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Girls vs boys maths challenge

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Reply 280
Original post by PhysicsKid
1/tan x (cot x)


if that is an equals sign then yes, but you need'nt use the definition cotx anyway, expressing it as 1/tanx will get you your desired result
(edited 10 years ago)
Original post by Jkn
Without using a calculator, prove or disprove that 398712+436512=447212\displaystyle 3987^{12}+4365^{12}=4472^{12}


3987^12 + 4365^12 = ((3987+4365)^2-(2x3987x4365))^6
Rounding both numbers:
We get (8000^2 - 32000000)^6 = 32000000^6 =approx.( 3x10^7)^6 = 729 x 10^42 = 7.3 x 10^44
4472^12 = approx. (5x10^3)^12 = 244142875 x 10^36 = approx 2.5x10^44
Even by constantly rounding up the values for 4472^12 and rounding down 3.2 to 3 for the other expression- which is magnified a lot given the indices involved, 4472^12 < 3987^12 +.4365^12 (the difference between the two is far too big even given the big effect rounding will have had for them to be equal).
Original post by Jkn
Without using a calculator, prove or disprove that 398712+436512=447212\displaystyle 3987^{12}+4365^{12}=4472^{12}


Spoiler



ETA: Hah, the difference is about 10^33 :P
(edited 10 years ago)
Reply 283
None of these questions make sense lmao
Original post by SamL94
None of these questions make sense lmao


I reply thus:


Ignorance exists in the map, not in the territory. If I am ignorant about a phenomenon, that is a fact about my own state of mind, not a fact about the phenomenon itself. A phenomenon can seem mysterious to some particular person. There are no phenomena which are mysterious of themselves. To worship a phenomenon because it seems so wonderfully mysterious, is to worship your own ignorance.


--Eliezer Yudkowsky
Original post by Robbie242
if that is an equals sign then yes, but you need'nt use the definition cotx anyway, expressing it as 1/tanx will get you your desired result


Obviously subbing in tan x = 1/tan x, meaning tan x must equal 1 but also:
tan x = 1/tan x
tan^2 x = 1
sin^2 x/cos^2 x = 1
1-cos^2 x/cos^2 x=1
1/cos^2 x - 1 = 1
1/cos^2 x = 0
Original post by Smaug123

Spoiler



ETA: Hah, the difference is about 10^33 :P


Could you explain that method please? :P
Reply 287
Prove that

[br]r=0nr(nr)=n2n1[br][br]\sum\limits_{r=0}^nr{n\choose r}=n2^{n-1}[br]
.
Reply 288
Original post by PhysicsKid
Obviously subbing in tan x = 1/tan x, meaning tan x must equal 1 but also:
tan x = 1/tan x
tan^2 x = 1
sin^2 x/cos^2 x = 1
1-cos^2 x/cos^2 x=1
1/cos^2 x - 1 = 1
1/cos^2 x = 0


Mistake in the bold, I assume you've taken the reciprocal here but you've forgetten that that just flips the numerator and denominator around, your cos2xcos^{2}x term has disappeared also your denominator is wrong you should've just said tanx=±1tanx=\pm 1

Hence you've reached a contradiction of 1=01=0

It should be with your method
1cos2x=cos2x1-cos^{2}x=cos^{2}x so 2cos2x=12cos^{2}x=1
cos2x=12cos^{2}x=\frac{1}{2}
cosx=±12cosx=\pm \frac{1}{\sqrt{2}} which yields x=45x=45 and x=45x=-45 since cos(θ)=cosθcos(-\theta)=cos\theta and other solutions in a given interval
(edited 10 years ago)
Original post by Robbie242
Mistake in the bold, I assume you've taken the reciprocal here but you've forgetten that that just flips the numerator and denominator around, your cos2xcos^{2}x term has disappeared also your denominator is wrong you should've just said tanx=±1tanx=\pm 1

Hence you've reached a contradiction of 1=01=0

It should be with your method
1cos2x=cos2x1-cos^{2}x=cos^{2}x so 2cos2x=12cos^{2}x=1
cos2x=12cos^{2}x=\frac{1}{2}
cosx=±12cosx=\pm \frac{1}{\sqrt{2}} which yields x=45x=45 and x=45x=-45 since cos(θ)=cosθcos(-\theta)=cos\theta and other solutions in a given interval


1-cos^2 x / cos^2 x = 1/cos^2 x - cos^2 x / cos^ 2 x = 1
sec^2 x - 1 = 1
sec^2 x = 2
I must be more careful with signs as I cancelled out the 1s! :redface:
Reply 290
Original post by PhysicsKid
1-cos^2 x / cos^2 x = 1/cos^2 x - cos^2 x / cos^ 2 x = 1
sec^2 x - 1 = 1
sec^2 x = 2
I must be more careful with signs as I cancelled out the 1s! :redface:


Yep precisely

haha don't worry, we all make silly mistakes from time to time :P
Original post by PhysicsKid
Could you explain that method please? :P


Theorem (Fermat's Little Theorem): if pp does not divide aa, then ap11(modp)a^{p-1} \equiv 1 \pmod{p} for any prime p.
ab(modp)a \equiv b \pmod{p} just means that there exists integer n such that ab=npa - b = n p.
The nicest proof I know uses group theory, but I can provide an easier proof if you like (a bit of a pain to type out, though).

I checked that both the numbers on the LHS don't have a factor 13, and that the number on the RHS does. Hence by Fermat's Little Theorem, the LHS is just 1+1 = 2 mod 13 [it can easily be verified that "mod" works fine when you add things together], while the RHS is 0 mod 13, because 13 divides the RHS, so there is an n such that 13n = RHS, and hence RHS = 0 (mod 13).
But if two numbers are equivalent to different things mod any number, then they can't be the same number. Hence not equal.
Reply 292
Original post by Bluebubbles123
Ok, so here's how I think it should work. Someone asks a maths question ( that they know the answer to) and the first person to answer it correctly scores a point for their genders team.

So who's gonna win?

First question : Work out the mean of these three numbers : √75, √75 and 6 divided by √3.

Come on girls!


Not a good idea, given a higher % of boys studying maths and maths related degrees, chances are more boys will be on the maths forum. The boys vs. girls I mean, not the challenge itself.

Edit: Oh well 15 pages, maybe someone has an overview whether I was right or not...
Original post by TomzOJO
LOOOL these don't seem like GCSE q's anymore, and can people explain how they got their answer more effectively for others that do not understand


Yeah, sorry - the thread was a bit derailed :frown:

How about:
(spoiler defines a "group")

Spoiler


The question is:
Show that the identity is unique in a group. That is, show that if there are two elements e,f such that e+a = a+e = f+a = a+f = a for all a in the set X, then e=f.

This may take you a while to get your head around, but it's a good exercise in exploring a definition. That skill is very much not taught at school - even at A-level - so don't be disheartened if you don't know what to do. Write things down and see if they help!
Good luck!
Reply 294
Original post by KeyFingot
Prove that

[br]r=0nr(nr)=n2n1[br][br]\sum\limits_{r=0}^nr{n\choose r}=n2^{n-1}[br]
.

Proof by induction.

Check for the case when n = 1.

r=01r(1r)=0(10)+1(11)=1=1×20 \displaystyle \sum \limits_{r = 0}^{1} r\binom{1}{r} = 0\binom{1}{0} + 1\binom{1}{1} = 1 = 1\times 2^0

So assume r=0nr(nr)=n2n1 \displaystyle \sum\limits_{r=0}^nr\binom{n}{r}=n2^{n-1}.

r=0n+1r(n+1r)=0(n+10)+1(n+11)+...+(n+1)(n+1n+1)=0(n0)+0(n0)+1(n1)+1(n1)+...+n(nn)+n(nn)=2×n2n1=n2n=n2(n+1)1 \displaystyle \sum\limits_{r=0}^{n+1}r\binom{n + 1}{r} = 0\binom{n + 1}{0} + 1\binom{n + 1}{1} + ... + (n+1)\binom{n + 1}{n + 1} = 0\binom{n}{0} + 0\binom{n}{0} + 1\binom{n}{1} + 1\binom{n}{1} + ... + n\binom{n}{n} + n\binom{n}{n} = 2\times n2^{n - 1} = n2^n = n2^{(n + 1) - 1}

Thereby completing the induction.
(edited 10 years ago)
Original post by Robbie242
Mistake in the bold, I assume you've taken the reciprocal here but you've forgetten that that just flips the numerator and denominator around, your cos2xcos^{2}x term has disappeared also your denominator is wrong you should've just said tanx=±1tanx=\pm 1

Hence you've reached a contradiction of 1=01=0

It should be with your method
1cos2x=cos2x1-cos^{2}x=cos^{2}x so 2cos2x=12cos^{2}x=1
cos2x=12cos^{2}x=\frac{1}{2}
cosx=±12cosx=\pm \frac{1}{\sqrt{2}} which yields x=45x=45 and x=45x=-45 since cos(θ)=cosθcos(-\theta)=cos\theta and other solutions in a given interval


If you take the reciprocal you get cosec^2 x = 2. Hence both sin:2 x and cos^2 x = 1/2. Hence both sin x and cos x = +/- 1/sqrt(2). Hence x = -45 or 45. (The cos x = -(cos -x) works for sin as well). That, with my other post is a full proof, isn't it?
Reply 296
Original post by PhysicsKid
If you take the reciprocal you get cosec^2 x = 2. Hence both sin:2 x and cos^2 x = 1/2. Hence both sin x and cos x = +/- 1/sqrt(2). Hence x = -45 or 45. (The cos x = -(cos -x) works for sin as well). That, with my other post is a full proof, isn't it?


Did you just say sin2x+cos2x=12sin^{2}x+cos^{2}x=\frac{1}{2} ? Remember the identity sin2x+cos2x=1sin^{2}x+cos^{2}x=1 so that is a false statement

Not quite, see from your beginning contradiction it went downhill from there

square rooting gives sin2x+cos2x\sqrt{sin^{2}x+cos^{2}x} not sinx+cosx=±12sinx+cosx=\pm \frac{1}{\sqrt{2}}
e.g. xa+xb\sqrt{x^{a}+x^{b}} is not equal to xa+xb\sqrt{x^{a}}+\sqrt{x^{b}} or even simpler 1+3\sqrt{1+3} is not equal to 1+3\sqrt{1}+\sqrt{3}

Even so you could see a contradiction from your initial statement anyway, plug 45 into sinx+cosx and it yields the answer 2\sqrt{2}

I would honestly just have left it at tanx=±1tanx=\pm 1 and then take the arctan (inverse) of both sides :redface:

Also its sin(θ)=sinθsin(-\theta)=-sin\theta and cos(θ)=cosθcos(-\theta)=cos\theta by definition sin is an odd function and cos is an even function

Unless you actually mean sin2x=12sin^{2}x= \frac{1}{2} and cos2x=12cos^{2}x= \frac{1}{2}
(edited 10 years ago)
Original post by Robbie242
Did you just say sin2x+cos2x=12sin^{2}x+cos^{2}x=\frac{1}{2} ? Remember the identity sin2x+cos2x=1sin^{2}x+cos^{2}x=1 so that is a false statement

Not quite, see from your beginning contradiction it went downhill from there

square rooting gives sin2x+cos2x\sqrt{sin^{2}x+cos^{2}x} not sinx+cosx=±12sinx+cosx=\pm \frac{1}{\sqrt{2}}
e.g. xa+xb\sqrt{x^{a}+x^{b}} is not equal to xa+xb\sqrt{x^{a}}+\sqrt{x^{b}} or even simpler 1+3\sqrt{1+3} is not equal to 1+3\sqrt{1}+\sqrt{3}

Even so you could see a contradiction from your initial statement anyway, plug 45 into sinx+cosx and it yields the answer 2\sqrt{2}

I would honestly just have left it at tanx=±1tanx=\pm 1 and then take the arctan (inverse) of both sides :redface:

Also its sin(θ)=sinθsin(-\theta)=-sin\theta and cos(θ)=cosθcos(-\theta)=cos\theta by definition sin is an odd function and cos is an even function

Unless you actually mean sin2x=12sin^{2}x= \frac{1}{2} and cos2x=12cos^{2}x= \frac{1}{2}

No I was saying that if sin^2 x / cos^2 x = 1, then cos^2 x/ sin^2 x= 1. 1-sin^2 x/sin^2x = 1. 1/sin^2 x - sin^2 x / sin^2 x = 1. cosec^2 x - 1 = 1. cosec^2 x = 2. Hence sin^2 x = 1/2 and sin x = 1/root 2 and so on. :smile:
Reply 298
Original post by PhysicsKid
No I was saying that if sin^2 x / cos^2 x = 1, then cos^2 x/ sin^2 x= 1. 1-sin^2 x/sin^2x = 1. 1/sin^2 x - sin^2 x / sin^2 x = 1. cosec^2 x - 1 = 1. cosec^2 x = 2. Hence sin^2 x = 1/2 and sin x = 1/root 2 and so on. :smile:


Fair enough but you've complicated it more than what was needed :tongue:
Original post by Smaug123
Theorem (Fermat's Little Theorem): if pp does not divide aa, then ap11(modp)a^{p-1} \equiv 1 \pmod{p} for any prime p.
ab(modp)a \equiv b \pmod{p} just means that there exists integer n such that ab=npa - b = n p.
The nicest proof I know uses group theory, but I can provide an easier proof if you like (a bit of a pain to type out, though).

I checked that both the numbers on the LHS don't have a factor 13, and that the number on the RHS does. Hence by Fermat's Little Theorem, the LHS is just 1+1 = 2 mod 13 [it can easily be verified that "mod" works fine when you add things together], while the RHS is 0 mod 13, because 13 divides the RHS, so there is an n such that 13n = RHS, and hence RHS = 0 (mod 13).
But if two numbers are equivalent to different things mod any number, then they can't be the same number. Hence not equal.


Ahh I see :biggrin: Simple yet elegant!

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