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Year 13 Maths Help Thread

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Original post by RDKGames
I'm not good at proofs of this sort, especially with rationaility involved, but I had a good midnight go at it and will look into the second part more tomorrow, will update here whenever I figure it out :smile:

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Original post by RDKGames
I'm not good at proofs of this sort, especially with rationaility involved, but I had a good midnight go at it and will look into the second part more tomorrow, will update here whenever I figure it out :smile:

Spoiler



If x x is rational then it can be expressed in the form a/b,a,bZ, gcd(a,b)=1 a/b, a,b \in \mathbb{Z}, \ \text{gcd} (a,b) =1 .
Then (a/b)347(a/b)2+291=0 (a/b)^3 -47(a/b)^2+291=0 and so clearly a347a2b+291b3=0 a^3-47a^2b+291b^3=0 .
Considering cases and changing the parity of a and b leads to LHS being odd but RHS is even so it's clearly not true and thus there is a contradiction. The only case that appears to work is where a and b are both even, but we assumed that a and b were co prime, so it leads to a contradiction again. Therefore no rational solutions.
Original post by B_9710
If x x is rational then it can be expressed in the form a/b,a,bZ, gcd(a,b)=1 a/b, a,b \in \mathbb{Z}, \ \text{gcd} (a,b) =1 .
Then (a/b)347(a/b)2+291=0 (a/b)^3 -47(a/b)^2+291=0 and so clearly a347a2b+291b3=0 a^3-47a^2b+291b^3=0 .
Considering cases and changing the parity of a and b leads to LHS being odd but RHS is even so it's clearly not true and thus there is a contradiction. The only case that appears to work is where a and b are both even, but we assumed that a and b were co prime, so it leads to a contradiction again. Therefore no rational solutions.


Damn this is exactly what I had in mind initially before it went away and couldn't get back to it while making sense of it. Thanks!
Original post by Ano123
[Note. No credit is given for using the rational roots theorem.]


Why not?
Original post by Zacken
Why not?


Too easy.
Reply 124
How can I calculate sinxdx\int |\sin x| dx (if it is even integrable)?
Original post by Palette
How can I calculate sinxdx\int |\sin x| dx (if it is even integrable)?


It's easy enough for definite integral a but I don't think there is an indefinite integral. But I'm not sure.
But if you want to find say absinx dx \int_a^b |\sin x| \ dx then it can be done.
If I(n)=0nsinx dx, n>0 I(n)= \int_0^n |\sin x| \ dx, \ n>0 , then
Unparseable latex formula:

\displaystyle I(n)= \[ \left\{\begin{array}{ll} 2 \lfloor{\frac{n}{\pi }} \rfloor +1-\cos n & \text{if }\lfloor \frac{n}{\pi } \rfloor \text{ is even} \\ 2\lfloor \frac{n}{\pi } \rfloor+1-\cos n & \text{if } \lfloor \frac{n}{\pi } \rfloor \text{ is odd} \\\end{array} \right. \]

.

i(m)=m0sinx dx, m<0 i(m)= \int_m^0 |\sin x| \ dx, \ m<0 , then
Unparseable latex formula:

\displaystyle i(m)= \[ \left\{\begin{array}{ll} 2 \lfloor{-\frac{m}{\pi }} \rfloor +1-\cos m & \text{if }\lfloor \frac{m}{\pi }+1 \rfloor \text{ is even} \\ 2\lfloor -\frac{m}{\pi } \rfloor+1-\cos n & \text{if } \lfloor \frac{m}{\pi }+1 \rfloor \text{ is odd} \\\end{array} \right. \]

.
I have no idea is there is a more 'proper' answer to this but this is what I came up with and verified it a few times.
Of course from what I got you could easily have I(m,n)=mnsinx dx, m<0,n>0I(m,n)= \int_m^n |\sin x| \ dx, \ m<0, n>0 and you would get 4 answers for the 4 different cases that arise according to the values relating to m and n.
Another question here, a supposed Diophantine equation, the question is
Prove that the equation x3+2y3=4z3 x^3+2y^3=4z^3 has no integer solutions (besides the trivial solution).
Deduce that the equation also has no rational solutions.
ALSO
Prove that the equation x4+2y4=4z4 x^4+2y^4=4z^4 has no non trivial rational solutions.
(edited 7 years ago)
Original post by Ano123
Another question here, a supposed Diophantine equation, the question is
Prove that the equation x3+2y3=4z3 x^3+2y^3=4z^3 has no integer solutions (besides the trivial solution).
Deduce that the equation also has no rational solutions.


Slightly beyond Y13 level :tongue:
Original post by Ano123
Another question here, a supposed Diophantine equation, the question is
Prove that the equation x3+2y3=4z3 x^3+2y^3=4z^3 has no integer solutions (besides the trivial solution).
Deduce that the equation also has no rational solutions.
ALSO
Prove that the equation x4+2y4=4z4 x^4+2y^4=4z^4 has no non trivial rational solutions.


Swear you can just quote FLT, if not, cba


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Reply 130
Original post by Ano123
Another question here, a supposed Diophantine equation, the question is
Prove that the equation x3+2y3=4z3 x^3+2y^3=4z^3 has no integer solutions (besides the trivial solution).
Deduce that the equation also has no rational solutions.


I found this french document online which answers the problem (I don't understand it at all, but I'll translate it if it helps. It might give you an idea on how to solve your other problem):

Translated (all credits go to the individuals who wrote the document http://lespascals.org/docs/Training-Maths.pdf ).
The problem is on the first page.
Spoiler, full solution:

Spoiler

(edited 7 years ago)
Original post by drandy76
Swear you can just quote FLT, if not, cba


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That's what I thought originally but I think FLT's easy way would only apply if the coefficients were all 1.
Original post by RDKGames
That's what I thought originally but I think FLT's easy way would only apply if the coefficients were all 1.


Just re read it, FLT looks to be valid for proving that no integer solutions exist but the infinite descent approach shows that there are no rational solutions either


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Self teaching myself FP3 so I'll post anything I don't understand here. Firstly, show that ln(1+sinx)=xx22+x36+...ln(1+sinx)=x-\frac{x^2}{2}+\frac{x^3}{6}+...but I keep getting -1/6 rather than +1/6 for the third term. Where am I going wrong here?

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Original post by RDKGames
Self teaching myself FP3 so I'll post anything I don't understand here. Firstly, show that ln(1+sinx)=xx22+x36+...ln(1+sinx)=x-\frac{x^2}{2}+\frac{x^3}{6}+...but I keep getting -1/6 rather than +1/6 for the third term. Where am I going wrong here?

Spoiler



You need to square the maclaurins expansion of sin2x sin^2x

The expansion of ln(1+x) ln(1+x) is valid for 1<x1-1<x\leq 1, but 1sinx1-1 \leq sinx\leq 1 perhaps.
(edited 7 years ago)
Original post by NotNotBatman
You need to square the maclaurins expansion of sin2x sin^2x

The expansion of ln(1+x) ln(1+x) is valid for 1<x1-1<x\leq 1, but 1sinx1-1 \leq sinx\leq 1 perhaps.


Thanks, I got it in the end.

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Q3: Surely that should say that it tends to positive infinity because the denominator tends to 0 as x goes to 0? x3ln(x) goes to 0 as well as x3 so it only makes sense?
ImageUploadedByStudent Room1469904946.144849.jpg
(edited 7 years ago)
Original post by RDKGames
Thanks, I got it in the end.

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Q3: Surely that should say that it tends to positive infinity because the denominator tends to 0 as x goes to 0? x3ln(x) goes to 0 as well as x3 so it only makes sense?
ImageUploadedByStudent Room1469904946.144849.jpg


x^k ln x tends to 0- when x tends to 0+.

So its essentially 1/-0 which is -infinity.
(edited 7 years ago)
Original post by Math12345
x^k ln x tends to 0- when x tends to 0+.

So its essentially 1/-0 which is -infinity.


Ah yeah that makes sense. Just checked the sketch and it baffled me as to the fact that it looks like it actually goes to positive infinity but then I looked down the graph and it had another section to it. Thanks.
Original post by RDKGames
Thanks, I got it in the end.

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Q3: Surely that should say that it tends to positive infinity because the denominator tends to 0 as x goes to 0? x3ln(x) goes to 0 as well as x3 so it only makes sense?
ImageUploadedByStudent Room1469904946.144849.jpg


No, take a small value of x. ln(x) is negative for x between 0 and 1.
Original post by RDKGames
...


Bit of a stupid question; the limit doesn't exist, so why do we care if it's negative or positive infinity? Anywho: lnx\ln x tends to negative infinity as xx tends to 0, so you can can see why the entire denominator tends to negative 0. (or rather, 0 approaching from the left)
(edited 7 years ago)

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