Euler's Method

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  • #1
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Homework Statement



I have some trouble understanding the following solved problem:

http://img808.imageshack.us/img808/3340/euler2.jpg [Broken]

This is the solutions they have given us:

[itex]E(32 \ steps) = 0.00097030[/itex]

[itex]E(64 \ steps) = 0.00003202[/itex]

32 steps has h=0.625

64 steps has h= 0.03125

The formula for effective order at stepsize h is

[itex]q= \frac{\ln |E(2h)|- \ln |E(h)|}{\ln 2}[/itex]

[itex]\therefore q (0.03125)= \frac{\ln |E(0.625)|- \ln |E(0.03125)|}{\ln 2} \approx 4.92[/itex]

The Attempt at a Solution



So I don't understand how they figured out that 32 steps has a step size of 0.625? What is the relationship between the stepsize and number of steps? :confused:

I used to think that step size and the number of steps were related like this:

[itex]h = \frac{1}{step \ size}[/itex]

But using this, for 32 steps I get a step size of h=1/32=0.03125, and for 64 steps I get h=1/64=0.0156. The only way I can get the correct answer is to multiply the denominator by 2, but why should I do that?

So, what's the problem with what I'm doing? Is my method wrong, or is that a typo in the given answers?
 
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Answers and Replies

  • #2
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The problem states that y(1) = -1, and the solutions are at t=3.
The step size would be the change in t divided by the number of steps.
 
  • #3
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The problem states that y(1) = -1, and the solutions are at t=3.
The step size would be the change in t divided by the number of steps.
But still I don't get the correct answer. The correct answer says:

32 steps has h=0.625

64 steps has h= 0.03125
And if I divide the change in t by the number of steps I get

2/32=0.0625

2/64=0.03125

How come for 32 steps I got a different answer?
 
  • #4
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I'm not sure. It's weird that the number they gave is off by a factor of 10, but the one for 64 is correct. Is it possible it's a typo?
 
  • #5
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Yes. But are you sure in order to find the step size from the number of steps we have to divide the change in t by the number of steps? If that's the correct method then I think that may be a typo.
 
  • #6
vela
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Yes, Villyer is correct, and that's a typo.
 
  • #7
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Okay thank you very much for the confirmation. And thanks a lot Villyer!!
 

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