Author Topic: 4U Maths Question Thread (Read 667570 times) Tweet Share

RuiAce · « **Reply #690 on:** October 21, 2016, 05:09:45 am »

Quote from: katherine123 on October 21, 2016, 03:47:04 am

how to do part ii)

$\text{Note that }\\ \begin{align*}&\quad (1+x)^n+(1-x)^n\\ &= \left[\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+\binom{n}{3}x^3+\binom{n}{4}x^4+\dots+\binom{n}{n}x^n\right]\\ &+\left[ \binom{n}{0}-\binom{n}{1}x+\binom{n}{2}x^2-\binom{n}{3}x^3+\binom{n}{4}x^4-\dots+\binom{n}{n}x^n\right]\\ &= 2\left[ \binom{n}{0}+\binom{n}{2}x^2+\binom{n}{4}x^4+\dots+\binom{n}{n}x^n\right]\end{align*}$
$\text{The reason why we know that the last term is positive is because }n\text{ is divisible by 4}\\ \text{i.e. }n\text{ is even}\\ \text{If }n\text{ were to be odd, the last term is }\binom{n}{n-1}x^{n-1}$
$\text{So letting }x=i\text{ and noting that }i^2=-1, i^4=1\\ (1+i)^n+(1-i)^n= 2\left[ \binom{n}{0}-\binom{n}{2}+\binom{n}{4}-\binom{n}{6}+\dots+\binom{n}{n}\right]$
________________________
$\text{That basically clears out the LHS. Now we look to the RHS.}\\ \text{Since }n\text{ is divisible by 4, we can establish an identity. Let }n=4k:\\ \text{Then this bit was basically covered above:}\\ \cos \frac{n\pi}{4} = \cos k\pi = (-1)^k = (-1)^\frac{n}{4}$
$\text{So upon equating:}\\ 2\left[\binom{n}{0}-\binom{n}{2}+\binom{n}{4}-\binom{n}{6}+\dots+\binom{n}{n}\right]=2(\sqrt{2})^n(-1)^{\frac{n}{4}}\\\text{Final result follows from just cancelling the 2.}$

karyn.mcdonald · « **Reply #691 on:** October 21, 2016, 10:02:27 am »

hello... just wondering if someone could help me...

I always struggle knowing whether to use radians or degrees in certain situations>? how do I distinguish coz i'd HATE to lose a mark just for not typing it into my calculator!!

thnx

RuiAce · « **Reply #692 on:** October 21, 2016, 10:11:07 am »

Quote from: karyn.mcdonald on October 21, 2016, 10:02:27 am

hello... just wondering if someone could help me...

I always struggle knowing whether to use radians or degrees in certain situations>? how do I distinguish coz i'd HATE to lose a mark just for not typing it into my calculator!!

thnx

You ALWAYS use radians unless you see that circle ^o floating in the air. Only when that circle is there do you use degrees.

Never assume degrees when there's no symbols - radians is the standard in mathematics.

kb123 · « **Reply #693 on:** October 25, 2016, 05:56:39 pm »

I don't understand this question.... Help please??

Show that | |z1|-|z2| | < |z1+z2|. State the condition for
equality to hold.

RuiAce · « **Reply #694 on:** October 25, 2016, 06:03:10 pm »

Quote from: kb123 on October 25, 2016, 05:56:39 pm

I don't understand this question.... Help please??

Show that | |z1|-|z2| | < |z1+z2|. State the condition for
equality to hold.

Is there a typo here. Equality never holds for what you gave and I know two possible questions you intended to give.

Edit actually it does hold if z2=0 but that makes me think the question has a typo more now

kb123 · « **Reply #695 on:** October 25, 2016, 07:48:10 pm »

Quote from: RuiAce on October 25, 2016, 06:03:10 pm

Is there a typo here. Equality never holds for what you gave and I know two possible questions you intended to give.

Edit actually it does hold if z2=0 but that makes me think the question has a typo more now

its just supposed to be less than or equal to, but otherwise there aren't any typos

birdwing341 · « **Reply #696 on:** October 25, 2016, 07:51:44 pm »

Quote from: RuiAce on October 25, 2016, 06:03:10 pm

Is there a typo here. Equality never holds for what you gave and I know two possible questions you intended to give.

Edit actually it does hold if z2=0 but that makes me think the question has a typo more now

Looks like equality holds if z2=-z1

RuiAce · « **Reply #697 on:** October 25, 2016, 07:53:43 pm »

Quote from: kb123 on October 25, 2016, 07:48:10 pm

its just supposed to be less than or equal to, but otherwise there aren't any typos

$\textbf{I highly doubt the authenticity of a minus sign in the LHS and plus in the RHS}$
$\text{Because since }|z_2| \ge 0\\ \begin{align*}2|z_2|&\ge 0\\ |z_1|+2|z_2|&\ge |z_1|\\ \therefore |z_1|-|z_2|&\le |z_1|+|z_2|\\ \therefore |z_1|-|z_2|&\le ||z_1|+|z_2||\\ \therefore ||z_1|-|z_2||&\le |z_1+z_2|\end{align*}$
$\text{i.e. this question is common sense.}$

Quote from: birdwing341 on October 25, 2016, 07:51:44 pm

Looks like equality holds if z2=-z1

Oh right, true.

Equality if z2=-z1 or any one of z1, z2 being equal to 0

kb123 · « **Reply #698 on:** October 25, 2016, 08:06:54 pm »

Quote from: RuiAce on October 25, 2016, 07:53:43 pm

$\textbf{I highly doubt the authenticity of a minus sign in the LHS and plus in the RHS}$
$\text{Because since }|z_2| \ge 0\\ \begin{align*}2|z_2|&\ge 0\\ |z_1|+2|z_2|&\ge |z_1|\\ \therefore |z_1|-|z_2|&\le |z_1|+|z_2|\\ \therefore |z_1|-|z_2|&\le ||z_1|+|z_2||\\ \therefore ||z_1|-|z_2||&\le |z_1+z_2|\end{align*}$
$\text{i.e. this question is common sense.}$ Oh right, true.

Equality if z2=-z1 or any one of z1, z2 being equal to 0

I'm sorry I only started ext 2 maths last week, could you explain how you got from the second last step to the last step?

RuiAce · « **Reply #699 on:** October 25, 2016, 08:20:19 pm »

Oh I see. Sorry, entirely my fault, must've been too hungry because I made a false assumption.
$\text{The correct way to prove these things is to treat it geometrically.}\\ \text{This is in general for anything that is, or is similar to the triangle inequality.}$
$\text{Draw a triangle with edges }|a|, |b|\text{ and }|a+b|\\ \text{Note that }|a+b|\text{ can be drawn by vector addition.}$
$\text{The triangle inequality asserts that }|a|+|b|\ge |a+b|\\ \text{This is because the sum of any two side lengths in a triangle is always greater than or equal to the third.}$
$\text{Let }a = z_2\text{ and }b=z_1-z_2\\ \text{Then }|z_2|+|z_1+z_2| \ge |z_1|\\ \text{i.e. }|z_1|-|z_2|\le |z_1+z_2|$
$\text{So we just plonk absolute valuse around it. On the RHS, note that double absolute values just revert back to single.}\\ ||z_1|-|z_2|| \le ||z_1+z_2|| = |z_1+z_2|$
$\text{An assumption was made without loss of generality that }|z_1| \le |z_2|\\ \text{If }|z_2| < |z_1|\text{ then the proof is identical, except}\\ \text{let }a=z_1\text{ and }b=z_2-z_1$

kb123 · « **Reply #700 on:** October 25, 2016, 08:50:42 pm »

Quote from: RuiAce on October 25, 2016, 08:20:19 pm

Oh I see. Sorry, entirely my fault, must've been too hungry because I made a false assumption.
$\text{The correct way to prove these things is to treat it geometrically.}\\ \text{This is in general for anything that is, or is similar to the triangle inequality.}$
$\text{Draw a triangle with edges }|a|, |b|\text{ and }|a+b|\\ \text{Note that }|a+b|\text{ can be drawn by vector addition.}$
$\text{The triangle inequality asserts that }|a|+|b|\ge |a+b|\\ \text{This is because the sum of any two side lengths in a triangle is always greater than or equal to the third.}$
$\text{Let }a = z_2\text{ and }b=z_1-z_2\\ \text{Then }|z_2|+|z_1+z_2| \ge |z_1|\\ \text{i.e. }|z_1|-|z_2|\le |z_1+z_2|$
$\text{So we just plonk absolute valuse around it. On the RHS, note that double absolute values just revert back to single.}\\ ||z_1|-|z_2|| \le ||z_1+z_2|| = |z_1+z_2|$
$\text{An assumption was made without loss of generality that }|z_1| \le |z_2|\\ \text{If }|z_2| < |z_1|\text{ then the proof is identical, except}\\ \text{let }a=z_1\text{ and }b=z_2-z_1$

after you let a=z2 and b = z1 - z2

how come you got I z1 + z2 I in the second line??

RuiAce · « **Reply #701 on:** October 25, 2016, 09:12:02 pm »

I must be having a really bad day.

I just did it by paper. I knew what I was doing but I articulated it wrong every single time. Terrible impression.

Seriously so sorry for the hassle.

kb123 · « **Reply #702 on:** October 26, 2016, 03:52:07 pm »

Quote from: RuiAce on October 25, 2016, 09:12:02 pm

I must be having a really bad day.

I just did it by paper. I knew what I was doing but I articulated it wrong every single time. Terrible impression.
(Image removed from quote.)
Seriously so sorry for the hassle.

All good, my teacher explained it for me today

RuiAce · « **Reply #703 on:** October 26, 2016, 08:57:21 pm »

Quote from: kb123 on October 26, 2016, 03:52:07 pm

All good, my teacher explained it for me today

Awesome (y)

Sorry again though; feel a bit bad having a new MX2 student suffer the stupidity invoked by my bad day yesterday.

kiwiberry · « **Reply #704 on:** October 29, 2016, 01:56:00 pm »

how do you do this 😁

if a and b are two non-zero complex numbers, show that if a/b = ik,
(conjugate of a)b + (conjugate of b)a = 0

sorry I'm not sure how to type the conjugate sign hahah

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