limit of series 1 A

**transgalactic** · January 18th 2011, 07:03 AM

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
?

**Ackbeet** · January 18th 2011, 07:12 AM

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.

**transgalactic** · January 18th 2011, 07:15 AM

yes its itterations and i need to find yn where
n goes to infinity
?

**Prove It** · January 18th 2011, 07:28 AM

Why can't you just solve the DE exactly? It's first order linear...

**Ackbeet** · January 18th 2011, 07:29 AM

So, just to be clear, you're setting up the following iteration scheme:

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

$y_{0}=1,$

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

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

Is there an initial condition or something I should know about?

**transgalactic** · January 18th 2011, 07:38 AM

yes y(0)=1
the whole question is about approximating

so i cant solve it like first order linear

**Ackbeet** · January 18th 2011, 07:46 AM

Ok. So y(0) = 1 implies that

$y_{0}=1,$

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

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

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

That's what you got, I believe. The sequence $\{y_{n}\}$ converges to $e^{x}.$ To convince yourself of that fact, just expand $e^{x}$ 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?

**transgalactic** · January 18th 2011, 08:17 AM

as much rigorous as you can give me
my orof is very strict
and i only started with converging serieses

**Ackbeet** · January 18th 2011, 08:20 AM

Try a proof by induction.

**transgalactic** · January 18th 2011, 08:24 AM

ok
what expressio for n=k to take
and what expression for n=k+1 to take

**Ackbeet** · January 18th 2011, 08:52 AM

Well, it looks like you want to show that, aside from the first term,

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

What ideas does that give you?

**transgalactic** · January 18th 2011, 09:08 AM

$y_{n+1}\displaystyle =1+\int (t+ \sum_{j=0}^{n}\frac{t^{j}}{j!})dt$
how to prove that it equals $\displaystyle y_{n}=\sum_{j=0}^{n+1}\frac{x^{j}}{j!}$

**Ackbeet** · January 18th 2011, 09:16 AM

An induction proof has two parts: the base case, and the inductive step. What's your base case?

**transgalactic** · January 18th 2011, 09:38 AM

aahh i understand now
thanks

**chisigma** · January 18th 2011, 09:39 AM

Originally Posted by transgalactic

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
?

We have the first order linear DE...

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

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

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

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

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

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

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

$\dots$

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

$\dots$

... so that is...

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

Kind regards

$\chi$ $\sigma$

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