Math 112 Lab 8: More with Series

Transcription

1 Manipulating Series Math 112 Lab 8: More with Series Various operations can be done to obtain a new series from an old (well-known) series: the basic operations of arithmetic, as well as substitution, differentiation, and integration. We will look at some examples of these operations in this section. The operations will be done with Taylor polynomials, the structures we studied in Lab 7. In the homework we will write the Taylor expansion of a product of two functions: cosÿsinÿx ' lnÿ1 x Example 1. Starting with some familiar series (such as sinÿx, cosÿx, e x,1/ÿ1 x, etc.) obtain the tenth degree Taylor polynomial for lnÿ1 x 2 cosÿcosÿx in powers of x (i.e., about x 0). Solution: We will begin with determining the series for lnÿ1 x 2. This series can be obtained by starting with the series for 1/Ÿ1 t. In case we can t recall the series expansion for this function, we can ask Maple to help us. f1 : taylor(1 / (1 t), t 0, 6); Maple determines the fifth degree Taylor polynomial as f1 : 1 " t t 2 " t 3 t 4 " t 5 OŸt 6. In order to determine the series for lnÿ1 t, we just need to find the antiderivative of the above series f1. We do this in Maple with the int command. f2 : int(f1, t); The sixth degree Taylor polynomial for lnÿ1 t is f2 : t " 1 2 t2 1 3 t3 " 1 4 t4 1 5 t5 " 1 6 t6 OŸt 7. (Note that Maple did not add the constant C to the antiderivative, which turns out to be ok since if we calculate C, it has the value 0.) Recall from Lab 6 we must now convert f2 to a polynomial before substituting t x 2. This is done in Maple with the statements: f2 : convert(f2, polynom); f3 : subs(t x ^2, f2); The resulting Taylor polynomial for lnÿ1 x 2 is x 2 " 1 2 x4 1 3 x6 " 1 4 x8 1 5 x10 " 1 6 x12 OŸx 14. Now we need to determine the series for cosÿcosÿx. We know the series for cosÿx. We can have Maple determine it for us up to the tenth term. g1 : taylor(cos(x), x 0, 11); The tenth degree Taylor polynomial for cosÿx is: 1

2 g1 : 1 " 1 2 x x4 " x x8 " x10 OŸx 11. To get cosÿcosÿx, we want to substitute g1 into a series for cosÿt. But it shouldn t be the series in powers of t because cosÿt is not near 0 when t is near 0, and OŸx 11 would not be small. Instead, we need to use the series for cosÿt in powers of t " 1, because cosÿ0 1. We will let g2 be the series expansion for cosÿt about t 1. g2 : taylor(cos(t), t 1, 11); We obtain g2as cosÿ1 " sinÿ1 Ÿt " 1 " 1 2 cosÿ1 Ÿt " sinÿ1 Ÿt " cosÿ1 Ÿt " 1 4 " sinÿ1 Ÿt " 1 5 O Ÿt " After converting g1 and g2 into polynomials, we can substitute t g1 into g2. g1 : convert(g1, polynom); g2 : convert(g2, polynom); g3 : subs(t g1, g2); The resulting series is a little overwhelming and contains many more terms than we are interested in. It is cosÿ1 " sinÿ1 A " 1 2 cosÿ1 A2 1 6 sinÿ1 A cosÿ1 A4 " sinÿ1 A5, where A " 1 2 x x4 " x x8 " x10. Since we are only interested in the first ten terms of this polynomial, we can use the taylor command to extract only the terms we want. g4 : taylor(g3, x 0, 11); The much less frightening Taylor polynomial for g4is g4 : ŸcosŸ1 1 2 sinÿ1 x2 " 24 1 sinÿ1 " 1 8 cosÿ1 x4 " sinÿ1 1/48 cosÿ1 x sinÿ cosÿ1 x8 " sinÿ1 " cosÿ1 x10 OŸx 12. Finally, we convert g4 to a polynomial and multiply it by f3, the twelfth degree Taylor polynomial for lnÿ1 x 2. Last, we extract the tenth degree polynomial for lnÿ1 x 2 cosÿcosÿx. The final Maple statements g4 : convert(g4, polynom); h1 : f3 * g4; finalpoly : taylor(h1, x 0, 11); convert(finalpoly, polynom); 2

3 produce the result cosÿ1 x sinÿ1 " 1 2 cosÿ1 x4 " 7 24 sinÿ cosÿ1 x sinÿ1 " 1/6 cosÿ1 x Now you ll be glad to know that the statement taylor(ln(1 x ^2) * cos(cos(x)), x 0, 11); cosÿ1 " sinÿ1 x10. produces the same result! But say you are stranded in the Australian Outback and don t have a handy copy of Maple available the good news is that you could derive this crazy polynomial on your own (given that you know the series expansions for your basic functions and rules of integration). Using Series to Solve Differential Equations Many differential equations (DE s) can t be solved explicitly in terms of simple familiar functions. So, one application of series approximation formulas, like the Maclaurin series, is in the solution of DE s. Example 2. Consider the differential equation y U x y with initial condition yÿ0 2. This DE cannot be solved by separating variables. So instead find a third degree polynomial that approximates the solution y X fÿx for x X 0. Solution: Let y fÿx be the solution to the DE: y U x y with initial condition yÿ0 2. Our goal will be to obtain an approximation formula of the form fÿx X a 0 a 1 x a 2 x 2 a 3 x 3. Thus we need to determine the numerical values of a 0, a 1, a 2, and a 3. From Lab 7, given fÿx X a 0 a 1 x a 2 x 2 a 3 x 3 is the Maclaurin series for y near x 0, recall that: a 0 fÿ0, a 1 f U Ÿ0, a 2 f UU Ÿ0 /2, a 3 f UUU Ÿ0 /6. 1) Find a 0. The initial condition yÿ0 2 means that Thus, at this point we know that a 0 fÿ0 2. fÿx X 2 a 1 x a 2 x 2 a 3 x 3. 2) Find a 1. Since fÿx must satisfy the DE y U x y, then f U Ÿx x fÿx. Thus, f U Ÿ0 0 fÿ Hence, a 2 2 and fÿx X 2 2x a 2 x 2 a 3 x 3. 3

4 3) Find a 2. We need a formula for computing f UU Ÿx since a 2 f UU Ÿ0 /2. We can obtain a formula for f UU Ÿx from item 2) above where we know f U Ÿx x fÿx. Differentiating both sides of this equation yields: f UU Ÿx 1 f U Ÿx. and so f UU Ÿ0 1 f U Ÿ Therefore we have a 2 3/2 and fÿx X 2 2x 3 2 x2 a 3 x 3. 4) Find a 3. We need to find f UUU Ÿx in order to determine a 3, since a 3 f UUU Ÿ0 /6. We can obtain such a formula from the fact that f UU Ÿx 1 f U Ÿx. Differentiating this equation yields f UUU Ÿx f UU Ÿx, and so f UUU Ÿ0 f UU Ÿ0 3 and a 3 3/6 1/2. Hence, fÿx X 2 2x 3 2 x2 1 2 x3. It is important to note that the above approximation is only valid for x X 0. In order to solve this DE in Maple, we just need to follow the process outlined above and proceed with the appropriate Maple commands. First define f as the third degree approximation polynomial. Since we know fÿ0 2, we can immediately substitute in 2 for a 0. f : a_0 a_1 * x a_2 * x ^2 a_3 * x ^3; f : subs(a_0 2, f); Since f satisfies the DE y U x y, then f satisfies the DE diff(f, x) x f. Call this DE Eq1, and substitute x 0 into Eq1 to determine the coefficient a 1. Eq1 : diff(f, x) x f; subs(x 0, Eq1); f : subs(a_1 2, f); Taking the derivative of diff(f, x) x f gives us an expression for calculating the second derivative f UU Ÿx and determining f UU Ÿ0. The first two statements determine the value for a 2, and the third statement substitutes the value of a 2 into f. Eq2 : diff(f, x$2) 1 diff(f, x); subs(x 0, Eq2); f : subs(a_2 3 / 2, f); Last, taking the derivative of diff(f, x$2) 1 diff(f, x) gives us an expression for calculating the third derivative f UUU Ÿx and determining f UUU Ÿ0. The final expression for f is then revealed. Eq3 : diff(f, x$3) diff(f, x$2); subs(x 0, Eq3); f : subs(a_3 1 / 2, f); Now that we have written out the step-by-step process to determine the approximating polynomial f, 4

5 we see that Maple actually will determine the exact solution for this differential equation. Use the dsolve command as demonstrated in Lab 6. DE1 : diff(y(x), x) x y(x); dsolve({de1, y(0) 2}, y(x)); y : -x * exp(x); We can determine the third degree Taylor polynomial for y to check that we obtained the correct approximation function fÿx. Also, we can plot the graphs of both the actual solution y and the approximation polynomial fÿx near x 0. The statements that do this are: taylor(y, x 0, 4); actualplot : plot(y, x , color red): cubicplot : plot(f, x , color blue): plots[display]({actualplot, cubicplot}); Example 3. Consider the DE y U xy 1 with initial condition yÿ0 1. Find an approximation formula of the form fÿx a 0 a 1 x a 2 x 2 a 3 x 3 for x X 0. [Hint: Finding f UU Ÿx and f UUU Ÿx will involve the product rule!] Solution: Define f to be a third degree polynomial again and follow the same process in Maple as discussed in Example 1. It is important to remember to use the product rule when determining the equations for f UU Ÿx and f UUU Ÿx. restart; f : a_0 a_1 * x a_2 * x ^2 a_3 * x ^3; f : subs(a_0 1, f); Eq1 : diff(f, x) x * f 1; subs(x 0, Eq1); f : subs(a_1 1, f); Eq2 : diff(f, x$2) x * diff(f, x) 1 * f; subs(x 0, Eq2); f : subs(a_2 1 / 2, f); Eq3 : diff(f, x$3) 1 * diff(f, x) x * diff(f, x$2) diff(f, x); subs(x 0, Eq3); f : subs(a_3 1 / 3, f); 5

6 These statements produce the approximation formula for f as: for x X 0. fÿx 1 x 1 2 x2 1 3 x3 We can also use Maple s dsolve command to solve the DE y U xy 1 with initial condition yÿ0 1: DE2 : diff(y(x), x) x * y(x) 1; dsolve({de2, y(0) 1}, y(x)); y : rhs(%); Maple returns a result for y with an unfamiliar erf expression in it: y : 1 2 = 2 erfÿ x 1 e 1/2x2. In Maple, erf(x) is defined as: erfÿx 2 ; x e "t 2 dt. = 0 Although erf(x) is an intimidating expression, we can still plot and evaluate the function y in Maple. To plot the actual solution and approximation polynomial, use the statements: actualplot : plot(y, x , color red): cubicplot : plot(f, x , color blue): plots[display]({actualplot, cubicplot}); The plots of y and its cubic approximation are displayed in Figure 1 below. The third degree Taylor polynomial for y about x 0 can also be obtained with the statement: taylor(y, x 0, 4); Figure 1: The graphs of the actual and approximate solutions to the DE y U xy 1 with initial condition yÿ0 1. 6

7 Figure1: The graphs of the actual and approximate solutions to the DE y U xy 1 with initial condition yÿ0 1. References Israel, Robert B. (2000), Calculus the Maple Way, Addison Wesley. 7

8 Lab 8 Homework Assignment: Due at the BEGINNING of the next lab class. Please come to class with your lab ready to hand in. Directions: Work the following problems using Maple, unless indicated otherwise. Print out your entire Maple worksheet and hand it in. What you hand in must show the exact Maple commands that you used to solve a given problem. 1. Use the well-known Maclaurin series for fÿx e x to obtain the seventh degree Maclaurin series for gÿx e 3x. [Hint: You will need to use the Maple subs command.] 2. Use a Taylor series to evaluate the following limit: lim xv0 x " arctanÿx x 3 [Hint: You can check your result with Maple s limit command. Go to the Maple help menu if you have forgotten how to use the limit command.] 3. Find the first three nonzero terms in the Maclaurin series for the function y e x lnÿ1 " x. Use the series expansions for e x and lnÿ1 " x. [Hint: Don t forget to use convert to rewrite your polynomials before performing any operations on them.] 4. Starting with well-known series, obtain the 9th degree Maclaurin polynomial for fÿx cosÿsinÿx lnÿ1 x about x Consider the DE y U 2y with initial condition yÿ0 1. (a) Find an approximation formula of the form for x X 0. fÿx a 0 a 1 x a 2 x 2 a 3 x 3 (b) Check your approximation from part (a) by using dsolve and then finding the third degree Taylor polynomial for that solution. Your answers for parts (a) and (b) should be the same! 6. Consider the DE with initial conditions yÿ0 1 and y U Ÿ0 0. (a) Find an approximation formula of the form y UU " xy U " y 0 fÿx a 0 a 1 x a 2 x 2 a 3 x 3 8

9 for x X 0. [Hint: Finding f UUU Ÿx will involve the product rule!) (b) Check your approximation from part (a) by using dsolve and then finding the third degree Taylor polynomial for that solution. Because of the two initial conditions, the dsolve command to solve this equation is: dsolve({diff(y(x), x$2) - x * diff(y(x), x) - y(x) 0, y(0) 1, D(y)(0) 0}, y(x)); 9