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

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???

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

#1 - e. Note stirling's approximation to n! is 閳?2锜簄)(n/e)^n. Taking the nth root of this, we get (2锜?^(1/(2n)) * 閳?n^(1/n) * n/e. Claerly, as n閳巻鍨? (2锜?^(1/(2n))閳?. Note that n^(1/n)=e^(ln n/n). The limit of ln n/n as n閳巻鍨?is 0, and e^0 is 1, and 閳? is 1, so as n閳巻鍨? 閳?n^(1/n) also approaches 1. Thus as n閳巻鍨? n!^(1/n) 閳姧/e, which mean that n/(n!^(1/n)) 閳?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+2i锜簁, where k is any integer. Thus, you might have x=0 (trivial, since anything to the 0=1), but also x=2i锜? x=4i锜? x=-3764i锜? etc.

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

yes

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

1-answer is 1

2-answer is 1^1,y^0

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

2. Yes. solutions are :

s = lim A / (B)^n ; A, |B| %26gt; 1, from R

n-%26gt;00

for instance s = (+/-) 0.0000000000000000000000..........

or 1 - 0.99999999999999 ....

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

something i dont understand

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

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.