Powers

Friday 25 June 2004 at 3:37 pm | In Articles | 3 Comments

I was asked recently why 2^0=1. Remember that if n is a positive whole number then 2^n= \underset{n \text{ times}}{\underbrace{2 \times 2 \times 2\times \dots \times 2}}. Clearly you can’t multiply 2 by itself 0 times :-?

The key, when extending properties of the number system, is to use definitions that work for every number. So, for example

    \dfrac{2^5}{2^2}=\dfrac{2 \times 2 \times 2 \; \times \! \not{2} \; \times \! \not{2}}{\not{2}\; \times \! \not{2}}=2^3=2^{5-2}

which gives you the rule that

    to divide powers you subtract the indices (the small superscripted numbers)

This leads to

    1=\dfrac{2^3}{2^3}=2^{3-3}=2^0

Similarly, a^0=1 for any positive real number.

What about zero powers of non-negative powers? 0^0 is a controversial case I have mentioned on 29 February (Q2.) and see Dr Math FAQ for more on this.

And if the number is negative? Great care is needed in this case. For example, using only real numbers, (-1)^{1/3}=-1 but (-1)^{1/2} is not a real number. The problem arises because the general definition of a power is given by a^x=e^{x\ln a} and \ln a is undefined if a is negative or 0. Using complex numbers (which helps with (-1)^{1/2}) just makes things more complex :-? – see Log of Complex Number

Misunderstanding

Wednesday 16 June 2004 at 4:23 pm | In Articles | 1 Comment

When teaching maths that you are familiar with, it is not easy to see why students struggle with it – indeed once the student has understood the problem, they can’t see why they had difficulty before! This means that you have to be very careful what you say in case an attentive student takes it literally.

How many times has a lecturer said “the integral of e is itself ” ? So this happens:

    Q. Find the value of \displaystyle \int_e^{e^2}\left(\frac{5}{x}+e\right)\ dx

    A. \displaystyle \int_e^{e^2}\left(\frac{5}{x}+e\right)\ dx=\left[5\ln |x| + e\right]_e^{e^2}=\dots=5

When the error was pointed out to the student, they responded with

    Why does e integrate to ex\;? e^x integrates to e^x and as e=e^1 surely the integral of e is also e\;?

How would you respond to this?

Generalisation of derivative

Sunday 6 June 2004 at 2:25 pm | In Articles | Post Comment

Inspired by a posting on S.O.S. Mathematics CyberBoard

Most students will be familiar with the definition of the derivative of a real-valued function of a real variable defined on some interval (a,b):

    If x \in (a,b) then f is differentiable at x if \displaystyle \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} exists and the limit is denoted f^{\prime}(x)

It is also clear that for this to make sense f must be defined at x (and of course it is a well-known consequence of the definition that f is also continuous at x). But what if f is defined on (a,b) but not at x, can we do anything then? Yes, we can define a pseudo-derivative \Lambda f of f provided f is defined on a neighbourhood of x:

    \displaystyle \Lambda f(x)=\lim_{h \to 0}\frac{f(x+h)-f(x-h)}{2h}

This pseudo-derivative has similar properties to the derivative and indeed it has the same values where f is differentiable but there are significant differences as the following exercises show:

  1. If f is differentiable at x show that \Lambda f(x)=f^{\prime}x)
  2. If f(x)=|x| show that \Lambda f(0) exists although f^{\prime}(0) does not
  3. If f(x)=\left\{\begin{array}{ll}x & \mbox{ if } x<0\\-x^2 & \mbox{ if } x \geq 0\end{array}\right. show that f has a local maximum at 0 but \Lambda f(0) \neq 0
  4. Suppose f is differentiable on (a,b), except at a point x_0 in (a,b), with f^{\prime}(x)\geq 0 for  x \neq x_0.
    If \Lambda (f)(x_0) exists show that \Lambda (f)(x)\geq0

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