network profile
| bio | website | |
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| location | ||
| age | ||
| visits | member for | 1 year |
| seen | yesterday | |
| stats | profile views | 13 |
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May 15 |
awarded | Caucus |
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Jun 19 |
awarded | Teacher |
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Jun 19 |
revised |
What is the usefulness of matrices? added 25 characters in body |
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Jun 19 |
answered | What is the usefulness of matrices? |
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May 31 |
awarded | Scholar |
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May 31 |
accepted | Refraction equation, quartic equation |
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May 31 |
comment |
Refraction equation, quartic equation I've tested it and it works great! Only 10-20 iterations of bisection search is enough. I could also improve the solution using Newton's method. |
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May 30 |
comment |
Refraction equation, quartic equation Right, converting the original equation into a polynomial introduces unwanted ambiguity. The polynomial can have as many as four roots while there's only one correct solution to the original problem. I just hoped that I could find a descent analytic solution but it turns out that a numerical method is the only way to solve the problem efficiently. |
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May 29 |
awarded | Commentator |
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May 29 |
comment |
Refraction equation, quartic equation Yes, but deriving $x$ from Snell's Law gives the same result - quartic equation. In fact, my problem is to find the point of refraction of light according to Snell's Law. |
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May 29 |
comment |
Refraction equation, quartic equation That's a good idea. Finding a root for $0<x<d$ is guaranteed, since $f(0)=-d^2h^2<0$ and $f(d)=d^2n^2w^2>0$ and $f$ is continous. |
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May 29 |
revised |
Refraction equation, quartic equation added 5 characters in body; edited tags |
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May 29 |
comment |
Refraction equation, quartic equation Edit to my previous comment: when starting with $x=0$, it converges to $-5.7$ (which means $n=-1.33$). The correct result can be obtained starting for example with $x=1$ or $x=12.5$. Do you have any idea how to choose the starting point which guarantees the convergence and the correct result? |
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May 29 |
comment |
Refraction equation, quartic equation Although it generally works well for $x=dh/(h+w)$, there is a problem for this. Starting with $x=25$ the result is getting close to $x=30$ while $x=30$ is not a root. The correct result can be obtained starting with $x=0$ which gives approx. $5.7$. Also I'm not sure if starting with $x=0$ always gives the correct result. |
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May 29 |
comment |
Refraction equation, quartic equation Thank you! I will check this method if it's accurate enough. |
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May 29 |
comment |
Refraction equation, quartic equation @joriki: Right, thanks for fixing it ;) |
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May 28 |
comment |
Refraction equation, quartic equation @Valentin: I need a solution for $n>1$ (or at least for $n=1.33$). I'd prefer to find an exact solution. |
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May 28 |
revised |
Refraction equation, quartic equation edited tags |
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May 28 |
awarded | Student |
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May 28 |
awarded | Editor |
