Mathematics Weblog

Problem

Monday 26 April 2004 at 4:53 pm | In Articles | 4 Comments

Firstly, show that:

$\log_{a^n}b^n=\log_{a}b$

Secondly, are there any other operations you can do to and and the $\log$ retains the same value? In other words:

Find all functions

$f,g:\mathbb{R^+}\mapsto \mathbb{R^+}$

such that

$\log_{f(a)}g(b)=\log_a b$

for all

$a,b\in\mathbb{R^+}$

Cancelling

Tuesday 20 April 2004 at 4:27 pm | In Articles | 1 Comment

Students love to cancel wherever they can, so much so that the book Comic Sections (now out of print) had the following joke:

The student law of universal cancellation: If the same symbol x occurs in any two different places on the one page it may be cancelled

So you get horrors like $\dfrac{\sin x}{n}=\dfrac{\text{si}\hspace{-1mm}\not{n}\hspace{1mm} x}{\not n}=$ six$$ 😮

And yet strange things can happen. It is true that $\frac{2666}{6665}=\frac{266}{665}=\frac{2}{5}$ .
What is remarkable is that you can have as many sixes as you like and cancel them as many times as you like so, for example, $\frac{26666666}{66666665}=\frac{266}{665}$ . Can you prove this is true? Can you find 3 other similar fractions?

Think about it!

Sunday 18 April 2004 at 5:07 pm | In Articles | 2 Comments

What is the value of $(x-a)(x-b)(x-c)\dots(x-z)$ ?

The LambertW Function

Friday 9 April 2004 at 2:35 pm | In Articles | 1 Comment

Many equations cannot be solved exactly without using special functions. For example, to solve 2^x=3 requires the use of the $\ln$ function (or similar). This function is sometimes defined in terms of an integral from which their properties can be deduced. Thus $\ln$ is defined by $\displaystyle \ln x=\int ^x _1 \dfrac{dt}{t}\text{ for } x>0$ and it is then clear that, for example, $\ln 1=0$

There are many equations that can only be solved in terms of newly-defined functions. One such function that isn’t all that well known is the LambertW function where $w\left(x\right)$ is defined as a solution (for ) of we^w=x . This allows you to solve equations like 2^x=x^8 which was asked about on the S.O.S. Mathematics CyberBoard

To solve 2^x=x^8 let $w=-\ln x$ so that $x=e^{-w}$ . Then

$2^x=x^8 \Longrightarrow 2^{e^{-w}}=e^{-8w} \Longrightarrow e^{-w}\ln 2=-8w \Longrightarrow -\frac{1}{8}\ln 2=we^w$

Thus $w=w(\frac{1}{8}\ln 2)$ and so $x=e^{-w}=-8\dfrac{w(\frac{1}{8}\ln 2)}{\ln 2}$ which is our answer.

Using tables or software this gives 1.100.

But hang on, is that the only solution? No, because 2^x<x ^8 for small values of and 2^x grows much faster than x^8 so 2^x>x^8 for large values of . Since both $x \mapsto 2^x$ and $x \mapsto x^8$ are continuous on $\mathbb{R}$ there is another value of for which 2^x=x^8 . A quick fiddle with a calculator gives $x\approx43.5$ .

Research into the LambertW function to find out how this other solution can be given in terms of this function.

Joke

Sunday 4 April 2004 at 2:25 pm | In Articles | Post Comment

Joi Ito’s Page contains a wonderful joke which I really have to repeat here.

$\displaystyle \lim_{x \to 8} \dfrac{1}{x-8}=\infty$

$\displaystyle \lim_{x \to 5} \dfrac{1}{x-5}= \rotatebox{90}{5}$

😀

Trig Ratios

Sunday 4 April 2004 at 10:55 am | In Articles | 1 Comment

A level syllabuses these days expect you to remember the exact values of sin cos and tan of certain angles. $\sin 30^\circ$ is easy enough as the calculator will give you the exact answer, but unless you know roughly what $\sin 60^\circ$ should be then the calculator will be no help.

But, help is at hand 😀 Memorising formulae is easier when there’s a pattern and the following table gives such a pattern.

$\begin{array}{|c|c|c|c|c|c|} \hline \theta & 0^\circ & 30^\circ & 45^{\circ} & 60^\circ & 90^\circ \\ \hline \begin{array}{c} \\ \sin \theta \\ \\ \end{array} & \dfrac{\sqrt{0}}{2} & \dfrac{\sqrt{1}}{2} & \dfrac{\sqrt{2}}{2} & \dfrac{\sqrt{3}}{2} & \dfrac{\sqrt{4}}{2} \\ \hline \begin{array}{c} \\ \cos \theta \\ \\ \end{array} & \dfrac{\sqrt{4}}{2} & \dfrac{\sqrt{3}}{2} & \dfrac{\sqrt{2}}{2} & \dfrac{\sqrt{1}}{2} & \dfrac{\sqrt{0}}{2} \\ \hline \end{array}$

Isn’t that amazing! I only came across this a few years ago but apparently it’s been around at least since the 1950’s.

What about tan? Since $\tan \theta = \dfrac{\sin \theta}{\cos \theta}$ you just divide a value from the second row by the one below it (but please not for $90^\circ$ ; see Tuesday 16 March).

Who Wants to be a Millionaire?

Saturday 3 April 2004 at 8:33 pm | In Articles | 1 Comment

On tonight’s programme the following question was asked:
Which is not a prime number?
The choices are

The sad thing is that the contestant said he gave up as he would only be guessing (a wrong answer would lose him a lot of money). 😥