leehuan
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First Year Mathematics B (Integration, Series, Discrete Maths & Modelling) (3 Viewers)
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Flop21
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Re: MATH1231/1241/1251 SOS Thread
2 questions about marking.
1. Will I lose marks if I just write down the limit in questions asking for 'does this series converge, if so find its limit'?
2. Will I lose marks if I don't show fully working when row reducing in questions where it's just a step, e.g. finding eigenvectors I can just sometimes do it my head and will just write the final form.
2 questions about marking.
1. Will I lose marks if I just write down the limit in questions asking for 'does this series converge, if so find its limit'?
2. Will I lose marks if I don't show fully working when row reducing in questions where it's just a step, e.g. finding eigenvectors I can just sometimes do it my head and will just write the final form.
leehuan
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Re: MATH1231/1241/1251 SOS Thread
But with 2 I also skip row reductions if it's that obvious. I think my tutor does as well.
I work around it by saying matrix*x = 0 therefore x = my answer
Well if you want to find the limit of a series, unless it's that obvious how do you plan to do it by inspection though
But with 2 I also skip row reductions if it's that obvious. I think my tutor does as well.
I work around it by saying matrix*x = 0 therefore x = my answer
leehuan
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Re: MATH1231/1241/1251 SOS Thread
Good thing is that they're like we WANT to give you marks. So they don't have mark allocations like in the HSC so that they aren't limiting themselves.
So I reckon if you jumped some steps but didn't jump too excessively to get the right answer you'd get it. (But of course if you jumped too many steps only to get the wrong answer well that's another story)
Good thing is that they're like we WANT to give you marks. So they don't have mark allocations like in the HSC so that they aren't limiting themselves.
So I reckon if you jumped some steps but didn't jump too excessively to get the right answer you'd get it. (But of course if you jumped too many steps only to get the wrong answer well that's another story)
Re: MATH1231/1241/1251 SOS Thread
Yeah that's correct.(for the same linearly independence qn)
Another friend wants to know if there's fault in what he did and I can't really communicate anything
Re: MATH1231/1241/1251 SOS Thread
. We generally define (for $z$ and $c$ complex and $z \neq 0$) $z^c$ as $e^{c \log z}$, where $\log$ is the complex logarithm and is \textsl{multivalued} (so powers are generally multivalued). For complex numbers $z = re^{i\theta}$ ($r > 0$, $\theta \in \mathbb{R}$), we define $\log z = \ln r + i\theta + 2\pi i n$ ($n\in \mathbb{Z}$), where $\ln $ is the usual real logarithm.$)
 = \color{red}{\frac{\pi}{2}}\color{black}{}\Big{)}$. Hence $i^i = e^{i\left(i\frac{\pi}{2} + 2\pi i n \right)} = e^{-\frac{\pi}{2} - 2\pi n}$, $n\in \mathbb{Z}$. If we take $n = 0$, this gives us the \textbf{principal value} (corresponding to the \textsl{principal logarithm} being used -- you can search online for more info on this) of $e^{-\frac{\pi}{2}}$.$)
I could probably research it up but I feel as though InteGrand's answers are more comprehensible...
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Drsoccerball
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Re: MATH1231/1241/1251 SOS Thread
Spending too much time on mathematical memes ?I could probably research it up but I feel as though InteGrand's answers are more comprehensible...
FritzTheCat
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Is everyone still doing this? I'd very much like to participate.
Here's a problem I've been stuck on for a little while. Any takers?
Here's a problem I've been stuck on for a little while. Any takers?
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sida1049
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Consider the equation \lambda_1.v_1 + ... \lambda_m.v_m = 0. We need to show that all coefficients are zero. To do this, take the dot product of v_i to both sides to show that for arbitrary i = 1, 2, ... , m, \lambda_i = 0 (by the property of orthogonality). And we're done.
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