C3 Iterative formula help

Hi,
Was wondering if anyone could help me, this is part of my revision.
'Use the iterative formula Xn+1 = ^5root(53-2xn) with a suitable starting value, to find the real root of the equation x^5 + 2x - 53 = 0
Show the result of each iteration, and give the root correct to 3 decimal places.'
(OCR June 07 3b)

In questions I've done, I think I've always been given the starting value. Is there a method to finding this?

Reply 1

ttoby

15

The mark scheme doesn't say anything about how to choose your starting value. So all I can suggest is pick a number that looks closeish to roughly where the root is and run the iteration on that. If your iteration diverges then just cross it all out and pick another number.

Reply 2

Jing_jing

14

I'd use the sign change rule to find a suitable starting value and use that

Reply 3

ghostwalker

17

I think it's clear that values of |x|<1 will not be a solution.

Since x^5 will be the dominant term involving x for values of |x| > 1, I'd look to a very rough solution to

x^5 = 53 as a starting point.

Original post by kej817

Hi,
Was wondering if anyone could help me, this is part of my revision.
'Use the iterative formula Xn+1 = ^5root(53-2xn) with a suitable starting value, to find the real root of the equation x^5 + 2x - 53 = 0
Show the result of each iteration, and give the root correct to 3 decimal places.'
(OCR June 07 3b)

In questions I've done, I think I've always been given the starting value. Is there a method to finding this?

Use the table of values button on your calculator and find the value of x^5 + 2x - 52 from 0 to 5 in steps of 1 (presumably you can see it must be positive). Look for a sign change. As this function is continuous there must be a root as it goes from negative to positive.

Reply 5

kej817

OP

5

Thanks everyone, I've done it now and it all seems to be converging to a root so it looks right (got 2.175 to 3 dp).
Woo :smile:

Reply 6

John taylor

1

Can you rearrange the following the way I did?
Show that -x^3 + 3x^2 - 1 = 0 can be written as x = square root of (1/3-x).
I did...
3x^2 = x^3 + 1 (thne divided x^3 + 1 by x^2)

so 3 = x - x^-2 (then took the x on to the other side)
3 - x = x^-2
3 - x = 1/x^2 so x^2 (3-x) = 1
so x^2 = (1/3-x)
so x = square root of (1/3-x)

Thanks :smile:

There are alternative methods by factorising but...i just did it this way!
If anythings wrong please say so...

Reply 7

ghostwalker

17

Original post by John taylor

Can you rearrange the following the way I did?
Show that -x^3 + 3x^2 - 1 = 0 can be written as x = square root of (1/3-x).
I did...
3x^2 = x^3 + 1 (thne divided x^3 + 1 by x^2)

so 3 = x - x^-2 (then took the x on to the other side)
3 - x = x^-2
3 - x = 1/x^2 so x^2 (3-x) = 1
so x^2 = (1/3-x)
so x = square root of (1/3-x)

Thanks :smile:

There are alternative methods by factorising but...i just did it this way!
If anythings wrong please say so...

You are told the iterative formula to use in the question. Setting that aside:

Doing it that way without further consideration is probably not the wisest choice with the "3-x" in the denominator. For a suitable starting point it does converge though.

C3 Iterative formula help

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