we looked at integrals which weren’t proper integrals because the limits of integration were infinite (either on one side, or both). This was one of the constraints we had on a well defined Riemann sum. The other constraint we had was that there were no infinite discontinuities in the integrand. Here we will show that sometimes we can indeed define an improper integral which does include such a discontinuity within the limits of integration.

Improper Integrals of the second kind: Infinite discontinuities in the integrand

We have seen what happens when we integrate $x$ from $\infty$ or to $-\infty$ . Sometimes it gives a convergent value and we can find the integral, sometimes the improper integral does not converge and it gives us an answer of $\pm\infty$ . Indeed sometimes it doesn’t blow up, but it just doesn’t converge – as in the integral of $\cos x$ over an infinite range.

Now we are going to look at what happens if the function itself has an infinite discontinuity in it and we want to integrate over this region, or include it at one of the limits of integration.…

By Jonathan Shock| July 28th, 2015|Courses, English, First year, Level: intermediate, MAM1000, Uncategorized, Undergraduate|3 Comments

UCT MAM1000 lecture notes part 6 (part i)

Some thoughts on exam preparation

It feels like the exams are an age away, but believe me, the coming weeks will fly by and before you know it the exams will be upon you. There are things that you can do now, which are going to make the period before the exams, and the exams themselves be a lot less stressful than they would be otherwise.

Part of the issue with MAM1000 is that it’s a big course, which covers a lot of material. It will have been many months since you studied limits by the time you get to the exam, so keeping your hand in with all the topics covered will make life much easier for you.

Of course you need to do the tutorials each week, but I recommend, starting now, going through a couple of old tuts each week. It shouldn’t take too long as you have done them before, and the more times you do them, the faster they will go.…

By Jonathan Shock| July 25th, 2015|Courses, English, First year, Level: intermediate, MAM1000, Uncategorized, Undergraduate|6 Comments

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