## MathSofiya 3 years ago 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 \[y_1=0.2\] \[y_2=0.392\] \[y_3=0.56224\] \[y_4=0.70232\]

1. MathSofiya

Do I continue until the numbers stop changing or can I stop at y(1.4)?

2. SmoothMath

It's only asking you to approximate y(1.4) so you don't need to go any further.

3. MathSofiya

Thank you sir.

4. SmoothMath

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

5. MathSofiya

Correct. My x0=1. I don't know why I added a 4th line. Sry

6. SmoothMath

You actually don't even need the third line.

7. MathSofiya

Yes I do. \[0+0.2(1-0)=0.2\] \[0.2+0.2(1.2-(0.2)(1.2))=0.392\] \[0.392+0.2(1.4-(1.4)(0.392))=0.56224\]

8. SmoothMath

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|

9. MathSofiya

Two steps. Got it.

10. MathSofiya

....?

11. SmoothMath

I was thinking about going into some simple explanation of how this approximation actually works.

12. MathSofiya

If you want to, I'd love to hear it.

13. SmoothMath

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.

14. MathSofiya

Ok, let's hear it.

15. MathSofiya

16. SmoothMath

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.

17. SmoothMath

|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.

18. MathSofiya

Understood. Now on to the steps I guess...

19. MathSofiya

The smaller the steps the greater the accuracy ;) Yes?

20. SmoothMath

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|

21. SmoothMath

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|

22. SmoothMath

And you can see how adding in more steps, adjusting the slope more often, results in a better approximation.

23. MathSofiya

Yep. Makes sense. Thank you Smoothie, I really appreciate it.

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