# iteration

• Sep 27th 2006, 06:49 PM
dan
iteration
im trying to write an iteration for x in x^x-y=0

i'v used x_(n+1)= x_n - (f(x_n)/f'(x_n))

maybe i'm doing somthing wrong...

so far the math i'v been doing latly has been all wrong..:p

dan
• Sep 27th 2006, 06:56 PM
ThePerfectHacker
Quote:

Originally Posted by dan
im trying to write an iteration for x in x^x-y=0

i'v used x_(n+1)= x_n - (f(x_n)/f'(x_n))

That looks correct to me.
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Only problem is that you are might be breaking the theorem. Meaning some theorem do not work because your problem does not satisfy the conditions for the theorem. For example, the fundamental theorem of calculus sometimes fails, not because it is wrong but if you use a discountinous function it will not work (those were the assumed conditions). In this algorithm I know there is a special condition your function needs to satisfy (besides for being differenciable on some open interval) I forget what it is. So even if you did everything correct you might not satify a special condition.
• Sep 28th 2006, 04:15 AM
Soroban
Hello, Dan!

Quote:

I'm trying to write an iteration for x in: . x^x - y = 0

I've used: .x_(n+1) .= .x_n - f(x_n)/f '(x_n)

It seems that you are seeking an x-intercept of: .y .= .x^x
. . (If I'm wrong, disregard this entire post.)

Is your derivative correct? . . . It's tricky.

Take logs: . ln y .= .ln(x^x) .= .x·ln(x)

Differentiate: . y'/y .= .x(1/x) + ln(x)

Then: . y' .= .y[(1 + ln(x)] .= .x^x[1 + ln(x)]

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .x^x
The interation becomes: . x_(n+1) . = . x_n - -----------------
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . x^x[1 + ln(x)]
. . . . . . . . . . . . . . . . . . . . . . . . .1
Therefore: .x_(n+1) . = . x_n - ------------
. . . . . . . . . . . . . . . . . . . . . . 1 + ln(x)

• Sep 28th 2006, 03:57 PM
dan
Quote:

Originally Posted by Soroban
Hello, Dan!

It seems that you are seeking an x-intercept of: .y .= .x^x
. . (If I'm wrong, disregard this entire post.)

Is your derivative correct? . . . It's tricky.

Take logs: . ln y .= .ln(x^x) .= .x·ln(x)

Differentiate: . y'/y .= .x(1/x) + ln(x)

Then: . y' .= .y[(1 + ln(x)] .= .x^x[1 + ln(x)]

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .x^x
The interation becomes: . x_(n+1) . = . x_n - -----------------
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . x^x[1 + ln(x)]
. . . . . . . . . . . . . . . . . . . . . . . . .1
Therefore: .x_(n+1) . = . x_n - ------------
. . . . . . . . . . . . . . . . . . . . . . 1 + ln(x)

that looks good ...

thanks

dan
• Sep 28th 2006, 05:00 PM
ThePerfectHacker
Quote:

Originally Posted by dan
that looks good ...

thanks

dan

I am curious does the iteration work now? Because it still might not as I said in the first post.