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Iteration question

coconut64

need help with b) as I font get how you can obtain the equation shown inthr question . This is as far as I get

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thanks

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Reply 1

Gregorius

Original post by coconut64

need help with b) as I font get how you can obtain the equation shown inthr question . This is as far as I get s

Subtract x/2 from both sides and then...

Reply 2

coconut64

Original post by Gregorius

Subtract x/2 from both sides and then...

But that gives you this : 7/x^2 - x/2 = 0 but the question shows that there should be a x instead of 0 and there should be a positive sign ...

Thanks

Reply 3

Gregorius

Original post by coconut64

But that gives you this : 7/x^2 - x/2 = 0 but the question shows that there should be a x instead of 0 and there should be a positive sign ...

Thanks

You need to remember that you are solving for p. The fact that you have stuck 7 there shows that you have done it! Your equation is equivalent to the original equation, as required.

Reply 4

coconut64

Original post by Gregorius

You need to remember that you are solving for p. The fact that you have stuck 7 there shows that you have done it! Your equation is equivalent to the original equation, as required.

That would be a guess though as I have - in my equation but in the question it is +

Also I am confused with what you meant by solving for p , there is no p ?

Thanks

Reply 5

NotNotBatman

Original post by coconut64

That would be a guess though as I have - in my equation but in the question it is +

Also I am confused with what you meant by solving for p , there is no p ?

Thanks

The way you can do this and this has always worked for me, is by working backwards and then swap it around, so

x = \frac{p}{x^2} +\frac{x}{2}

Then I would want to multiply both sides by 2x^2 to get

2x^3 = 2p +x^3

then

x^3 = 2p

so you can find what p is, then you can use that and form the equation, I hope I haven't shown too much here.

Reply 6

Notnek

Original post by NotNotBatman

The way you can do this and this has always worked for me, is by working backwards and then swap it around, so

x = \frac{p}{x^2} +\frac{x}{2}

Then I would want to multiply both sides by 2x^2 to get

2x^3 = 2p +x^3

then

x^3 = 2p

so you can find what p is, then you can use that and form the equation, I hope I haven't shown too much here.

Great advice. Working backwards for these questions can be very useful.

Reply 7

coconut64

Original post by notnek

Great advice. Working backwards for these questions can be very useful.

If I am to construct the equation, why can I do so? It is not working out for me; I am struggling to construct the same equation as the one shown in the question. How do I do it not considering the backward method? Thanks

Reply 8

Notnek

Original post by coconut64

NotNotBatman has shown you how to work backwards. Quote him if you don't understand part of his post.

Reply 9

coconut64

Original post by notnek

NotNotBatman has shown you how to work backwards. Quote him if you don't understand part of his post.

I understand his method but I am just wondering if you know how to construct the equation without using the backward method. Cheers

Reply 10

NotNotBatman

Original post by coconut64

I understand his method but I am just wondering if you know how to construct the equation without using the backward method. Cheers

Do the backwards method backwards...
that's how you would construct it, but it can be difficult to see.

Reply 11

Notnek

Original post by coconut64

I understand his method but I am just wondering if you know how to construct the equation without using the backward method. Cheers

Sorry I misread you post.

From the working in your first post, try adding

\frac{x}{2}

to both sides. It's possible that Gregorius meant this instead of subtracting from both sides.

(edited 7 years ago)

Reply 12

RDKGames

Study Forum Helper

Original post by coconut64

I understand his method but I am just wondering if you know how to construct the equation without using the backward method. Cheers

Divide both sides by

x^2

then do

\frac{14}{x^2}=\frac{p}{x^2} + \frac{x}{2}

and solve for p. Flows nicely this way.

Reply 13

Gregorius

Original post by coconut64

Also I am confused with what you meant by solving for p , there is no p ?

The question says that you should show that <something> can be written in the form <something else involving p> where p is to be found. You are "solving for p" in the sense that you have to find p so that the identity is true.

Others in this thread have suggested ways to approach this question, so I will just note that when you get a question phrased in this way, you have to read the question setter's "code" - questions written in this form are giving you the hint to start from the end (with the expression involving p) and work back to the original expression, finding a value of p that makes it work.

But I also notice that you ask the excellent question:

I understand his method but I am just wondering if you know how to construct the equation without using the backward method.

The question setter had to construct the equation involving p without knowing in advance the form of that expression - how did he or she do it? The answer lies in the surrounding context of the question. You are being asked to construct iteration schemes that will converge to numerical solutions to equations. The equation setter has one great advantage over you - they know the conditions under which such iteration schemes actually converge to the right answer! Therefore they can look for equations of the right form that do as they want - there is a degree of trial and error in this sort of construction, but once you know roughly what you're looking for, things tend to drop out relatively easily.

However, in answering the question, you don't necessarily know the direction that you need to set off in - hence you should use the "working backwards from the form of the answer" method (as suggested by the "code" of the question).

Reply 14

coconut64

So it is okay to start using the the equation given in this question, even in exams as I don't really see this backward method in mark scheme normally. Thanks

Reply 15

DFranklin

Original post by coconut64

So it is okay to start using the the equation given in this question, even in exams as I don't really see this backward method in mark scheme normally. Thanks

Yes.