does any body know how to write an iteration to compute y in y-log(y)=x ?

thank's,

dan

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- Jul 10th 2006, 07:28 AMdany-log(y)=x
does any body know how to write an iteration to compute y in y-log(y)=x ?

thank's,

dan - Jul 10th 2006, 08:08 AMCaptainBlackQuote:

Originally Posted by**dan**

RonL - Jul 10th 2006, 09:09 AMThePerfectHacker
(I assume your question is on Newton's Method cuz that is what you asked the other time).

You want to solve,

Which is equivalent to,

The iteration sequence is,

Where,

Thus,

Thus,

Thus,

Simplfy,

More simplfying,

And yet one more time, (watch those signs :eek: )

Finally we have,

- Jul 10th 2006, 10:00 AMdany-log(y)=x
Ok ,

I think you might have lost me… I suppose the a with the little n (How do you write that anway?) is your initial value but you lost me on how to compute that.

Sorry about my slow mind :confused: .

dan - Jul 10th 2006, 10:14 AMCaptainBlackQuote:

Originally Posted by**dan**

in terms of , so you get a sequence:

,

which with luck will converge to the required solution.

RonL - Jul 10th 2006, 12:47 PMdany-log(y)=x
so...

a_0 =1-1/x ???? I think I'm lost. I need a way to calulate a_0 in relation to x...I think :confused: I gues they should't let people like me on calculus forums :rolleyes:

dan - Jul 10th 2006, 01:02 PMThePerfectHackerQuote:

Originally Posted by**dan**

- Jul 10th 2006, 01:12 PMdany-log(y)=x
Just any old random guess?? I know that some iterations have a formula for making that guess. But if x = 10 I could just guess 14 and it would still approach??

Dan - Jul 10th 2006, 01:30 PMCaptainBlackQuote:

Originally Posted by**dan**

iteration as given by PH it appears to diverge with , or !

It must be a mistake in PH's algebra as the iteration (which is what PH had before he tried simplifying it):

does converge with , to after two iterations.

RonL - Jul 10th 2006, 02:15 PMJameson
PH: Some slight errors in your algebra.

Starting from here:

For the next step you wrote:Quote:

Originally Posted by**ThePerfectHacker**

And factoring: - Jul 10th 2006, 02:33 PMThePerfectHacker
I realized.

- Jul 10th 2006, 06:09 PMmalaygoelQuote:

Originally Posted by**dan**

You are given x

Guess any value of a(a is iterating variable, should be greater than x)

Find

You will get some value

this is your next value of a

Substitute it again in the above expression.

After some steps it will give the answer(when you the same value of a for two times)

Note: I am not sure about this idea, please someone check it.

Keep Smiling

Malay - Jul 10th 2006, 07:36 PMCaptainBlackQuote:

Originally Posted by**malaygoel**

seems to converge faster at least near x=10. - Jul 11th 2006, 04:22 AMmalaygoelQuote:

Originally Posted by**malaygoel**

It works

You could take any starting value of a

Keep Smiling

Malay - Jul 11th 2006, 04:39 AMCaptainBlackQuote:

Originally Posted by**malaygoel**

faster disguised the fact that I knew it converged for reasonable starting

values and values for x.

Try a_0=0.000001 with x=10 :p

RonL