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 a and b 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$ :o

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 w) 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 x and 2^x grows much faster than x^8 so 2^x>x^8 for large values of x. Since both x \mapsto 2^x and x \mapsto x^8 are continuous on \mathbb{R} there is another value of x 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.

    After explaining to a student through various lessons and examples that \displaystyle \lim_{x \to 8} \dfrac{1}{x-8}=\infty
    I tried to check if she really understood that, so I gave her a different example. This was the result:
    \displaystyle \lim_{x \to 5} \dfrac{1}{x-5}=
\rotatebox{90}{5}

:D

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 :D 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

  1. 41
  2. 43
  3. 45
  4. 47

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). :cry:

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