Euler's method

  • Thread starter anb
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  • #1
anb
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Hi all. I've been having a bit of trouble just understanding Euler's method, which I need to know (and probably should have known for a while now) for a question on an assignment. I'm going to provide an example with confusing wording (at least to me) and show what I've done to try to understand it. I think I get it but I want to be absolutely sure.

Homework Statement



So for the question we have (dx/dt) = f(x) = -x with initial condition x(0) = 1. The first part asks us to give an exact solution to x(1). I believe the solution is (e^-1) and I'm not having trouble with this part (but if that's incorrect then maybe that's my problem - I reasoned that you could just separate variables and integrate both sides, giving -lnx=t+K for some constant K. x=1 at t=0 so K=0 since ln1=0. So then if t=1 and K=0, lnx=-1 which means x=e^-1).

The second part says "using Euler's method with step size D=1 [delta was used in the text but I can't seem to get a "Delta" symbol working], estimate x(1) numerically ... then repeat using D=(10^-n) for n=1,2,3..."

Homework Equations


The Attempt at a Solution



Now here's what I did. The text tells us that to approximate solutions with euler's method we use the recursion xn+1=xn+f(xn)*D. So I take x0=1 since x(0)=1 is the initial condition. Therefore f(x0)=-1 since f(x)=-x, and with D=t that gives us the approximation x1=0 (because this is such a poor approximation I assume I may have gone wrong here, ie maybe I just don't understand which values to plug where).

Now when it asks us to repeat with different step sizes D=(10^-n), I'm not sure if it means that for approximating xn+1 we should use D=(10^-n), or if the text is asking us to repeat using that step size for the approximation of x1. The latter doesn't really make sense to me, because that would lead to an even worse approximation for all n>0. My problem though is that if I repeat the incursion for n>0, I'm pretty sure I get
xn+1=0 for all n>1 since the approximation x1=0, so x2=x1+f(x1)*D=0 no matter what D is. I realize that this could still be the intended way of solving the problem, it just doesn't seem very nice because the actual step size doesn't matter and I'm used to having to use every part of a problem to solve it (and in this case the step size seems like a red herring).

Keep in mind I've never worked with euler's method before (shock), and I may just be over-thinking the problem
 

Answers and Replies

  • #2
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It is not strange that you get a bad answer when you only do one step.

What you are supposed to do next is do the whole thing over but this time having a step size of [tex] 10^{-1} [/tex].
This means that you need to step 10 times to get to [tex] x = 1[/tex]. This should give you a better approximation. Remember in every step you need to update both the x-value and the function value.

With n = 2 you do a hundred steps. If you know any programming you can automate this process (I don't think your teacher wants you to do it manually a hundred times).
 
  • #3
anb
3
0
It is not strange that you get a bad answer when you only do one step.

What you are supposed to do next is do the whole thing over but this time having a step size of [tex] 10^{-1} [/tex].
This means that you need to step 10 times to get to [tex] x = 1[/tex]. This should give you a better approximation. Remember in every step you need to update both the x-value and the function value.

With n = 2 you do a hundred steps. If you know any programming you can automate this process (I don't think your teacher wants you to do it manually a hundred times).
Oh, I think I understand now - so for a step size of 1/10 I should first get x1=9/10, which means at the next step I compute 9/10 + (1/10)f(9/10), and so on with the result of each step, ten times? (So for the step after I'd do 81/100 + (1/10)f(81/100), etc.?) After trying this out, I am getting a much better approximation of e^-1 after the tenth step.
 

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