Mathematics 116 HWK 25a Solutions 8.6 p610

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1 Mathematics 6 HWK 5a Solutions 8.6 p6 Problem, 8.6, p6 Fin a power series representation for the function f() = etermine the interval of convergence. an Solution. Begin with the geometric series = n +... = vali for <. Multiply by, istributing the, to fin f() = = n = also vali for <. So the interval of convergence is (,). n n+ = n. n= Problem 7, 8.6, p6 Fin a power series representation for the function f() = + an etermine the interval of convergence. Solution. First o a little rewriting, by iviing both numerator an enominator by : f() = + = + = + Now use the series representation + = ( ) n n +... = ( ) n n (vali for < ) with replace by. (So the new result will be vali for <.) This gives us f() = + = [ + ( ) ( ) ( ) n ( )n +...] = ( ) n ( )n (vali for < ) which we can rewrite as f() = n ( )n n = ( ) n n n+ (vali for < ). Since the conition < is equivalent to the conition <, the interval of convergence is (, ). Page of 5 A. Sontag December,

2 Math 6 HWK 8.6 p6 Solns continue Problem, 8.6, p6 (a). Use ifferentiation to fin a power series representation for f() = (+). What is the raius of convergence? (b). Use part (a) to fin a power series for f() = (+) 3. (c). Use part (b) to fin a power series for f() = (+). 3 Solution. (a). The key is to realize that ( ) + = ( + ) which gives, in succession, In sigma notation, this woul be ( + ) = ( ) + ( + ) = ( ( ) n n +...) ( + ) = ( n( ) n n +...) ( + ) = n( ) n+ n +... ( + ) = + = ( ) n n = ( ) n n n = ( ) n (n + ) n The geometric series use to represent + has raius of convergence R =. We know that ifferentiation oes not change the raius of convergence for a power series (though it can change the interval). Therefore the ifferentiate series use to represent (+) has raius of convergence R =. Solution. (b). The iea here is the same as for part (a). We know ( + ) = ( + ) 3 so, using, part (a), we have ( + ) 3 = ( + ) Page of 5 A. Sontag December,

3 Math 6 HWK 8.6 p6 Solns continue ( + ) 3 = [ ( ) n+ n n +...] ( + ) 3 = [ ( ) n+ n(n ) n +...] In sigma notation, this woul be ( + ) = ( + ) 3 = n(n ) ( ) n+ n +... n(n ) ( ) n n +... ( + ) 3 = ( ) n (n + ) n = [( )n (n + ) n ] ( + ) 3 = n+ n(n + ) ( ) n or ( + ) 3 = n (n + )(n + ) ( ) n By the same reasoning as in part (a), the raius of convergence is R =. Solution. (c). We nee only multiply the series from (b) by, istributing the. This gives ( + ) 3 = [ ( + ) 3 = ( + ) 3 = n (n + )(n + ) ( ) n = n(n ) ( ) n n +...] n(n ) ( ) n n +...] n (n + )(n + ) ( ) n+ or ( + ) 3 = n= n (n )n ( ) n Once again, the raius of convergence is R =. The reasoning has to o with Theorem 8(i), which tells us that a constant multiple of a convergent series is convergent. [To be more specific, we can reason as follows. For each in the interval of convergence for the series from part (b), we can Page 3 of 5 A. Sontag December,

4 Math 6 HWK 8.6 p6 Solns continue use Theorem 8(i) to conclue that the series in (c) converges for that particular. So we on t lose convergence. But we can t gain convergence either. For suppose is some nonzero number outsie the original interval of convergence an suppose that multiplying by the series in part (b) were to yiel a convergent series. Then multiplying by the series in part (c) shoul also yiel a convergent series, which is a contraiction.] Problem, 8.6, p6 Evaluate the inefinite integral as a power series. + Solution. + = [ + ( ) + ( ) ( ) n +...] + = [ ( ) n n +...] 5 = C n ( )n 3 n = ( ) ( ) n = ( ) n n = C + ( ) n n+ n + Note that these representations are vali for <. arctan Problem 3, 8.6, p6 Evaluate the inefinite integral as a power series. Solution. Since arctan = ( )n n n + we have arctan = ( )n n n arctan = [ ( )n n +...] n + arctan = C ( )n n+ (n + ) +... Page of 5 A. Sontag December,

5 Math 6 HWK 8.6 p6 Solns continue arctan = arctan = arctan ( ) n n n + = = ( ) n n+ n + ( ) n n n + ( ) n n n + = C + ( ) n n+ (n + ) Note that the raius of convergence is R =, the same as for the Maclaurin series for arctan. Problem 5, 8.6, p6 Use a power series to approimate the efinite integral with error < Solution. Using the same technique as for problem, we can obtain the following representation for the inefinite integral: + 5 = 6 = C n ( )n 6 5n Therefore. ]. [ + = n ( )n 6 5n (.)6 =. + (.) (.) ( ) n (.)5n+ 5n Since this alternating series meets the criteria for the Alternating Series Error Estimate (ignoring the alternating sign, the terms ecrease an have limit zero), we nee only use enough terms so that the first term we on t use has absolute value at most.5. A bit of ecimal arithmetic or a few keystrokes on the calculator tell us that we nee to use two terms but two will be enough. With the esire accuracy, we have. (.) = Page 5 of 5 A. Sontag December,