ATAR Notes: Forum

Uni Stuff => Science => Faculties => Mathematics => Topic started by: Odette on April 07, 2008, 12:52:49 pm

Title: Maths Questions ^_^
Post by: Odette on April 07, 2008, 12:52:49 pm: Ok since I seem to have a few questions regarding maths, from now on i'll post them up in this thread.
Here's one i had a bit of trouble with:
1.Given that [x^3 - 3x^2 - 4x +16 = (well its actually 3 -) (x-2) (x+3) (x-c) + Px + Q], Find the values of P, Q and c.
Oh and what does the 3 -s mean? (Hope that made sense)
Thanks in advance :) ^_^

Edit: by the 3 -s i mean ≡ Lol
Title: Re: Maths Questions ^_^
Post by: AppleXY on April 07, 2008, 03:37:02 pm: Umm. Maybe you should word your question better.

Did you mean (well actually 3 - s) ? :)

But I don't know where the 3 - s came from, unless if it's a factor.
Title: Re: Maths Questions ^_^
Post by: Odette on April 07, 2008, 04:07:02 pm: No i meant ≡ Lol just couldnt find the symbol for it =P
Title: Re: Maths Questions ^_^
Post by: Odette on April 07, 2008, 04:27:11 pm: Oh new question...
2.a) Three of the factors of x^4 + ax^3 + bx^2 + x +c are x, x+1 and x-1. Find a,b and c...
b) Write down an expression of the form x^3 + px^2 + qx + r which gives a remainder 4 when divided by x, x-2 or x+3.

Thanks ^_^
Title: Re: Maths Questions ^_^
Post by: /0 on April 07, 2008, 05:06:10 pm: Quote from: Odette on April 07, 2008, 12:52:49 pm
Ok since I seem to have a few questions regarding maths, from now on i'll post them up in this thread.
Here's one i had a bit of trouble with:
1.Given that [x^3 - 3x^2 - 4x +16 = (well its actually 3 -) (x-2) (x+3) (x-c) + Px + Q], Find the values of P, Q and c.
Oh and what does the 3 -s mean? (Hope that made sense)
Thanks in advance :) ^_^

Edit: by the 3 -s i mean ≡ Lol

The $\equiv$ (\equiv) symbol means that the LHS = RHS for all x. It is called an identity. It is unlike an equation in that you can't solve for x.

An easy way is to expand the RHS and group terms:

1

$x^3-3x^2-4x+16 \equiv (x^3-cx^2+x^2-cx-6x+6c) + Px+Q$

$x^3 - 3x^2 - 4x+16 \equiv x^3 +(1-c)x^2 + (P-c-6)x+Q+6c$

Then you get simultaneous equations:

$-3=1-c$

$-4=P-c-6$

$16=Q+6c$

And you can solve :)
Title: Re: Maths Questions ^_^
Post by: Odette on April 07, 2008, 05:29:33 pm: Oh thank you! ^_^ I get it now lolz + 1 coming your way
Title: Re: Maths Questions ^_^
Post by: AppleXY on April 08, 2008, 05:58:49 pm: yeah, $\equiv$ means congruent to or "equal", as DB0 said, "identity".
Title: Re: Maths Questions ^_^
Post by: Odette on April 24, 2008, 10:45:10 pm: Ok here's one i've started to solve but didnt quite get the answer for...
(1 + tanx ) ^2
I ended up with 1 + 2tanx + tan ^2 x ...
How do you simplify that even further?
Title: Re: Maths Questions ^_^
Post by: Collin Li on April 24, 2008, 10:46:36 pm: $\sec^2x + 2\tan x$ , since $1+\tan^2x = \sec^2x$

$= \sec x(\sec x + 2\sin x)$

That's all I can think of.
Title: Re: Maths Questions ^_^
Post by: Odette on April 24, 2008, 10:50:25 pm: ahh thanks ! i didnt know 1 + tan ^2 x = sec ^2 x (i'm new to this lolz)

Oh another question i was a little confused about:
sin ( pi/2 - x) [secx - cosx]

P.S i dont know how to use latex or whatever its called.
Title: Re: Maths Questions ^_^
Post by: Collin Li on April 24, 2008, 10:58:25 pm: Quote from: Odette on April 24, 2008, 10:50:25 pm
ahh thanks ! i didnt know 1 + tan ^2 x = sec ^2 x (i'm new to this lolz)

It comes from $\sin^2x + \cos^2x = 1$ , and dividing both sides by $\cos^2x$ . :)

Quote from: Odette on April 24, 2008, 10:50:25 pm
Oh another question i was a little confused about:
sin ( pi/2 - x) [secx - cosx]

P.S i dont know how to use latex or whatever its called.

You can simplify $\sin\left(\frac{\pi}{2}-x\right)$ into $\cos x$ . This can be shown geometrically by drawing any right-angled triangle with an angle $x$ . The other angle must be $\frac{\pi}{2} - x$ , since a triangle has angles adding up to $\pi$ . I've only briefly outlined this, so if you don't get it, ask about it.

This should now yield: $\cos x(\sec x - \cos x) = 1 - \cos^2x = \sin^2x$

LaTeX is pretty straightforward. Here are some links, give it a go - you learn best by experimenting:
http://vcenotes.com/forum/index.php/topic,864.0.html
http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php
Title: Re: Maths Questions ^_^
Post by: Odette on April 24, 2008, 11:18:23 pm: Ah does that have anything to do with the sine and cosine graphs being similar except one is moved to the right pi/2 units? (i dunno something the lecturer mentioned today)

And thanks again ^_^ I appreciate it
Title: Re: Maths Questions ^_^
Post by: Collin Li on April 24, 2008, 11:21:31 pm: Quote from: Odette on April 24, 2008, 11:18:23 pm
Ah does that have anything to do with the sine and cosine graphs being similar except one is moved to the right pi/2 units? (i dunno something the lecturer mentioned today)

Yeah, that's another way you can think of it. You should try the right-angled triangle method though, it's rather nice.
Title: Re: Maths Questions ^_^
Post by: Odette on April 24, 2008, 11:22:52 pm: Quote from: coblin on April 24, 2008, 11:21:31 pm
Quote from: Odette on April 24, 2008, 11:18:23 pm
Ah does that have anything to do with the sine and cosine graphs being similar except one is moved to the right pi/2 units? (i dunno something the lecturer mentioned today)

Yeah, that's another way you can think of it. You should try the right-angled triangle method though, it's rather nice.

Ok, um any link for that one? I've never heard of that method, so i'm not too sure what you're talking about.
Title: Re: Maths Questions ^_^
Post by: Odette on April 24, 2008, 11:39:00 pm: Ah never mind i did some research and found out what you were talking about :)
Thanks again coblin =) it's pretty straight forward now. Oh and i'll practice using LaTeX from now on :)
Title: Re: Maths Questions ^_^
Post by: Odette on April 27, 2008, 01:19:36 pm: Ok i'm a little confused about this, so any help will be appreciated... Oh and an explanation of where they came from would be good..
Deduce the formula in terms of t
$\sin \theta$ = $\frac {2t} {1 + t^2}$
$\cos \theta$ = $\frac {1-t^2} {1+t^2}$
Where t = $\tan \frac{\theta} {2}$

P.S this is my first time using LaTeX
Title: Re: Maths Questions ^_^
Post by: Mao on April 27, 2008, 03:54:05 pm: what is the question asking....? =\
Title: Re: Maths Questions ^_^
Post by: Odette on April 27, 2008, 04:32:27 pm: Oops lol i just realised its already in terms of t.... haha um, my question really should be where did this formula come from?
Title: Re: Maths Questions ^_^
Post by: dcc on April 27, 2008, 04:46:12 pm: $\sin \theta = \dfrac{2t}{1 + t^2}$

where $t = \tan \dfrac{\theta}{2}$

RHS:

$\dfrac{2t}{1+t^2} = \dfrac{2\tan \dfrac{\theta}{2}}{1 + \tan^2 \dfrac{\theta}{2}}$

$\implies \dfrac{2\dfrac{\sin \dfrac{\theta}{2}}{\cos \dfrac{\theta}{2}}}{\mbox{sec}^2 \dfrac{\theta}{2}}$ (since $1 + \tan^2x = sec^2x$ )

$\implies 2\dfrac{\sin \dfrac{\theta}{2}}{\cos \dfrac{\theta}{2}} \cdot \cos^2 \dfrac{\theta}{2}$ (since $\dfrac{1}{\mbox{sec}^2x} = \cos^2 x$ )

$\implies 2\sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2} = \sin \theta$ (since $\sin 2u = 2\sin u \cos u$ )
Title: Re: Maths Questions ^_^
Post by: Odette on April 27, 2008, 04:50:21 pm: Thanks, but i'm completely lost
Title: Re: Maths Questions ^_^
Post by: Mao on April 27, 2008, 04:50:41 pm: so, you are given that:

$\tan\left(\frac{\theta}{2}\right) = t$

that is, $\tan\left(\frac{\theta}{2}\right) = \frac{Opposite}{Adjacent} = \frac{t}{1}$

if we represent this in a triangle:

(http://obsolete-chaos.wikispaces.com/space/showimage/odette_tan.gif)

the hypotenuse, by pythagoras's theorem, is $\sqrt{1+t^2}$

sine is opposite/hypotenuse, so

$\sin \left( \frac{\theta}{2}\right) = \frac{t}{\sqrt{1+t^2}}$

cosine is then

$\cos \left( \frac{\theta}{2}\right) = \frac{1}{\sqrt{1+t^2}}$

to find $\sin(\theta)$ , we use the double angle formula:

$\sin(\theta) = \sin\left(2\cdot \frac{\theta}{2}\right)$

$\sin(\theta) = 2\cdot \sin\left(\frac{\theta}{2}\right) \cdot \cos\left(\frac{\theta}{2}\right)$

$\sin(\theta) = 2 \cdot \frac{t}{\sqrt{1+t^2}} \cdot \frac{1}{\sqrt{1+t^2}}$

$\sin(\theta) =\frac{2t}{1+t^2}$

$\cos(\theta) = \cos\left(2\cdot \frac{\theta}{2}\right)$

$\cos(\theta) = 2\cdot \cos^2\left(\frac{\theta}{2}\right) -1$

$\cos(\theta) = \frac{2}{1+t^2}-1$

$\cos(\theta) = \frac{2-\left(1+t^2\right)}{1+t^2}$

$\cos(\theta) = \frac{1-t^2}{1+t^2}$
Title: Re: Maths Questions ^_^
Post by: dcc on April 27, 2008, 05:03:17 pm: $\cos \theta &= \dfrac{1 - t^2}{1 + t^2}$

$\begin{align*} RHS &= \dfrac{1 - t^2}{1 + t^2}\\ &= \dfrac{1 - \tan^2 \dfrac{\theta}{2}}{1 + \tan^2 \dfrac{\theta}{2}}\\ &= \dfrac{1 - \tan^2 \dfrac{\theta}{2}}{\mbox{sec}^2 \dfrac{\theta}{2}}\\ &= (1 - \tan^2 \dfrac{\theta}{2})(\cos^2 \dfrac{\theta}{2})\\ &= \cos^2 \dfrac{\theta}{2} - \cos^2 \dfrac{\theta}{2} \cdot \tan^2 \dfrac{\theta}{2}\\ &= \cos^2 \dfrac{\theta}{2} - \sin^2 \dfrac{\theta}{2}\\ &= \dfrac{\cos \theta + 1}{2} - \left(\dfrac{1 - \cos \theta}{2}\right) \mbox{ because $\cos \theta = 2\cos^2 \dfrac{\theta}{2} - 1$ and $\cos \theta = 1 - 2\sin^2 \dfrac{\theta}{2}$}\\ &= \dfrac{2\cos \theta}{2} = \cos \theta\\ \end{align*}$
Title: Re: Maths Questions ^_^
Post by: Odette on April 27, 2008, 05:10:24 pm: OH I GET IT!! YAY!! THANK YOU MAO AND DCC ;D ;D ;D :D :D :D
Title: Re: Maths Questions ^_^
Post by: Odette on April 28, 2008, 01:14:13 pm: Ok new set of questions and new topic :D COMPLEX NUMBERS!! YAY
Here are some of the questions I need help with...
1.Let z = 4 + 8i and $\zeta$ = 8 + 6i. Evaluate each of the following, giving your answer in the form x + yi.
a) z $\zeta$
b)z + 3 $\overline {\zeta}$
c) $\frac {z} {\zeta}$
d) $\frac {z + \zeta} {\zeta}$

2. Using z from the previous question, evaluate z $^5$ , giving your answer in the form x + yi

Thanks again to whoever helps :)
Title: Re: Maths Questions ^_^
Post by: Odette on April 28, 2008, 01:19:44 pm: Oh and if anyone could possibly explain the concept of complex numbers that would be greatly appreciated, because the lecturer didn't explain it clearly and this is the first time i've been introduced to complex numbers.
Title: Re: Maths Questions ^_^
Post by: Odette on April 28, 2008, 09:40:12 pm: Anyone? Please ^_^ Dcc? Mao? Anyone? I really need to know this and i just dont get it :( :'(
Title: Re: Maths Questions ^_^
Post by: Mao on April 28, 2008, 09:46:42 pm: sorry, i would, except i have bloody english SAC 2morrow (and the next two days)

2 english SACs in the same week is seriously not cool =(
Title: Re: Maths Questions ^_^
Post by: AppleXY on April 28, 2008, 10:05:00 pm: Yeah same. Would've helped you but I have other stuphs :(.

It's pretty simple, do you have a ref. text?
Title: Re: Maths Questions ^_^
Post by: Odette on April 28, 2008, 10:06:25 pm: No just an exercisebook lolz
Title: Re: Maths Questions ^_^
Post by: Odette on April 28, 2008, 10:13:28 pm: That's ok, um i guess i'll have to figure it out somehow :( :'(
Title: Re: Maths Questions ^_^
Post by: dcc on April 28, 2008, 10:40:40 pm: Complex Numbers were created because mathematicians were getting annoyed at equations like $x^2 + 1 = 0$ (which of course has no solution in the reals). They are an extension on the real numbers which includes a new unit called the 'imaginary unit'. Mathematicians define this 'imaginary unit' as $i^2 = -1 \implies i = \sqrt{-1}$ .

All complex numbers ( $z \in C$ ) take the form:

$z = x + yi \mbox{ }x,y \in R$ , where i is this imaginary unit i discussed earlier.

ok, the main operations to know are:
Real part of a Complex Number:
Say you had a complex number $z = 3 + 2i$ , the REAL component of this is the part of z which is not proceeded by an I, so you can say:
$Re(z) = 3$

More generally: if $z = a + bi$ , then

$Re(z) = a$

Imaginary Part of a Complex Number:
The imaginary component of a complex number is the bit which is proceeded by an i (which is why it is called the imaginary part):

If $z = 9 + 66i$ , then:

$Im(z) = 66$

More generally: if $z = a + bi$ , then:

$Im(z) = b$

Complex Conjugate:
The complex conjugate of a complex number is given by changing the sign of the imaginary part. Graphically it represents a reflection in the $Re(z)$ axis (but thats on argand diagrams, so you might not of seen them).

If you have: $z = 3 - 2i$
then the complex conjugate of z is:

$\overline{z} = 3 + 2i$ (just change the negative to positive, or positive to negative)

Note that this little bar: $\overline{z}$ indicates a complex conjugate and is achieved in latex by typing \overline{z}

Addition:
$(a + bi) + (c + di) = (a + c) + (b + d)i$ (i.e. to add two complex numbers, you add the real & imaginary bits of each number)

Subtraction:
$(a + bi) - (c + di) = (a - c) + (b - d)i$ (as you would expect)

Multiplication::
Multiplication is much like expanding brackets (it IS expanding brackets) but you have to remember to simplify numbers etc:
$(a + bi)(c + di) = ac + adi + bci + bdi^2 = ac + (ad + bc)i + bd(-1) \mbox{ (as $i^2 = -1$)}\\ \implies (a + bi)(c + di) = (ac - bd) + (ad + bc)i$

Division:
Division is weird, so it requires a bit of thought. I'll give you an example just straight up:

Say you wanted to find: $\dfrac{3 + 2i}{1 + 5i}$

The first step is to multiply both top and bottom lines by the complex conjugate of the bottom line. This is done because if you multiply a complex number with its conjugate, you generally get a real number, which simplifies the process immensely.

$\dfrac{3 + 2i}{1 + 5i} = \dfrac{(3 + 2i)(1 - 5i)}{(1 + 5i)(1 - 5i)} = \dfrac{3 - 15i + 2i - 10i^2}{1 - 5i + 5i - 25i^2} = \dfrac{13 - 13i}{1 + 25} = \dfrac{13 - 13i}{26} = \dfrac{1}{2} - \dfrac{1}{2}i$

In General:

$\dfrac{a + bi}{c + di} = \dfrac{(a + bi)(c - di)}{c^2 + d^2} = \dfrac{ac - adi + bci + bd}{c^2 + d^2} = \dfrac{(ac + bd) + (bc - ad)i}{c^2 + d^2}$

That should be all you need for complex numbers.
Title: Re: Maths Questions ^_^
Post by: Odette on April 28, 2008, 10:45:49 pm: Ok thanks Dcc, I think I get it now, so I'll have a go at the questions myself, then I'll post them up to see if I'm on the right track.
Title: Re: Maths Questions ^_^
Post by: Odette on April 28, 2008, 11:40:13 pm: Ok here goes nothing lol....
1 a) (4 + 8i) (8 + 6i) = (4 $\times$ 8 - 8 $\times$ 6) + (4 $\times$ 6 + 8 $\times$ 8)i
= (32 - 48) + (24 + 64)i
= -16 + 88i

b) (4 + 8i) + 3(8 - 6i)= (4 + 8) + 3(8 - 6)i
= 12 + 24i -18i
= 12 + 6i (Not too sure about this one)

c) $\frac {4 + 8i} {8 + 6i}$ = $\frac {(4 \times 8 + 8 \times 6) + (8 \times 8 - 4 \times 6)i} {8^2 + 6^2}$
= $\frac {(32+48) + (64 - 24)i} {64 + 36}$
= $\frac {80 + 40i} {100}$
= $\frac {4} {5}$ + $\frac {2} {5}$ i

d) $\frac{(4 + 8i) + (8+6i)} {8 + 6i}$ = $\frac {(32 - 24) + (24 + 64)i} {8 + 6i}$
= $\frac {8 + 88i} {8+6i}$
= $\frac {(64 + 528) + (704 - 48)i} {64 +36}$
= $\frac {592 + 656i} {100}$
= 5 $\frac {23}{25}$ + 6 $\frac{14} {25}$ i

2. (4 + 8i) $^5$ = 41984 - 38912i (can't be bothered typing it all out for this one)

Please let me know if these are right or wrong, thanks :)
Title: Re: Maths Questions ^_^
Post by: Odette on April 29, 2008, 08:46:30 pm: Anyone? Please ^_^
Oh never mind then, i'll have to figure it out myself, thanks for the explanation dcc.
Title: Re: Maths Questions ^_^
Post by: Odette on May 15, 2008, 06:47:31 pm: Ok i really need help with something... evaluate (4 + 8i) ^1/2 using de moivre's theorem, I have the answer but it's in a+bi form, I want the full answer but my calculator wont give me the answer >.< I have a TI-89, would anyone be able to tell me how to get it into decimal form, so that its just one answer rather than in a+bi form?
Thanks
Title: Re: Maths Questions ^_^
Post by: AppleXY on May 15, 2008, 06:53:19 pm: Press the green diamond button and "ENTER" to obtain an approximate value for any calculation :)
Title: Re: Maths Questions ^_^
Post by: Odette on May 15, 2008, 06:55:10 pm: Thank you ^_^
It didn't work >.<
Title: Re: Maths Questions ^_^
Post by: Odette on May 15, 2008, 07:09:18 pm: Oh and another question (8+6i) ^1/3 ... I keep getting the same answer and it's wrong.
Here's what i did:
r = sqrt (8^2 + 6^2)
r= sqrt (100)
r= 10

$\theta$ = tan^-1 (6/8)
= 0.64

z^1/3= 10 ^1/3 (cos (1/3 * 0.64) + i sin (1/3 *0.64) )
= 10 ^1/3 (cos (0.22) + i sin (0.22))
= 2.15 e ^0.22i
= 2.10 + 0.46i
Title: Re: Maths Questions ^_^
Post by: dcc on May 15, 2008, 07:15:41 pm: Solutions for $z^2 = 4 + 8i$ are:

$z = \pm \left[2\sqrt{1 + 2i}\right]$

But that is not a question which favours the use of de'moivre.
Title: Re: Maths Questions ^_^
Post by: Odette on May 15, 2008, 07:20:39 pm: So my answer is wrong? -is confused-

Oh and another question, how exactly to you print a graph you've sketched using you calculator?
Sorry I don't use my calculator much.

Nevermind I used a graphing program ^_^
Title: Re: Maths Questions ^_^
Post by: Mao on May 15, 2008, 08:25:59 pm: Quote from: Odette on May 15, 2008, 07:09:18 pm
Oh and another question (8+6i) ^1/3 ... I keep getting the same answer and it's wrong.
Here's what i did:
r = sqrt (8^2 + 6^2)
r= sqrt (100)
r= 10

$\theta$ = tan^-1 (6/8)
= 0.64

z^1/3= 10 ^1/3 (cos (1/3 * 0.64) + i sin (1/3 *0.64) )
= 10 ^1/3 (cos (0.22) + i sin (0.22))
= 2.15 e ^0.22i
= 2.10 + 0.46i
why is this wrong?

there are two more solutions, is that what you meant?

remember:

$|z|^n cis(n\theta) = |w| cis (\phi + 2k\pi)$ , where k is an integer.
Title: Re: Maths Questions ^_^
Post by: Odette on May 15, 2008, 09:46:41 pm: Huh? What's k? I don't get it >.<
Title: Re: Maths Questions ^_^
Post by: /0 on May 15, 2008, 11:22:26 pm: Quote from: dcc on April 28, 2008, 10:40:40 pm
Complex Numbers were created because mathematicians were getting annoyed at equations like $x^2 + 1 = 0$ (which of course has no solution in the reals). They are an extension on the real numbers which includes a new unit called the 'imaginary unit'. Mathematicians define this 'imaginary unit' as $i^2 = -1 \implies i = \sqrt{-1}$ .

Has anyone defined the log of a negative number? :D
Title: Re: Maths Questions ^_^
Post by: Mao on May 15, 2008, 11:27:39 pm: $e^{\pi \cdot i} = cis(\pi)= -1$

:P
Title: Re: Maths Questions ^_^
Post by: AppleXY on May 16, 2008, 09:11:07 pm: Quote from: Odette on May 15, 2008, 09:46:41 pm
Huh? What's k? I don't get it >.<

k is an integer, like 1,2,3...
Title: Re: Maths Questions ^_^
Post by: Odette on May 16, 2008, 09:50:15 pm: what's k in my questions? lolz
Title: Re: Maths Questions ^_^
Post by: /0 on May 16, 2008, 09:56:12 pm: Since sine and cosine have period $2\pi$ , if you add $2\pi$ to their angle, the results stays the same. You can visualise this by graphing a standard sine graph. Every $2\pi$ , the same y-value occurs. So plugging in different k's just gives you a bunch of angles differing by $2\pi$ , which, when evaluated, will give you the same result.
Title: Re: Maths Questions ^_^
Post by: Odette on May 17, 2008, 09:14:54 pm: Quote from: DivideBy0 on May 16, 2008, 09:56:12 pm
Since sine and cosine have period $2\pi$ , if you add $2\pi$ to their angle, the results stays the same. You can visualise this by graphing a standard sine graph. Every $2\pi$ , the same y-value occurs. So plugging in different k's just gives you a bunch of angles differing by $2\pi$ , which, when evaluated, will give you the same result.

This is what i have to say to that: Huh? >.< -feels stupid-
Title: Re: Maths Questions ^_^
Post by: /0 on May 18, 2008, 03:16:53 pm: Think about the unit circle, the angle measure all the way around is $360^{\circ}=2\pi^c$ . Since $\sin{\theta}$ is equal to the y-coordinate at any time and $\cos{\theta}$ is equal to the x-coordinate at any time, if you add $2\pi$ radians onto an angle, all the point $(\cos{\theta},\sin{\theta})$ will do is move around the circle again and end up at the same point. If you haven't encountered this before it's not easy to explain without a diagram, so there's more info here: http://www.sparknotes.com/testprep/books/sat2/math2c/chapter9section6.rhtml
Title: Re: Maths Questions ^_^
Post by: Odette on May 18, 2008, 03:24:50 pm: Hmm i still dont know what k is, example perhaps? (sorry)