Iterative formula

V = (2x^2 + x^3)^{0.5}

An attempt is made to find the value of x when V = 10, show that a possible iterative formula to find x is:

x_{n+1} = \frac{10}{2+x_n}

If I did:

[br]10 = \sqrt {x^2 (2+x)}[br]10 = \sqrt {x^2} \times \sqrt {2+x}[br]\dfrac {10}{ \sqrt{2+x}} = x[br][br]

I guess I'm on the right lines but how do I replace the x with xn and xn+1?

Thanks!

V = (2x^2 + x^3)^{0.5}

An attempt is made to find the value of x when V = 10, show that a possible iterative formula to find x is:

x_{n+1} = \frac{10}{2+x_n}

If I did:

[br]10 = \sqrt {x^2 (2+x)}[br]10 = \sqrt {x^2} \times \sqrt {2+x}[br]\dfrac {10}{ \sqrt{2+x}} = x[br][br]

I guess I'm on the right lines but how do I replace the x with xn and xn+1?

Thanks!

So your new value of x is given by

\frac{10}{\sqrt(2+x)}

so you have

x_{n+1}=\frac{10}{\sqrt(2+x_n)}

Original post by brianeverit

So your new value of x is given by

\frac{10}{\sqrt(2+x)}

so you have

x_{n+1}=\frac{10}{\sqrt(2+x_n)}

Aah okay so I just needed to show that xn+1 is the next value and xn the previous by placing them in the appropriate place?

Aah okay so I just needed to show that xn+1 is the next value and xn the previous by placing them in the appropriate place?

Yes

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