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    (Original post by Indeterminate)
    That's the solution I was looking for

    The complex analysis route is really quite messy in this case.
    I thought so, i dont do any complex analysis as of yet as I am not at Uni yet! But the square at the bottom and the limts made me think of that sub.


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    Problem 551
    A,B are fixed points on a circle, not diametrically opposite each other.P is a variable point of the circle and Q is the point diametrically opposite P. Find the Locus of the point of intersection of AP and BQ.

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    I am new to this but I had a quick question: Why is it possible to add the two integrals when the first one has the 'dx' operator and the second has the 'du' operator on it ?
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    (Original post by -Gifted-)
    I am new to this but I had a quick question: Why is it possible to add the two integrals when the first one has the 'dx' operator and the second has the 'du' operator on it ?
    ∫ f(x) dx = ∫ f(u) du = ∫ f(t) dt = ∫ f(k) dk etc
    as the limits are the same they represent the same number
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    (Original post by -Gifted-)
    I am new to this but I had a quick question: Why is it possible to add the two integrals when the first one has the 'dx' operator and the second has the 'du' operator on it ?
    Such things known as dummy variables in calculus. As TeeEm has phrased.
    You can see some as transformations aswell ie x->-x etc


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    (Original post by TeeEm)
    ∫ f(x) dx = ∫ f(u) du = ∫ f(t) dt = ∫ f(k) dk etc
    as the limits are the same they represent the same number
    Thanks, makes sense now
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    (Original post by physicsmaths)
    Problem 551
    A,B are fixed points on a circle, not diametrically opposite each other.P is a variable point of the circle and Q is the point diametrically opposite P. Find the Locus of the point of intersection of AP and BQ.

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    Solution 551

    The locus is a circle through A and B centered on the intersection of the tangents to the original circle at A and B.
    There are two cases to consider, but I'll write a short argument, the details can be added.
    Let the point of intersection of AP and BQ be T.
    We know that as Q varies, the angle AQB stays constant.
    We also know that the angle QAP, and thus QAT are right angles, as QAP lies in a semicircle.
    Thus in the triangle QAT, the remaining angle QTA is constant, thus BTA is constant and the locus is a circle.

    Now let the the intersection of the tangents to the original circle through A and B meet at S and let O be the center of the original circle.
    ASB = 180 - AOB = 180 - 2AQB = 2(90 - AQB) and let T lie outside the original circle (if it doesn't in your sketch, swap the positions of P and Q), so ATB = 90 - AQB (considering the triangle AQT), this and symmetry implies that S is the center of the locus circle.
    (I haven't done this last part in a particularly beautiful way, but notice that it implies that if we used the locus circle as the generating circle, keeping A and B constant, the locus generated will be the original circle, and notice also that the two circles intersect each other at right angles.)
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    Problem 552***

    Let  \alpha, \beta \in [0,\pi] and k > 0

    By considering the Maclaurin series of \ln(1-z), or otherwise, show that

    \displaystyle \sum_{k=1}^{\infty} \dfrac{x^k \cos(k\alpha)}{k} = -\dfrac{1}{2} \ln \left(x^2 - 2x \cos(\alpha) + 1\right)

    for |x| < 1

    Show that

    \displaystyle \int_{0}^{\infty}\dfrac{1}{x} \ln \left(\dfrac{x^2 + 2kx \cos \beta + k^2}{x^2 + 2kx \cos \alpha + k^2}\right) \ dx = \alpha^2 - \beta^2
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    (Original post by Indeterminate)
    Problem 552***

    Let  \alpha, \beta \in [0,\pi] and k > 0

    By considering the Maclaurin series of \ln(1-z), or otherwise, show that

    \displaystyle \sum_{k=1}^{\infty} \dfrac{x^k \cos(k\alpha)}{k} = -\dfrac{1}{2} \ln \left(x^2 - 2x \cos(\alpha) + 1\right)

    for |x| < 1

    Show that

    \displaystyle \int_{0}^{\infty} \ln \left(\dfrac{x^2 + 2kx \cos \beta + k^2}{x^2 + 2kx \cos \alpha + k^2}\right) \ dx = \alpha^2 - \beta^2
    What level of mathematics is required for this?
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    (Original post by EricPiphany)
    Solution 551

    The locus is a circle through A and B centered on the intersection of the tangents to the original circle at A and B.
    There are two cases to consider, but I'll write a short argument, the details can be added.
    Let the point of intersection of AP and BQ be T.
    We know that as Q varies, the angle AQB stays constant.
    We also know that the angle QAP, and thus QAT are right angles, as QAP lies in a semicircle.
    Thus in the triangle QAT, the remaining angle QTA is constant, thus BTA is constant and the locus is a circle.

    Now let the the intersection of the tangents to the original circle through A and B meet at S and let O be the center of the original circle.
    ASB = 180 - AOB = 180 - 2AQB = 2(90 - AQB) and let T lie outside the original circle (if it doesn't in your sketch, swap the positions of P and Q), so ATB = 90 - AQB (considering the triangle AQT), this and symmetry implies that S is the center of the locus circle.
    (I haven't done this last part in a particularly beautiful way, but notice that it implies that if we used the locus circle as the generating circle, keeping A and B constant, the locus generated will be the original circle, and notice also that the two circles intersect each other at right angles.)
    Sorry, completely missed this, I will try find my solution tonight, If not I will write it out. It was a beautfiul problem.


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    (Original post by qwertzuiop)
    What level of mathematics is required for this?
    I'd say undergrad

    The connoisseurs amongst the A-level students could do it, though
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    (Original post by Indeterminate)
    Problem 552***

    Let  \alpha, \beta \in [0,\pi] and k > 0

    By considering the Maclaurin series of \ln(1-z), or otherwise, show that

    \displaystyle \sum_{k=1}^{\infty} \dfrac{x^k \cos(k\alpha)}{k} = -\dfrac{1}{2} \ln \left(x^2 - 2x \cos(\alpha) + 1\right)

    for |x| < 1

    Show that

    \displaystyle \int_{0}^{\infty} \ln \left(\dfrac{x^2 + 2kx \cos \beta + k^2}{x^2 + 2kx \cos \alpha + k^2}\right) \ dx = \alpha^2 - \beta^2
    I'm confused about the integral in the second part, because in the first part you prove a formula for |x| < 1, but the integral ranges from 0 to infinity. Also, when I substitute x = ku, the result seems to imply that the integral is not independent on the value of k.
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    (Original post by EricPiphany)
    I'm confused about the integral in the second part, because in the first part you prove a formula for |x| < 1, but the integral ranges from 0 to infinity. Also, when I substitute x = ku, the result seems to imply that the integral is not independent on the value of k.
    Hint:

    After substituting u=x/k, you can split the integral into 2: the first over (0,1) and the second over (1,\infty)

    If you can find away around the latter, then the result will be applicable
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    Problem 553 (*)

    Here's a nice, easy to start with problem

    Let a_{1}, a_{2},... be a sequence of positive integers such that every (positive) integer appears in the sequence only once. Suppose further that  n divides  S_{n} = a_{1} + ... + a_{n} . Show that such a sequence exist.
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    (Original post by Blazy)
    Problem 553 (*)

    Here's a nice, easy to start with problem

    Let a_{1}, a_{2},... be a sequence of positive integers such that every (positive) integer appears in the sequence only once. Suppose further that  n divides  S_{n} = a_{1} + ... + a_{n} . Show that such a sequence exist.
    Solution 553

    We construct a sequence that works. Start with  a_1=1 then let  a_2 be some large number such that we may have  a_3=2 , i.e  a_2 \equiv 3 \pmod{6} . Now let  a_4 be some even larger number compared to  a_2 such that we may have  a_5=3 , and such a number exists by CRT. Hence, we keep repeating this, alternating the sequence between large number and the sequence  \{ 1,2,3... \} , missing out a part of the sequence if that number has already been used as one of our large numbers, and hence, we have constructed a sequence that works.
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    (Original post by Renzhi10122)
    Solution 553

    We construct a sequence that works. Start with  a_1=1 then let  a_2 be some large number such that we may have  a_3=2 , i.e  a_2 \equiv 5 \pmod{6} . Now let  a_4 be some even large number compared to  a_2 such that we may have  a_5=3 , and such a number exists by CRT. Hence, we keep repeating this, alternating the sequence between large number and the sequence  \{ 1,2,3... \} , missing out a part of the sequence if that number has already been used as one of our large numbers, and hence, we have constructed a sequence that works.
    I have never studied the Chinese remainder theorem so this question may be stupid, but I'm trying to understand your construction.
    I understand that a_{2n-1}=n so a_{1}=1, a_{2}=k&gt;1 but then a_{2k-1}=k=a_{2} so k appears twice in the sequence.
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    (Original post by EricPiphany)
    I have never studied the Chinese remainder theorem so this question may be stupid, but I'm trying to understand your construction.
    I understand that a_{2n-1}=n so a_{1}=1, a_{2}=k&gt;1 but then a_{2k-1}=k=a_{2} so k appears twice in the sequence.
    Would that not be missed out though, I think that is what is happening.


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    (Original post by EricPiphany)
    I have never studied the Chinese remainder theorem so this question may be stupid, but I'm trying to understand your construction.
    I understand that a_{2n-1}=n so a_{1}=1, a_{2}=k&gt;1 but then a_{2k-1}=k=a_{2} so k appears twice in the sequence.
    I've just used CRT to justify the existence of such a large number with the property I want. We then miss it out in our sequence as physicsmaths said.
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    (Original post by Renzhi10122)
    I've just used CRT to justify the existence of such a large number with the property I want. We then miss it out in our sequence as physicsmaths said.
    OK, I'm still struggling with the construction though. If a_1=1, a_2=6n-1 then S_2=6n, S_3=6n+2 which isn't divisible by 3.
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    (Original post by EricPiphany)
    OK, I'm still struggling with the construction though. If a_1=1, a_2=6n-1 then S_2=6n, S_3=6n+2 which isn't divisible by 3.
    Woops... I meant =3 (mod 6)... sorry, I'll edit that.
 
 
 
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