what is the limit of the series yn
The limit of that sequence is as you should be able to recognize the MacLaurin series for
However, the derivative of is , not x+ y.
If you are trying to use Picard's method to solve this equation, as Ackbeet suggested, then you are doing it wrong.
From dy/dx= x+y we get dy= (x+y)dx so that . Of course, since we don't know y but we can approximate it. If we are given y(0)= 1 (which you don't say but is implied by "y0= 1") as a first approximation, assume y(x)= 1 for all x. Then dy/dx= x+1 so that dy= (x+1)dx and [tex]y= (1/2)x^2+ x+ C[tex] and y(0)= C= 1 gives , as you have.
But then, and, again, gives which is NOT what you have. I expect you forgot the added x.
Now, and gives .
One more time: and gives .
I think I would conjecture now that
and I would attempt to prove that by "induction"
When n= 0, that says that y0= 1 which is true.
Now assume that, for some k, so that and gives
just as we wish.
Jan 21st 2011, 04:25 AM
Originally Posted by Ackbeet
Ok. So y(0) = 1 implies that
That's what you got, I believe. The sequence converges to To convince yourself of that fact, just expand in a Taylor series about 0, and you'll see that it's the exact same series as you're getting here.