anonymous
  • anonymous
how does eulers method of approximation work?
Mathematics
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schrodinger
  • schrodinger
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nowhereman
  • nowhereman
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.
anonymous
  • anonymous
and implicit?
amistre64
  • amistre64
so instead of going to x=0 you just step down a bit?

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nowhereman
  • nowhereman
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.
anonymous
  • anonymous
oh ok, thanks man
amistre64
  • amistre64
elevation at the starting point..whats that mean? go abit in that direction and take the end point..... whats that mean?
amistre64
  • amistre64
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nowhereman
  • nowhereman
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 \[f(t_0 + h) \approx y_0 + h\cdot F(t_0, y_0)\]
amistre64
  • amistre64
ahhh... that sounds more like integration methods.... unless im mistaken :)
nowhereman
  • nowhereman
Well yes, solving initial-value problems / differential equations is a generalization of integration.
amistre64
  • amistre64
i was thinking more along the lines of the Newton stuff in the books finding roots and what nots. ok
nowhereman
  • nowhereman
Hehe, that explains the confusion ;-)

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