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Theo P
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Post Number: 235
 Posted on Friday, 13 September, 2013 - 05:54 pm:

Hi. I've been doing some revision of A Level Maths using the 'Mathematics Workbook' sent to me from the university. I've collected a few assorted questions / things I didn't quite understand.
Sorry to post them all together, but I didn't want to start loads of threads!

Series Expansions
What is the meaning of 'well-behaved' in the context of functions?
The relevant question was to find the series expansion of $\frac{1-\cos^2{x}}{x^2}$ which is $1 - \frac{x^2}{3} + \cdots$.
I guess the point they're getting at is that it looks like the function is undefined at x=0, but it actually isn't?!

Trig
What does $\tan^{-1}{a} + \tan^{-1}{b} = \tan^{-1}{\left(\frac{a+b}{1-ab}\right)} + n\pi$ mean?
It looks like something to do with adding angles, but I'm not sure how to interpret it.
I drew a diagram of my thoughts ...

Conic Sections
Can anyone recommend somewhere I could read up on conic sections? We didn't really cover them as such in A-level, and I'm finding them quite confusing.
How can I find the directrix / focus / eccentricity of a conic section, given the equation for it?
I was thinking that I could compare to a known form. Or more satisfying would be a method directly from their definition?

I found this question particularly tricky:
Show $r^{-1} = k\cos{\theta} + m$ can describe any conic section.
What I did was to fiddle about with it to get it into cartesian form ...
$1 = kr\cos{\theta} + mr$
$1 = kx + m\sqrt{x^2 + y^2}$
$x^2 \left(m^2 - k^2\right) + my^2 = 1-2kx$
Then considering $k=m$ gives $y^2 = 1/k - 2x$ which is a parabola.
If $m>k$ the coefficient of $x^2$ is positive, so it looks like an ellipse.
If $m<k$ then the opposite is true, so it's a hyperbola.
However, this all looks very messy and the forms aren't exactly the same as the standard ones:
$(x-x_0)^2 / a^2 + y^2 / b^2 = 1$ as opposed to simply $x^2 / a^2 + y^2 / b^2 = 1$, for example.
Is there a neater way to show it? Also, how would I go about classifying the conic sections - finding focus, etc?

Parametric Differentiation
Show that if $x=a\cos{\theta}$, $y=b\sin{\theta}$ then $\frac{d^2y}{dx^2} < 0$ for $y>0$.
This question was under parametric differentiation, and I tried doing it parametrically but got lost and converted to cartesian instead.
How would one go about doing it parametrically? I found myself wanting to use second-order equivalents of $1 / \frac{dx}{dy} = \frac{dy}{dx}$ and the chain rule, which aren't allowed!
This is what I did ...
Converting to cartesian: $x^2/a^2 + y^2/b^2 = 1$
Differentiating implicitly: $2x/a^2 + dy/dx 2y/b^2 = 0$
Same again: $2/a^2 + (\frac{dy}{dx}^2 \frac{2}{b^2} + \frac{d^2y}{dx^2} \frac{2y}{b^2} = 0$
Rearranging: $\frac{d^2y}{dx^2} = -\frac{b^2}{2y}\left(\frac{2}{a} + \frac{2}{b^2}\left(\frac{dy}{dx}\right)^2\right)$
which is <0 for y>0.

Differential Equations / SHM
The first DE is $\frac{d^2y}{dx^2} + k^2 y = 0$.
I'm told to re-write it as $z \frac{dx}{dy} + k^2 y = 0$ where $z = \frac{dy}{dx}$.
It follows that $z = \pm \sqrt{c^2 - k^2 y^2}$ (with c a constant of integration).
We then have that $\frac{dy}{dx} = \pm \sqrt{c^2 - k^2 y^2}$ and can solve by reciprocating to find x in terms of y.
$x = \pm \frac{1}{k} \sin^{-1}{\frac{ky}{c}} + c_2$
Re-arranging gives $y = R \sin{k\left(\pm x - x_0 \right) }$ where $R = c/k$ and $x_0$ is a constant (of integration).
The answers give it without the plus/minus sign - is that because the $x_0$ takes account of it?

The second part is very similar except for a sign-change: $\frac{d^2y}{dx^2} - k^2 y = 0$.
The solution comes out as $y = R \sinh{k\left(x-x_0 \right)}$, and the question then asks if, in this case, it is the general solution?
The presence of the question makes me doubtful, but I don't see why it wouldn't be.
Also, by expanding the sinh, you can transform the solution to the form $y = Ae^{kx} + Be^{-kx}$ which looks right for a general solution.
Is the first (sin) solution general too, or not?

Quixote (Tim King)
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Post Number: 204
 Posted on Friday, 13 September, 2013 - 09:31 pm:

Let's do these one at a time, in the order that you posted them.

'Well behaved' is a nonspecific term. In my opinion, if we say that an object is 'well-behaved' we mean that 'it has enough nice properties for us to draw some interesting conclusions about it.'

For example, a general function from $R$ to $R$ needn't be well-behaved. For example, consider the function defined by $f(x)=0$ when $x \neq 1$ and $f(1)=1$.

This has the disturbing property that $\int_0^2 f(x) dx =0$, even though $f(x) \geq 0$ for all $x$, and $f$ isn't identically zero. If we don't want such counter-intuitive things to be true, we might restrict our study to that of 'continuous' functions - those that have no 'jumps'.

As another example, when studying Taylor functions and the like, we need to state the values of $x$ for which our series expansion is valid. For example, the GP sum formula says that
$\frac{1}{1-x}=1+x+x^2+x^3...$
We can substitute $x=2$ this formula to 'prove' that
$-1=1+2+4+8+16...$
which is obviously rubbish. We say that the sum on the right is only well-behaved (or, in precise language, 'converges') when $|x|<1$

In your example of $f(x)=\frac{1-cos(x)^2}{x^2}$, we can (correctly) assume that your series expansion is well-behaved for $x=0$ to deduce that, when $x$ is close to zero, $f(x)$ is close to 1.

In fact, if we extend the definition of $f$ to include zero as well, by specifying that $f(0)=1$, it can be shown that we get a continuous function which is differentiable for every value of $x$, including zero. This is certainly well-behaved.

Do post back if you want to discuss anything in this post. Otherwise, I guess we can move on to trig :P
Theo P
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Post Number: 236
 Posted on Friday, 13 September, 2013 - 10:07 pm:

I think that I understand the examples you've made up. They make sense. I still don't really get the original one, though.

Why can you correctly assume the series expansion is valid for $x=0$? I can see that if I were to carry on finding terms in the expansion, it would just be a polynomial - is that the justification?

And why do you switch from the precise $x=0$ to saying 'when $x$ is close to zero, $f(x)$ is close to 1.'? If $f$ is already 'well-behaved for $x=0$', then why do we need to extend the definition to include zero in the domain by specifying that $f(0)=1$?
Or does having a potential division by zero automatically exclude any offending values in the domain? What about $x^2 / x$?

How can it be shown that you get a continuous function? I plotted the graph on the computer, and it looks continuous but that's hardly justification!
Daniel Fretwell
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Post Number: 1981
 Posted on Friday, 13 September, 2013 - 10:37 pm:

At A-level maths the "it looks about right" ethos works...I guess the purpose of these questions is to show you that actually you need to be more rigorous if you are to study for a degree in maths.

So the point is you cannot actually evaluate $f(0)$ directly, the function isn't "well defined" there...but you CAN evaluate it's series expansion at $x=0$. The question is, will it give something meaningful?

Of course in this case comparing both sides at $x=0$ is ridiculous (one side of the equation doesn't even exist there), however if you look at values close to $0$ then you find that both sides DO give the same value.

What you are really doing by plugging $x=0$ into the series expansion is getting the "most likely" value of $f(0)$ based on its surrounding values. You are getting the value that you would expect to find if $f$ were continuous at $x=0$. If you look at your graph you see that at $x=0$ the function looks like it SHOULD have value $1$ yet we know it doesn't.

Rigorously what you are getting when plugging in $x=0$ is the limit of $f(x)$ as $x$ tends to $0$. There is a subtle difference between this and $f(0)$ which you have probably never met before since at A-level most functions you use are continuous everywhere.

I guess the purpose of this question is to stop you thinking that series expansions (and other things) are just tools for plugging in random values, they are more delicately defined than that.
Theo P
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Post Number: 237
 Posted on Friday, 13 September, 2013 - 11:02 pm:

Okay so ...
$\frac{1-\cos^2{x}}{x^2}$ is not defined for $x = 0$ and never will be. However, by using a series expansion (or trying small values), we can show that as $x$ tends to zero, the value tends to 1. To make $f$ 'well behaved', we can then specify that $f(0)=1$.
Is that right?

The wording in the workbook makes it sound like $f$ is 'well behaved' even without defining $f(0)=1$.
"The interesting thing about the function in part (v) is that the series shows it is 'well-behaved' at $x=0$, despite appearances."
Quixote (Tim King)
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Post Number: 205
 Posted on Saturday, 14 September, 2013 - 01:49 am:

"Why can you correctly assume the series expansion is valid for x=0?"

What I meant here was, 'Let's assume that the series expansion is valid for numbers close to x=0. In fact, while we cannot prove it at this stage, this assumption is correct.' By substituting in x=0 we obtain $lim_{x \rightarrow 0} f(x)=1$.

Non-rigourous chit-chat coming up: Most of the time, a series expansion about a point $a$ will approximate f(x) very well provided x is close to a. There are a couple of functions which don't work. If we let $f(x)=exp(1/x^2)$ and $f(0)=0$ then the Taylor series for $f$ about $x=0$ is identically zero so doesn't converge to the function anywhere apart from the initial point of $x=0$, but such perverse examples are rare and 99% of the time a Taylor expansion will provide a good local approximation to the function.

"How can it be shown that you get a continuous function?"

Wait until your first year at uni, or see below.

Fact 1: any 'nice' function that you know, such as cos(x) and x^2 are continuous.

Fact 2: When you multiply continuous functions or divide them (provided you aren't dividing by zero) what you get is still continuous.

Fact 3: If you have a function of the form $f(x)=\frac{P(x)}{Q(x)}$ where $lim_{x \rightarrow 0}P(x)=lim_{x \rightarrow 0}Q(x)=0$, then $lim_{x \rightarrow 0} f(x)=\frac{P'(0)}{Q'(0)}$ (This is called 'L'hopital's rule')

The combination of facts 1 and 2 tell us that $\frac{1-cos^2(x)}{x^2}$ is continuous for all nonzero $x$. Fact 3 tells us that $lim_{x \rightarrow 0} f(x)=1$, so it follows that if we extend the definition of $f$ to include zero (at present, we have not specified the value of f at zero) such that f(0)=1, the extension of $f$ is continuous over $R$. (I haven't actually defined what continuous means, so you have to take this on faith a little bit, but if you accept the 'facts' above, what I have written here is essentially a rigourous proof.)

"Okay so ... 1−cos2xx2 is not defined for x=0 and never will be. However, by using a series expansion (or trying small values), we can show that as x tends to zero, the value tends to 1. To make f 'well behaved', we can then specify that f(0)=1. "

Correct. If I define $h(x)=x/x$ then I have not defined $f(0)$, because 0/0 is undefined. However you can see that $h(x)$ is 'really' just the constant function that sends everything to 1, so $h$ has the following 'natural extension' which is well-behaved:

$h(x)=x/x$ for $x \neq 0$ and h(0)=1.

In your example, you have defined $f(x)=\frac{1-cos^2(x)}{x^2}$, but this on its own doesn't define $f(0)$. However, we can use the Taylor series to show that $lim_{x \rightarrow 0} f(x)=1$ and hence that $f$ has a 'natural extension' to the real line, which is
$f(0)=1$ and $f(x)=\frac{1-cos^2(x)}{x^2}$ for $x \neq 0$.
(As I have said, we are assuming (at this stage I cannot actually prove it) that the Taylor expansion does in fact provide a good local approximation to $f$. Most of the time, this assumption is ok.)

The remark in the workbook is just saying 'f(x) looks at first sight horrible at $x=0$ because you are dividing by zero. However, it turns out that this function doesn't exhibit bad behaviour by 'blowing up' at $x=0$ (compare with $g(x)=1/x^2$) but 'behaves nicely' by tending to 1.

Sorry for the long post!
Theo P
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Post Number: 238
 Posted on Saturday, 14 September, 2013 - 09:07 am:

Fantastic - thank you Tim and Daniel.

... trig? :D
Quixote (Tim King)
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Post Number: 206
 Posted on Sunday, 15 September, 2013 - 09:39 pm:

"What does it mean to say that $\tan^{-1}{a} + \tan^{-1}{b} = \tan^{-1}{\left(\frac{a+b}{1-ab}\right)} + n\pi$?

I suppose it's the $+n \pi$ that's confusing you. Let's work this through one at a time.

i) What general property does a real-valued function need in order for it to have to have an inverse?

ii) Show that $f: R \rightarrow R, \space x \mapsto tan(x)$ doesn't have an inverse, but that the restriction of $f$ to a domain which you should specify (there is more than one possible choice for the domain - choose what you consider to be the 'most natural') does have an inverse.

Your answer to part (ii) gives you a way of defining $tan^{-1}(x)$ for any real number $x$.

Now let $a$ and $b$ be real such that $ab \neq 1$. Then:

Is it true that $\tan^{-1}{a} + \tan^{-1}{b} = \tan^{-1}{\left(\frac{a+b}{1-ab}\right)}$?

Hopefully considering the above will tell you why there's a $+ n\pi$ in the identity which you were asked to prove.
Theo P
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Post Number: 239
 Posted on Sunday, 15 September, 2013 - 10:05 pm:

I think I may have misinterpreted the question "What exactly does part (v) mean?" I thought it was asking to give some deeper meaning to the equation, but it might just be 'explain the $+n\pi$'.

"What general property does a real-valued function need in order for it to have to have an inverse?"
It needs to be a bijection?

"Show that $f: R \rightarrow R, \space x \mapsto tan(x)$ doesn't have an inverse, but that the restriction of $f$ to a domain which you should specify (there is more than one possible choice for the domain - choose what you consider to be the 'most natural') does have an inverse."
tan is a periodic function, with period $\pi$. E.g. $\tan(\pi / 4) = \tan(5 \pi / 4) = 1$. The restriction of the domain to $[-\pi /2, \pi /2]$ makes sense and is probably most natural.

It's not true that $\tan^{-1}{a} + \tan^{-1}{b} = \tan^{-1}{\left(\frac{a+b}{1-ab}\right)}$.
Does the $+n\pi$ just allow adjustment of the answer so that the restricted domain doesn't throw things off by a factor of pi?
I think it should be read as 'for some integer n' as opposed to 'for all n'? I can only think of examples where it's off by $\pm \pi$.
Quixote (Tim King)
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Post Number: 207
 Posted on Monday, 16 September, 2013 - 06:44 pm:

Precisely. If $a$ and $b$ are real with $ab \neq 1$, then define
$P=tan^{-1}(a)+tan^{-1}(b)$
$Q=tan^{-1}\frac{a+b}{1-ab}$

Then it is easy to show that $tan(P)=tan(Q)$, and so it follows immediately that $P=Q+n \pi$ for some integer $n$.

However, it is also clear that $-\pi/2 < Q <\pi/2$ and $-\pi<P<\pi$ so you can only ever be off by $\pi$, as you suggest.

I don't know much about conics so can somebody else discuss this? If nobody replies, I'll do my best.
Theo P
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Post Number: 240
 Posted on Monday, 16 September, 2013 - 11:33 pm:

Thanks. That all makes sense.

We could move onto Parametric Differentiation or D.E.s temporarily, if you like?
Quixote (Tim King)
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Post Number: 208
 Posted on Tuesday, 17 September, 2013 - 04:45 pm:

For parametric differentiation, rather than trying to 'remember' second-order derivative formulae, it's better to derive them yourself, using the first-order chain rule.

Here's a hint:

If $x$ and $y$ both depend on $t$, then it is possible to express any formula involving $x$ and $y$ in terms of $t$ only. In particular, we can express $\frac{dy}{dx}$ as a function of $t$.

Now let $F(t)$ be the function given by $F(t)=\frac{dy}{dx}$.

Then to find $\frac{d^2y}{dx^2}$, you need to differentiate $F$ with respect to $x$. How can you do this?

Theo P
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Post Number: 241
 Posted on Tuesday, 17 September, 2013 - 05:52 pm:

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(F(t)\right) = F'(t) . \frac{dt}{dx} = F'(t) / \frac{dx}{dt}$
In this case, $x=a\cos{\theta}$ and $y=b\sin{\theta}$ so $\frac{dx}{d\theta} = -a \sin{\theta}$ and $\frac{dy}{d\theta} = b \cos{\theta}$.
So $\frac{dy}{dx} = -\frac{b}{a} \cot{\theta}$.
Then $\frac{d^2y}{dx^2} = \frac{d}{d\theta} (-\frac{b}{a} \cot{\theta}) / \left(-a \sin{\theta}\right) = -\frac{b}{a^2} \text{cosec}^3{\theta} = -\frac{b^2 \text{cosec}^2{\theta}}{a^2 y} < 0$ for $y > 0$.
If that's right, it's much neater.
Quixote (Tim King)
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Post Number: 209
 Posted on Tuesday, 17 September, 2013 - 07:11 pm:

Correct. For differential equationss, the $x_0$ and $R$ do take into account the plus or minus. Can you make this precise? (Hint: $sinh(-x)=-sinh(x)$.)

I will have a think about the hyperbolic thing, but I can't see why it wouldn't be the general solution either.
Theo P
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Post Number: 242
 Posted on Tuesday, 17 September, 2013 - 07:21 pm:

Great, I thought that was the case with the DEs. And yes, I can see where it comes from with sin(-x)=-sin(x) and sinh(-x) = -sinh(x).

Now, I wonder if there are any conic experts snooping around here ... ?
kmsquared
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Post Number: 821
 Posted on Tuesday, 17 September, 2013 - 07:42 pm:

I wouldn't claim to be a conic expert, but if I wanted to derive an equation for a general conic in polar coordinates I'd make the origin a focus and a line x=a the directrix and let the eccentricity be e. Then you can transform what you get into the form given in the question.

Kate
Theo P
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Post Number: 243
 Posted on Thursday, 19 September, 2013 - 01:08 pm:

If the origin is the focus and $x=a$ is the directrix, then to satisfy the property "distance from focus is e times distance from directrix", we have $r = e \left( a - r\cos{\theta} \right)$. Rearranging gives $r^{-1} = a^{-1} \cos{\theta} + \left(ae\right)^{-1}$.
The form in the question is $r^{-1} = k \cos{\theta} +m$ so we have the same form with $k = a^{-1}$ and $m = \left( ae \right )^{-1}$.
So this shows that the given form can describe any conic section.

How can one do transformations with polar coordinates - for example a translation or rotation?
I can only think how to do transformations with Cartesian coordinates. The Cartesian form of the above equation is rather messy - I make it to be: $$\dfrac{\left(1-e^2\right)^2}{\left(ae\right)^2} \left(x + \dfrac{ae^2}{\left(1-e^2\right)}\right)^2 + \dfrac{1-e^2}{\left(ae\right)^2} y^2 = 1$$
I guess I could compare coefficients with that, but surely there's a nicer way?!

Can anyone recommend a book or online notes that covers the basics on conics and would answer these sorts of questions?
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Post Number: 249
 Posted on Thursday, 19 September, 2013 - 08:31 pm:

Hehe, I thought that this would be the least useful chapter in MODA (covering both the Cartesian and polar forms, focus-directrix property, reflector property and applications):

http://cp4space.files.wordpress.com/2012/11/moda-ch13.pdf

I guess you've proved me wrong.
Theo P
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Post Number: 246
 Posted on Sunday, 22 September, 2013 - 02:15 pm:

Sorry, I kept meaning to have a proper look at this, but only just got round to it.

I'm stuck on the first question ...
What do you mean by 'Prove that the directrix is the polar of the focus'?
What is the polar of something? What are projective and affine transformations?

I think this might be a bit too high-level for me, with talk of non-degenerate quadratic curves, affine transformed circles, Steiner's porism, and quoted IMO questions.

I was looking more for a basic text on conics. Perhaps my current understanding will suffice for now.
Kiran Ashtekar
Frequent poster

Post Number: 102
 Posted on Sunday, 20 October, 2013 - 09:52 pm:

About the conic sections questions :

I have a few queries myself, which I am asking about in a separate thread. I hope the responses to those may throw some (indirect) light on the q.s asked here by Theo P.
Theo P
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Post Number: 248
 Posted on Monday, 21 October, 2013 - 09:30 am:

In the end, I managed to work through some of the first bit of Adam's MODA chapter. It cleared up a few things. I'll probably come back to it when conics crop up (I think in Dynamics & Relativity, next term).

I'll be sure to look at your thread too, Kiran, to see if there are any useful replies - thanks for the heads-up!