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Numerical
Approximation of Zeroes:  
 You may have noticed that always using the midpoint of the interval generated some pretty long numbers. My computations weren't too hard, because I was making intelligent use of my calculator. But there were times when it was pretty obvious that I could have used better, and easier, x-values, and still gotten the same answer. If you're working with a calculator that doesn't allow for the symbolic calculations of the polynomial for a given value of x, then choosing x-values with fewer decimal places could be especially handy. So let's find the polynomial's zero again, but this time we'll go to the trouble of picking our the x-values, instead of just mindlessly always using the midpoint.
Since the zero is closer to x = 1, I'll try x = 1.4. x = 1.4: y = 1.92224 This gets me a lot closer to the axis, but this y-value is positive so the zero is still lower. I'll try x = 1.2: x = 1.2: y = –1.398369 That's nearly as negative
as the last y-value
was positive. Okay, so the zero is above x
= 1.2. I'll try
x
= 1.3: x = 1.3: y = 0.00993 Oooo; that's pretty good. Okay, so I'm really close to the polynomial's zero, but I'm still a little high. I'll edge down by hundredths and try x = 1.29: x = 1.29: y = –0.151005035... That took me further from the axis, and below. So the zero is higher, and probably a lot closer to 1.300 than to 1.290. So I'll try x = 1.297: x = 1.297: y = –0.0388891527... This y-value is closer to zero, but I'm still too low, and x = 1.300 got me a bit closer still, so I'll try x = 1.299: x = 1.299: y = –0.006394647... I'm getting closer, but I'm still too low. I've noticed that 1.3000 was not quite as close to the zero as was 1.2990, but the polynomial also seems to grow faster after the zero than before it, so I'll split the difference and see what I get: Copyright © Elizabeth Stapel 2005-2011 All Rights Reserved x = 1.2995: y = 0.001761215... Okay, now I'm too high again, but I'm getting pretty close. Lemme edge down a bit: x =
1.2994: y = 0.000129009...
Finally! I've managed to bracket the zero between two x-values that have the same first three decimal places. That's all the accuracy I need, so I'm done. x = 1.299 As you can see, by picking your own x-values, you can simplify and shorten your computations. But you do have to pay attention to what you're doing, or you could end up picking lots of fairly useless values and taking nearly as long as you would by using the midpoint process. Either way, though, numerically approximating zeroes isn't terribly complicated; it's just long and tedious. If you don't have a graphing calculator, just a regular calculator, you may have to do a little more preliminary work on your own (if they don't tell you "look between here and here for the zero"), but the general process will still be the same.
The first thing I have to do is figure out the general area of the zero. Without a graph and without limits being given to me, I'll have to use what I know of polynomials to judge the values I get. I'll try a few values in the vicinity of the origin: x =
0: y = –9.1
(I'm
below the axis) Bingo! I've found a sign change, so the zero is somewhere between x = –3 and x = –2. Also, the polynomial is above the axis at the lower x-value, so I know the graph goes down as it crosses the axis. Then a negative y-value will mean that my x-value was too high, and a positive y-value will mean that my x-value was too low. This will tell me in which direction I need to move. Since y = –2.5 (at x = –2) is a lot closer to zero than is y = 36 (at x = –3), the zero must be a lot closer to x = –2, so I'll test some values in the tenths: x =
–2.1:
y = –0.693469
(I'm
getting closer to the axis; I'll move down a bit) Okay, now I know the zero is between x = –2.2 and x = –2.1, and it looks like it's closer to x = –2.1. I'll work my way down into the hundredths: x =
–2.13:
y = –0.0870641...
(I'm
really close, but I'm still above the axis; I'll move down) The sign change tells me that the zero is bracketed by –2.14 and –2.13, with the polynomial being closer to the axis at –2.13. I'll test values in the thousandths: x =
–2.133:
y = –0.02469719...
(I've
already crossed the axis, so I need to back up) Now I have the zero bracketed between –2.135 and –2.134, numbers that have the same first two decimal places, so I'm done. x = –2.13 If you're writing a computer program or graphing-calculator program to find numerical approximations, use the midpoint process: it's mindless and rote, which is what you need for an automated process like a program. If your text specifies the method you should use, you should stick with that. Otherwise, use whichever method you prefer. Just allow yourself some time, and be sure to write down your steps carefully. About the only problem you can really have with this process would come from copying down a number wrong. As long as you're neat and don't rush, you should do fine. << Previous Top | 1 | 2 | Return to Index
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