Show that
\lim_{n \to \infty} \frac{n!}{a^{2n+1}} = \infty \qquad a \in \mathbb{N}
Here is my try:
Take a_n = a^{2n+1}/n! . Since
\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{a^{2(n+1)+1}/\,(n+1)!}{a^{2n+1}/\,n!} = \lim_{n \to \infty} \frac{a^{2}}{n+1} =...
Define a sequence
A_n(r) = \int_{-1}^1(1-x^2)^n \cos(rx)\, dx, \qquad n \in \mathbb{N}, r \in \mathbb{R}.
Prove that
A_n(r) = \frac{n!}{r^{2n+1}}[P_n(r)\sin r - Q_n(r)\cos r]
where P_n and Q_n are two polynomials with integer coefficients. What is the degree of P_n and of Q_n...
HallsofIvy, thanks.
2x_1=0 and 0=0 => x_1=0, x_2 could be any number
0=0 and -2x_2=0 => x_2=0, x_1 could be any number
I just try another solution.
Rewrite the equations:
(x_1cos\alpha + x_2sin\alpha) - x_1 = 0
(x_1sin\alpha - x_2cos\alpha) - x_2 = 0
Suppose x_1 = cos\frac{\alpha}{2}...
Let me try...
Solve equation 1:
x_2 = \frac{-x_1(cos\alpha - 1)}{sin\alpha}
Substitute it to the second:
x_1sin\alpha + \frac{-x_1(cos\alpha - 1)}{sin\alpha}(-cos\alpha-1) = 0
x_1sin\alpha + \frac{x_1(cos^2\alpha - 1)}{sin\alpha} = 0
2x_1sin\alpha = 0
What is the solutions for...