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. (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 n th 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.
I already know the mathematical answers but I would appreciate if it could be explained in words for each question. Please try to explain as simply as possible. Thank you very much!
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.
(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 n th 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. 1Taken from OpenStax Calc
Trending now
This is a popular solution!
Step by step
Solved in 6 steps with 5 images
