Girls vs boys maths challenge Watch

Robbie242
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#301
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#301
(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.
Fair enough but you've complicated it more than what was needed
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PhysicsKid
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#302
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(Original post by Smaug123)
Theorem (Fermat's Little Theorem): if p does not divide a, then a^{p-1} \equiv 1 \pmod{p} for any prime p.
a \equiv b \pmod{p} just means that there exists integer n such that a - 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 Simple yet elegant!
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Felix Felicis
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#303
(Original post by KeyFingot)
Prove that



\sum\limits_{r=0}^nr{n\choose r}=n2^{n-1}
.
(Original post by 0x2a)
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Proof by induction.

Check for the case when n = 1.

 \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  \displaystyle \sum\limits_{r=0}^nr\binom{n}{r}  =n2^{n-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.
I don't like induction

Consider

\displaystyle \begin{aligned} f(x) = (x+1)^{n} = \sum_{r=0}^{n} \binom{n}{r} x^{r} \implies f'(1) & = \sum_{r=0}^{n} \binom{n}{r} \cdot r  \\ & = n \cdot (x+1)^{n-1} \bigg|_{x = 1} \\ & = n \cdot 2^{n-1} \end{aligned}

as required.
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PhysicsKid
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#304
(Original post by Robbie242)
Fair enough but you've complicated it more than what was needed
I always do! :P
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Smaug123
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(Original post by Felix Felicis)
I don't like induction

Consider

\displaystyle \begin{aligned} f(x) = (x+1)^{n} = \sum_{r=0}^{n} \binom{n}{r} x^{r} \implies f'(1) & = \sum_{r=0}^{n} \binom{n}{r} \cdot r  \\ & = n \cdot (x+1)^{n-1} \bigg|_{x = 1} \\ & = n \cdot 2^{n-1} \end{aligned}

as required.
Yeah, this was the method I used - it's so much nicer
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Jkn
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#306
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#306
(Original post by PhysicsKid)
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).
Your initial factorization is incorrect. Also, you have not justified that your rounding did not create the error.
(Original post by Smaug123)
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It's false, by either one of FLT or FLT. In the former case, it's just blindingly obvious; in the latter, we take p=13; then the LHS is 1+1 mod 13, while the RHS is 0 (since only 4472 is divisible by 13).


ETA: Hah, the difference is about 10^33 :P
Interestingly though, they match in the first 10 decimal places as well as the last decimal place. The accuracy is in the region of 10^{-8}% It's a 'near-solution'. There was another similar one in the background of the Simpsons episode where homer goes behind a bookshelf and ends up in another dimension (and then ends up in the real world) along with loads of other things like P=NP. :lol:
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Smaug123
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(Original post by Jkn)
Interestingly though, they match in the first 10 decimal places as well as the last decimal place. The accuracy is in the region of 10^{-8}% It's a 'near-solution'. There was another similar one in the background of the Simpsons episode where homer goes behind a bookshelf and ends up in another dimension (and then ends up in the real world) along with loads of other things like P=NP. :lol:
Ah, I vaguely remember that episode from my dim and distant past
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PhysicsKid
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#308
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#308
(Original post by Jkn)
Your initial factorization is incorrect. Also, you have not justified that your rounding did not create the error.

Interestingly though, they match in the first 10 decimal places as well as the last decimal place. The accuracy is in the region of 10^{-8}% It's a 'near-solution'. There was another similar one in the background of the Simpsons episode where homer goes behind a bookshelf and ends up in another dimension (and then ends up in the real world) along with loads of other things like P=NP. :lol:
Fair enough for your second point but a^2 + b^2 = (a+b)^2 - 2ab
a^12 + b^12 = (a^2 + b^2)^6, or not?:confused: So a^2+ b^2 may be written as:
( (a+b)^2 - 2ab)^6. No?
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Smaug123
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(Original post by PhysicsKid)
Fair enough for your second point but a^2 + b^2 = (a+b)^2 - 2ab
a^12 + b^12 = (a^2 + b^2)^6, or not?:confused: So a^2+ b^2 may be written as:
( (a+b)^2 - 2ab)^6. No?
No. That requires (a^2+b^2)^6 = a^{12}+b^{12}, which implies that (1+1)^6 = 1+1 - clearly rubbish.
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PhysicsKid
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#310
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#310
(Original post by Smaug123)
No. That requires (a^2+b^2)^6 = a^{12}+b^{12}, which implies that (1+1)^6 = 1+1 - clearly rubbish.
Right. I was going to use the binomial expansion but thought that was a shortcut- out of interest, what is (a^2 + b^2)^6?
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Smaug123
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(Original post by PhysicsKid)
Right. I was going to use the binomial expansion but thought that was a shortcut- out of interest, what is (a^2 + b^2)^6?
Binomial expansion:
a^{12} + 6 a^{10} b^2 + 15 a^8 b^4 + 20 a^6 b^6 + 15 a^4 b^8 + 6 a^2 b^{10} + b^{12}
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PhysicsKid
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#312
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#312
(Original post by Smaug123)
Binomial expansion:
a^{12} + 6 a^{10} b^2 + 15 a^8 b^4 + 20 a^6 b^6 + 15 a^4 b^8 + 6 a^2 b^{10} + b^{12}
Thanks So just the normal expansion- not like (k+3x)^7 or something similar?
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Felix Felicis
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#313
(Original post by PhysicsKid)
Thanks So just the normal expansion- not like (k+3x)^7 or something similar?
Always like that, (a+b)^{n} \not\equiv a^{n} + b^{n}
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KeyFingot
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(Original post by Felix Felicis)
I don't like induction

Consider

\displaystyle \begin{aligned} f(x) = (x+1)^{n} = \sum_{r=0}^{n} \binom{n}{r} x^{r} \implies f'(1) & = \sum_{r=0}^{n} \binom{n}{r} \cdot r  \\ & = n \cdot (x+1)^{n-1} \bigg|_{x = 1} \\ & = n \cdot 2^{n-1} \end{aligned}

as required.
Obviously both methods are valid, but I prefer this one. It's neater
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PhysicsKid
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#315
(Original post by Felix Felicis)
Always like that, (a+b)^{n} \not\equiv a^{n} + b^{n}
<br />
<br />
Yet (a^b)^c = a^bc. It's so ANNOYING!
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Robbie242
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#316
(Original post by PhysicsKid)
<br />
<br />
Yet (a^b)^c = a^bc. It's so ANNOYING!
This is because this is only term it influences, if there is an addition or subtracting going on then you need to take account of all terms that's like saying (x+1)^{2}=x^{2}+1^{2} which is clearly rubbish
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Kvothe the Arcane
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#317
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#317
(Original post by Felix Felicis)
Always like that, (a+b)^{n} \not\equiv a^{n} + b^{n}
when n \neq 1
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tohaaaa
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Maths question: A-Level
(should be simple)

A curve, y=f(x) passes through point (0,4) and is such that f'(x) = 8x3 + 4x - 2

Find f(x)
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0x2a
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#319
(Original post by tohaaaa)
Maths question: A-Level
(should be simple)

A curve, y=f(x) passes through point (0,4) and is such that f'(x) = 8x3 + 4x - 2

Find f(x)
 \displaystyle \int (8x^3 + 4x - 2) dx = 2x^4 + 2x^2 - 2x + c

 2(0)^4 + 2(0)^2 - 2(0) + c = 4 \Rightarrow c = 4

 \therefore f(x) = 2x^4 + 2x^2 - 2x + 4
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tohaaaa
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#320
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#320
(Original post by 0x2a)
 \displaystyle \int (8x^3 + 4x - 2) dx = 2x^4 + 2x^2 - 2x + c

 2(0)^4 + 2(0)^2 - 2(0) + c = 4 \Rightarrow c = 4

 \therefore f(x) = 2x^4 + 2x^2 - 2x + 4
If I hadn't wasted my thumbs up, I would have given you one

Yep, fully correct!
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