In this project, we use the Maclaurin polynomials for f(x) = e x to proved that e is irrational. The proof relies on supposing that e is rational and arriving at a contradiction. Therefore, in the following steps, we suppose e = r/s for some integers r and s where s 6= 0. The work needs to be typed up neatly and submitted via Canvas as either a Word document or as a PDF. If you want to include pictures of your work, make sure it is neat and legible. I would prefer you use the equation editor to try to make the work as neat as possible (1) Write the Maclaurin polynomials p0(x), p1(x), p2(x), p3(x), and p4(x) for e x . Evaluate p0(1), p1(1), p2(1), p3(1), and p4(1) to estimate e. (2) Let Rn(x) denote the remainder when using pn(x) to estimate e x . Therefore, Rn(x) = e x − pn(x) and, in particular, Rn(1) = e − pn(1). Assuming that e = r s for integers r and s 6= 0, evaluate R0(1), R1(1), R2(1), R3(1), and R4(1). (3) Using the results from (2), show that for each remainder R0(1), R1(1), R2(1), R3(1), and R4(1), we can find an integer k such that kRn(1) is an integer for n = 0, 1, 2, 3, 4. (4) Write down the formula for the nth Maclaurin polynomial pn(x) for e x and the corresponding remainder Rn(x). Show that sn!Rn(1) is an integer. (5) Use Taylor’s theorem to write down an explicit formula for Rn(1). Conclude that Rn(1) 6= 0 and, therefore, sn!Rn(1) 6= 0. (6) Use Taylor’s theorem to find an estimate on Rn(1). Use this estimate combined with the result from (5) to show that |sn!Rn(1)| < se n+1 . Conclude that if n is large enough, then |sn!Rn(1)| < 1. Therefore, sn!Rn(1) is an integer with a magnitude less than 1. Thus, sn!Rn(1) = 0. But, from (5), we know that sn!Rn(1) 6= 0. We have arrived at a contradiction and, consequently, the original supposition that e is rational must be false.
In this project, we use the Maclaurin polynomials for f(x) = e x to proved that e is irrational. The proof relies on supposing that e is rational and arriving at a contradiction. Therefore, in the following steps, we suppose e = r/s for some integers r and s where s 6= 0. The work needs to be typed up neatly and submitted via Canvas as either a Word document or as a
(1) Write the Maclaurin polynomials p0(x), p1(x), p2(x), p3(x), and p4(x) for e x . Evaluate p0(1), p1(1), p2(1), p3(1), and p4(1) to estimate e.
(2) Let Rn(x) denote the remainder when using pn(x) to estimate e x . Therefore, Rn(x) = e x − pn(x) and, in particular, Rn(1) = e − pn(1). Assuming that e = r s for integers r and s 6= 0, evaluate R0(1), R1(1), R2(1), R3(1), and R4(1).
(3) Using the results from (2), show that for each remainder R0(1), R1(1), R2(1), R3(1), and R4(1), we can find an integer k such that kRn(1) is an integer for n = 0, 1, 2, 3, 4.
(4) Write down the formula for the nth Maclaurin polynomial pn(x) for e x and the corresponding remainder Rn(x). Show that sn!Rn(1) is an integer.
(5) Use Taylor’s theorem to write down an explicit formula for Rn(1). Conclude that Rn(1) 6= 0 and, therefore, sn!Rn(1) 6= 0.
(6) Use Taylor’s theorem to find an estimate on Rn(1). Use this estimate combined with the result from (5) to show that |sn!Rn(1)| < se n+1 . Conclude that if n is large enough, then |sn!Rn(1)| < 1. Therefore, sn!Rn(1) is an integer with a magnitude less than 1. Thus, sn!Rn(1) = 0. But, from (5), we know that sn!Rn(1) 6= 0. We have arrived at a contradiction and, consequently, the original supposition that e is rational must be false.
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