1873_solutions

# D n1 it follows from exercise 2 that at least one of

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Unformatted text preview: c c 1 which occurs when c |1 c |  x  c  |1 c |. If c  1 then the latter inequality says that 2c 1  x  1 and if c  1 the inequality says that 1  x  2c 1. 3. Does a power series have to have the same interval of convergence as its derived series? x n1 and of x n . The first of these intervals No. Look at the intervals of convergence of n 1 contains the number 1 but the second does not. 4. Suppose that f and g are given functions, that c is a given number and that r  0. Suppose that both of the functions f and g are the sums of their Taylor series at every number x in the interval c r, c  r . Suppose finally that there exists a number   0 such that f x  g x whenever x belongs to the interval c , c   . c−r c−δ c+r c+δ c Prove that f x  g x for every number x in the interval c r, c  r . The given information tells us that fn c gn c for every nonnegative integer n. 5. True or false: If f and g have derivatives or all orders in a neighborhood of a number c and if f n c  g for every nonnegative integer n then we have f x  g x for every number x sufficiently close to c. The statement if false. Define f to be the function introduced in this example and g to be the constant function 0. 6. n c Use Scientific Notebook to calculate a variety of nth partial sums of the Maclaurin series of the function f defined by the equation x x f x  e sin6 1x for every real number x. Then use Scientific Notebook to plot some of these nth partial sums with the graph of f on the interval 2, 2 and explore the accuracy of these partial sums as approximation so f on the interval. Some Exercises on the Series Expansion of exp 1. Prove that if c and x are any real numbers then Ý exp x  n0 ec x n! c n. Solution: exp x  exp c exp x Ý  ec n0 c Ý 1x n! c n  n0 ec x n! 2. Prove that Þ 0 e x dx  1 Ý 2 From the fact that 374 n0 1n 2n  1 n! c n. Ý e x2 1 n x 2n n!  n0 1 n x 2n n! and from the fact that the series 0, 1 , we see that Þ0 1 e x 2 dx  converges bounded (in fact, uniformly) in x on the inteval Þ0 1 Ý  n0 Ý 1 n x 2n dx n! n0 1 n x 2n dx  n! Þ0 1 Ý 1n 2n  1 n! n0 3. Show that even before we have showed that the function f defined in the proof of the above theorem is the exponential function exp we could have seen from the binomial theorem and Cauchy’s theorem on products of series (click here to see it) that for all numbers x and y we have f x f y  f xy . Solution: Suppose that x and y are real numbers. Ý Ý xm m! fxfy  m0 n0 yn n! From Cauchy’s theorem on products of series we know that the latter product is Ý n n0 j0 j xn j y n j ! j! Ý n  n0 Ý 1 n! j0 n n! xn jyj j !j! n  n0 1 n! j0 n xn jyj j and from the binomial theorem we know that Ý n0 4. Ý n 1 n! j0 n xn jyj j  n0 1 xy n! n  f xy . a. Prove that Ý e n0 1. n! b. Prove that if m is any positive integer then the number m m! j0 1 j! is an integer and prove that Ý 0  m! jm1 1  1. j! c. Prove that if m is any positive integer...
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## This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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