492 reputation
1310
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location Wellington, New Zealand
age
visits member for 2 years, 7 months
seen 2 hours ago

I am an undergraduate computer science and mathematics student in New Zealand. My fields of interest are computer graphics, in particular the physics of light transport, and to some extent cryptography, as well as programming and software development in general.


Jul
22
comment What's wrong with my aproach to solving this equation with multiple logarithms?
@Deepak I thought the brackets made it explicit enough. What would be ambiguous is e.g. $\log a^b$, which is where people either specify $\log (a^b)$, or $(\log a)^b$, or $\log(a)^b$ (or, equivalently, $\log^b a$).
Jul
13
comment Proving 7n+5 is never a cubic number?
Another interesting variant on this: we know $x^6 \equiv 1 \pmod{7}$, ignoring the case that $x$ is divisible by 7 for now, thus $x^3 \equiv \pm 1 \pmod{7}$ because $7$ is prime. But $5 \not \equiv \pm 1 \pmod{7}$.
Jul
5
comment How to distinguish walking on a sphere or on a torus?
@Martijn Staple it to his tail? (sorry Omnomnomnom...)
Jul
2
awarded  Curious
Jun
26
comment Subtraction by addition
@Joker_vD I still don't know what is so wrong with the traditional carry/borrow method that motivates educators to constantly come up with convoluted and crippled "algorithms" to perform such a simple task.
Jun
24
comment Liouville's theorem and holomorphic function
Thank you! $\left . \right. $
Jun
24
accepted Liouville's theorem and holomorphic function
Jun
24
asked Liouville's theorem and holomorphic function
Jun
22
comment Better Divisibility by 8
Good chunk sizes are in my opinion 16, 40, 200, $10^n$ for $n > 2$ (that last part is why you can ignore all but the last three digits). You will probably come to choose your own favorite chunks as you do this more often, and the operation will become second nature (you almost certainly already have mechanical algorithms burnt into your brain for division by 3, 9, and perhaps 11)
Jun
21
comment Find the last two digits of $9^{{9}^{9}}$
The Chinese remainder theorem might help, but this looks efficient enough already. You have a typo, by the way, the last equation should be mod 100 :)
Jun
13
comment Quotient group of $\mathbb{Z} \times \mathbb{Z}$
Ah! got it, it's clear now. Thanks again for your help!
Jun
13
comment Quotient group of $\mathbb{Z} \times \mathbb{Z}$
Thanks, that makes sense, the Smith normal form seems very handy. Just one question, to conclude that $(\mathbb{Z} \oplus \mathbb{Z}) / (\mathbb{Z} \oplus 2 \mathbb{Z}) = \mathbb{Z} / 2 \mathbb{Z}$ can you just cancel out $\mathbb{Z}$? I thought of the third isomorphism theorem but it isn't that and I don't think I've seen arithmetic done on direct products like this before.
Jun
13
accepted Quotient group of $\mathbb{Z} \times \mathbb{Z}$
Jun
13
comment Quotient group of $\mathbb{Z} \times \mathbb{Z}$
@DonAntonio Thanks, I think I understand the algorithm given in your document, it's really quite elegant. I have one more question but I will post it on your answer below...
Jun
13
comment Quotient group of $\mathbb{Z} \times \mathbb{Z}$
@DonAntonio No, we haven't studied it. Does it provide a systematic way of classifying such quotient groups?
Jun
13
comment Quotient group of $\mathbb{Z} \times \mathbb{Z}$
Nice approach for showing $(a, b)$ can only fall in two equivalence classes, I know about using algebra to rewrite things that way but occasionally get muddled up! Thanks!
Jun
13
comment Quotient group of $\mathbb{Z} \times \mathbb{Z}$
@AWertheim Well for instance if there was a better way to find the generators of the quotient group than completely working it out and then finding the elements which generate it?
Jun
13
revised Quotient group of $\mathbb{Z} \times \mathbb{Z}$
changed homomorphism to additive notation
Jun
13
asked Quotient group of $\mathbb{Z} \times \mathbb{Z}$
May
27
accepted Complex contour integral and partial fractions