In this project. we use the Macburin polynomials for e x to prove that e is irrational. The proof relies on supposing that e is rational and arriving a a contradiction. Therefore, in the following steps, we suppose e = r/s for some integers r and s where s ≠ 0. 6. Use Taylor’s theorem to find an estimate on R n (1). Use this estimate combined with the result from part 5 to show that | s n ! R n | < s e n + 1 . Conclude that if n is large enough then | s n ! R n | < 1 . Therefore. sn ! R n (l) is an integer with magnitude less than 1. Thus, . sn ! R n (1) = 0. But from part 5, we know that sn ! R n (l) ? 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 Macburin polynomials for e x to prove that e is irrational. The proof relies on supposing that e is rational and arriving a a contradiction. Therefore, in the following steps, we suppose e = r/s for some integers r and s where s ≠ 0. 6. Use Taylor’s theorem to find an estimate on R n (1). Use this estimate combined with the result from part 5 to show that | s n ! R n | < s e n + 1 . Conclude that if n is large enough then | s n ! R n | < 1 . Therefore. sn ! R n (l) is an integer with magnitude less than 1. Thus, . sn ! R n (1) = 0. But from part 5, we know that sn ! R n (l) ? 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 Macburin polynomials for exto prove that e is irrational. The proof relies on supposing that e is rational and arriving a a contradiction. Therefore, in the following steps, we suppose e = r/s for some integers r and s where s ≠ 0.
6. Use Taylor’s theorem to find an estimate on Rn(1). Use this estimate combined with the result from part 5 to show that
|
s
n
!
R
n
|
<
s
e
n
+
1
. Conclude that if n is large enough then
|
s
n
!
R
n
|
<
1
. Therefore. sn!Rn(l) is an integer with magnitude less than 1. Thus, .sn!Rn(1) = 0. But from part 5, we know that sn!Rn(l) ? 0. We have arrived at a contradiction, and consequently, the original supposition that e is rational must be false.
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