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anonymous

  • 5 years ago

how does eulers method of approximation work?

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  1. nowhereman
    • 5 years ago
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    There are two kinds of Euler approximations called explicit and implicit. The explicit is a bit faster but the implicit one is much more stable. What you do is approximating the solution to an initial-value problem by piece-wise affine functions. So explicit Euler is the easiest approximation method you can think of. Just calculate the elevation at the starting point, go a bit in that direction and take the resulting end point as the new starting point. Then repeat that step.

  2. anonymous
    • 5 years ago
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    and implicit?

  3. amistre64
    • 5 years ago
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    so instead of going to x=0 you just step down a bit?

  4. nowhereman
    • 5 years ago
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    Don't what you mean amistre... For implicit in each step you set up a system of linear equations which contains the elevation at the starting and at the end point.

  5. anonymous
    • 5 years ago
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    oh ok, thanks man

  6. amistre64
    • 5 years ago
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    elevation at the starting point..whats that mean? go abit in that direction and take the end point..... whats that mean?

  7. amistre64
    • 5 years ago
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  8. nowhereman
    • 5 years ago
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    You have given a initial value problem \[f'(t) = F(t, f(t)),\; f(t_0) = y_0\] So the starting point is (t_0, y_0). Also you must choose a step size h, which then gives you the precision. The elevation at the starting point is \[f'(t_0) = F(t_0, y_0)\] So going in that direction you get the approximation \[f(t_0 + h) \approx y_0 + h\cdot F(t_0, y_0)\]

  9. amistre64
    • 5 years ago
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    ahhh... that sounds more like integration methods.... unless im mistaken :)

  10. nowhereman
    • 5 years ago
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    Well yes, solving initial-value problems / differential equations is a generalization of integration.

  11. amistre64
    • 5 years ago
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    i was thinking more along the lines of the Newton stuff in the books finding roots and what nots. ok

  12. nowhereman
    • 5 years ago
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    Hehe, that explains the confusion ;-)

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