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MathSofiya Group Title

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 \[y_1=0.2\] \[y_2=0.392\] \[y_3=0.56224\] \[y_4=0.70232\]

  • 2 years ago
  • 2 years ago

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  1. MathSofiya Group Title
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    Do I continue until the numbers stop changing or can I stop at y(1.4)?

    • 2 years ago
  2. SmoothMath Group Title
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    It's only asking you to approximate y(1.4) so you don't need to go any further.

    • 2 years ago
  3. MathSofiya Group Title
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    Thank you sir.

    • 2 years ago
  4. SmoothMath Group Title
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    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

    • 2 years ago
  5. MathSofiya Group Title
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    Correct. My x0=1. I don't know why I added a 4th line. Sry

    • 2 years ago
  6. SmoothMath Group Title
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    You actually don't even need the third line.

    • 2 years ago
  7. MathSofiya Group Title
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    Yes I do. \[0+0.2(1-0)=0.2\] \[0.2+0.2(1.2-(0.2)(1.2))=0.392\] \[0.392+0.2(1.4-(1.4)(0.392))=0.56224\]

    • 2 years ago
  8. SmoothMath Group Title
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    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|

    • 2 years ago
  9. MathSofiya Group Title
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    Two steps. Got it.

    • 2 years ago
  10. MathSofiya Group Title
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    ....?

    • 2 years ago
  11. SmoothMath Group Title
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    I was thinking about going into some simple explanation of how this approximation actually works.

    • 2 years ago
  12. MathSofiya Group Title
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    If you want to, I'd love to hear it.

    • 2 years ago
  13. SmoothMath Group Title
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    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.

    • 2 years ago
  14. MathSofiya Group Title
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    Ok, let's hear it.

    • 2 years ago
  15. MathSofiya Group Title
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    I posted this question earlier and asked what the bigger picture was. He simply gave a link to a wikipedia page.

    • 2 years ago
  16. SmoothMath Group Title
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    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.

    • 2 years ago
  17. SmoothMath Group Title
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    |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.

    • 2 years ago
  18. MathSofiya Group Title
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    Understood. Now on to the steps I guess...

    • 2 years ago
  19. MathSofiya Group Title
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    The smaller the steps the greater the accuracy ;) Yes?

    • 2 years ago
  20. SmoothMath Group Title
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    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|

    • 2 years ago
  21. SmoothMath Group Title
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    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|

    • 2 years ago
  22. SmoothMath Group Title
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    And you can see how adding in more steps, adjusting the slope more often, results in a better approximation.

    • 2 years ago
  23. MathSofiya Group Title
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    Yep. Makes sense. Thank you Smoothie, I really appreciate it.

    • 2 years ago
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