What dose ${x_{n}}$ ${x_n}$ converges ? when $x_{1} = \frac{5}{2}, 5 x_{n + 1} = x_{n}^{2} + 6$ $x_1=5/2,5x_(n+1)=x_n^2+6$

Question

What dose ${x_{n}}$ ${x_n}$ converges ? when $x_{1} = \frac{5}{2}, 5 x_{n + 1} = x_{n}^{2} + 6$ $x_1=5/2,5x_(n+1)=x_n^2+6$

1 Answer

Impact of this question

1263 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-07T10:00:57

See below.

Explanation:

This is a nonlinear difference equation. Normally it is hard to analyze convergence in this case. So we will make some basic convergence considerations.

If $x_{n}$ $x_n$ converges to $x^{\circ}$ $x^@$ then

$5 x^{\circ} = {(x^{\circ})}^{2} + 6$ $5x^@=(x^@)^2+6$ and solving for $x^{\circ}$ $x^@$ we have

$x^{\circ} = {2, 3}$ $x^@ = {2,3}$

Checking for $x = 2 + δ$ $x = 2+delta$ we conclude that $x^{\circ} = 2$ $x^@=2$ is an stable attraction point for $- \infty < δ < 1$ $-oo < delta < 1$ and $x^{\circ} = 3$ $x^@ = 3$ is an inestable attraction point. Resuming, for $- \infty < x_{1} < 3$ $-oo < x_1 < 3$ the sequence converges to $2$ $2$ otherwise is diverges.

Science

Math

Humanities

... and beyond

What dose ${x_{n}}$ ${x_n}$ converges ? when $x_{1} = \frac{5}{2}, 5 x_{n + 1} = x_{n}^{2} + 6$ $x_1=5/2,5x_(n+1)=x_n^2+6$

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

What dose {x_n}{xn}{x_n} converges ? when x_1=5/2,5x_(n+1)=x_n^2+6x1=52,5xn+1=x2n+6x_1=5/2,5x_(n+1)=x_n^2+6

1 Answer

Explanation:

Related questions

Impact of this question

What dose ${x_{n}}$ ${x_n}$ converges ? when $x_{1} = \frac{5}{2}, 5 x_{n + 1} = x_{n}^{2} + 6$ $x_1=5/2,5x_(n+1)=x_n^2+6$