makes the function pass through (1, 0). 
| 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? |  |
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!
 |  |  |
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