Are calculators allowed in the 4u test
HSC 2017: MX2 -- EX1 -- ECO -- BUS --
Goal: Finance/Law/Maths/
Prove De Moivre's Theorem by Mathematical Induction where |z|=1
2016 Preliminary
Maths Extension • English Advanced • Biology • Legal Studies • Physics • PDHPE
2017 HSC
Maths Extension 2 • English Advanced • Biology • Physics • PDHPE
Atar aim: 97
Desired Course: Physiotherapy
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
For the first part about the limit:
The best I could come up with was that (based on what my teacher said) is that all exponential functions dominate over all polynomial functions which dominate over all log functions. Therefore, based on the second line of the statement, because both g(t) (the function g denotes that left hand expression just before log(t)) and f(t) = t^a both tend to 0, and f(t) dominates over log(t), it must follow that the limit is true because g(t) also tends to 0 and takes precedence over the log function.
Hopefully that made sense...
I'm still trying to practice on latex - I got the second part but I don't think it technically uses parts...
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
HSC 2017
Physics, Chemistry, Biology, English Advanced, Japanese Continuers
HSC 2016
Mathematics Extension 1, Mathematics Extension 2
And I probably did something wrong.
HSC 2017
Physics, Chemistry, Biology, English Advanced, Japanese Continuers
HSC 2016
Mathematics Extension 1, Mathematics Extension 2
You really don't need to split based upon parity. It only makes things worse.
You haven't considered
enough.
Side note your conclusion is wrong.
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
Here's a nice anecdote.
"One day Gauss' teacher asked his class to add together all the numbers from 1 to 100, assuming that this task would occupy them for quite a while. He was shocked when young Gauss, after a few seconds thought, wrote down the answer 5050. The teacher couldn't understand how his pupil had calculated the sum so quickly in his head, but the eight year old Gauss pointed out that the problem was actually quite simple.
He had added the numbers in pairs - the first and the last, the second and the second to last and so on, observing that 1+100=101, 2+99=101, 3+98=101, ...so the total would be 50 lots of 101, which is 5050"
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
Damn it...I was doing that
Binomial Question.pdf
I think I got it, but it looks dodgy...
For odd n, I don't think your S_2 is an integer. I think you don't need that subtraction by 1.
Also note that you can evaluate the sums C(n,k)^2 and kC(n,k)^2 combinatorially. (This takes less effort than other methods of computation imo.)
These sums amount to:
A: The number of ways of choosing a team of n people out of 2n people, n of whom are in year 12 and n of whom are in year 11.
This can be done in C(2n,n) ways.
B: Same as above, but also selecting one of the year 12 students to be team captain.
This can be done in nC(2n,n)/2 ways. (The number of ways of choosing a team and specifying a captain is clearly nC(2n,n). By symmetry, this captain will be in year 12 exactly half of the time.)
Currently studying:
PhD (Pure Mathematics) at ANU
Note that the above yields an end answer of something like
Busy prepping for a meeting right now but paradoxica can confirm if that is what he was looking for.
Currently studying:
PhD (Pure Mathematics) at ANU
Yeah...I realized my mistake. My bad...
For z^5-1=0, the roots are 1, w^2-w^4, and
z^5-1=(z-1)(z-cis2pi/5)(z-cis-2pi/5)(z-cis4pi/5)(z-cis-4pi/5) How do you get back to the LHS from expanding the RHS?
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