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Math 104 - Calculus I. Part 8 Power series, Taylor series. Series. First .. a review of what we have done so far: 1. We examined series of constants and learned that we can say everything there is to say about geometric and telescoping series.
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Math 104 - Calculus I Part 8 Power series, Taylor series
Series • First .. a review of what we have done so far: • 1. We examined series of constants and learned that we can say everything there is to say about geometric and telescoping series. • 2. We developed tests for convergence of series of constants. • 3. We considered power series, derived formulas and other tricks for finding them, and know them for a few functions. • 4. We used the ratio tests to determine intervals on which power series converge, and use the other tests to check convergence at the endpoints of the intervals.
Geometric series • a/(1-r) = a + ar + ar2 + ar3 + ... • provided |r|< 1. • We often use partial fractions to detect telescoping series, for which we can calculate explicitly the partial sums sn
Tests for convergence for series of constants • Fundamental divergence test (nth term must go to zero for convergence to be possible) • Integral test • Comparison and limit comparison tests • Ratio test • Root test • Alternating series test
Power series • f(x) = a0 + a1 x + a2 x2 + a3 x3 + ... • where an = f(n)(0)/n! • We know the series for ex, sin(x), cos(x), • 1/(1-x), and a few related functions.
Convergence of power series • Before we get too excited about finding series, let's make sure that, at the very least, the series converge. • Next week, we'll deal with the question of whether they converge to the function we expect. But for now, we'll assume that if they converge, they converge to the function they "came from". • (Strictly speaking, this is not always true -- but it is true for a large class of functions, which includes nearly all the ones encountered in basic science and mathematics. This fact was not fully appreciated until the early part of the twentieth century.) • Fortunately, most of the question of whether power series converge is answered fairly directly by the ratio test.
Recall that... • for a series of constants , we have that the series converges (absolutely) if the the limit of the absolute value of is less than one, diverges if • the limit is greater than one, and the test is indeterminate if the limit equals one. • To use the ratio test on power series, just leave the x there and calculate the limit for each value of x. This will give an inequality that x must satisfy in order for the series to converge.
Your turn... • Calculate the series for the function sin(x) and determine for which x the series converges.
OK, your turn... A. -1 < x < 1 B. -2 < x <2 C. 1/2 <x < 1/2 D. -2 <x <2 E. -1/2 <x <1/2
One more... A. -1 < x < 1 B. -1 <x <1 C. 1 <x <1 D. -1 <x <1 E. 0 <x <1
From these examples, • ...it should be apparent that power series converge for values of x in an interval that is centered at zero, i.e., an interval of the form[-a, a] , (-a, a], [-a, a) or (-a, a) (where amight be either zero or infinity). The interval is called the interval of convergenceand the number ais called the radius of convergence.
Let’s go back To finding series of functions:
Try this... • Take the derivative of the series for sin(x) to get
Integrate both sides of the geometric series from 0 to x to get:
Start from the geometric series again... 2 And substitute x for x everywhere it appears to get
A challenge to think about... • How to get the other one from previously
Application of Series • 1. Limits: Series give a good idea of the behavior of functions in the neighborhood of 0: • We know for other reasons that • We could do this by series:
This can be used on complicated limits... • Calculate the limit:A. 0 • B. 1/6 • C. 1 • D. 1/12 • E. does not exist
Application of series (continued) • 2. Approximate evaluation of integrals: Many integrals that cannot be evaluated in closed form (i.e., for which no elementary anti-derivative exists) can be approximated using series (and we can even estimate how far off the approximations are). • Example:Calculate to the nearest 0.001.
According to Maple... • The last series is an alternating series with decreasing terms. We need to find the first one that is less than 0.0005 to ensure that the error will be less than 0.001. According to Maple: • evalf(1/(7*factorial(3))), evalf(1/(9*factorial(4))),evalf( 1/(11*factorial(5))); • evalf(1/(13*factorial(6))); .02380952381, .004629629630, .0007575757576 .0001068376068
Keep going... • So it's enough to go out to the 5! term. We do this as follows: • Sum((-1)^n/((2*n+1)*factorial(n)),n=0..5) = sum((-1)^n/((2*n+1) *factorial(n)),n=0..5); • evalf(%); .7467291967=.7467291967
and finally... • So we get that to the nearest thousandth. • Again, according to Maple, the actual answer (to 10 places) is • evalf(int(exp(-x^2),x=0..1)); .74669241330
Try this... • Sum the first four nonzero terms to approximate • A. 0.7635 • B. 0.5637 • C. 0.3567 • D. 0.6357 • E. 0.6735
Application of series (cont.) • 3. Differential equations : Another important application of series is to find "solutions" to differential equations (when other methods fail). • Earlier, we found the series for based on the differential equation it satisfies (y ' = y ), together with the initial condition y(0)=1.
Airy functions • The equation y '' + xy = 0 has no "elementary" solution (the solutions of this equation are called "Airy functions", named after a famous British Royal Astronomer). But we can find series for the solution that satisfies y(0)=1, y '(0)=0, as follows: • From the initial conditions, we can assume that the series for y(x) begins:
Airy (continued) • Then the series for series:
So... We work our way from left to right -- since y’’+ xy is supposed to be zero, every coefficient must be zero. Therefore:
What is the denominator of the x term of this series? 9 • A. 720 • B. 1080 • C. 1440 • D. 4440 • E. 12960
Application of series (cont.) • 4. Algebraic equations depending on a parameter ("regular perturbations"). • Consider the equation . We could calculate the solutions of this equation using the quadratic formula, but it will be instructive to think of the two solutions (x) as functions of the parameter a. • We will find the power series of one of these functions. • Keep in mind that we are thinking of x as a function of a (and not the other way around)! • Write x = f(a). • First, if a=0, there are two roots, x = 2 and x = -2. We'll calculate the series for the larger root, so f(0)=2.
Graph • To see how good the approximation is, we plot the "actual" solution (in blue) and our approximation (in red) on the same graph
And…. • What is the beginning of the series for the other root?
A max/min problem • During World War II, the blood of thousands of recruits had to be tested for various diseases. The incidence of the diseases was quite low (less than one per hundred). Assume that the probability that a random person has the disease is p (for some p between 0 and 1, but much closer to 0 than to 1). Let q = 1 - p be the probability that the person does not have the disease. • Because the tests were costly, someone had the bright idea to pool several (x) samples at a time. If the pooled sample tested negative, then all x samples were negative. But if the pooled sample tested positive, then each individual sample must be tested again.
Undaunted... we try expanding the terms in powers of p. We know that ln(1-p) = -p + ... and If we substitute this much into the equation, we get: The solution of this is (which is at least consistent with the limit!). What we have here is an example of an asymptotic series, or, a series of powers other than whole numbers. We can calculate more coefficients by expanding the equation in higher powers of p.
