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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- Jan 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

? - Jan 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. - Jan 18th 2011, 07:15 AMtransgalactic
yes its itterations and i need to find yn where

n goes to infinity

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

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

$\displaystyle \displaystyle y_{n+1}(x)=\int(x+y_{n}(x))\,dx,$ right? If so, I don't get what you get. I'm seeing

$\displaystyle y_{0}=1,$

$\displaystyle y_{1}=\displaystyle\int(x+1)\,dx=x+\frac{x^{2}}{2} ,$

$\displaystyle \displaystyle y_{2}=\int\left(x+x+\frac{x^{2}}{2}\right)\,dx=x^{ 2}+\frac{x^{3}}{6},\dots$

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

the whole question is about approximating

so i cant solve it like first order linear - Jan 18th 2011, 07:46 AMAckbeet
Ok. So y(0) = 1 implies that

$\displaystyle y_{0}=1,$

$\displaystyle y_{1}=\displaystyle\int(x+1)\,dx=1+x+\frac{x^{2}}{ 2},$

$\displaystyle y_{2}=\displaystyle\int\left(x+1+x+\frac{x^{2}}{2} \right)\,dx=1+x+x^{2}+\frac{x^{3}}{6},$

$\displaystyle y_{3}=\displaystyle\int\left(x+1+x+x^{2}+\frac{x^{ 3}}{6}\right)dx=1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6 }+\frac{x^{4}}{24},\dots$

That's what you got, I believe. The sequence $\displaystyle \{y_{n}\}$ converges to $\displaystyle e^{x}.$ To convince yourself of that fact, just expand $\displaystyle e^{x}$ 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? - Jan 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 - Jan 18th 2011, 08:20 AMAckbeet
Try a proof by induction.

- Jan 18th 2011, 08:24 AMtransgalactic
ok

what expressio for n=k to take

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

$\displaystyle \displaystyle y_{n}=\sum_{j=0}^{n+1}\frac{x^{j}}{j!}.$

What ideas does that give you? - Jan 18th 2011, 09:08 AMtransgalactic
$\displaystyle y_{n+1}\displaystyle =1+\int (t+ \sum_{j=0}^{n}\frac{t^{j}}{j!})dt$

how to prove that it equals $\displaystyle \displaystyle y_{n}=\sum_{j=0}^{n+1}\frac{x^{j}}{j!}$ - Jan 18th 2011, 09:16 AMAckbeet
An induction proof has two parts: the base case, and the inductive step. What's your base case?

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

thanks

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

$\displaystyle \displaystyle y^{'} = y + x\ \ ,\ y(0)=1$ (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...

$\displaystyle \displaystyle y(x)= \sum_{n=0}^{\infty} \frac{y^{(n)} (0)}{n!}\ x^{n}$ (2)

But the $\displaystyle y^{(n)} (0)$ can be derived from (1) as follows...

$\displaystyle \displaystyle y^{(0)} (0) = 1$

$\displaystyle \displaystyle y^{(1)}(x)= y+x \implies y^{(1)}(0)= 1$

$\displaystyle \displaystyle y^{(2)}(x) = 1 + y^{(1)}(x) = 1+x+y \implies y^{(2)}(0)= 2$

$\displaystyle \displaystyle y^{(3)}(x) = 1+x+y \implies y^{(3)}(0)= 2$

$\displaystyle \dots$

$\displaystyle \displaystyle y^{(n)}(x) = 1+x+y \implies y^{(n)}(0)= 2$

$\displaystyle \dots$

... so that is...

$\displaystyle \displaystyle y(x)= 1 + x + 2\ \sum_{n=2}^{\infty} \frac{x^{n}}{n!} = 2\ e^{x} -1 -x$ (3)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$