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MathSofiya

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

  • one year ago
  • one year ago

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

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

    • one year ago
  3. MathSofiya
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    Thank you sir.

    • one year ago
  4. SmoothMath
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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

    • one year ago
  5. MathSofiya
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    Correct. My x0=1. I don't know why I added a 4th line. Sry

    • one year ago
  6. SmoothMath
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    You actually don't even need the third line.

    • one year ago
  7. MathSofiya
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    Yes I do. \[0+0.2(1-0)=0.2\] \[0.2+0.2(1.2-(0.2)(1.2))=0.392\] \[0.392+0.2(1.4-(1.4)(0.392))=0.56224\]

    • one year ago
  8. SmoothMath
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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|

    • one year ago
  9. MathSofiya
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    Two steps. Got it.

    • one year ago
  10. MathSofiya
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    ....?

    • one year ago
  11. SmoothMath
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    I was thinking about going into some simple explanation of how this approximation actually works.

    • one year ago
  12. MathSofiya
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    If you want to, I'd love to hear it.

    • one year ago
  13. SmoothMath
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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.

    • one year ago
  14. MathSofiya
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    Ok, let's hear it.

    • one year ago
  15. MathSofiya
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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.

    • one year ago
  16. SmoothMath
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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.

    • one year ago
  17. SmoothMath
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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.

    • one year ago
  18. MathSofiya
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    Understood. Now on to the steps I guess...

    • one year ago
  19. MathSofiya
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    The smaller the steps the greater the accuracy ;) Yes?

    • one year ago
  20. SmoothMath
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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|

    • one year ago
  21. SmoothMath
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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|

    • one year ago
  22. SmoothMath
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    And you can see how adding in more steps, adjusting the slope more often, results in a better approximation.

    • one year ago
  23. MathSofiya
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    Yep. Makes sense. Thank you Smoothie, I really appreciate it.

    • one year ago
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