[Mukade]

So, so, we're starting a maths club to help all suffering from any maths related traums/injuries!
Please feel free to post any maths related question here, and I'll try my best to answer them!

[RicocheT]

How did Wolf come to the conclusion that 50+20 = 60???

Quick Comments

	syfSoundwave:	*10 mins later* "...Did you say 40 or 50?"
	TtSYF.tRoy:	Made me giggle for a long time. Good post Undone
	mGGNemesis:	The banes blew up 10 marines making it to 60

[Malice]

How much wood could a wood chuck chuck if a wood chuck could chuck wood?

[Mukade]

the answer you seek is 7

[RVLiveR]

x=y
x^2=xy
x^2-y^2=xy-y^2
(x+y)(x-y)=y(x-y)
x+y=y
1+1=1

7=((1+1)+(1+1))+((1+1)+1)
7=(1+1)+(1+1)
7=1+1
7=1

Wood chuck only chucks one wood if a wood chuck could chuck wood.

[AxS.pinealan]

Seems legit lol
I wonder if there is anyone who cant notice the mistake in this proof

Originally Posted by sRLiveR

x=y
x^2=xy
x^2-y^2=xy-y^2
(x+y)(x-y)=y(x-y)
x+y=y
1+1=1

7=((1+1)+(1+1))+((1+1)+1)
7=(1+1)+(1+1)
7=1+1
7=1

Wood chuck only chucks one wood if a wood chuck could chuck wood.

[Mukade]

Dividing by zero in the first? Not even sure how the second one is a thing

[x5_Namakaye]

1+1=1 makes me cry

Quick Comments

mGGNemesis:

1 ht + 1 ht = 1 archon sounds perfectly legit :D

[syfPhoenix]

Originally Posted by sRLiveR

7=((1+1)+(1+1))+((1+1)+1)
7=(1+1)+(1+1)
7=1+1
7=1

Wood chuck only chucks one wood if a wood chuck could chuck wood.

Dafuq?!?!

10char

[AxS.Sorathez]

Originally Posted by sRLiveR

x=y
x^2=xy
x^2-y^2=xy-y^2
(x+y)(x-y)=y(x-y)
x+y=y
1+1=1

7=((1+1)+(1+1))+((1+1)+1)
7=(1+1)+(1+1)
7=1+1
7=1

Wood chuck only chucks one wood if a wood chuck could chuck wood.

The mistake in this proof is step number 5.
There you divide by (x-y) but x=y, therefore (x-y)=0

Dividing by zero. Tut tut tut

Dividing by zero in the first? Not even sure how the second one is a thing

The second one is a thing because he established that 1+1 = 1

Extending on that:
If we assume that 1+1=1, then 2=1, and therefore (by subtracting 1 from each side) 1=0.

Now:
Barack Obama has no leafy green top, but 1=0, therefore he has 1 leafy green top.
The wavelength of a photon emitted from Obama is (x)nm.
Because 1=0 we can say that
1x=0x
so
1x=0
Adding 585 to both sides
1x+585=585

But since 1=0,
We can show that
585=0 by multiplying both sides by 585.
Therefore:
x=585.

And as such, the photons that radiate from Barack Obama have a wavelength of 585nm, and he therefore appears bright orange.

Obama's waist is ycm wide, and similarly to above, we can show that y=0cm.

Therefore, Obama tapers to a point.

Now that we have shown that he has a leafy green top, is bright orange and that he tapers to a point, I can conclude that Barack Obama is a carrot.

Quick Comments

	syfPhoenix:	This^^
	mGGNemesis:	I got lost somewhere when he mentioned Obama
	Mukade:	Ahhhh, right, I thought the second one was separate

[IcedTea]

What is the answer to the ultimate question of life, the universe and everything?

Quick Comments

Asrathiel:

Duh, 42

[Xormentor]

I remember my first Math C(extension) class in grade 10, we learnt .999(recurring)=1 exactly.

http://www.debate.org/debates/.9999-repeating-1/1/

[RVLiveR]

mm this is probably nobody and way out of topic but is anyone here representing their school in SEAMC, coming up in ~2 weeks or so?

[Bash]

[FaDeHealZ]

I did last year, but can't go this year, have piano exam

[|Erasmus|]

http://imgs.xkcd.com/verizon_billing.mp3

[Mukade]

Originally Posted by IcedTea

What is the answer to the ultimate question of life, the universe and everything?

The answer is on the interval 0<x<everything else

[mGGPrometheus]

yay math thread!
Obligatory:

Quick Comments

	mGGDrGooSe:	I question the idea tha the mathematicians would look up to see anyone in the first place.
	Mukade:	Lolololol, nice
	MazEi:	xkcd!

[Meatex]

maths is silly :P

Quick Comments

Mukade:

My god, he's right!

[Seffy]

Calculus 101 pls.

Quick Comments

Mukade:

Can do :)

[faithHunter]

I now realize how mathemathically less inclined I am compared to the rest of you guys now after reading through this thread.

I don't even get half the terms that you are using :P

Quick Comments

	mGGNemesis:	lol poor faith; dw :)
	RVLiveR:	it's okay, everything is equal to 1 if 1+1=1
	AxS.Sorathez:	If 1+1=1 then anything and everything is true :D

[AxS.pinealan]

Hey I got a trigonometry question for you guys:
Use mathematical induction to prove that for any positive integer n,
sinx+sin3x+...+sin(2n-1)x= (1-cos2nx)/2sinx
where sinx =/= 0

[Mukade]

Originally Posted by pinealan

Hey I got a trigonometry question for you guys:
Use mathematical induction to prove that for any positive integer n,
sin x+sin 3x+...+sin(2n-1)x= (1-cos2nx)/2sin x
where sin x =/= 0

I'll use the indentity (1) (sin x)^2= (1-cos 2x)/2
because (sin nx)^2/(sin x) is easier to work with.
and also use
(2) (sin A sin B=1/2(cos(A-B)-cos(A+B))


Set n=1

lhs=sinx
rhs=(sin x)^2/sin x
=sin x
=lhs
true for n=1

Assume n=k
sin x+ sin 3x+...+sin(2k-1)x=(sin kx)^2/(sin x)

For n=k+1

lhs= ((sin kx)^2)/sin x+sin(2k+1)x
=(((sin kx)^2)+sin(2k+1)x sin x)/sin x
=(((sin kx)^2)+1/2(cos(x-2kx-x)-cos(2kx+2x))/sin x (2)
=(((sin kx)^2)+1/2(cos(-2kx)-cos(2kx+2x))/sin x
=(((sin kx)^2)+1/2(cos(2kx)-cos(2kx+2x))/sin x
=((((sin kx)^2)+1/2(1-2(sin kx)^2-(1-2(sinkx+x))^2))/sin x
=((((sin kx)^2)+1/2(2(sin kx+x))^2-2(sin kx)^2))/sin x
=((sin kx)^2+(sin kx+x))^2-(sin kx)^2)/sin x
=(sin(k+1)x)/sin x
=rhs

Therefore, true for all n.

[Mukade]

Admittedly, I've done this question before and still have the textbook worked solution >.>

Quick Comments

syfPhoenix:

Knew it was too good to be true. :P

[AuXCoolpuff]

I don't want to live on this planet any more.

Quick Comments

	Mukade:	Lololol
	mGGNemesis:	lolol

[|Erasmus|]

You Tube

You Tube

Quick Comments

Mukade:

This is amazing

[Zepph]

I could fap to this thread.

Maths .

Quick Comments

	Maynarde:	NERD
	TheGentleman:	@Maynerd: And don't you forget it!!
	RVLiveR:	zepph is my new best friend~ ^.^

[TheGentleman]

I know this probably belongs in the puns thread but... I'm so glad this thread exists.

I wish you were sine^2 and I was cosine^2 so together we could be 1

Quick Comments

	Mukade:	Cute as bro :P
	AxS.pinealan:	lol nice pun, never have thought of that before
	AxS.Sorathez:	I love it :D You can always be my Cos^2
	mGGNemesis:	awwww nice <3

[AxS.pinealan]

Well, frankly, I pulled that question from my textbook too :P

[AxS.Sorathez]

Originally Posted by Mukade

I'll use the indentity (1) (sin x)^2= (1-cos 2x)/2
because (sin nx)^2/(sin x) is easier to work with.
and also use
(2) (sin A sin B=1/2(cos(A-B)-cos(A+B))


Set n=1

lhs=sinx
rhs=(sin x)^2/sin x
=sin x
=lhs
true for n=1

Assume n=k
sin x+ sin 3x+...+sin(2k-1)x=(sin kx)^2/(sin x)

For n=k+1

lhs= ((sin kx)^2)/sin x+sin(2k+1)x
=(((sin kx)^2)+sin(2k+1)x sin x)/sin x
=(((sin kx)^2)+1/2(cos(x-2kx-x)-cos(2kx+2x))/sin x (2)
=(((sin kx)^2)+1/2(cos(-2kx)-cos(2kx+2x))/sin x
=(((sin kx)^2)+1/2(cos(2kx)-cos(2kx+2x))/sin x
=((((sin kx)^2)+1/2(1-2(sin kx)^2-(1-2(sinkx+x))^2))/sin x
=((((sin kx)^2)+1/2(2(sin kx+x))^2-2(sin kx)^2))/sin x
=((sin kx)^2+(sin kx+x))^2-(sin kx)^2)/sin x
=(sin(k+1)x)/sin x
=rhs

Therefore, true for all n.

I had quite the love/hate relationship with mathematical induction in High School...

[TheGentleman]

Originally Posted by pinealan

Well, frankly, I pulled that question from my textbook too :P

Now now... we won't do your homework for you!

[|Erasmus|]

And here's what we all want math to be:

+ Show +

Quick Comments

ToRSpartaz:

MAAAAAAAAAAAAAAAATH

[ToRSpookToR]

[6:50:21 AM] Nick "ToRSpartaz" Simpson: 10 + - 11 is 21

Quick Comments

Mukade:

Pure genius

[ToRSpartaz]

In my defense, I only woke up a few minutes before that comment...

Quick Comments

mGGNemesis:

lolol

[mGGNemesis]

I this thread

Quick Comments

Mukade:

:)

[ToRSpartaz]

Love Math, hate proofs T_T, proofs make me die a little inside

Quick Comments

	ToRLuneth:	So does basic arithmetic it seems <3
	ToRSmotPokingFish:	this^^

[Antel0pe]

Wait, so i can scan sheets of my homework in, and people will just answer it yeh? ^.^
I approve of this thread

Quick Comments

	|Erasmus|:	no.
	ToRsupapower:	no.
	Mukade:	I might. I promise I won't answer them wrong
	mGGPrometheus:	I probably would
	RVLiveR:	mm koreans asking for help at math..
	mGGNemesis:	anyone could... though I'm not sure about trusting the answers ;)

[RVLiveR]

yo guys for anybody who will be at the coming SEAMC you should greet me ^.^ I'll be in the JIS A-Team with two koreans so I'll stand out :P

[Antel0pe]

Originally Posted by Antelopes

*sRLiveR: *
mm koreans asking for help at math..

Wow, that stereotype T_T.
I'm Wood League in Math ;_;

[AxS.pinealan]

Originally Posted by Antelopes

Wow, that stereotype T_T.
I'm Wood League in Math ;_;

You mean Korean wood league? Thats NA Masters

Quick Comments

	Mukade:	Lolololol
	syfPhoenix:	I lol'd a little.

[faithHunter]

I think this image is relevant:

[mGGMueX]

I have one.

Derive the commutativity of addition from the 4 addition axioms closure, associativity, zero, inverses.

[SayfT]

zero would be a + 0 = a
associativity should be (a+b) + c = a + (b+c)
inverses would be a + -a = -a + a

and closure is escaping me for the moment ><

[AxS.pinealan]

what is commutativity -__-''
looks like so really basic stuff when i searched on wiki

[SayfT]

how stuff is added, at least thats how I understand it, multiplicativity (is that even a word?!) would be how stuff is multiplied, or the axioms that multiplication is based on.


fyi: I could be wrong :P

[Pox]

Commutativity is the fact that a+b=b+a for all a,b.

Quick Comments

	RVLiveR:	not if 1+1=1 (see proof above)
	FaDeBadger:	That "proof" is false

[Mukade]

An operation is commutative if a x b = b x a where x is some operation. Like, you can say addition is commutative for a,b e C.

[mGGPrometheus]

I don't think you can derive commutativity from only closure, associativity, identity and inverses.

For example consider the general linear group consisting of nxn matrices with nonzero determinant.
- It is closed (multiplying two matrices with nonzero determinant results in a matrix with nonzero determinant)
- It is associative (matrix multiplication is associative)
- It has an identity
- It has inverses
And yet any two matrices in the general linear group need not commute.

[mGGMueX]

Originally Posted by mGGTitan

I don't think you can derive commutativity from only closure, associativity, identity and inverses.

For example consider the general linear group consisting of nxn matrices with nonzero determinant.
- It is closed (multiplying two matrices with nonzero determinant results in a matrix with nonzero determinant)
- It is associative (matrix multiplication is associative)
- It has an identity
- It has inverses
And yet any two matrices in the general linear group need not commute.

Matrix multiplication is not commutative but addition is.

It is definitely possible to derive from the 4 stated axioms.

[Pox]

Originally Posted by mGGMueX

Matrix multiplication is not commutative but addition is.

This is exactly the point - both (GL,*) and (R,+) are groups (i.e. satisfy the four given axioms) but (GL,*) is not commutative - so the group axioms are not enough to prove that (R,+) is commutative, you need some extra fact.

[Mukade]

Commutativity is just a property. Of course everything isn't commutative, one being matrix multiplication, normal division, etc.

[ToRSpookToR]

shama lama ding dong

[Mukade]

As for deriving commutativity from the peano axioms, that's actually fairly tricky, but it can be done. The trick is to prove associativity first with induction, and then a cute little proof for commutativity follows (that I don't remember).

[mGGMueX]

You guys failed the test... I was hoping to get some help revising this

I'll post the proof later if noone can figure it out.

[Mukade]

Originally Posted by mGGMueX

You guys failed the test... I was hoping to get some help revising this

I'll post the proof later if noone can figure it out.

I have it in a textbook, lol

[Mukade]

For addition anyway, but the other operations are similarly easy.

[Pox]

I'll give anyone $100 for a singular solution to the 3+1-dimensional navier-stokes equations with a smooth initial condition. It's, uh, my homework.

Quick Comments

AxS.pinealan:

I dont even know what is navier-stokes equation

[Mukade]

Originally Posted by Av.ToRPox

I'll give anyone $100 for a singular solution to the 3+1-dimensional navier-stokes equations with a smooth initial condition. It's, uh, my homework.

Sorry dude, beyond me.

[mGGPrometheus]

Originally Posted by mGGMueX

You guys failed the test... I was hoping to get some help revising this

I'll post the proof later if noone can figure it out.

I'd like to see that... from what I've learnt in group theory, any group with those 4 axioms can either be commutative or noncommutative (in fact those 4 axioms are the defining properties of a group)

Btw for anyone who likes calculus, try this simple integration problem =P

Quick Comments

Zepph:

Oh man...memories. 1/cosx is secx? so its something like log secx - tanx + C ? gosh i hardly remember

[AxS.pinealan]

Originally Posted by mGGTitan

Btw for anyone who likes calculus, try this simple integration problem =P

lnlsecx+tanxl + C
thats an identity in my textbook which I am supposed to memorize =P

[Mukade]

ln (tan x + sec x) +C, lol, ez

[mGGPrometheus]

Originally Posted by pinealan

lnlsecx+tanxl + C
thats an identity in my textbook which I am supposed to memorize =P

Yep, although I never thought that was the obvious solution. An alternative solution is to multiply the top and bottom by cos(x) and use the substitution u = sin(x).

[AxS.pinealan]

Well, my solution is to multiply the top and botton by (secx+tanx)

[FlashVeno]

3+100÷5(2)

is brackets the same as times? if it is then its 43
but if you follow bodmas brackets are first, so its 13
wolfram alpha says 43..what do you guys think?

[|Erasmus|]

you evaluate the (2) first, not the 5 x 2 cause that's not what's in brackets.

wolfram alpha is 100% correct.

Quick Comments

mGGAequitas:

wolframalpha is always correct :)

[TheGentleman]

I thought I'd do this graphically for you cause it looks better ^^



I find it's best to look at the equation written in fractions like so:



Because it can also be written like so:



At this point in time it's up to you how you solve the equation. You can either multiply the Numerator of each fraction and Denominator of each fraction by each other to get this:



OR you can resolve each fraction first and then multiply the numbers left behind by each other like so:



Either way the answer is the same. The important thing to note is that 100 is the numerator and 5 is the denominator of fraction one while 2 is a fraction on its own. 5 * 2 is multiplying the denominator of one fraction by the numerator of the other and is the reason why 13 is not the correct answer.

i.e. is the problem.

I hoped this helps

+ [old content... now irrelevance] +

3+100÷5(2)

expands to look like this:

3 + ((100 / 5) * 2)
= 3 + (20 * 2)
= 43

OR

3 + ((100 * 2) / 5)
= 3 + (200 / 5)
= 43

in short yes... brackets indicate multiplication unless otherwise specified.

Edit:

Originally Posted by |Erasmus|

you evaluate the (2) first, not the 5 x 2 cause that's not what's in brackets.

Actually... the 2 is in brackets on its own so technically the stuff within the brackets is already resolved and it just indicates multiplication. Then it's simple BOMDAS... considering Multiplication and Division are considered equal it doesn't matter whether you perform 100 / 5 first or 100 * 2 first... just as long as you don't do 5 * 2 because 5 is a denominator not a numerator.

P.S. I read eras' message wrong confusing me ^^ still all information is good information.

[|Erasmus|]

yeah... that's what i said... it's just (2) inside the brackets, so only the 2 is evaluated by bomdas. so you don't do 5x2. which gives the alternative (wrong) answer 13

Quick Comments

TheGentleman:

ah right... nvm... I just tripped over the terminology that was used <.< My bad.

[FlashVeno]

thanks now i know xD
i thought it was 13 before lol

[syfPhoenix]

So much of the stuff in this thread has gone over my head... :S im screwed when i get back to school T_T

Quick Comments

eCKo`Tazerenix:

Most of it is uni level anyway

[AxS.pinealan]

Was working on some public exam past papers when I stumbled upon this:
Intergrate

I have no idea how to do this T.T

Edit: thx for telling me that link didnt work, i'll just write it here
Int sec^5(x) dx

Quick Comments

	|Erasmus|:	link no work for me...
	mGGPrometheus:	^
	Mukade:	Not for me either. You know how much I love doing your maths problems for you :P

[mGGPrometheus]

Originally Posted by pinealan

Was working on some public exam past papers when I stumbled upon this:
Intergrate

I have no idea how to do this T.T

Edit: thx for telling me that link didnt work, i'll just write it here
Int sec^5(x) dx

As in the previous problem, you can multiply the top and bottom by the same factor:

The result follows by u-substitution.

[ToRSpartaz]

Reading some of these calc equations makes me just want to study again >< not a good sign O_o

Quick Comments

FaDeBadger:

This, loved it in High school

[AxS.pinealan]

er, I still cant work out the answer... do I have to use integration by part?
What I now have is
Not sure if it is correct. May I ask for a bit more help?

[Mukade]

That's a really hard problem pinealan. I think you have to use the reduction formula for sec^n (x), no idea what it is though, sorry.

[Mukade]

http://en.wikipedia.org/wiki/Integral_of_secant_cubed That should have them!

[faithHunter]

Kinda off topic but can someone explain to me what exactly is a Tesseract? The youtube videos I watched said that its a "cube inside of a cube" but when you rotate it it gets really really weird and incomprehensible IMHO.

[AxS.pinealan]

the thing i hate about integration is that there are tons of possible representations of the answer, and I have no clue if mine is a correct one.
Another answer that I just worked out with Mukade's help:

Quick Comments

Zepph:

Is that your handwriting? It's so ridiculously neat I could eat it.

[Delraich]

Something I found recently that might be interesting for you to solve (cipher codes). The solutions are on the web, so if you want a challenge don't Google the answer. Try and solve them in less than 44 hours without cheating

I will start with the easiest to the hardest:

1.
UG LMIZ KPIZTMA,

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GWCZA AQVKMZMTG,

VQKPWTIA

2.
TF KLHY UPJOVSHZ

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Quick Comments

Zepph:

Challenge accepted!

[AxS.pinealan]

Originally Posted by Delraich

Something I found recently that might be interesting for you to solve (cipher codes). The solutions are on the web, so if you want a challenge don't Google the answer. Try and solve them in less than 44 hours without cheating

One hour passed, decoded the first three, lets see if I can get the rest.

[AxS.pinealan]

Originally Posted by pinealan

*Zepph: *
Is that your handwriting? It's so ridiculously neat I could eat it.

Sorry for the double post, but yes, thats my handwriting. Well I took my time to make it as neat as possible

[mGGPrometheus]

Originally Posted by pinealan

er, I still cant work out the answer... do I have to use integration by part?
What I now have is
Not sure if it is correct. May I ask for a bit more help?

After letting u = sin(x), du = cos(x)dx, the integral becomes . Then you just need to use partial fractions.



Since this must be true for all u, we can try different values of u. Letting u = -1, we see that D = 1/8. Letting u = 1, A = 1/8. Letting u = 0, A+B+C+D+E+F = 1, so B+C+E+F = 3/4, so B = 3/4 - C - E - F. Then you expand the RHS so you get a cubic in 'u' and equate the terms on both sides to solve for the other constants. In the end you get



And ^ can be integrated pretty easily. I didn't want to type the working out because it's quite a messy problem

[eCKo`Tazerenix]

Originally Posted by faithHunter

Kinda off topic but can someone explain to me what exactly is a Tesseract? The youtube videos I watched said that its a "cube inside of a cube" but when you rotate it it gets really really weird and incomprehensible IMHO.

Basically if a square is a 2D 'square' and a cube is a 3D 'square', a tesseract is a 4D 'square'. The thing to understand about dimensions is a dimension is defined as being perpendicular to all other dimensions. Now in 2D or 3D space we can easily visualize this but we can't in 4D space, because we live in 3D space. When people say a tesseract is a cube inside a cube with the verticies joined, thats not actually a tesseract, its the shadow of a tesseract viewed in 3D space, much like you can draw the shadow of a cube in 2D space. Scientists just sort of added a dimension and did some analysis and thinking to figure out what the 3D shadow would look like. Since we can't actually visualize a tesseract, scientists or mathematicians try and make visualizations for it with computers, with stupid videos and gifs of a tesseract rotating, but these dont actually mean anything because to grasp it you need to see 4D space which is impossible for 3D species.

Quick Comments

AxS.pinealan:

simply put, its not something that ANYONE can exactly explain

[Mukade]

Originally Posted by mGGTitan

After letting u = sin(x), du = cos(x)dx, the integral becomes . Then you just need to use partial fractions.



Since this must be true for all u, we can try different values of u. Letting u = -1, we see that D = 1/8. Letting u = 1, A = 1/8. Letting u = 0, A+B+C+D+E+F = 1, so B+C+E+F = 3/4, so B = 3/4 - C - E - F. Then you expand the RHS so you get a cubic in 'u' and equate the terms on both sides to solve for the other constants. In the end you get



And ^ can be integrated pretty easily. I didn't want to type the working out because it's quite a messy problem

Good working o.0 I wish I'd thought of that.

[mGGPrometheus]

Oh it's just a useful trick to deal with integrals of sec^n(x)dx and cosec^n(x) where n is odd. With cosec^n(x) you would multiply the top and bottom by sin(x)

[AxS.pinealan]

Looks like a neat technic that I can use in my exams. Thx Titan!
However, I still wanna know if this is a valid answer. I got this answer after doing integration by part twice.

[mGGPrometheus]

The only sure way to know is take the derivative! It's really close but I think it's

integration by parts is a pretty sweet way to do this problem. It probably requires less work than partial fractons

Quick Comments

AxS.pinealan:

btw, how do you make those images of the answers?

[AxS.pinealan]

Oh right, I forgot the 3 when you integrate that sec^3x. Thx a lot Titan!

Quick Comments

mGGPrometheus:

http://www.codecogs.com/latex/eqneditor.php

[AxS.pinealan]

More maths is always better! Enough of calculus for the week, so can anyone help me with this parametric equation?

[Antel0pe]

Help anyone?
doing a GCSE math book on Geometry and my tuition teacher presumed I learned trigonometry ;;

I'm kinda lost T_T

[AxS.pinealan]

Originally Posted by Antelopes

Help anyone?
doing a GCSE math book on Geometry and my tuition teacher presumed I learned trigonometry ;;

For (a), use the cosine rule

Then in (b), since you know the angle POQ, you can multiple it with the radius to get the arc length. After that i guess you know how to do the perimeter.

As for (c), you need to know the area of the segment PQ. To do that, you first calculate the area of the larger part of sector POQ, which you can do so by multiplying pir^2 with the reflex angle in radians. Then, find out the area of the triangle POQ. That, you can use the sine rule.
Then you just add them up. wala.

I hope that you can understand my messy working

Quick Comments

Antel0pe:

thanks~ ^-^ with this, and some workings i did with the teacher, i think i can solve it~

[mGGNemesis]

stuff here is actually getting pretty serious

[mGGPrometheus]

Originally Posted by pinealan

More maths is always better! Enough of calculus for the week, so can anyone help me with this parametric equation?

Solve for t so that t = -1 + 1/x. Then sub that into the second equation. From there's it should just be a matter of simplifying fractions.

[Seffy]

Hey guys,

My friend sent me this picture cos she thought I was good at Mathematics.. clearly I suck LOL so I'm seeing if anyone here have any idea as to what these are, as my friend or myself have no idea what these formulae are. Thanks in advance!

Quick Comments

AxS.pinealan:

lol what are these things, got no idea what they are

[mGGPrometheus]

Top left is Fermat's Last Theorem, which is a famous problem in mathematics that was recently solved by Andrew Wiles.

Top right some kind of Fourier Series in quantum mechanics. In QM you can completely describe objects using different kinds of 'basis states', such a position, momentum, energy, spin etc. In some cases knowledge of one set of states allows you to know another set of states, thanks to the uncertainty principle. I'm not quite sure what it's saying but I think the general gist that one side of the equation is one state (e.g position), the other is another state (e.g momentum), and the equation lets you transform between the two.

Bottom left is an (Inverse) Fourier Transform. If you have a signal whose amplitude varies with time, you can analyze what frequencies it's made up of using a Fourier transform. For example, say you have a piano note being played. The time-signal will be some kind of oscillating wave. Apply a Fourier transform and you can see the different sound frequencies it's made up of. One of the frequencies you see will be the actual note, however you may also see higher frequencies contributing to the timbre of the instrument.

Bottom right is some kind of inner product in quantum mechanics (similar to a dot product except with functions instead of vectors).

Dunno what the middle is

Quick Comments

	Seffy:	Thanks Titan! Helped my friend a lot. :)
	AxS.pinealan:	0.0 you are so good at maths, and you even know about QM stuff

[eCKo`Tazerenix]

Let ε < 0.