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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
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    Assume x=\alpha is an integer solution to the above cubic.

    Therefore {\alpha}^3=47{\alpha}^2-291

    In the event \alpha is even; then {\alpha}^3 and {\alpha}^2 are always even. However, 47{\alpha}^2 is even, and 291 is odd, therefore even-odd is always odd. even\not=odd therefore we reach a contradiction.


    In the event \alpha is odd; then {\alpha}^3 and {\alpha}^2 are always odd. However, 47{\alpha}^2 is odd, and 291 is odd, therefore odd-odd is always even. odd\not=even therefore we reach a contradiction yet again.

    Thus we prove there are no integer solutions by means of contradiction. QED.



    ...and I'm dry on this one for now
    (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
    Spoiler:
    Show


    Assume x=\alpha is an integer solution to the above cubic.

    Therefore {\alpha}^3=47{\alpha}^2-291

    In the event \alpha is even; then {\alpha}^3 and {\alpha}^2 are always even. However, 47{\alpha}^2 is even, and 291 is odd, therefore even-odd is always odd. even\not=odd therefore we reach a contradiction.


    In the event \alpha is odd; then {\alpha}^3 and {\alpha}^2 are always odd. However, 47{\alpha}^2 is odd, and 291 is odd, therefore odd-odd is always even. odd\not=even therefore we reach a contradiction yet again.

    Thus we prove there are no integer solutions by means of contradiction. QED.



    ...and I'm dry on this one for now
    If  x is rational then it can be expressed in the form  a/b, a,b \in \mathbb{Z}, \ \text{gcd} (a,b) =1 .
    Then  (a/b)^3 -47(a/b)^2+291=0 and so clearly  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.
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    (Original post by B_9710)
    If  x is rational then it can be expressed in the form  a/b, a,b \in \mathbb{Z}, \ \text{gcd} (a,b) =1 .
    Then  (a/b)^3 -47(a/b)^2+291=0 and so clearly  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!
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    (Original post by Ano123)
    [Note. No credit is given for using the rational roots theorem.]
    Why not?
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    (Original post by Zacken)
    Why not?
    Too easy.
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    How can I calculate \int |\sin x| dx (if it is even integrable)?
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    (Original post by Palette)
    How can I calculate \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  \int_a^b |\sin x| \ dx then it can be done.
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    If  I(n)= \int_0^n |\sin x| \ dx, \ n>0 , then  \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)= \int_m^0 |\sin x| \ dx, \ m<0 , then  \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)= \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.
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    Another question here, a supposed Diophantine equation, the question is
    Prove that the equation  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  x^4+2y^4=4z^4 has no non trivial rational solutions.
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    (Original post by Ano123)
    Another question here, a supposed Diophantine equation, the question is
    Prove that the equation  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
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    (Original post by Ano123)
    Another question here, a supposed Diophantine equation, the question is
    Prove that the equation  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  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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    (Original post by Ano123)
    Another question here, a supposed Diophantine equation, the question is
    Prove that the equation  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:
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    The infinite descent is the principle by which there are no strictly decreasing series of positive integers.
    We use this principle to prove that there are no solutions to certain problems involving integers: if based on a solution, we are able to fabricate another one stricly smaller but still in integers, then we can repeat this indefinitely, so the initial problem has no solution.

    Exercice: Solve in integers the equation: x^3+2y^3=4z^3

    Solution:

    Suppose that (x,y,z) is a solution to the equation, x, y and z being integers.

    Then:
    x^3+2y^3=4z^3 \Rightarrow x^3=2(2z^3-y^3)
    So x is even.

    We pose x', an integer such that x=2x'
    Then:
    8x'^3+2y^3=4z^3
    So:
    4x'^3+y^3=2z^3
    Which can be rewritten as:
    y^3=2(z^3-2x'^3)

    So y is even. We pose y', an integer such that y=2y'
    Then:
    4x'^3+8y'^3=2z^3
    So:
    2x'^3+4y'^3=z^3
    Which can be rewritten as:
    z^3=2(x'^3+2y'^3)

    So z is even. We pose z', an integer such that z=2z'
    Then:
    2x'^3+4y'^3=8z'^3
    Which can be rewritten as:
    x'^3+2y'^3=4z'^3

    We have deduced from the solution (x,y,z) a new solution (x',y',z'), stricly smaller than the first.
    Thus, we have constructed an infinite serie of solutions entirely decreasing, which is contradictory.
    The equation has no solution.


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    (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.
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    (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)=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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    ln(1+sinx)=sinx-\frac{sin^2x}{2}+\frac{sin^3x}{3  }+...

    =sinx(1-\frac{sinx}{2}+\frac{sin^2x}{3}+  ...)

    =[x-\frac{x^3}{6}+...][1-\frac{1}{2}(x-\frac{x^3}{6}+...)+...]

    =x-\frac{1}{2}x-\frac{1}{6}x^3+...

    ...and the -1/6 comes from the -1/6 in the first bracket multiplied by 1 in the other.
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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)=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:
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    ln(1+sinx)=sinx-\frac{sin^2x}{2}+\frac{sin^3x}{3  }+...

    =sinx(1-\frac{sinx}{2}+\frac{sin^2x}{3}+  ...)

    =[x-\frac{x^3}{6}+...][1-\frac{1}{2}(x-\frac{x^3}{6}+...)+...]

    =x-\frac{1}{2}x-\frac{1}{6}x^3+...

    ...and the -1/6 comes from the -1/6 in the first bracket multiplied by 1 in the other.
    You need to square the maclaurins expansion of  sin^2x

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

    The expansion of  ln(1+x) is valid for -1<x\leq 1, but -1 \leq sinx\leq 1 perhaps.
    Thanks, I got it in the end.

    -

    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?
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    (Original post by RDKGames)
    Thanks, I got it in the end.

    -

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

    So its essentially 1/-0 which is -infinity.
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    (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.
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    (Original post by RDKGames)
    Thanks, I got it in the end.

    -

    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?
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    No, take a small value of x. ln(x) is negative for x between 0 and 1.
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    (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: \ln x tends to negative infinity as x tends to 0, so you can can see why the entire denominator tends to negative 0. (or rather, 0 approaching from the left)
 
 
 
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