Picture related allows to approximate the square root of any

>>8302521
What the fuck? Rn+1 will always be bigger than Rn are you retarded?

Like, what makes you think that making a number bigger will make its square root smaller?

File: example.png (7KB, 396x452px) Image search: [Google]

7KB, 396x452px

>>8302530
huh? x=144

>>8302521
Let 0<x<1

easy enough to show, assuming
[math]R_i[/math] converges (which it does), just sub [math]R_ \infty[/math] into your relation,
and solve for [math]x[/math]

File: 22222222222.png (3KB, 241x176px)

3KB, 241x176px

>>8302545
x=0.8

>>8302553
I don't know that much math

>>8302564
it's elementary Algebra:
[math]R_ \infty= \frac{1}{2} \left (R_ \infty+\frac{x}{R_ \infty} \right)[/math]
multiply by [math]2R{ \infty}[/math]:
[math]2R_{ \infty}^2=R_{ \infty}^2+x[/math]
subtract [math]R_{ \infty}^2[/math] from each side, and Bob's yer uncle

>>8302598
[math] R_ \infty= \frac{1}{2} \left (R_ \infty-1+\frac{x}{R_ \infty-1} \right) [/math] tho

>>8302598
And what does that prove?

>>8302621
wat

>>8302598
>infinity minus 1
keke, implying infinity is even a logically coherent mathematical entity

>>8302624
It proves that if the sequence converges, it must converge to sqrt(x), as this is the only fixed point.

It remains to be shown that the sequence converges.

>>8302521
If R_i > sqrt(x) then R_{i+1} < R_i.

because if R_i > sqrt(x) then 1 > x/(R_i)^2.

and 1 > 1/2(1 + x/(R_i)^2)

and so R_i > 1/2(R_i + x/R_i)

and so R_i > R_{i+1}.

Note for all x and y,

(y - sqrt(x))^2 > 0

and so y^2 - 2ysqrt(x) + x > 0

y^2 + x > 2ysqrt(x)

(y^2 + x)/2y > sqrt(x)

1/2(y + x/y) > sqrt(x)

R_i > sqrt(x), for all i != 1

so you are right OP. R_i < r_{i+1} only when i = 1.

>>8302708
>infinity
[math]R_{ \infty}[/math] is a number, you imbecile.

>>8302784
[math]If R_i > sqrt(x) then R_{i+1} < R_i. \\

because if R_i > sqrt(x) then 1 > x/(R_i)^2. \\

and 1 > 1/2(1 + x/(R_i)^2) \\

and so R_i > 1/2(R_i + x/R_i) \\

and so R_i > R_{i+1}. \\

Note for all x and y, \\

(y - sqrt(x))^2 > 0 \\

and so y^2 - 2ysqrt(x) + x > 0 \\

y^2 + x > 2ysqrt(x) \\

(y^2 + x)/2y > sqrt(x) \\

1/2(y + x/y) > sqrt(x) \\

R_i > sqrt(x), for all i != 1 \\

so you are right OP. R_i < r_{i+1} only when i = 1.[/math]