Start with a sine wave (shown at right). We have to give it the right period, and by now weve learned that doing a shrink by a factor of makes the function pass through (1, 0).
Now let's try a term with a frequency twice that of the first. We don't know what the coefficient is, so let's use n with a range from 0 to 2. Unfortunately, nothing helps.

It's OK to have
a coefficient of
zero, so we move on to the third component.
Here things are looking better.
That value of 0.4 was pretty good, but if we play around with n some more, it looks like .33 might be better. So we substitute the value as one-third. We go on to the fourth component (which looks terrible, so we assume its coefficient is zero) and then the fifth.
Pretty good--and 0.2 (the value of n in the picture) is about one-fifth. Now we're getting somewhere, and a pattern emerges. But all the even coefficients are zero. On some reflection,that makes sense. Why?

The pattern suggests that the square wave function is

which turns out to be correct. We can write this using summation notation:

To see how the function will look as you add and remove terms from the series, download this and drag the n slider. To do this, you will need to configure NuCalc Web Helper as a helper application.

How did we make the summation sign? For that secret, you have to buy the book!

page backhomepage forward

© 1996 Key Curriculum Press

[Privacy Policy] [Terms of Use]

Home || The Math Library || Quick Reference || Search || Help 

© 1994- The Math Forum at NCTM. All rights reserved.