meta user
network profile
| bio | website | xcrypt-devblog.blogspot.be |
|---|---|---|
| location | Belgium | |
| age | 21 | |
| visits | member for | 1 year, 6 months |
| seen | Feb 5 at 16:39 | |
| stats | profile views | 17 |
I have always been fascinated by games. At the age of 18, I started to get into game programming with C++ and DirectX. I have been working on various little game projects since then.
I decided to take a game development course, but I soon found out how misleading those courses are. In the meantime, I got more and more interested in game physics. At the age of 21 I went to Ghent University, taking a BSc mathematics course intended to be followed by a MSc mathematical computer science. My goal is to create my own game physics engine in C++ and to become a game physics programmer.
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Feb 5 |
awarded | Informed |
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Feb 3 |
comment |
insight on lemma on linear independence and spanning set @atricolf because we take the $a_{ij}$ from the $w_i = \sum_j^n{a_{ij}v_j}$, then we choose the unkowns $c_i$ such that the system of linear equations holds. In the end we don't assume that $\sum_{i=1}^{n+1}c_iw_i = 0$. We just end up with $0$ because of the simplification. This is where I seemed to go wrong for hours. |
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Feb 3 |
accepted | insight on lemma on linear independence and spanning set |
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Feb 3 |
comment |
insight on lemma on linear independence and spanning set I think you're assuming $\sum_{i=1}^{n+1}{c_ia_{ij}} = 0$ because we looked at the system of linear equations above. But it never stated $\sum_{i=1}^{n+1}{c_ia_{ij}} = 0$. If we assume that we take exactly those $c_i, a_{ij}$ such that the system holds, it makes a little more sense. Hmm, I think I'm starting to understand now. |
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Feb 3 |
comment |
insight on lemma on linear independence and spanning set I don't understand why $\sum_{i=1}^{n+1}c_ia_{ij} = 0$. I also don't understand why $c_1,\ldots,c_{n+1}$ has atleast one $c_i \neq 0$. I understand all the other steps. |
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Feb 3 |
revised |
insight on lemma on linear independence and spanning set added 392 characters in body |
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Feb 3 |
revised |
insight on lemma on linear independence and spanning set added 392 characters in body |
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Feb 3 |
asked | insight on lemma on linear independence and spanning set |
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Jan 26 |
accepted | Question on polynomial rings |
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Jan 26 |
comment |
Question on polynomial rings Oh, that makes sense. Thanks |
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Jan 26 |
comment |
Question on polynomial rings @K.Stm. I'm not sure I haven't seen factor rings yet. I think it's not closed under multiplication because all elements in the set are of degree $\leq$ 1. When you multiply two monomials, there is no closure. |
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Jan 26 |
asked | Question on polynomial rings |
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Jan 24 |
comment |
How much time is too much (to put into a single problem)? Thank you - I really like your answer. I am also having trouble with this and certainly going to give this a try. |
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Jan 16 |
comment |
Mean value theorem implies first fundamental theorem of calculus? @dwarandae No in our book both these theorems get proven before the first fundamental theorem of calculus |
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Jan 16 |
comment |
Mean value theorem implies first fundamental theorem of calculus? Ah @gnometorule that will most likely be the reason. Variables with same names but different meanings get so confusing... Thanks. But maybe your comment should be an answer :) |
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Jan 16 |
comment |
Mean value theorem implies first fundamental theorem of calculus? @dwarandae what is fftc? |
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Jan 16 |
asked | Mean value theorem implies first fundamental theorem of calculus? |
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Jan 12 |
accepted | limit of absolute values |
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Jan 12 |
comment |
limit of absolute values Because then you generalized all cases with 1 equation/expression |
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Jan 12 |
comment |
limit of absolute values Hmm, this looks correct, but I was hoping for something that would steer me in the direction of |x|/x |
