i have dy/dx=x+y

y0=1

y1=1+x+(x^2)/2

y2=1+x+(x^2)/2+(x^3)/6

what is the limit of the series yn

?

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- January 18th 2011, 07:03 AMtransgalacticlimit of series 1 A
i have dy/dx=x+y

y0=1

y1=1+x+(x^2)/2

y2=1+x+(x^2)/2+(x^3)/6

what is the limit of the series yn

? - January 18th 2011, 07:12 AMAckbeet
Two comments:

1. It looks an awful lot like the exponential series.

2. Theoretically, assuming certain smoothness conditions are met, which it looks like they are, the series should converge to the solution of the DE. You're essentially doing Picard iterations there, if I'm not mistaken. - January 18th 2011, 07:15 AMtransgalactic
yes its itterations and i need to find yn where

n goes to infinity

? - January 18th 2011, 07:28 AMProve It
Why can't you just solve the DE exactly? It's first order linear...

- January 18th 2011, 07:29 AMAckbeet
So, just to be clear, you're setting up the following iteration scheme:

right? If so, I don't get what you get. I'm seeing

Is there an initial condition or something I should know about? - January 18th 2011, 07:38 AMtransgalactic
yes y(0)=1

the whole question is about approximating

so i cant solve it like first order linear - January 18th 2011, 07:46 AMAckbeet
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.

How rigorous a proof does this require? - January 18th 2011, 08:17 AMtransgalactic
as much rigorous as you can give me

my orof is very strict

and i only started with converging serieses - January 18th 2011, 08:20 AMAckbeet
Try a proof by induction.

- January 18th 2011, 08:24 AMtransgalactic
ok

what expressio for n=k to take

and what expression for n=k+1 to take - January 18th 2011, 08:52 AMAckbeet
Well, it looks like you want to show that, aside from the first term,

What ideas does that give you? - January 18th 2011, 09:08 AMtransgalactic

how to prove that it equals - January 18th 2011, 09:16 AMAckbeet
An induction proof has two parts: the base case, and the inductive step. What's your base case?

- January 18th 2011, 09:38 AMtransgalactic
aahh i understand now

thanks

:) - January 18th 2011, 09:39 AMchisigma
We have the first order linear DE...

(1)

... and we have 'forgotten' the procedure to solve it... no problem because we can use an alternative approach!...

Let's suppose that y(x) is analytic in x=0 so that is...

(2)

But the can be derived from (1) as follows...

... so that is...

(3)

Kind regards