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I have

$ \frac{dy}{dx} = y^2, y(0) = y_0 $

I have solved this as

$y = \frac{y_0}{1 - x y_0}$

Which has the Taylor expansion

$ y_0+y_0^2 x+y_0^3 x^2+y_0^4 x^3+y_0^5 x^4+ ...$

However, when I perform Picard iteration, I get:

$Iteration 0: y = y_0 $

$Iteration 1: y = y_0 + \int y_0^2 = y_0 + y_0^2x$

$Iteration 2: y = y_0 + \int (y_0 + y_0^2x)^2 = y_0 + y_0^2x + y_0^3x^2 + \mathbb{\frac{y_0^4x^3}{3}}$

Everything is of the right order, there is just a factor of 1/3 at the end, where am I going wrong?

Thanks

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Do one more iteration... –  David Mitra Jan 6 '12 at 12:35
    
Ok so I did another iteration and I am still left with terms at the end off by a factor. However, the $x^3$ term is now correct. Can I then deduce that by taking limits all terms do this? Thanks –  T. Kiley Jan 6 '12 at 12:45
1  
I don't think you're even guaranteed that $any\ iteration$ contains terms that are exactly the first few terms of the Taylor expansion of the solution. Of course, the iterates converge to the solution on some set. –  David Mitra Jan 6 '12 at 12:54

1 Answer 1

up vote 3 down vote accepted

I see nothing wrong. A Picard iteration does not necessarily give you the first few terms of the Taylor expansion of the solution exactly. It only gives you approximate solutions.

If you do one more interationfor your problem, however, you'll see it contains the forth term.

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