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# Rearrangement watch

1. Hey guys could u tell me how to rearrange this. There are 2 rearrangements and ive figured out one of them but cant get the other one cus when i draw the graph its weird lol

so the equation is 0=5/x^(1/2)+x-7

ive got x= 7-5/x^(1/2)

and need help on the other one

and also guys what does -1<g'(x)<1 mean does tht mean if u put in your starting point and its below 0 the process will converge to a root?

thnxs
2. (Original post by chuck111)
so the equation is 0=5/x^(1/2)+x-7
If I've understood you right: try taking the +x and -7 to the other side, take the reciprocal of both sides, multiply both sides by 5, and square.

(Original post by chuck111)
and also guys what does -1<g'(x)<1 mean does tht mean if u put in your starting point and its below 0 the process will converge to a root?
It just means the derivative of g is always between -1 and 1, in general. Are you talking about some specific iterative process?
3. (Original post by generalebriety)
If I've understood you right: try taking the +x and -7 to the other side, take the reciprocal of both sides, multiply both sides by 5, and square.

It just means the derivative of g is always between -1 and 1, in general. Are you talking about some specific iterative process?
yer im tlaking about the iterative poocess f(x)=0 into g(x) and i need to know about the magnitude of g'x which involves knowing about -1<g'x<1
4. (Original post by chuck111)
yer im tlaking about the iterative poocess f(x)=0 into g(x) and i need to know about the magnitude of g'x which involves knowing about -1<g'x<1
There are hundreds of iterative processes. Which one?
5. (Original post by generalebriety)
There are hundreds of iterative processes. Which one?
i dunno our teacher jsut said the rearrangement one thts all she said. I know theres newton raphson but in the one were doing it is like when a starting value is chosen a vertical line is drawn from the x-axis to the curve (y=g (x). This line is then continued to the y=x line and from there it is then drawn back to the curve which is the new starting point
6. (Original post by generalebriety)
There are hundreds of iterative processes. Which one?
From the sounds of it, he means this one. (Standard C3, if you can remember that far back )
7. (Original post by generalebriety)
If I've understood you right: try taking the +x and -7 to the other side, take the reciprocal of both sides, multiply both sides by 5, and square.
I did tht and got −0.5√((7−x)/5) is tht correct?
8. (Original post by generalebriety)
There are hundreds of iterative processes. Which one?
The one where .

Massive OT for everyone else, but if you've done the contraction mapping theorem, the significance of should become clear...
9. (Original post by DFranklin)
The one where .

Massive OT for everyone else, but if you've done the contraction mapping theorem, the significance of should become clear...
hey i was just wondeirng if my rearrangment for 0=5/x^(1/2)+x-7

is correct: x=−0.5√((7−x)/5)
10. hey i was just wondeirng if my rearrangment for 0=5/x^(1/2)+x-7

is correct: x=−0.5√((7−x)/5)

plz i need to know really quickly cus i think i might have dun my c/wk wrong and have to hand it in today (((
11. Is it supposed to be for an iterative procedure?
12. (Original post by stevencarrwork)
Is it supposed to be for an iterative procedure?
yes
is it correct??
13. (Original post by stevencarrwork)
Is it supposed to be for an iterative procedure?
it looks like the rearrangement method
14. Where did the 0.5 come from?

Is the original equation 0 = (5/SQRT(X)) + X - 7

In which case, surely you just have X = 7 - (5/(SQRT(X))
15. (Original post by stevencarrwork)
Where did the 0.5 come from?

Is the original equation 0 = (5/SQRT(X)) + X - 7

In which case, surely you just have X = 7 - (5/(SQRT(X))
yes its 0 = (5/SQRT(X)) + X - 7

and ive already got tht rearrangment 7 - (5/(SQRT(X)) and i need to know if the one i asked for is correct.
16. (Original post by stevencarrwork)
In which case, surely you just have X = 7 - (5/(SQRT(X))
He's already found that, he's looking for a second rearrangement (which you wouldn't have known unless you'd read the near identical thread he posted yesterday).

I've now merged the two threads.

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