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March 19, 2024, 09:54:17 pm

Author Topic: 4U Maths Question Thread  (Read 658459 times)  Share 

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envisagator

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Re: 4U Maths Question Thread
« Reply #1935 on: July 16, 2018, 10:29:25 pm »
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Fairly sure you can continue with this question once I give you what to assume.
[tex]\text{You know that }u_{k+1} = 4u_k - 3u_{k-1}. Is this statement just subbing in k+1 into the 'fact' that shows how each term in the relation is created?, and do you choose to sub in k+1 because you know you are going to use that to prove in the final step??. Thank You!!
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RuiAce

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Re: 4U Maths Question Thread
« Reply #1936 on: July 16, 2018, 10:35:22 pm »
+3
Fairly sure you can continue with this question once I give you what to assume.
[tex]\text{You know that }u_{k+1} = 4u_k - 3u_{k-1}.
Is this statement just subbing in k+1 into the 'fact' that shows how each term in the relation is created?, and do you choose to sub in k+1 because you know you are going to use that to prove in the final step??. Thank You!!
Yeah well you're meant to prove it holds when \(n=k+1\) so you're trying to prove that \(u_{k+1} = 2+3^{k+1} \).

But the recurrence relation was defined to be \(u_n = 4u_{n-1} - 3u_{n-2}\). Of course you can assume what you've defined it to be in the first place
« Last Edit: July 16, 2018, 10:47:06 pm by RuiAce »

kaustubh.patel

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Re: 4U Maths Question Thread
« Reply #1937 on: July 16, 2018, 11:44:39 pm »
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hey Rui a bit of help please with some harder inequalities.

RuiAce

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Re: 4U Maths Question Thread
« Reply #1938 on: July 17, 2018, 06:46:18 am »
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hey Rui a bit of help please with some harder inequalities.
What are the conditions on a, b and c?

kaustubh.patel

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Re: 4U Maths Question Thread
« Reply #1939 on: July 17, 2018, 11:05:57 am »
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Oh sorry a, b and c are all positive integers

RuiAce

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Re: 4U Maths Question Thread
« Reply #1940 on: July 17, 2018, 03:51:54 pm »
+3
Without any precursor assumptions to build on I really could not find any intuitive way of doing this, so I just resorted to working backwards.



« Last Edit: July 18, 2018, 07:33:25 am by RuiAce »

envisagator

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Re: 4U Maths Question Thread
« Reply #1941 on: July 18, 2018, 04:38:25 pm »
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Hi Rui, is relying on Cauchy's inequalitity AM>= GM too much to prove inequalities a bad habit or should i considering using proofs previously done and expanding on it to proof another result, idk if you understand by what i meant. Plus, when I state i have used it is just saying AM>= GM fine??. Thanks in advance!!
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RuiAce

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Re: 4U Maths Question Thread
« Reply #1942 on: July 19, 2018, 02:16:56 am »
+2
Hi Rui, is relying on Cauchy's inequalitity AM>= GM too much to prove inequalities a bad habit or should i considering using proofs previously done and expanding on it to proof another result, idk if you understand by what i meant. Plus, when I state i have used it is just saying AM>= GM fine??. Thanks in advance!!
Your questions at Terry Lee coaching are kinda rigged so that you have no choice but to use them. In the actual HSC you'll only need it if you prove it beforehand. Preferably, yeah, always use what you've already proven to prove new stuff.

kaustubh.patel

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Re: 4U Maths Question Thread
« Reply #1943 on: July 19, 2018, 04:47:44 pm »
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Hey rui please some help with polynomials
solve 4x^5+x =0 over the complex field
« Last Edit: July 19, 2018, 04:53:33 pm by kaustubh.patel »

RuiAce

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Re: 4U Maths Question Thread
« Reply #1944 on: July 19, 2018, 05:53:26 pm »
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Hey rui please some help with polynomials
solve 4x^5+x =0 over the complex field
\begin{align*}4x^5+x&=0\\ x(4x^4+1)&=0\\ x(2x^2+i)(2x^2-i)&=0\end{align*}
\[ \text{The obvious solution is }x=0\\ \text{We need to then solve }\boxed{x^2 = \pm \frac{i}{2}}\\ \text{which can be done using any one of the usual approaches.} \]
Rewrite \(x = r \text{cis}\theta\), so that \(x^2 = \frac{i}{2}\) can be rewritten as \( r^2 \text{cis}2\theta = \frac12 \text{cis} \frac\pi2 \). This gives the solutions \(x = \frac{1}{\sqrt2}\text{cis}\frac\pi4\), \(x=\frac{1}{\sqrt2}\text{ cis }\left(-\frac{5\pi}{4}\right) \). Then do the same thing for \(x^2 = -\frac{i}{2} \).

Alternatively, just solve \(4x^4+1=0\), i.e. \(x^4 = -\frac14\), using the same approach as above.

yammy

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Re: 4U Maths Question Thread
« Reply #1945 on: July 22, 2018, 02:02:30 am »
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Hey Rui,
Can you please help me with question 3 and 4?

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Re: 4U Maths Question Thread
« Reply #1946 on: July 22, 2018, 07:38:06 am »
+1
Hey Rui,
Can you please help me with question 3 and 4?

With 3, sketch your locus first before you do anything with it.. with q4 you have to use by parts
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RuiAce

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Re: 4U Maths Question Thread
« Reply #1947 on: July 22, 2018, 07:52:51 am »
+4
Hey Rui,
Can you please help me with question 3 and 4?
Sketch solution for Q3: That region isn't even gonna be a region. It's literally just gonna be a tiny segment of the ray. First, verify that the ray \( \arg z = \frac\pi3\) and the circle \( |z|=2\) intersect at \(z = 2 \text{cis} \frac\pi3\). (Recall from my lecture that we only need to consider the boundary, which is why we don't care too much about the interior of the circle; \(|z|<2\).)

Then, because you know the locus of \(z\), you also know the locus of \(z-i\). Reason being that \(-i\) is just a translation downwards by one unit.

Since the endpoints of the locus of \(z\) were at \(0\) with an open circle and \(2\text{cis}\frac\pi3\) with a closed circle, the endpoints of the locus of \(z-i\) are at \(-i\) and \(2\text{cis} \frac\pi3-i\). So you just have to find the arguments of these two complex numbers to know the extremities. Of course, you will need to expand out that cis into Cartesian form beforehand.


I will look at Q4 later, but my guess would to be start with \(U_n = \int \frac{dx}{(x^2+1)^{n-1}} \) and then take the IBP suggestion, differentiating \( \frac{1}{(x^2+1)^{n-1}} \) and integrating \(1\).
« Last Edit: July 22, 2018, 08:37:24 am by RuiAce »

RuiAce

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Re: 4U Maths Question Thread
« Reply #1948 on: July 24, 2018, 11:12:23 am »
+2
Sorry, forgot to back to this. But I couldn't get what they got. Their a) is obviously wrong, because there's no reason for the power in the bottom to be \(n\), but it looks like everything in b) is inverted as well..
\begin{align*}U_{n-1} &= \int \frac{dx}{(x^2+1)^{n-1}}\\ &= \frac{x}{(x^2+1)^{n-1}}-\int x\cdot \frac{2(-n+1)x}{(x^2+1)^n}\\ &= \frac{x}{(x^2+1)^{n-1}}-\int \frac{2x^2(1-n)}{(x^2+1)^n}\,dx \end{align*}

Subject to some minor computational error

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Re: 4U Maths Question Thread
« Reply #1949 on: July 24, 2018, 08:00:49 pm »
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Hi Rui, for the inequality proof question posted as an image below, I've been looking at it for a while, just cant see the actually logical reason to why the part highlighted in the box is true. Plus, when i do some questions, is it 'normal' just to sorta guess how some of your facts are bigger or smaller than another fact such that when you combine them you can get easily get to the desired result. I dont know whether that makes sense, also is it necessary to write equality iff - bla bla bla?? Thank You in advance, sorry for it being a mouthful to take in
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