HELP! question involving the irrationality of e (1 Viewer)

unholybuddha · Jul 11, 2010

i can do parts a),b) and c) but can't do part d)

the upside down ! signs are < signs.. i'm not really good at using latex

unholybuddha · Jul 11, 2010

Affinity · Jul 12, 2010

unholybuddha said:
$For n = 1,2,3,..., let s_{\text{n}} = 1 + \sum_{r=1}^{n} \frac{1}{r!}$
$$(a) Prove by mathematical induction that e - s_{\text{n}} = e\int_{0}^{1}\frac{x^n}{n!}e^{ -x}dx$
$$(b) From (a), deduce that 0 < e - s_{\text{n}} < \frac{3}{(n+1)!}$ $ for n = 1,2,3,.... . [Remember that e < 3 and e^{ -x} \leq 1 for$ $ x\geq 0.$
$$(c) By using the inequality of (b), deduce that (e - s_{\text{n}})n! $ $ is not an integer for n=2,3,4,...$
$(d) Show that there cannot exist positive integers p and q such that e = p/q$

Consider the case when n = q for the result in (c).

unholybuddha · Jul 12, 2010

can u elaborate a bit more please? i can't see where to go..
thanx for ur help anyway

Affinity · Jul 13, 2010

if e = p/q then (e-s_q)*q! = (q-1)!*p - q! * s_n
which is an integer (q! * s_n ----> Check term by term)
Which contradicts (c), so you can't have e = p/q for some p,q

unholybuddha · Jul 13, 2010

ahhh got it now!
thnx for the help

HELP! question involving the irrationality of e (1 Viewer)

unholybuddha

Member

unholybuddha

Member

Affinity

Active Member

unholybuddha

Member

Affinity

Active Member

unholybuddha

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)