Two Math quizzes........good luck?

Question

1-find lim n/((n!)^(1/n)) at n approachs infinity???
2-the equation exp(x)=1 can be expressed as an infinity degree polynomial equation if we used McLorin seires to substute exp(x)
......does it mean that the equation exp(x)=1 has infinite solutions rather than zero?!!!if yes what are they???

Answer 1

#1 - e. Note stirling's approximation to n! is √(2πn)(n/e)^n. Taking the nth root of this, we get (2π)^(1/(2n)) * √(n^(1/n) * n/e. Claerly, as n→∞, (2π)^(1/(2n))→1. Note that n^(1/n)=e^(ln n/n). The limit of ln n/n as n→∞ is 0, and e^0 is 1, and √1 is 1, so as n→∞, √(n^(1/n) also approaches 1. Thus as n→∞, n!^(1/n) →n/e, which mean that n/(n!^(1/n)) → e.

#2 - Actually, yes, though it's not because of the infinite degree polynomial, but rather because of Euler's identity, e^(ix) = cos x + i sin x. The soutions are of the form x=0+2iπk, where k is any integer. Thus, you might have x=0 (trivial, since anything to the 0=1), but also x=2iπ, x=4iπ, x=-3764iπ, etc.

Answer 2

I wanted to remark on problem 2 a bit. Note that exp(x) can be written as an "infinity degree" power series, yet the equation exp(x)=0 has no solution (even complex ones). It's an interesting exercise (with a computer) to see what happens to the zeroes of the finite MacLaurin polynomials of exp(x), as the degrees go up.

So you'd start with 1,1+x,1+x+x^2/2,1+x+x^2/2+x^3/3!, etc., which have 0, 1, 2, 3, etc. roots.

Answer 3

2. Yes. solutions are :

s = lim A / (B)^n ; A, |B| > 1, from R
n->00

for instance s = (+/-) 0.0000000000000000000000..........
or 1 - 0.99999999999999 ....

Answer 4

1-answer is 1
2-answer is 1^1,y^0

Answer 5

yes

Answer 6

something i dont understand