The Picard Iteration to find a series of Functions converging towards a solution I want to find a series of functions converging to the solution of $\displaystyle \frac{dx}{dt}=2x$ with $x(0)=x_0$
But, I was to do this using the Picard-Lindolf iteration and am completely stuck on how to use it!! Please could someone please explain?
I know that you do several iterations and then use induction, but how do you iterate?
EDIT:
Arrghh! Been stuck on this for 2 Hours now, I have the iteration x_5 (t) = x_0(1 + 2t + 2t^2 + 4/3t^3 + 3/4^t^4 ..+ but i dont see a pattern with this series at all! 
I want to obviously prove by induction this series is of the solutions but i have no idea how to write it in series form, not sure if its impossible or just too late in the night! If someone could help me/guide me towards the the series form i would be very grateful! Many thanks
 A: For your problem, we are given:
$\tag 1 \displaystyle \frac{dx}{dt}=2x, ~~x(0)=x_0$
The solution to $(1)$ is: 
$$x(t) = x_0 e^{2t}.$$
The Picard-Lindelof Iteration is given by:
$$\tag 2 \displaystyle x_0(t) = x_0, ~~x_{n+1}(t) = x_0 + \int^t_{t_0} f(s, x_n(s))ds$$
For $(1)$, we have: $f(s, x_n(s)) = 2s$ and using $(2)$, yields:


*

*$\displaystyle x(0) = x_0$

*$\displaystyle x_1(t) = x_0 + \int^t_{t_0} f(s, x_0(s))ds = x_0 + \int^t_{0} 2(x_0) ds = x_0 + 2x_0t$

*$\displaystyle x_2(t) = x_0 + \int^t_{t_0} f(s, x_1(s))ds = x_0 + \int^t_{0} 2(x_0 + 2x_0 s) ds = x_0 + 2x_0 t (t+1)$

*$\displaystyle x_3(t) = x_0 + \int^t_{t_0} f(s, x_2(s))ds = x_0 + \int^t_{0} 2(x_0 + 2x_0 s (s+1)) ds = x_0 + \frac{2}{3} x_0t(2t^2+3t+3)$

*$\displaystyle \ldots$

*See a pattern for $x_n(t)$?
Now, lets experiment by choosing $x_0 = 1$, and plot the four solutions $\{x(t), x_0, x_1(t), x_2(t), x_3(t)\}$ using Wolfram Alpha.
Does it converge in a suitable interval and diverge for other values of $t$? 
I'll leave that up to you because your question implies you understand how to do induction to see what is going on.
Update
If you expand the series for $e^{2t}$, we get:
$$e^{2t} = 1 + 2t + 2t^2 + \frac{4}{3}t^3 + \frac{2}{3}t^4 + \ldots$$
Now, look at my successive $x_n(t)$, and what do you notice?
Got it? (Note, you made an error in your calc in comments.)
Update 2
$$x(t) = x_0e^{2t} = x_0(1 + 2t + 2t^2 + \frac{4}{3}t^3 + \frac{2}{3}t^4 + \ldots) = x_0 \sum_{k=0}^\infty \frac{(2t)^{k}}{t!}$$


*

*$\displaystyle x(0) = x_0 = x_0(1)$

*$\displaystyle x_1(t) = x_0 + 2x_0t = x_0(1+2t)$

*$\displaystyle x_2(t) = x_0 + 2x_0 t (t+1) = x_0(1 + 2t + 2t^2)$

*$\displaystyle x_3(t) = x_0 + \frac{2}{3} x_0t(2t^2+3t+3) = x_0(1 + 2t + 2t^2 + \frac{4t^3}{3})$

*$\ldots$
Now compare the $x(t)$ solution to the $x_n(t)$ and what do you see.
Regards
