anonymous
  • anonymous
Use Euler's method with step size 0.2 to estimate y(1.4), where y(x) is the solution of the initial-value problem y'=x-xy, y(1)=0 \[y_1=0.2\] \[y_2=0.392\] \[y_3=0.56224\] \[y_4=0.70232\]
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
Do I continue until the numbers stop changing or can I stop at y(1.4)?
anonymous
  • anonymous
It's only asking you to approximate y(1.4) so you don't need to go any further.
anonymous
  • anonymous
Thank you sir.

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anonymous
  • anonymous
What x value did you begin at? 1. At a step size of .2, that means I'll need to do two steps to get to 1.4
anonymous
  • anonymous
Correct. My x0=1. I don't know why I added a 4th line. Sry
anonymous
  • anonymous
You actually don't even need the third line.
anonymous
  • anonymous
Yes I do. \[0+0.2(1-0)=0.2\] \[0.2+0.2(1.2-(0.2)(1.2))=0.392\] \[0.392+0.2(1.4-(1.4)(0.392))=0.56224\]
anonymous
  • anonymous
I'm pretty confident that you don't. You KNOW the y value when x is 1. That's given. Your step size is .2, and each time you do your fancy-pants calculation, it gives you an approximation for what y is when x is .2 higher than before. |dw:1343082671327:dw|
anonymous
  • anonymous
Two steps. Got it.
anonymous
  • anonymous
....?
anonymous
  • anonymous
I was thinking about going into some simple explanation of how this approximation actually works.
anonymous
  • anonymous
If you want to, I'd love to hear it.
anonymous
  • anonymous
Like, if you want to understand it instead of just plugging things in, it's simple enough that you would be able to get what you are actually doing.
anonymous
  • anonymous
Ok, let's hear it.
anonymous
  • anonymous
I posted this question earlier and asked what the bigger picture was. He simply gave a link to a wikipedia page.
anonymous
  • anonymous
So the idea is that you have a function that looks like y' = suchandsuch For a given starting point, you can use that function to find out what the first derivative is at that point. The first derivative gives us slope, or rate of change. So sweet, we know what the slope of the function is at that point. Sure, the slope of the function is CHANGING, but maybe it's not changing that fast. That's where the approximation business comes in. We figure out what the slope is, assume that the slope is constant, at least for a little bit, and we see where the function would end up.
anonymous
  • anonymous
|dw:1343083148521:dw| So here's a simple example. The top line is the actual curve where the slope is steadily increasing. The bottom line is our rough approximation where we assume that the slope stays the same.
anonymous
  • anonymous
Understood. Now on to the steps I guess...
anonymous
  • anonymous
The smaller the steps the greater the accuracy ;) Yes?
anonymous
  • anonymous
Okay, yeah. So a SUPER simple approximation would be to say "Let's assume that the slope doesn't change at all." And I get something like this where the slope might actually change quite a lot and I get way off: |dw:1343083474058:dw|
anonymous
  • anonymous
Or, I could add in some steps. Basically the idea is that I occasionally stop and account for the slope changing. So instead of going the whole distance under the assumption of no slope change, I stop a little bit in and go "What's a good approximation for the slope at this point" I use the same y' = function that I used before|dw:1343083628908:dw|
anonymous
  • anonymous
And you can see how adding in more steps, adjusting the slope more often, results in a better approximation.
anonymous
  • anonymous
Yep. Makes sense. Thank you Smoothie, I really appreciate it.

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