CAPÍTULO 2 FUNDAMENTOS TEÓRICOS
2.1 MARCO TEÓRICO
2.1.3 Inteligencias múltiples
4. When this happens, declare a to be a root of f (x) to within the desired accuracy. Let us redo our calculations above, using this procedure. Note that for our f (x) = x2− 2 we have f′(x) = 2x and hence f (a)
f′(a) evaluates to
a2
−2
2a . Thus it makes sense to
calculate a nice table as follows: 2
a y = a2− 2 y′ = 2a y y′ a − y y′ 1.0 −1.0 2.0 −0.50000 1.5000 1.5000 0.25000 3.0 0.083333 1.4167 1.4167 0.0070389 2.8334 0.0024843 1.4142 1.4142 −0.00003836 2.8284 −0.000013562 1.4142
10.2
Limitations of the Newton’s Method
Even though it is easy to understand and nice to implement, the above method can fail for several reasons. We give these below to help the reader appreciate why one should not accept the results blindly.
In spite of all the listed limitations, the method works pretty well for most rea- sonable functions.
1. Case of no real roots!. If we start with a function like f (x) = x2 + 2, then
we know that it has no real roots. What would happen to our method?
The method will keep you wandering about the number line, leading to no value! You should try this!
2. Reaching the wrong root. Usually you expect that if you start near a potential root, you should land into it. It depends on the direction in which the tangent line carries you.
2In this example, the formula which changes a to a −a2−2
2a simplifies to 1 2(a +
2
a). This has a
particularly pleasant description. Whichever side of √2 we take our positive a, we can see that 2 a
lies on the other side and their average is exactly 12(a + 2a). This is easily seen to be closer to √2 by observing them as points on the number line. Thus, we know for sure that we are clearly getting closer and closer to our√2. We can even estimate the new accuracy as half the distance between a and 2
a. This particular method was probably well known since ancient times and leads to excellent
approximations of any desired accuracy with rational numbers. Techniques for approximating other square roots using rational numbers by special methods are also scattered in ancient mathematical works. Here is a sequence of approximations to√2 using this simple formula but keeping the answers as rational numbers, rather than decimals.
1, 3/2,17 12, 577 408, 665857 470832, 886731088897 627013566048.
You will find the second to the sixth commonly used in ancient literature. The fourth is accurate to 5 places. The fifth is accurate to 11 places and the sixth to 23 places. The seventh one, when calculated, will come out correct to 48 places!
3. Not reaching a root even when you have them! Consider f (x) = x3− 5x.
Then f′(x) = 3x2 − 5. Then the starting value of a = 1 takes you to
1 − ff (1)′(1) = 1 − −4
−2 = −1. Now the value a = −1 leads to
−1 − ff (−1)′
(−1) = −1 − 4 −2 = 1.
Thus you will be in a perpetual cycle between −1 and 1, never finding any of the roots 0,√5, −√5. The reader should observe that taking other values of a does the trick. For example a = 2 takes you thru:
2 → 2.2857 → 2.2376 → 2.2361. This is the √5 to 5 decimal places.
Other starting values will take you to other roots.
4. Horizontal tangent. If at any stage of the process, the derivative becomes zero, then the value of the new “a” is undefined and we get stuck. For more complicated functions, the derivative may not even be defined! Often changing the starting value takes care of this.
5. Accuracy problems. In addition to the above, there are possible problems with roundoff errors of calculations both in the calculation of the derivative and the function. There are also problems of wild jumps in values if the derivative becomes rather small and so yy′ gets very large!
6. Conclusion. We have only sketched the method. An interested reader can pursue it further to study the branch of mathematics called Numerical Analysis, which studies practical calculation techniques to ensure desired accuracy.
Chapter 11
Summation of series.
We now discuss a simple application of elementary polynomials which can be used to build elegant useful formulas.
Given any function f (x) defined for all positive integers, by a series we mean a sum:
f (1) + f (2) + · · · + f(n) where n is some positive integer.
In usual books on this subject, a compact notation is presented and it goes like this: Pn
i=1f (i) is a short hand notation for the sum of all values obtained by setting
i = 1, i = 2, · · · , i = n in succession. Thus, if f (x) = x2, then we get:
5 X i=1 f (i) = 5 X i=1 i2 = 12+ 22+ 32+ 42 + 52 = 55.
The symbol P stands for the Greek letter “S” to suggest a sum. The reader should note that the particular “index” i in the expression does not affect the final answer, i.e. 5 X i=1 f (i) = 5 X j=1 f (j).
(Just expand each and see!)
It is possible to create manipulation techniques for this summation symbol, but we refrain from getting into it. Our aim, in this chapter, is to build formulas for such summations for some convenient functions. We have actually avoided the summation notation, choosing to write out the long form instead. The reader is advised to try and convert our statements to the summation notation for practice and efficiency.
11.1
Application of polynomials to summation for-
mulas.
Consider any polynomial expression g(x) and define a new polynomial f (x) = g(x) − g(x − 1).
Consider the following sequence of substitutions.
x f (x) = g(x) − g(x − 1) 1 f (1) = g(1) − g(0) 2 f (2) = g(2) − g(1) · · · ·
i f (i) = g(i) − g(i − 1) · · · ·
n f (n) = g(n) − g(n − 1)
If we add up the f -column and equate it to the sum of the last column, we see that most values of g cancel leaving us with a very simple formula
f (1) + f (2) + · · · + f(i) + · · · + f(n) = g(n) − g(0).
This gives us very efficient and clever ways for adding up a series of terms. For example, let g(x) = x. Then f (x) = 1 and we get the trivial result
f (1) + · · · + f(n) = 1 + · · · + 1 = g(n) − g(0) = n − 0 = n. If we take g(x) = x2+ x, then f (x) = 2x and we get the identity
f (1) + · · · + f(n) = 2 + 4 + · · · + 2n = g(n) − g(0) = (n2+ n) − (0) = n(n + 1). Division of both sides by 2 gives the well known
Arithmetic series formula: 1 + 2 + · · · n = n(n + 1)
2 .
Thus to add up values of a gives f (x) all we have to do is to come up with a cleverly constructed g(x) such that f (x) = g(x) − g(x − 1). The best strategy is to try lots of simple g(x) and learn to combine.
Here is a table from simple calculations. g(x) g(x) − g(x − 1) x 1 x2 2x − 1 x3 3x2− 3x + 1 x4 4x3− 6x2+ 4x − 1 x5 5x4− 10x3+ 10x − 5x + 1